├── .gitignore ├── PDE01.ipynb ├── PDE02.ipynb ├── PDE03.ipynb ├── README.md ├── ch01.ipynb ├── ch02.ipynb ├── ch03.ipynb ├── ch04.ipynb ├── ch05.ipynb ├── ch07.ipynb ├── ch08.ipynb ├── ch09.ipynb ├── ch10.ipynb ├── ch11.ipynb ├── ch12.ipynb ├── codes ├── .gitignore ├── ch02_code1.ipynb ├── ch02_code2.ipynb ├── ch03_code1.ipynb ├── ch03_code2.ipynb ├── ch03_code3.ipynb ├── ch05_code1.ipynb ├── ch05_code2.ipynb ├── ch05_code3.ipynb ├── ch08_code1.ipynb ├── ch08_code2.ipynb ├── ch08_code3.ipynb ├── ch08_code4.ipynb ├── ch08_code5.ipynb ├── ch08_code6.ipynb └── phase_portrait.ipynb ├── exams ├── .gitignore ├── Quiz_01.ipynb ├── Quiz_02_ch01_02.ipynb ├── Quiz_02_ch01_02.slides.html ├── README.md ├── figs │ ├── 2023_final_2_B_2.jpg │ ├── fig_j0.ipynb │ ├── fig_j0.pdf │ ├── fig_j0.png │ ├── mid_2019_2_A_1.png │ └── mid_2019_2_B_1.png ├── final_2019_2_A.ipynb ├── final_2019_2_B.ipynb ├── final_2019_2_code1.ipynb ├── final_2019_2_code2.ipynb ├── final_2019_2_code3.ipynb ├── final_2020_1.ipynb ├── final_2020_2_A.ipynb ├── final_2020_2_B.ipynb ├── final_2022_1_A.ipynb ├── final_2022_1_A_solution.ipynb ├── final_2022_1_B.ipynb ├── final_2022_1_B_solution.ipynb ├── final_2022_2_A.ipynb ├── final_2022_2_A_code.ipynb ├── final_2022_2_A_solution.ipynb ├── final_2022_2_B.ipynb ├── final_2022_2_B_solution.ipynb ├── final_2023_1_A.ipynb ├── final_2023_1_A_solution.ipynb ├── final_2023_1_B.ipynb ├── final_2023_1_B_solution.ipynb ├── final_2023_1_solution_5.ipynb ├── final_2023_2_A.html ├── final_2023_2_A.ipynb ├── final_2023_2_B.html ├── final_2023_2_B.ipynb ├── mid_2019_1.ipynb ├── mid_2019_2_A.ipynb ├── mid_2019_2_B.ipynb ├── mid_2020_1.ipynb ├── mid_2021_2_A.ipynb ├── mid_2022_1_A.ipynb ├── mid_2022_1_A_solution.ipynb ├── mid_2022_1_B.ipynb ├── mid_2022_1_B_solution.ipynb ├── mid_2022_2_A.ipynb ├── mid_2022_2_A_solution.ipynb ├── mid_2022_2_B.ipynb ├── mid_2022_2_B_solution.ipynb ├── mid_2023_1_A.ipynb ├── mid_2023_1_A_solution.ipynb ├── mid_2023_1_B.ipynb ├── mid_2023_1_B_solution.ipynb ├── mid_2023_2_A.ipynb └── mid_2023_2_B.ipynb ├── exercises ├── ch01_exer02.ipynb ├── ch01_exer03.ipynb ├── ch02_exer02.ipynb ├── ch02_exer07.ipynb ├── ch04_exer06.ipynb ├── ch10_exer02.ipynb ├── ch10_exer03.ipynb ├── ch10_exer04.ipynb ├── ch10_exer05.ipynb └── figs │ ├── ch01_exer01.png │ ├── ch01_exer03.png │ └── ch04_exer06.png ├── figures ├── .gitignore ├── Fourier2.jpg ├── ch01_figure01.png ├── ch01_figure02.png ├── ch01_figure03.png ├── ch02_figure01.png ├── ch02_figure02.png ├── ch02_figure03.png ├── ch02_figure04.png ├── ch02_figure05.png ├── ch03_figure01.png ├── ch03_figure02.png ├── ch03_figure03.png ├── ch03_figure04.png ├── ch04_ex_.png ├── ch04_figure01.png ├── ch04_figure02.png ├── ch04_figure03.png ├── ch07_figure00.png ├── ch07_figure01.png ├── ch07_figure02.png ├── ch07_figure03.png ├── ch07_figure04.png ├── ch07_figure05.png ├── ch07_figure06.png ├── ch07_figure07.png ├── ch07_figure08.png ├── ch07_figure09.png ├── ch07_figure10.png ├── ch07_figure11.png ├── ch07_figure12.png ├── ch07_figure13.png ├── ch07_figure14.png ├── ch07_figure15.png ├── ch07_figure16.png ├── ch07_figure17.png ├── ch08_figure01.png ├── ch08_figure02.png ├── ch08_figure03.png ├── ch08_figure04.png ├── ch08_figure05.png ├── ch08_figure06.png ├── ch08_figure07.png ├── ch09_centripetal_force.png ├── ch09_figure01.png ├── ch09_figure02.png ├── ch09_figure03.png ├── ch09_figure04.png ├── ch09_figure05.png ├── ch09_figure06.png ├── ch09_figure07.png ├── ch09_figure08.png ├── ch09_figure09.png ├── ch09_figure10.png ├── ch09_figure11.png ├── ch09_figure12.png ├── ch09_figure13.png ├── ch09_figure14.png ├── ch09_figure15.png ├── ch09_figure16.png ├── ch09_figure17.png ├── ch09_figure18.png ├── ch09_figure19.png ├── ch09_figure20.png ├── ch09_figure21.png ├── ch09_figure22.png ├── ch09_figure23.png ├── ch09_figure24.png ├── ch09_figure25.png ├── ch09_figure26.png ├── ch09_figure27.png ├── ch09_figure28.png ├── ch09_figure29.png ├── ch09_figure30.png ├── ch09_figure31.png ├── ch09_figure32.png ├── ch09_figure33.png ├── ch09_figure34.png ├── ch09_figure35.png ├── ch09_figure36.png ├── ch09_figure37.png ├── ch09_figure38.png ├── ch09_figure39.png ├── ch09_figure40.png ├── ch09_figure41.png ├── ch09_figure42.png ├── ch09_figure43.png ├── ch09_figure44.png ├── ch09_figure45.png ├── ch09_figure46.png ├── ch09_figure47.png ├── ch11_figure01.png ├── ch11_figure02.png ├── ch11_figure03.png ├── ch11_figure04.png ├── ch11_figure05.png ├── ch11_figure06.png ├── ch11_figure07.png ├── ch11_figure08.png ├── ch12_figure01.png ├── ch12_figure02.png ├── ch12_figure03.png ├── fig_ex7_1.jpg ├── my_computer.jpg ├── pde01_figure01.jpg ├── pde01_figure02.jpg ├── pde01_figure03.jpg ├── pde01_figure04.jpg ├── pde01_figure05.jpg ├── pde01_figure06.jpg ├── pde01_figure07.jpg ├── pde01_figure08.jpg ├── pde01_figure09.jpg ├── pde01_figure10.jpg ├── pde01_figure11.jpg ├── pde01_figure12.png ├── pde01_figure13.jpg ├── pde02_figure01.jpg ├── pde02_figure02.jpg ├── pde02_figure03.jpg ├── pde02_figure04.jpg ├── pde02_figure05.jpg ├── pde02_figure06.jpg ├── pde02_figure07.jpg ├── pde02_figure08.jpg ├── pde02_figure09.jpg ├── pde02_figure10.jpg ├── pde02_figure11.jpg ├── pde02_figure12.jpg ├── pde02_figure13.jpg ├── pde02_figure14.jpg ├── pde02_figure15.jpg ├── pde02_figure16.jpg ├── pde02_figure17.jpg ├── pde02_figure18.jpg ├── pde02_figure19.jpg ├── pde02_figure20.jpg ├── pde02_figure21.jpg ├── pde02_figure22.jpg ├── pde02_figure23.jpg ├── pde02_figure24.jpg ├── pde02_figure25.jpg ├── pde02_figure26.jpg ├── pde02_figure27.jpg ├── pde02_figure28.jpg ├── pde03_figure01.jpg ├── pde03_figure02.jpg ├── pde03_figure03.jpg ├── pde03_figure04.jpg ├── pde03_figure05.jpg ├── pde03_figure06.jpg ├── pde03_figure07.jpg ├── pde03_figure08.jpg ├── pde03_figure09.jpg ├── pde03_figure10.jpg ├── pde03_figure11.jpg ├── pde03_figure12.jpg └── pde03_figure13.jpg └── supplements ├── .gitignore ├── Green's_Function.ipynb ├── Laplace_Equation_in_Sphere.ipynb ├── LegendrePolynomials_IntegralRelations.ipynb ├── Poisson_Equation_in_Sphere.ipynb ├── README.md ├── RootLocus.ipynb ├── hydrogen.ipynb └── hydrogen_code.ipynb /.gitignore: -------------------------------------------------------------------------------- 1 | .ipynb_checkpoints/ -------------------------------------------------------------------------------- /README.md: -------------------------------------------------------------------------------- 1 | ## Engineering Mathematics Lecture Notes 2 | 3 | * [Introduction to Differential Equations](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/ch01.ipynb) 4 | * [First-Order Differential Equations](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/ch02.ipynb) 5 | * [Higher-Order Differential Equations](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/ch03.ipynb) 6 | * [The Laplace Transform](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/ch04.ipynb) 7 | * [Series Solutions of Linear Differential Equations](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/ch05.ipynb) 8 | * [Vectors](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/ch07.ipynb) 9 | * [Matrices](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/ch08.ipynb) 10 | * [Systems of Linear Differential Equations](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/ch10.ipynb) 11 | * [Systems of Nonlinear Differential Equations](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/ch11.ipynb) 12 | --- 13 | * [Vector Calculus](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/ch09.ipynb) 14 | * [Orthogonal Functions and Fourier Series](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/ch12.ipynb) 15 | * [PDE - Parabolic-Type Problems](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/PDE01.ipynb) 16 | * [PDE - Hyperbolic-Type Problems](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/PDE02.ipynb) 17 | * [PDE - Elliptic-Type Problems](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/PDE03.ipynb) 18 | --- 19 | * [Supplements](https://github.com/SeoulTechPSE/EngMath/blob/master/supplements/README.md) 20 | --- 21 | 22 | ## References 23 | 24 | * D.G. Zill and W.S. Wright, **Advanced Engineering Mathematics**, *7th ed.*, 2020. 25 | * S.J. Farlow, **Partial Differential Equations for Scientists and Engineers**, *Reprint ed.*, 1993. 26 | * H.F. Weinberger, **A First Course in Partial Differential Equations with Complex Variables and Transform Methods**, 1965. 27 | -------------------------------------------------------------------------------- /codes/.gitignore: -------------------------------------------------------------------------------- 1 | .ipynb_checkpoints/ -------------------------------------------------------------------------------- /codes/ch02_code1.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": {}, 7 | "outputs": [], 8 | "source": [ 9 | "%matplotlib inline\n", 10 | "\n", 11 | "import numpy as np\n", 12 | "import matplotlib.pyplot as plt\n", 13 | "plt.style.use('ggplot')\n", 14 | "\n", 15 | "nx, ny = 0.25, 0.25\n", 16 | "x = np.arange(-3, 3, nx)\n", 17 | "y = np.arange(-3, 3, ny)\n", 18 | "\n", 19 | "X, Y = np.meshgrid(x, y)\n", 20 | "\n", 21 | "dy = (X*X -Y*Y)\n", 22 | "dx = np.ones(dy.shape)\n", 23 | "\n", 24 | "plt.figure(figsize=(10, 10))\n", 25 | "plt.quiver(X, Y, dx, dy, color='red')\n", 26 | "plt.xticks(np.arange(-3, 3.5))\n", 27 | "plt.yticks(np.arange(-3, 3.5))\n", 28 | "plt.xlabel('x')\n", 29 | "plt.ylabel('y')\n", 30 | "plt.title(r'$\\frac{dy}{dx}=x^2-y^2$')\n", 31 | "plt.show()" 32 | ] 33 | } 34 | ], 35 | "metadata": { 36 | "kernelspec": { 37 | "display_name": "Python 3 (ipykernel)", 38 | "language": "python", 39 | "name": "python3" 40 | }, 41 | "language_info": { 42 | "codemirror_mode": { 43 | "name": "ipython", 44 | "version": 3 45 | }, 46 | "file_extension": ".py", 47 | "mimetype": "text/x-python", 48 | "name": "python", 49 | "nbconvert_exporter": "python", 50 | "pygments_lexer": "ipython3", 51 | "version": "3.9.13" 52 | } 53 | }, 54 | "nbformat": 4, 55 | "nbformat_minor": 4 56 | } 57 | -------------------------------------------------------------------------------- /codes/ch02_code2.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": {}, 7 | "outputs": [], 8 | "source": [ 9 | "%matplotlib inline\n", 10 | "\n", 11 | "import numpy as np\n", 12 | "import matplotlib.pyplot as plt\n", 13 | "plt.style.use('ggplot')\n", 14 | "\n", 15 | "x0 = 2.0\n", 16 | "y0 = 4.0\n", 17 | "xf = 2.5\n", 18 | "\n", 19 | "h = 0.1\n", 20 | "n = int((xf -x0)/h) +1\n", 21 | "\n", 22 | "x = np.linspace(x0, xf, n)\n", 23 | "y = np.zeros(n)\n", 24 | "\n", 25 | "y[0] = y0\n", 26 | "for i in range(1, n):\n", 27 | " y[i] = y[i -1] +h*(0.1 *np.sqrt(y[i -1]) +0.4 *x[i -1]**2)\n", 28 | " print('x = %3.1f, y = %5.4f' % (x[i], y[i]))\n", 29 | "\n", 30 | "plt.figure(figsize=(10, 10))\n", 31 | "plt.plot(x, y,'o')\n", 32 | "plt.xlabel(\"$x$\")\n", 33 | "plt.ylabel(\"$y$\")\n", 34 | "plt.title(\"Euler's method: $h=0.1$\")\n", 35 | "\n", 36 | "plt.show()" 37 | ] 38 | } 39 | ], 40 | "metadata": { 41 | "kernelspec": { 42 | "display_name": "Python 3 (ipykernel)", 43 | "language": "python", 44 | "name": "python3" 45 | }, 46 | "language_info": { 47 | "codemirror_mode": { 48 | "name": "ipython", 49 | "version": 3 50 | }, 51 | "file_extension": ".py", 52 | "mimetype": "text/x-python", 53 | "name": "python", 54 | "nbconvert_exporter": "python", 55 | "pygments_lexer": "ipython3", 56 | "version": "3.9.13" 57 | } 58 | }, 59 | "nbformat": 4, 60 | "nbformat_minor": 4 61 | } 62 | -------------------------------------------------------------------------------- /codes/ch03_code2.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": { 7 | "slideshow": { 8 | "slide_type": "slide" 9 | } 10 | }, 11 | "outputs": [], 12 | "source": [ 13 | "import numpy as np\n", 14 | "from scipy.integrate import solve_ivp\n", 15 | "import matplotlib.pyplot as plt\n", 16 | "plt.rcParams.update({'font.size': 16})\n", 17 | "\n", 18 | "def func(t, y):\n", 19 | " return [y[1], -y[0] -y[0]**3 -y[1]]\n", 20 | "\n", 21 | "tf = 14\n", 22 | "sol1 = solve_ivp(func, [0, tf], [-3, 4], t_eval = np.linspace(0, tf, 200))\n", 23 | "sol2 = solve_ivp(func, [0, tf], [0, -8], t_eval = np.linspace(0, tf, 200))" 24 | ] 25 | }, 26 | { 27 | "cell_type": "code", 28 | "execution_count": null, 29 | "metadata": { 30 | "slideshow": { 31 | "slide_type": "fragment" 32 | } 33 | }, 34 | "outputs": [], 35 | "source": [ 36 | "plt.figure(figsize=(12, 8))\n", 37 | "plt.plot(sol1.t, sol1.y[0], 'b-', label=r'$x_0=-3,\\; \\dot{x}_0=4$')\n", 38 | "plt.plot(sol2.t, sol2.y[0], 'r-', label=r'$x_0=0,\\; \\dot{x}_0=-8$')\n", 39 | "plt.axis((0, tf, -3, 3))\n", 40 | "plt.xlabel('t')\n", 41 | "plt.ylabel('x')\n", 42 | "plt.legend()\n", 43 | "plt.title(r'$\\frac{d^2x}{dt^2} +\\frac{dx}{dt} +x +x^3=0$')\n", 44 | "plt.show()" 45 | ] 46 | } 47 | ], 48 | "metadata": { 49 | "celltoolbar": "Slideshow", 50 | "kernelspec": { 51 | "display_name": "Python 3 (ipykernel)", 52 | "language": "python", 53 | "name": "python3" 54 | }, 55 | "language_info": { 56 | "codemirror_mode": { 57 | "name": "ipython", 58 | "version": 3 59 | }, 60 | "file_extension": ".py", 61 | "mimetype": "text/x-python", 62 | "name": "python", 63 | "nbconvert_exporter": "python", 64 | "pygments_lexer": "ipython3", 65 | "version": "3.9.13" 66 | } 67 | }, 68 | "nbformat": 4, 69 | "nbformat_minor": 4 70 | } 71 | -------------------------------------------------------------------------------- /codes/ch03_code3.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": { 7 | "slideshow": { 8 | "slide_type": "slide" 9 | } 10 | }, 11 | "outputs": [], 12 | "source": [ 13 | "import numpy as np\n", 14 | "from scipy.integrate import solve_ivp\n", 15 | "import matplotlib.pyplot as plt\n", 16 | "plt.rcParams.update({'font.size': 22})\n", 17 | "\n", 18 | "k1, k2, m1, m2 = 6, 4, 1, 1\n", 19 | "\n", 20 | "def func(t, y):\n", 21 | " return [y[1], -(k1 +k2)/m1 *y[0] +k2/m1 *y[2], y[3], k2/m2 *y[0] -k2/m2 *y[2]]\n", 22 | "\n", 23 | "tf = 14\n", 24 | "sol = solve_ivp(func, [0, tf], [0, 1, 0, -1], t_eval = np.linspace(0, tf, 200))" 25 | ] 26 | }, 27 | { 28 | "cell_type": "code", 29 | "execution_count": null, 30 | "metadata": { 31 | "slideshow": { 32 | "slide_type": "fragment" 33 | } 34 | }, 35 | "outputs": [], 36 | "source": [ 37 | "plt.figure(figsize=(12, 8))\n", 38 | "plt.plot(sol.t, sol.y[0], 'b-', label='$x_1$') \n", 39 | "plt.plot(sol.t, sol.y[2], 'r-', label='$x_2$')\n", 40 | "plt.axis((0, tf, -0.6, 0.6))\n", 41 | "plt.legend()\n", 42 | "plt.xlabel('t')\n", 43 | "plt.ylabel('x')\n", 44 | "plt.title('Coupled Spring/Mass Systems')\n", 45 | "plt.show()" 46 | ] 47 | } 48 | ], 49 | "metadata": { 50 | "celltoolbar": "Slideshow", 51 | "kernelspec": { 52 | "display_name": "Python 3 (ipykernel)", 53 | "language": "python", 54 | "name": "python3" 55 | }, 56 | "language_info": { 57 | "codemirror_mode": { 58 | "name": "ipython", 59 | "version": 3 60 | }, 61 | "file_extension": ".py", 62 | "mimetype": "text/x-python", 63 | "name": "python", 64 | "nbconvert_exporter": "python", 65 | "pygments_lexer": "ipython3", 66 | "version": "3.9.13" 67 | } 68 | }, 69 | "nbformat": 4, 70 | "nbformat_minor": 4 71 | } 72 | -------------------------------------------------------------------------------- /codes/ch05_code1.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": {}, 7 | "outputs": [], 8 | "source": [ 9 | "%matplotlib inline\n", 10 | "import matplotlib.pyplot as plt\n", 11 | "plt.style.use('ggplot')\n", 12 | "\n", 13 | "import numpy as np\n", 14 | "from scipy.special import jv, yv" 15 | ] 16 | }, 17 | { 18 | "cell_type": "code", 19 | "execution_count": null, 20 | "metadata": {}, 21 | "outputs": [], 22 | "source": [ 23 | "fig = plt.figure(figsize=(12, 12))\n", 24 | "\n", 25 | "ax1 = fig.add_subplot(211)\n", 26 | "\n", 27 | "x = np.linspace(0, 20, 200)\n", 28 | "for m in range(5):\n", 29 | " y = jv(m, x)\n", 30 | " ax1.plot(x, y, label=r'$J_%d(x)$' % m)\n", 31 | "\n", 32 | "ax1.axis((0, 20, -0.6, 1)) \n", 33 | "\n", 34 | "ax1.set_title(r'$J_m(x), m=0,1,2,3,4$')\n", 35 | "#ax1.set_xlabel('x')\n", 36 | "ax1.set_ylabel('y')\n", 37 | "ax1.legend()\n", 38 | "\n", 39 | "ax2 = fig.add_subplot(212)\n", 40 | "\n", 41 | "for m in range(5):\n", 42 | " y = yv(m, x)\n", 43 | " ax2.plot(x, y, label=r'$Y_%d(x)$' % m)\n", 44 | "\n", 45 | "ax2.axis((0, 20, -3, 1)) \n", 46 | "\n", 47 | "ax2.set_title(r'$Y_m(x), m=0,1,2,3,4$')\n", 48 | "ax2.set_xlabel('x')\n", 49 | "ax2.set_ylabel('y')\n", 50 | "ax2.legend()\n", 51 | "\n", 52 | "plt.show()" 53 | ] 54 | } 55 | ], 56 | "metadata": { 57 | "kernelspec": { 58 | "display_name": "Python 3 (ipykernel)", 59 | "language": "python", 60 | "name": "python3" 61 | }, 62 | "language_info": { 63 | "codemirror_mode": { 64 | "name": "ipython", 65 | "version": 3 66 | }, 67 | "file_extension": ".py", 68 | "mimetype": "text/x-python", 69 | "name": "python", 70 | "nbconvert_exporter": "python", 71 | "pygments_lexer": "ipython3", 72 | "version": "3.9.13" 73 | } 74 | }, 75 | "nbformat": 4, 76 | "nbformat_minor": 4 77 | } 78 | -------------------------------------------------------------------------------- /codes/ch05_code2.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": {}, 7 | "outputs": [], 8 | "source": [ 9 | "%matplotlib inline\n", 10 | "import matplotlib.pyplot as plt\n", 11 | "plt.style.use('ggplot')\n", 12 | "\n", 13 | "import numpy as np\n", 14 | "from scipy.special import jv, spherical_jn" 15 | ] 16 | }, 17 | { 18 | "cell_type": "code", 19 | "execution_count": null, 20 | "metadata": {}, 21 | "outputs": [], 22 | "source": [ 23 | "fig = plt.figure(figsize=(12, 6))\n", 24 | "ax = fig.add_subplot(111)\n", 25 | "\n", 26 | "x = np.linspace(np.finfo(np.float32).eps, 10, 200)\n", 27 | "for n in range(5):\n", 28 | " y = spherical_jn(n, x)\n", 29 | " #y = jv(n +0.5, x)\n", 30 | " ax.plot(np.append(0, x), np.append(0, y), label=r'$j_{%d}(x)$' % n)\n", 31 | "\n", 32 | "ax.axis((0, 10, -0.4, 1.0)) \n", 33 | "\n", 34 | "ax.set_title(r'$j_{n}(x), n=0,1,2,3,4$')\n", 35 | "ax.set_xlabel('x')\n", 36 | "ax.set_ylabel('y')\n", 37 | "ax.legend()\n", 38 | "\n", 39 | "plt.show()" 40 | ] 41 | } 42 | ], 43 | "metadata": { 44 | "kernelspec": { 45 | "display_name": "Python 3 (ipykernel)", 46 | "language": "python", 47 | "name": "python3" 48 | }, 49 | "language_info": { 50 | "codemirror_mode": { 51 | "name": "ipython", 52 | "version": 3 53 | }, 54 | "file_extension": ".py", 55 | "mimetype": "text/x-python", 56 | "name": "python", 57 | "nbconvert_exporter": "python", 58 | "pygments_lexer": "ipython3", 59 | "version": "3.9.13" 60 | } 61 | }, 62 | "nbformat": 4, 63 | "nbformat_minor": 4 64 | } 65 | -------------------------------------------------------------------------------- /codes/ch05_code3.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": {}, 7 | "outputs": [], 8 | "source": [ 9 | "%matplotlib inline\n", 10 | "import matplotlib.pyplot as plt\n", 11 | "plt.style.use('ggplot')\n", 12 | "\n", 13 | "import numpy as np\n", 14 | "from scipy.special import eval_legendre" 15 | ] 16 | }, 17 | { 18 | "cell_type": "code", 19 | "execution_count": null, 20 | "metadata": {}, 21 | "outputs": [], 22 | "source": [ 23 | "fig = plt.figure(figsize=(8, 8))\n", 24 | "ax = fig.add_subplot(111)\n", 25 | "\n", 26 | "x = np.linspace(-1, 1, 100)\n", 27 | "for n in range(6):\n", 28 | " y = eval_legendre(n, x)\n", 29 | " ax.plot(x, y, label=r'$P_{%d}(x)$' % n)\n", 30 | "\n", 31 | "ax.axis((-1.05, 1.05, -1.05, 1.05)) \n", 32 | "\n", 33 | "ax.set_title(r'$P_n(x), n=0,1,\\cdots,5$')\n", 34 | "ax.set_xlabel('x')\n", 35 | "ax.set_ylabel('y')\n", 36 | "ax.legend(loc='lower left', bbox_to_anchor=(1, 0))\n", 37 | "\n", 38 | "plt.show()" 39 | ] 40 | } 41 | ], 42 | "metadata": { 43 | "kernelspec": { 44 | "display_name": "Python 3 (ipykernel)", 45 | "language": "python", 46 | "name": "python3" 47 | }, 48 | "language_info": { 49 | "codemirror_mode": { 50 | "name": "ipython", 51 | "version": 3 52 | }, 53 | "file_extension": ".py", 54 | "mimetype": "text/x-python", 55 | "name": "python", 56 | "nbconvert_exporter": "python", 57 | "pygments_lexer": "ipython3", 58 | "version": "3.9.13" 59 | } 60 | }, 61 | "nbformat": 4, 62 | "nbformat_minor": 4 63 | } 64 | -------------------------------------------------------------------------------- /codes/ch08_code1.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": {}, 7 | "outputs": [], 8 | "source": [ 9 | "import pprint\n", 10 | "import numpy as np\n", 11 | "\n", 12 | "def power_method(A, n_iteration):\n", 13 | " # Ideally choose a random vector\n", 14 | " # To decrease the chance that our vector is orthogonal to the eigenvector\n", 15 | " k = np.random.rand(A.shape[1])\n", 16 | "\n", 17 | " for _ in range(n_iteration):\n", 18 | " # calculate the matrix-by-vector product Ak\n", 19 | " k_1 = np.dot(A, k)\n", 20 | "\n", 21 | " # calculate the norm\n", 22 | " k_1_norm = np.linalg.norm(k_1)\n", 23 | "\n", 24 | " # re normalize the vector\n", 25 | " k = k_1 /k_1_norm\n", 26 | "\n", 27 | " return k\n", 28 | "\n", 29 | "def Rayleigh_quotient(A, k):\n", 30 | " # calculate Rayleigh quotient\n", 31 | " lambda_ = np.dot(np.dot(A, k), k) /np.dot(k, k)\n", 32 | " \n", 33 | " return lambda_\n", 34 | "\n", 35 | "A = np.array([[4, 2], [3, -1]])\n", 36 | "k_1 = power_method(A, 10)\n", 37 | "lambda_1 = Rayleigh_quotient(A, k_1)\n", 38 | "\n", 39 | "print('A =')\n", 40 | "pprint.pprint(A)\n", 41 | "print('\\nDominant Eigenvector:', np.around(k_1, 3))\n", 42 | "print('Dominant Eigenvalue:', np.around(lambda_1, 3))" 43 | ] 44 | } 45 | ], 46 | "metadata": { 47 | "kernelspec": { 48 | "display_name": "Python 3", 49 | "language": "python", 50 | "name": "python3" 51 | }, 52 | "language_info": { 53 | "codemirror_mode": { 54 | "name": "ipython", 55 | "version": 3 56 | }, 57 | "file_extension": ".py", 58 | "mimetype": "text/x-python", 59 | "name": "python", 60 | "nbconvert_exporter": "python", 61 | "pygments_lexer": "ipython3", 62 | "version": "3.7.5" 63 | } 64 | }, 65 | "nbformat": 4, 66 | "nbformat_minor": 4 67 | } 68 | -------------------------------------------------------------------------------- /codes/ch08_code2.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": {}, 7 | "outputs": [], 8 | "source": [ 9 | "import pprint\n", 10 | "import numpy as np\n", 11 | "\n", 12 | "def power_method(A, n_iteration):\n", 13 | " # Ideally choose a random vector\n", 14 | " # To decrease the chance that our vector is orthogonal to the eigenvector\n", 15 | " k = np.random.rand(A.shape[1])\n", 16 | "\n", 17 | " for _ in range(n_iteration):\n", 18 | " # calculate the matrix-by-vector product Ak\n", 19 | " k_1 = np.dot(A, k)\n", 20 | "\n", 21 | " # calculate the norm\n", 22 | " k_1_norm = np.linalg.norm(k_1)\n", 23 | "\n", 24 | " # re normalize the vector\n", 25 | " k = k_1 /k_1_norm\n", 26 | "\n", 27 | " return k\n", 28 | "\n", 29 | "def Rayleigh_quotient(A, k):\n", 30 | " # calculate Rayleigh quotient\n", 31 | " lambda_ = np.dot(np.dot(A, k), k) /np.dot(k, k)\n", 32 | " \n", 33 | " return lambda_" 34 | ] 35 | }, 36 | { 37 | "cell_type": "code", 38 | "execution_count": null, 39 | "metadata": {}, 40 | "outputs": [], 41 | "source": [ 42 | "def deflation_method(A, n_iteration):\n", 43 | " \n", 44 | " n = A.shape[1]\n", 45 | "\n", 46 | " K = np.zeros((n, n))\n", 47 | " Lambda = np.zeros(n)\n", 48 | " \n", 49 | " k0 = np.zeros(n)\n", 50 | " L0 = 0\n", 51 | " for i in range(n):\n", 52 | " A = A -L0 *np.outer(k0, k0)\n", 53 | " k0 = power_method(A, n_iteration)\n", 54 | " L0 = Rayleigh_quotient(A, k0)\n", 55 | " K[i, :] = k0\n", 56 | " Lambda[i] = L0\n", 57 | " \n", 58 | " return K, Lambda \n", 59 | "\n", 60 | "# symmetric matrix\n", 61 | "A = np.array([[1, 2, -1], [2, 1, 1], [-1, 1, 0]])\n", 62 | "\n", 63 | "K, Lambda = deflation_method(A, 20) \n", 64 | "\n", 65 | "print('A =')\n", 66 | "pprint.pprint(A)\n", 67 | " \n", 68 | "print('\\nk_1 =', np.around(K[0, :], 3))\n", 69 | "print('k_2 =', np.around(K[1, :], 3))\n", 70 | "print('k_3 =', np.around(K[2, :], 3))\n", 71 | "\n", 72 | "print('\\nlambda_1 =', np.around(Lambda[0], 3))\n", 73 | "print('lambda_2 =', np.around(Lambda[1], 3))\n", 74 | "print('lambda_3 =', np.around(Lambda[2], 3))" 75 | ] 76 | } 77 | ], 78 | "metadata": { 79 | "kernelspec": { 80 | "display_name": "Python 3", 81 | "language": "python", 82 | "name": "python3" 83 | }, 84 | "language_info": { 85 | "codemirror_mode": { 86 | "name": "ipython", 87 | "version": 3 88 | }, 89 | "file_extension": ".py", 90 | "mimetype": "text/x-python", 91 | "name": "python", 92 | "nbconvert_exporter": "python", 93 | "pygments_lexer": "ipython3", 94 | "version": "3.7.5" 95 | } 96 | }, 97 | "nbformat": 4, 98 | "nbformat_minor": 4 99 | } 100 | -------------------------------------------------------------------------------- /codes/ch08_code3.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": {}, 7 | "outputs": [], 8 | "source": [ 9 | "import pprint\n", 10 | "import numpy as np\n", 11 | "\n", 12 | "def inverse_power_method(A, n_iteration):\n", 13 | " # Ideally choose a random vector\n", 14 | " # To decrease the chance that our vector is orthogonal to the eigenvector\n", 15 | " k = np.random.rand(A.shape[1])\n", 16 | "\n", 17 | " for _ in range(n_iteration):\n", 18 | " # calculate the matrix-by-vector product Ak\n", 19 | " w = np.linalg.solve(A, k)\n", 20 | "\n", 21 | " # calculate the norm\n", 22 | " w_norm = np.linalg.norm(w)\n", 23 | "\n", 24 | " # re normalize the vector\n", 25 | " k = w /w_norm\n", 26 | "\n", 27 | " return k\n", 28 | "\n", 29 | "def Rayleigh_quotient(A, k):\n", 30 | " # calculate Rayleigh quotient\n", 31 | " lambda_ = np.dot(np.dot(A, k), k) /np.dot(k, k)\n", 32 | " \n", 33 | " return lambda_\n", 34 | "\n", 35 | "A = np.array([[4, 2], [3, -1]])\n", 36 | "k_2 = inverse_power_method(A, 10)\n", 37 | "lambda_2 = Rayleigh_quotient(A, k_2)\n", 38 | "\n", 39 | "print('A =')\n", 40 | "pprint.pprint(A)\n", 41 | "print('\\nEigenvector =', np.around(k_2, 3))\n", 42 | "print('Eigenvalue of Least Magnitude =', np.around(lambda_2, 3))" 43 | ] 44 | } 45 | ], 46 | "metadata": { 47 | "kernelspec": { 48 | "display_name": "Python 3", 49 | "language": "python", 50 | "name": "python3" 51 | }, 52 | "language_info": { 53 | "codemirror_mode": { 54 | "name": "ipython", 55 | "version": 3 56 | }, 57 | "file_extension": ".py", 58 | "mimetype": "text/x-python", 59 | "name": "python", 60 | "nbconvert_exporter": "python", 61 | "pygments_lexer": "ipython3", 62 | "version": "3.7.5" 63 | } 64 | }, 65 | "nbformat": 4, 66 | "nbformat_minor": 4 67 | } 68 | -------------------------------------------------------------------------------- /codes/ch08_code4.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": {}, 7 | "outputs": [], 8 | "source": [ 9 | "import pprint\n", 10 | "import numpy as np\n", 11 | "\n", 12 | "def lu_factor(A):\n", 13 | " \"\"\"\n", 14 | " LU factorization with partial pivoting\n", 15 | " \n", 16 | " PA = LU \n", 17 | " P(permutation), L(unit Lower triangular) and U(upper triangular) \n", 18 | " \n", 19 | " Return P, L, U\n", 20 | " \"\"\"\n", 21 | " n = A.shape[0] \n", 22 | " U = A.copy()\n", 23 | " P = np.identity(n)\n", 24 | " L = np.identity(n)\n", 25 | "\n", 26 | " for k in range(n -1):\n", 27 | "\n", 28 | " # Partial Pivoting\n", 29 | " max_row_index = np.argmax(abs(U[k:n,k])) +k\n", 30 | " P[[k,max_row_index]] = P[[max_row_index,k]]\n", 31 | " U[[k,max_row_index]] = U[[max_row_index,k]]\n", 32 | "\n", 33 | " # LU\n", 34 | " L[k+1:,k] = U[k+1:,k] /U[k,k]\n", 35 | " U[k+1:,k] = 0.0\n", 36 | " U[k+1:,k+1:] -= np.tensordot(L[k+1:,k], U[k,k+1:], axes=0)\n", 37 | "\n", 38 | " return P, L, U\n", 39 | "\n", 40 | "A = np.array([[7, 3, -1, 2], [3, 8, 1, -4], [-1, 1, 4, -1], [2, -4, -1, 6]], dtype='float64')\n", 41 | "P, L, U = lu_factor(A)\n", 42 | "\n", 43 | "print('A ='); pprint.pprint(A)\n", 44 | "print('\\nP ='); pprint.pprint(P)\n", 45 | "print('\\nL ='); pprint.pprint(L)\n", 46 | "print('\\nU ='); pprint.pprint(U)" 47 | ] 48 | } 49 | ], 50 | "metadata": { 51 | "kernelspec": { 52 | "display_name": "Python 3", 53 | "language": "python", 54 | "name": "python3" 55 | }, 56 | "language_info": { 57 | "codemirror_mode": { 58 | "name": "ipython", 59 | "version": 3 60 | }, 61 | "file_extension": ".py", 62 | "mimetype": "text/x-python", 63 | "name": "python", 64 | "nbconvert_exporter": "python", 65 | "pygments_lexer": "ipython3", 66 | "version": "3.7.5" 67 | } 68 | }, 69 | "nbformat": 4, 70 | "nbformat_minor": 4 71 | } 72 | -------------------------------------------------------------------------------- /codes/ch08_code5.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": 8, 6 | "metadata": {}, 7 | "outputs": [ 8 | { 9 | "name": "stdout", 10 | "output_type": "stream", 11 | "text": [ 12 | "A =\n", 13 | "array([[ 7., 3., -1., 2.],\n", 14 | " [ 3., 8., 1., -4.],\n", 15 | " [-1., 1., 4., -1.],\n", 16 | " [ 2., -4., -1., 6.]])\n", 17 | "\n", 18 | "b = array([1., 2., 3., 4.])\n", 19 | "\n", 20 | "x = array([-1.27619048, 1.87619048, 0.57142857, 2.43809524])\n" 21 | ] 22 | } 23 | ], 24 | "source": [ 25 | "import pprint\n", 26 | "import numpy as np\n", 27 | "\n", 28 | "def lu_factor(A):\n", 29 | " \"\"\"\n", 30 | " LU factorization with partial pivoting\n", 31 | " \n", 32 | " PA = LU \n", 33 | " P(permutation), L(unit Lower triangular) and U(upper triangular) \n", 34 | " \n", 35 | " Return P, L, U\n", 36 | " \"\"\"\n", 37 | " n = A.shape[0] \n", 38 | " U = A.copy()\n", 39 | " P = np.identity(n)\n", 40 | " L = np.identity(n)\n", 41 | "\n", 42 | " for k in range(n -1):\n", 43 | "\n", 44 | " # Partial Pivoting\n", 45 | " max_row_index = np.argmax(abs(U[k:n,k])) +k\n", 46 | " P[[k,max_row_index]] = P[[max_row_index,k]]\n", 47 | " U[[k,max_row_index]] = U[[max_row_index,k]]\n", 48 | "\n", 49 | " # LU\n", 50 | " L[k+1:,k] = U[k+1:,k] /U[k,k]\n", 51 | " U[k+1:,k] = 0.0\n", 52 | " U[k+1:,k+1:] -= np.tensordot(L[k+1:,k], U[k,k+1:], axes=0)\n", 53 | "\n", 54 | " return P, L, U\n", 55 | "\n", 56 | "def ufsub(L, b):\n", 57 | " \"\"\" Unit row oriented forward substitution \"\"\"\n", 58 | " y = b.copy()\n", 59 | " for i in range(1, L.shape[0]):\n", 60 | " y[i] -= np.dot(L[i,:i], y[:i])\n", 61 | " return y\n", 62 | "\n", 63 | "def bsub(U, y):\n", 64 | " \"\"\" Row oriented backward substitution \"\"\"\n", 65 | " x = y.copy()\n", 66 | " x[-1] /= U[-1,-1]\n", 67 | " for i in range(U.shape[0] -2, -1, -1): \n", 68 | " x[i] -= np.dot(U[i,i+1:], x[i+1:])\n", 69 | " x[i] /= U[i,i]\n", 70 | " return x \n", 71 | " \n", 72 | "A = np.array([[7, 3, -1, 2], [3, 8, 1, -4], [-1, 1, 4, -1], [2, -4, -1, 6]], dtype='float64')\n", 73 | "b = np.array([1, 2, 3, 4], dtype='float64')\n", 74 | "\n", 75 | "P, L, U = lu_factor(A)\n", 76 | "Pb = np.matmul(P, b)\n", 77 | "\n", 78 | "y = ufsub(L, Pb)\n", 79 | "x = bsub(U, y)\n", 80 | "\n", 81 | "print('A ='); pprint.pprint(A)\n", 82 | "print('\\nb =', end=' '); pprint.pprint(b)\n", 83 | "print('\\nx =', end=' '); pprint.pprint(x)" 84 | ] 85 | } 86 | ], 87 | "metadata": { 88 | "kernelspec": { 89 | "display_name": "Python 3", 90 | "language": "python", 91 | "name": "python3" 92 | }, 93 | "language_info": { 94 | "codemirror_mode": { 95 | "name": "ipython", 96 | "version": 3 97 | }, 98 | "file_extension": ".py", 99 | "mimetype": "text/x-python", 100 | "name": "python", 101 | "nbconvert_exporter": "python", 102 | "pygments_lexer": "ipython3", 103 | "version": "3.7.5" 104 | } 105 | }, 106 | "nbformat": 4, 107 | "nbformat_minor": 4 108 | } 109 | -------------------------------------------------------------------------------- /codes/ch08_code6.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": {}, 7 | "outputs": [], 8 | "source": [ 9 | "%matplotlib inline\n", 10 | "import matplotlib.pyplot as plt\n", 11 | "import numpy as np\n", 12 | "from scipy.optimize import curve_fit\n", 13 | "\n", 14 | "np.random.seed(1970)\n", 15 | "\n", 16 | "def func(x, a, b):\n", 17 | " return a *np.exp(-b*x)\n", 18 | "\n", 19 | "# Define the data to be fit with some noise\n", 20 | "\n", 21 | "xdata = np.linspace(0, 4, 50)\n", 22 | "ydata = func(xdata, 2.5, 1.3) +0.1 *np.random.normal(size=xdata.size)\n", 23 | "\n", 24 | "plt.plot(xdata, ydata, 'ro', label='data')\n", 25 | "\n", 26 | "# Fit for the parameters a and b of the function func:\n", 27 | "\n", 28 | "popt, pcov = curve_fit(func, xdata, ydata)\n", 29 | "\n", 30 | "plt.plot(xdata, func(xdata, *popt), 'b-',\n", 31 | " label='fit: a=%5.3f, b=%5.3f' % tuple(popt))\n", 32 | "\n", 33 | "plt.xlabel('x')\n", 34 | "plt.ylabel('y')\n", 35 | "plt.legend()\n", 36 | "\n", 37 | "plt.show()" 38 | ] 39 | } 40 | ], 41 | "metadata": { 42 | "kernelspec": { 43 | "display_name": "Python 3", 44 | "language": "python", 45 | "name": "python3" 46 | }, 47 | "language_info": { 48 | "codemirror_mode": { 49 | "name": "ipython", 50 | "version": 3 51 | }, 52 | "file_extension": ".py", 53 | "mimetype": "text/x-python", 54 | "name": "python", 55 | "nbconvert_exporter": "python", 56 | "pygments_lexer": "ipython3", 57 | "version": "3.7.5" 58 | } 59 | }, 60 | "nbformat": 4, 61 | "nbformat_minor": 4 62 | } 63 | -------------------------------------------------------------------------------- /codes/phase_portrait.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": 1, 6 | "metadata": {}, 7 | "outputs": [], 8 | "source": [ 9 | "%matplotlib inline\n", 10 | "import numpy as np\n", 11 | "import matplotlib.pyplot as plt" 12 | ] 13 | }, 14 | { 15 | "cell_type": "code", 16 | "execution_count": null, 17 | "metadata": {}, 18 | "outputs": [], 19 | "source": [ 20 | "def phase_portrait(width=6):\n", 21 | " \n", 22 | " xl = np.arange(-width, width, 0.1)\n", 23 | " yl = np.arange(-wisth, width, 0.1)\n", 24 | " x, y = np.meshgrid(xl, yl)" 25 | ] 26 | } 27 | ], 28 | "metadata": { 29 | "kernelspec": { 30 | "display_name": "Python 3", 31 | "language": "python", 32 | "name": "python3" 33 | }, 34 | "language_info": { 35 | "codemirror_mode": { 36 | "name": "ipython", 37 | "version": 3 38 | }, 39 | "file_extension": ".py", 40 | "mimetype": "text/x-python", 41 | "name": "python", 42 | "nbconvert_exporter": "python", 43 | "pygments_lexer": "ipython3", 44 | "version": "3.7.5" 45 | } 46 | }, 47 | "nbformat": 4, 48 | "nbformat_minor": 4 49 | } 50 | -------------------------------------------------------------------------------- /exams/.gitignore: -------------------------------------------------------------------------------- 1 | .ipynb_checkpoints/ -------------------------------------------------------------------------------- /exams/Quiz_01.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/Quiz_01.ipynb)" 8 | ] 9 | }, 10 | { 11 | "cell_type": "markdown", 12 | "metadata": {}, 13 | "source": [ 14 | "## Quiz " 15 | ] 16 | }, 17 | { 18 | "cell_type": "markdown", 19 | "metadata": {}, 20 | "source": [ 21 | "Find the solution of the initial-boundary value problem:\n", 22 | "\n", 23 | ">$\\begin{align*}\n", 24 | " &u_t = u_{xx} +f(x,t), \\;\\; 0 \\leq x \\leq L, \\;\\; 0 \\leq t \\\\ \n", 25 | " &u(0,t) = 0\\\\ \n", 26 | " &u(L,t) = 0\\\\ \n", 27 | " &u(x,0) = \\phi(x)\n", 28 | "\\end{align*}$" 29 | ] 30 | }, 31 | { 32 | "cell_type": "markdown", 33 | "metadata": {}, 34 | "source": [ 35 | "**Solution**" 36 | ] 37 | }, 38 | { 39 | "cell_type": "markdown", 40 | "metadata": {}, 41 | "source": [ 42 | ">$\\begin{align*}\n", 43 | " &\\;\\;\\;\\;\\Downarrow {\\scriptstyle \\text{Superposition Principle}} \\\\\n", 44 | " &v_t = v_{xx}, \\;\\; 0 \\leq x \\leq L, \\;\\; 0 \\leq t \\\\ \n", 45 | " &v(0,t) = 0\\\\ \n", 46 | " &v(L,t) = 0\\\\ \n", 47 | " &v(x,0) = \\phi(x) \\\\\n", 48 | " &\\;\\;\\;\\;+ \\\\\n", 49 | " &w_t = w_{xx} +f(x,t), \\;\\; 0 \\leq x \\leq L, \\;\\; 0 \\leq t \\\\ \n", 50 | " &w(0,t) = 0\\\\ \n", 51 | " &w(L,t) = 0\\\\ \n", 52 | " &w(x,0) = 0\\\\\n", 53 | " &\\;\\;\\;\\;\\Downarrow\\\\\n", 54 | " &u(x,t) = v(x,t) +w(x,t)\\\\ \n", 55 | "\\end{align*}$" 56 | ] 57 | }, 58 | { 59 | "cell_type": "markdown", 60 | "metadata": {}, 61 | "source": [ 62 | "where\n", 63 | "\n", 64 | ">$\\begin{align*}\n", 65 | " v(x,t) &= \\sum_{n=1}^\\infty a_n \\exp\\left[ -\\left(\\frac{n\\pi}{L}\\right)^2t\\right]\\sin\\frac{n\\pi x}{L}, \\;\\;a_n = \\frac{2}{L} \\int_0^L \\phi(x) \\sin \\frac{n\\pi x}{L}\\,dx \\\\\n", 66 | " w(x,t) &= \\sum_{n=1}^\\infty W_n(t) \\sin\\frac{n\\pi x}{L}, \\;\\; \n", 67 | " W_n(t) =\\int_0^t \\exp \\left[-\\left( \\frac{n\\pi}{L} \\right)^2 (t -\\tau) \\right] f_n(\\tau) \\,d\\tau, \\;\\;\n", 68 | " f_n(t) = \\frac{2}{L} \\int_0^L f(x, t) \\sin \\frac{n\\pi x}{L}\\,dx\n", 69 | "\\end{align*}$" 70 | ] 71 | } 72 | ], 73 | "metadata": { 74 | "kernelspec": { 75 | "display_name": "Python 3", 76 | "language": "python", 77 | "name": "python3" 78 | }, 79 | "language_info": { 80 | "codemirror_mode": { 81 | "name": "ipython", 82 | "version": 3 83 | }, 84 | "file_extension": ".py", 85 | "mimetype": "text/x-python", 86 | "name": "python", 87 | "nbconvert_exporter": "python", 88 | "pygments_lexer": "ipython3", 89 | "version": "3.7.5" 90 | } 91 | }, 92 | "nbformat": 4, 93 | "nbformat_minor": 4 94 | } 95 | -------------------------------------------------------------------------------- /exams/Quiz_02_ch01_02.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": { 6 | "slideshow": { 7 | "slide_type": "skip" 8 | } 9 | }, 10 | "source": [ 11 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/Quiz_02_ch01_02.ipynb)" 12 | ] 13 | }, 14 | { 15 | "cell_type": "markdown", 16 | "metadata": { 17 | "slideshow": { 18 | "slide_type": "slide" 19 | } 20 | }, 21 | "source": [ 22 | "## Quiz - March 31, 2020" 23 | ] 24 | }, 25 | { 26 | "cell_type": "markdown", 27 | "metadata": { 28 | "slideshow": { 29 | "slide_type": "slide" 30 | } 31 | }, 32 | "source": [ 33 | "**1.** $\\text{ }$ $y= -x \\ln |x| +cx\\;$ is a one-parameter family of solutions to the DE $x \\frac{dy}{dx}=y-x$. Which of the following is the largest interval of existence for a solution to the IVP $y(2)=-7$" 34 | ] 35 | }, 36 | { 37 | "cell_type": "markdown", 38 | "metadata": { 39 | "slideshow": { 40 | "slide_type": "fragment" 41 | } 42 | }, 43 | "source": [ 44 | "**Solution**\n", 45 | "\n", 46 | "\\begin{align*}\n", 47 | " &\\;\\;\\;\\;\\Downarrow {\\scriptstyle \\text{Standard form, } x\\,>\\,0 \\text{ or } x\\,<\\,0} \\\\\n", 48 | " &\\frac{dy}{dx} -\\frac{1}{x} y=-1 \\\\ \n", 49 | " &\\;\\;\\;\\;\\Downarrow {\\scriptstyle \\text{Initial condition, } y(2)=-7} \\\\\n", 50 | " c &= \\ln 2 -\\frac{7}{2}, \\;\\; x\\,>\\, 0\n", 51 | "\\end{align*}" 52 | ] 53 | }, 54 | { 55 | "cell_type": "markdown", 56 | "metadata": { 57 | "slideshow": { 58 | "slide_type": "slide" 59 | } 60 | }, 61 | "source": [ 62 | "**2.** $\\text{ }$ Classify the following differential equation by order and linearity.\n", 63 | "\n", 64 | "$$\\ddot{x} -\\left(1 -(\\dot{x})^3 \\right)\\dot{x} +x = 2$$" 65 | ] 66 | }, 67 | { 68 | "cell_type": "markdown", 69 | "metadata": { 70 | "slideshow": { 71 | "slide_type": "fragment" 72 | } 73 | }, 74 | "source": [ 75 | "**Solution** $\\text{ }$ Second-order nonlinear" 76 | ] 77 | }, 78 | { 79 | "cell_type": "markdown", 80 | "metadata": { 81 | "slideshow": { 82 | "slide_type": "slide" 83 | } 84 | }, 85 | "source": [ 86 | "**3.** $\\text{ }$ A solution to an initial value problem is always defined on $(-\\infty, \\infty)$" 87 | ] 88 | }, 89 | { 90 | "cell_type": "markdown", 91 | "metadata": { 92 | "slideshow": { 93 | "slide_type": "fragment" 94 | } 95 | }, 96 | "source": [ 97 | "**Solution** $\\text{ }$ False" 98 | ] 99 | }, 100 | { 101 | "cell_type": "markdown", 102 | "metadata": { 103 | "slideshow": { 104 | "slide_type": "slide" 105 | } 106 | }, 107 | "source": [ 108 | "**4.** $\\text{ }$ Which of the following is an example of a separable differential equation?\n", 109 | "\n", 110 | "1. $\\displaystyle x \\tan^{-1} y \\frac{dy}{dx} = e^{x+y}$ \n", 111 | "2. $y'=\\sin xy$\n", 112 | "3. $\\displaystyle (1 +x^2) \\frac{dy}{dx}=\\sqrt{x+y}$\n", 113 | "4. $\\displaystyle y'=\\frac{1}{x +y}$" 114 | ] 115 | }, 116 | { 117 | "cell_type": "markdown", 118 | "metadata": { 119 | "slideshow": { 120 | "slide_type": "fragment" 121 | } 122 | }, 123 | "source": [ 124 | "**Solution** $\\text{ }$ 1" 125 | ] 126 | }, 127 | { 128 | "cell_type": "markdown", 129 | "metadata": { 130 | "slideshow": { 131 | "slide_type": "slide" 132 | } 133 | }, 134 | "source": [ 135 | "**5.** $\\text{ }$ The general solution of the differential equation $5\\frac{dy}{dx}+30y=6\\;$ is" 136 | ] 137 | }, 138 | { 139 | "cell_type": "markdown", 140 | "metadata": { 141 | "slideshow": { 142 | "slide_type": "fragment" 143 | } 144 | }, 145 | "source": [ 146 | "**Solution**\n", 147 | "\n", 148 | "\\begin{align*}\n", 149 | " &\\;\\;\\;\\;\\Downarrow {\\scriptstyle \\text{Standard form}} \\\\\n", 150 | " & \\frac{dy}{dx}+6y=\\frac{6}{5} \\\\\n", 151 | " &\\;\\;\\;\\;\\Downarrow {\\scriptstyle \\text{Mutiply by the integral factor, } e^{6x} } \\\\\n", 152 | " &\\frac{d}{dx}\\left(e^{6x}y \\right)=\\frac{6}{5}e^{6x} \\\\ \n", 153 | " &\\;\\;\\;\\;\\Downarrow \\\\\n", 154 | " y &= c e^{-6x} +\\frac{1}{5}\n", 155 | "\\end{align*}" 156 | ] 157 | }, 158 | { 159 | "cell_type": "markdown", 160 | "metadata": { 161 | "slideshow": { 162 | "slide_type": "slide" 163 | } 164 | }, 165 | "source": [ 166 | "**6.** Which of the following is a homogeneous differential equation?\n", 167 | "\n", 168 | "1. $\\displaystyle \\frac{dy}{dx} = \\frac{2x +7y}{7x +2y}$\n", 169 | "1. $(y^2 +x)\\, dx = (x +y^2)\\, dy$\n", 170 | "1. $\\displaystyle \\frac{dy}{dx} +x^2 y = xy^2$\n", 171 | "1. $(x^2 +y^2)\\,dy -x^4 \\,dx=0$" 172 | ] 173 | }, 174 | { 175 | "cell_type": "markdown", 176 | "metadata": { 177 | "slideshow": { 178 | "slide_type": "fragment" 179 | } 180 | }, 181 | "source": [ 182 | "**Solution** $\\text{ }$ 1" 183 | ] 184 | }, 185 | { 186 | "cell_type": "markdown", 187 | "metadata": { 188 | "slideshow": { 189 | "slide_type": "slide" 190 | } 191 | }, 192 | "source": [ 193 | "**7.** Which of the following differential equations is a Bernoulli equation?\n", 194 | "\n", 195 | "1. $3\\, dy =\\sin x \\,(y^2 -y) \\,dx$\n", 196 | "1. $(x^2 +y^2) \\,dy -2xy \\,dx = 0$\n", 197 | "1. $e^y\\, dx = 5 -\\sin y \\,dy$\n", 198 | "1. $\\displaystyle\\frac{dy}{dx}=\\ln y$" 199 | ] 200 | }, 201 | { 202 | "cell_type": "markdown", 203 | "metadata": { 204 | "slideshow": { 205 | "slide_type": "fragment" 206 | } 207 | }, 208 | "source": [ 209 | "**Solution** $\\text{ }$ 1" 210 | ] 211 | }, 212 | { 213 | "cell_type": "markdown", 214 | "metadata": { 215 | "slideshow": { 216 | "slide_type": "slide" 217 | } 218 | }, 219 | "source": [ 220 | "**8.** Find the critical point(s) of the differential equation\n", 221 | "\n", 222 | "$$\\frac{dV}{dt}=V\\sqrt{c + mV}$$\n", 223 | "\n", 224 | "where $c$ and $m$ are nonzero constants." 225 | ] 226 | }, 227 | { 228 | "cell_type": "markdown", 229 | "metadata": { 230 | "slideshow": { 231 | "slide_type": "fragment" 232 | } 233 | }, 234 | "source": [ 235 | "**Solution** $\\text{ }$ $\\displaystyle 0, \\;-\\frac{c}{m}$" 236 | ] 237 | }, 238 | { 239 | "cell_type": "markdown", 240 | "metadata": { 241 | "slideshow": { 242 | "slide_type": "slide" 243 | } 244 | }, 245 | "source": [ 246 | "**9.** Rewrite the linear differential equation $\\displaystyle \\cos x \\,dy +y\\,\\sin x \\,dx = dx$ in the standard form of a linear differential equation" 247 | ] 248 | }, 249 | { 250 | "cell_type": "markdown", 251 | "metadata": { 252 | "slideshow": { 253 | "slide_type": "fragment" 254 | } 255 | }, 256 | "source": [ 257 | "**Solution** $\\text{ }$ $\\displaystyle \\frac{dy}{dx} +\\tan x\\, y = \\sec x$" 258 | ] 259 | }, 260 | { 261 | "cell_type": "markdown", 262 | "metadata": { 263 | "slideshow": { 264 | "slide_type": "slide" 265 | } 266 | }, 267 | "source": [ 268 | "**10.** Find the value of $k$ so that the differential equation\n", 269 | "\n", 270 | "$$\\left(8yt^5 +kt^2 e^{ty}\\right) \\,dy = \\left(\\frac{2}{t} -20 y^2 t^4 -ktye^{ty} -5e^{ty}\\right) \\,dt$$\n", 271 | "\n", 272 | "is exact." 273 | ] 274 | }, 275 | { 276 | "cell_type": "markdown", 277 | "metadata": { 278 | "slideshow": { 279 | "slide_type": "fragment" 280 | } 281 | }, 282 | "source": [ 283 | "**Solution** $\\text{ }$ $k=5$" 284 | ] 285 | } 286 | ], 287 | "metadata": { 288 | "celltoolbar": "Slideshow", 289 | "kernelspec": { 290 | "display_name": "Python 3", 291 | "language": "python", 292 | "name": "python3" 293 | }, 294 | "language_info": { 295 | "codemirror_mode": { 296 | "name": "ipython", 297 | "version": 3 298 | }, 299 | "file_extension": ".py", 300 | "mimetype": "text/x-python", 301 | "name": "python", 302 | "nbconvert_exporter": "python", 303 | "pygments_lexer": "ipython3", 304 | "version": "3.7.5" 305 | } 306 | }, 307 | "nbformat": 4, 308 | "nbformat_minor": 4 309 | } 310 | -------------------------------------------------------------------------------- /exams/README.md: -------------------------------------------------------------------------------- 1 | ## EngMath Quizs. 2 | 3 | * [2019_2_Quiz](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/Quiz_01.ipynb) 4 | 5 | ## EngMath Exams. 6 | 7 | * [2019_1_중간고사](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2019_1.ipynb) 8 | * [2020_1_중간고사](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2020_1.ipynb) 9 | * [2022_1_중간고사 A](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2022_1_A_solution.ipynb) 10 | * [2022_1_중간고사 B](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2022_1_B_solution.ipynb) 11 | * [2023_1_중간고사 A](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2023_1_A_solution.ipynb) 12 | * [2023_1_중간고사 B](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2023_1_B_solution.ipynb) 13 | --- 14 | * [2020_1_기말고사](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/final_2020_1.ipynb) 15 | * [2022_1_기말고사 A](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/final_2022_1_A_solution.ipynb) 16 | * [2022_1_기말고사 B](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/final_2022_1_B_solution.ipynb) 17 | * [2023_1_기말고사 A](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/final_2023_1_A_solution.ipynb) 18 | * [2023_1_기말고사 B](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/final_2023_1_B_solution.ipynb) 19 | * [2023_1_기말고사 5번 풀이 상세](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/final_2023_1_solution_5.ipynb) 20 | --- 21 | * [2019_2_중간고사_A](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2019_2_A.ipynb) 22 | * [2019_2_중간고사_B](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2019_2_B.ipynb) 23 | * [2021_2_중간고사 A](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2021_2_A.ipynb) 24 | * [2022_2_중간고사 A](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2022_2_A_solution.ipynb) 25 | * [2022_2_중간고사 B](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2022_2_B_solution.ipynb) 26 | --- 27 | * [2019_2_기말고사_A](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/final_2019_2_A.ipynb) 28 | * [2019_2_기말고사_B](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/final_2019_2_B.ipynb) 29 | * [2020_2_기말고사_A](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/final_2020_2_A.ipynb) 30 | * [2020_2_기말고사_B](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/final_2020_2_B.ipynb) 31 | * [2022_2_기말고사 A](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/final_2022_2_A_solution.ipynb) 32 | * [2022_2_기말고사 B](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/final_2022_2_B_solution.ipynb) 33 | --- 34 | -------------------------------------------------------------------------------- /exams/figs/2023_final_2_B_2.jpg: -------------------------------------------------------------------------------- https://raw.githubusercontent.com/SeoulTechPSE/EngMath/f7fe2b6bb0d6b8843a7c386fef6f2f4c748a0911/exams/figs/2023_final_2_B_2.jpg -------------------------------------------------------------------------------- /exams/figs/fig_j0.pdf: 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-------------------------------------------------------------------------------- https://raw.githubusercontent.com/SeoulTechPSE/EngMath/f7fe2b6bb0d6b8843a7c386fef6f2f4c748a0911/exams/figs/mid_2019_2_B_1.png -------------------------------------------------------------------------------- /exams/final_2019_2_code1.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": {}, 7 | "outputs": [], 8 | "source": [ 9 | "%matplotlib inline\n", 10 | "import numpy as np\n", 11 | "from scipy.special import jv, jn_zeros\n", 12 | "import matplotlib.pyplot as plt\n", 13 | "from matplotlib import animation, rc\n", 14 | "rc('animation', html='html5')" 15 | ] 16 | }, 17 | { 18 | "cell_type": "code", 19 | "execution_count": null, 20 | "metadata": {}, 21 | "outputs": [], 22 | "source": [ 23 | "fig = plt.figure(figsize=(6, 4))\n", 24 | "ax = plt.axes(xlim=(0, 1), ylim=(-1.2, 1.2))\n", 25 | "\n", 26 | "ax.set_xticks([0, 0.2, 0.4, 0.6, 0.8, 1.0])\n", 27 | "ax.set_yticks([-1.2, -0.6, 0.0, 0.6, 1.2])\n", 28 | "ax.set_xlabel('$r$')\n", 29 | "ax.set_ylabel('$u(r,t)$')\n", 30 | "\n", 31 | "plt.close()" 32 | ] 33 | }, 34 | { 35 | "cell_type": "code", 36 | "execution_count": null, 37 | "metadata": {}, 38 | "outputs": [], 39 | "source": [ 40 | "line, = ax.plot([], [], lw=2)\n", 41 | "def init():\n", 42 | " line.set_data([], [])\n", 43 | " return (line,)\n", 44 | "\n", 45 | "mm = 2\n", 46 | "alpha_0m =jn_zeros(0, mm)\n", 47 | "\n", 48 | "def u_sol(x, t): \n", 49 | "\n", 50 | " y = np.cos(alpha_0m[1] *t) *jv(0, alpha_0m[1] *x)\n", 51 | " \n", 52 | " return y\n", 53 | " \n", 54 | "def animate(t):\n", 55 | " \n", 56 | " xx = np.linspace(0, 1, 50) \n", 57 | " uu = u_sol(xx, t)\n", 58 | " \n", 59 | " line.set_data(xx, uu)\n", 60 | " ax.legend([f't = {t:.2f}'])\n", 61 | " return (line,)\n", 62 | "\n", 63 | "tt = list(np.linspace(0, 6.0 *np.pi, 300))\n", 64 | "anim = animation.FuncAnimation(fig, animate, init_func=init, frames=tt, interval=300, blit=True)" 65 | ] 66 | } 67 | ], 68 | "metadata": { 69 | "kernelspec": { 70 | "display_name": "Python 3 (ipykernel)", 71 | "language": "python", 72 | "name": "python3" 73 | }, 74 | "language_info": { 75 | "codemirror_mode": { 76 | "name": "ipython", 77 | "version": 3 78 | }, 79 | "file_extension": ".py", 80 | "mimetype": "text/x-python", 81 | "name": "python", 82 | "nbconvert_exporter": "python", 83 | "pygments_lexer": "ipython3", 84 | "version": "3.8.12" 85 | } 86 | }, 87 | "nbformat": 4, 88 | "nbformat_minor": 4 89 | } 90 | -------------------------------------------------------------------------------- /exams/final_2019_2_code2.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": {}, 7 | "outputs": [], 8 | "source": [ 9 | "%matplotlib inline\n", 10 | "import numpy as np\n", 11 | "from scipy import integrate\n", 12 | "import matplotlib.pyplot as plt\n", 13 | "from matplotlib import animation, rc\n", 14 | "rc('animation', html='html5')" 15 | ] 16 | }, 17 | { 18 | "cell_type": "code", 19 | "execution_count": null, 20 | "metadata": {}, 21 | "outputs": [], 22 | "source": [ 23 | "fig = plt.figure(figsize=(6, 4))\n", 24 | "ax = plt.axes(xlim=(0, 10), ylim=(-1.2, 1.2))\n", 25 | "\n", 26 | "ax.set_xticks([0.0, 2.0, 4.0, 6.0, 8.0, 10.0])\n", 27 | "ax.set_yticks([-1.2, -0.6, 0.0, 0.6, 1.2])\n", 28 | "ax.set_xlabel('$x$')\n", 29 | "ax.set_ylabel('$u(x,t)$')\n", 30 | "\n", 31 | "plt.close()" 32 | ] 33 | }, 34 | { 35 | "cell_type": "code", 36 | "execution_count": null, 37 | "metadata": {}, 38 | "outputs": [], 39 | "source": [ 40 | "line, = ax.plot([], [], lw=2)\n", 41 | "def init():\n", 42 | " line.set_data([], [])\n", 43 | " return (line,)\n", 44 | "\n", 45 | "def u_sol(x, t): \n", 46 | " f = lambda tau : np.sin(t -tau) *x/(2*np.sqrt(np.pi *np.power(tau, 3))) *np.exp(-x*x/(4.0 *tau)) \n", 47 | " y, _ = integrate.quad(f, 0, t)\n", 48 | " \n", 49 | " return y\n", 50 | " \n", 51 | "def animate(t):\n", 52 | " \n", 53 | " xx = np.linspace(0, 10, 300) \n", 54 | " uu = np.zeros_like(xx)\n", 55 | " \n", 56 | " for i, x in enumerate(xx):\n", 57 | " uu[i] = u_sol(x, t)\n", 58 | " \n", 59 | " line.set_data(xx, uu)\n", 60 | " ax.legend([f't = {t:.2f}'])\n", 61 | " return (line,)\n", 62 | "\n", 63 | "tt = list(np.linspace(np.finfo(float).eps, 6.0 *np.pi, 30))\n", 64 | "anim = animation.FuncAnimation(fig, animate, init_func=init, frames=tt, interval=600, blit=True)" 65 | ] 66 | } 67 | ], 68 | "metadata": { 69 | "kernelspec": { 70 | "display_name": "Python 3", 71 | "language": "python", 72 | "name": "python3" 73 | }, 74 | "language_info": { 75 | "codemirror_mode": { 76 | "name": "ipython", 77 | "version": 3 78 | }, 79 | "file_extension": ".py", 80 | "mimetype": "text/x-python", 81 | "name": "python", 82 | "nbconvert_exporter": "python", 83 | "pygments_lexer": "ipython3", 84 | "version": "3.7.5" 85 | } 86 | }, 87 | "nbformat": 4, 88 | "nbformat_minor": 4 89 | } 90 | -------------------------------------------------------------------------------- /exams/final_2019_2_code3.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "code", 5 | "execution_count": null, 6 | "metadata": {}, 7 | "outputs": [], 8 | "source": [ 9 | "%matplotlib inline\n", 10 | "import numpy as np\n", 11 | "import matplotlib.pyplot as plt\n", 12 | "\n", 13 | "term_no = 10\n", 14 | "def u_sol(r, theta, term_no):\n", 15 | " \n", 16 | " u = 1.0 /np.pi +0.5 *r *np.sin(theta)\n", 17 | " for m in range(1, term_no):\n", 18 | " u -= 2.0 /np.pi *np.power(r, 2*m) /(np.power(2*m, 2) -1) *np.cos(2*m *theta)\n", 19 | "\n", 20 | " return u\n", 21 | "\n", 22 | "azimuths = np.radians(np.linspace(0, 360, 180))\n", 23 | "zeniths = np.linspace(0, 1, 50)\n", 24 | "\n", 25 | "r, theta = np.meshgrid(zeniths, azimuths)\n", 26 | "usol = u_sol(r, theta, term_no)\n", 27 | "\n", 28 | "fig, ax = plt.subplots(figsize=(10, 5), subplot_kw=dict(projection='polar'))\n", 29 | "cs = ax.contourf(theta, r, usol, 50)\n", 30 | "cbar = fig.colorbar(cs)\n", 31 | "\n", 32 | "#plt.show()" 33 | ] 34 | } 35 | ], 36 | "metadata": { 37 | "kernelspec": { 38 | "display_name": "Python 3", 39 | "language": "python", 40 | "name": "python3" 41 | }, 42 | "language_info": { 43 | "codemirror_mode": { 44 | "name": "ipython", 45 | "version": 3 46 | }, 47 | "file_extension": ".py", 48 | "mimetype": "text/x-python", 49 | "name": "python", 50 | "nbconvert_exporter": "python", 51 | "pygments_lexer": "ipython3", 52 | "version": "3.7.5" 53 | } 54 | }, 55 | "nbformat": 4, 56 | "nbformat_minor": 4 57 | } 58 | -------------------------------------------------------------------------------- /exams/final_2022_1_A.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "id": "a26af8d5", 6 | "metadata": {}, 7 | "source": [ 8 | "## Engineering Mathematics I (161006-21001) - Final Exam.\n", 9 | "May 31, 2022" 10 | ] 11 | }, 12 | { 13 | "cell_type": "markdown", 14 | "id": "f492a544", 15 | "metadata": {}, 16 | "source": [ 17 | "**1[25].** The equation\n", 18 | "\n", 19 | "> $\\displaystyle (1 - x^2)y'' -xy' +\\alpha^2 y=0,$\n", 20 | "\n", 21 | " where $\\alpha$ is a constant, is called the *Chebyshev* equation.\n", 22 | " \n", 23 | " (a)$~$ Compute two linearly independent series solutions for $|x|<1$\n", 24 | " \n", 25 | " (b)$~$ Show that for every non-negative integer $\\alpha=n$ there is a polynomial solution of degree $n$. When appropriately normalized these are called the *Chebyshev polynomials*\n", 26 | "$~$\n", 27 | "\n", 28 | "$~$" 29 | ] 30 | }, 31 | { 32 | "cell_type": "markdown", 33 | "id": "67560fc1", 34 | "metadata": {}, 35 | "source": [ 36 | "**2[25].** $~$ Find the general solution of the differential equation on the interval $(0, \\infty)$\n", 37 | "\n", 38 | "> $\\displaystyle x^2 y'' +2x y'+\\left(\\alpha^2x^2 -n(n+1)\\right)y=0, \\;\\;$ $n$ is an integer \n", 39 | "\n", 40 | "Hint) $~$Use the substitution $\\displaystyle y(x)=\\frac{1}{\\sqrt{x}}z(x)$\n", 41 | "$~$\n", 42 | "\n", 43 | "$~$" 44 | ] 45 | }, 46 | { 47 | "cell_type": "markdown", 48 | "id": "717470bb", 49 | "metadata": {}, 50 | "source": [ 51 | "**3[10].** $~$ Find the parametric equation for the line through the indicated point that is parallel to the given planes\n", 52 | "\n", 53 | "> $\\begin{align*}\n", 54 | " x + y -4z &= 2 \\\\ \n", 55 | " 2x - y +z &= 10; \\;\\;(5, 6, -12) \n", 56 | "\\end{align*}$ \n", 57 | "\n", 58 | "$~$\n", 59 | "\n", 60 | "$~$" 61 | ] 62 | }, 63 | { 64 | "cell_type": "markdown", 65 | "id": "8d4ac247", 66 | "metadata": {}, 67 | "source": [ 68 | "**4[15].** $~$In 3-space, a plane through the origin can be written as $S= \\{ (x,y,z)\\;|\\; ax+by+cz=0, \\;\\;a, b, \\text{ and }c\\;\\;\\text{are real numbers}\\}$. Show that $S$ is a subspace of $\\mathbb{R}^3$\n", 69 | "\n", 70 | "$~$\n", 71 | "\n", 72 | "$~$" 73 | ] 74 | }, 75 | { 76 | "cell_type": "markdown", 77 | "id": "f3cb2812", 78 | "metadata": {}, 79 | "source": [ 80 | "**5[25].** $~$Consider the matrix\n", 81 | "\n", 82 | "> $\\mathbf{A}=\\begin{pmatrix}\n", 83 | " 3 & 0 & 1\\\\ \n", 84 | " 0 & 1 & 0\\\\ \n", 85 | " 1 & 0 & 1\n", 86 | "\\end{pmatrix}$\n", 87 | "\n", 88 | "(a) Find an orthogonal matrix $\\mathbf{P}$ that diagonalizes $\\mathbf{A}$ and the diagonal matrix $\\mathbf{D}$\n", 89 | "\n", 90 | "(b) Calculate the eigenvalues and eigenvectors of the matrix\n", 91 | "\n", 92 | "> $3\\mathbf{A}^2 +4\\mathbf{A}^{-1} -7\\mathbf{I}$" 93 | ] 94 | } 95 | ], 96 | "metadata": { 97 | "kernelspec": { 98 | "display_name": "Python 3 (ipykernel)", 99 | "language": "python", 100 | "name": "python3" 101 | }, 102 | "language_info": { 103 | "codemirror_mode": { 104 | "name": "ipython", 105 | "version": 3 106 | }, 107 | "file_extension": ".py", 108 | "mimetype": "text/x-python", 109 | "name": "python", 110 | "nbconvert_exporter": "python", 111 | "pygments_lexer": "ipython3", 112 | "version": "3.8.12" 113 | } 114 | }, 115 | "nbformat": 4, 116 | "nbformat_minor": 5 117 | } 118 | -------------------------------------------------------------------------------- /exams/final_2022_1_B.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "id": "a26af8d5", 6 | "metadata": {}, 7 | "source": [ 8 | "## Engineering Mathematics I (161006-21002) - Final Exam.\n", 9 | "May 31, 2022" 10 | ] 11 | }, 12 | { 13 | "cell_type": "markdown", 14 | "id": "f492a544", 15 | "metadata": {}, 16 | "source": [ 17 | "**1[25].** The equation\n", 18 | "\n", 19 | "> $\\displaystyle y'' -2xy' +2\\alpha y=0,$\n", 20 | "\n", 21 | " where $\\alpha$ is a constant, is called the *Hermite* equation.\n", 22 | " \n", 23 | " (a)$~$ Find two linearly independent solutions for $-\\infty < x <\\infty$\n", 24 | " \n", 25 | " (b)$~$ Show that there is a polynomial solution of degree $n$, in case $\\alpha=n$ is a non-negative integer\n", 26 | "$~$\n", 27 | "\n", 28 | "$~$" 29 | ] 30 | }, 31 | { 32 | "cell_type": "markdown", 33 | "id": "67560fc1", 34 | "metadata": {}, 35 | "source": [ 36 | "**2[25].** $~$ Find the general solution of the differential equation on the interval $(0, \\infty)$\n", 37 | "\n", 38 | "> $\\displaystyle x y'' +y=0$ \n", 39 | "\n", 40 | "Hint) $~$Use the substitution $\\displaystyle y(x)=\\sqrt{x}u(x)$, $~2\\sqrt{x}=z$\n", 41 | "\n", 42 | "$~$\n", 43 | "\n", 44 | "$~$" 45 | ] 46 | }, 47 | { 48 | "cell_type": "markdown", 49 | "id": "717470bb", 50 | "metadata": { 51 | "slideshow": { 52 | "slide_type": "slide" 53 | } 54 | }, 55 | "source": [ 56 | "**3[10].** $~$ Find the parametric equation for the line of intersection of the given planes\n", 57 | "\n", 58 | "> $\\begin{align*}\n", 59 | " 5x - 4y -9z &= 8 \\\\ \n", 60 | " x + 4y +3z &= 4 \n", 61 | "\\end{align*}$ \n", 62 | "\n", 63 | "\n", 64 | "$~$\n", 65 | "\n", 66 | "$~$" 67 | ] 68 | }, 69 | { 70 | "cell_type": "markdown", 71 | "id": "8d4ac247", 72 | "metadata": {}, 73 | "source": [ 74 | "**4[15].** $~$Determine whether the set of vectors $S= \\{ \\left \\langle a_1, 0, a_3 \\right \\rangle\\;|\\; \\;a_1, a_3 \\;\\text{are real numbers}\\}$ is a subspace of $\\mathbb{R}^3$ by defining appropriate addition and scalar multiplication\n", 75 | "\n", 76 | "$~$\n", 77 | "\n", 78 | "$~$" 79 | ] 80 | }, 81 | { 82 | "cell_type": "markdown", 83 | "id": "f3cb2812", 84 | "metadata": {}, 85 | "source": [ 86 | "**5[25].** $~$Identify the given conic section. Graph\n", 87 | "\n", 88 | "> $16x^2 + 24xy + 9y^2 +2x -y=0$\n", 89 | "\n", 90 | "$~$\n", 91 | "\n", 92 | "$~$" 93 | ] 94 | } 95 | ], 96 | "metadata": { 97 | "kernelspec": { 98 | "display_name": "Python 3 (ipykernel)", 99 | "language": "python", 100 | "name": "python3" 101 | }, 102 | "language_info": { 103 | "codemirror_mode": { 104 | "name": "ipython", 105 | "version": 3 106 | }, 107 | "file_extension": ".py", 108 | "mimetype": "text/x-python", 109 | "name": "python", 110 | "nbconvert_exporter": "python", 111 | "pygments_lexer": "ipython3", 112 | "version": "3.8.12" 113 | } 114 | }, 115 | "nbformat": 4, 116 | "nbformat_minor": 5 117 | } 118 | -------------------------------------------------------------------------------- /exams/final_2022_2_A.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/final_2022_2_B.ipynb)" 8 | ] 9 | }, 10 | { 11 | "cell_type": "markdown", 12 | "metadata": {}, 13 | "source": [ 14 | "## Engineering Mathematics II (161007-21001) - Final Exam.\n", 15 | "December 14, 2022" 16 | ] 17 | }, 18 | { 19 | "cell_type": "markdown", 20 | "metadata": {}, 21 | "source": [ 22 | "**1[20].** $~$ Find the solution of\n", 23 | "\n", 24 | "> $\n", 25 | "\\begin{align*}\n", 26 | " u_t &= u_{xx} + \\sin \\lambda_1 x, \\;\\;0 where $\\lambda_1$ is the first root of the equation $\\tan \\lambda = -\\lambda$" 34 | ] 35 | }, 36 | { 37 | "cell_type": "markdown", 38 | "metadata": {}, 39 | "source": [ 40 | "**2[20].** $~$ Solve the semi-infinite string problem\n", 41 | "\n", 42 | "> $\n", 43 | "\\begin{align*}\n", 44 | " u_{tt} &= u_{xx}, \\;\\;0 Draw the solution for various values of time" 52 | ] 53 | }, 54 | { 55 | "cell_type": "markdown", 56 | "metadata": {}, 57 | "source": [ 58 | "**3[20].** $~$ Solve the damped vibrating-string problem\n", 59 | "\n", 60 | "> $\n", 61 | "\\begin{align*}\n", 62 | " u_{tt}=c^2 &u_{xx} -\\beta u_t, \\;\\;0 $\n", 78 | "\\begin{align*}\n", 79 | " u_t = \\alpha &u_{xx} +b u +f(x,t), \\;\\; 0$\n", 94 | "\\begin{align*}\n", 95 | " \\nabla^2 u &= 0, \\;\\; 1= 0:\n", 78 | "# uu[i] = 0.5*(f_i(ch1) +f_i(ch2))\n", 79 | "# else:\n", 80 | "# uu[i] = 0.5*(f_i(ch1) -f_i(-ch2))\n", 81 | " \n", 82 | " line.set_data(xx, uu)\n", 83 | " return (line,)" 84 | ] 85 | }, 86 | { 87 | "cell_type": "code", 88 | "execution_count": null, 89 | "id": "5e2c9410", 90 | "metadata": {}, 91 | "outputs": [], 92 | "source": [ 93 | "tt = list(np.linspace(0, 10, 100))\n", 94 | "anim = animation.FuncAnimation(fig, animate, init_func=init, frames=tt, interval=250, blit=True)" 95 | ] 96 | } 97 | ], 98 | "metadata": { 99 | "kernelspec": { 100 | "display_name": "Python 3 (ipykernel)", 101 | "language": "python", 102 | "name": "python3" 103 | }, 104 | "language_info": { 105 | "codemirror_mode": { 106 | "name": "ipython", 107 | "version": 3 108 | }, 109 | "file_extension": ".py", 110 | "mimetype": "text/x-python", 111 | "name": "python", 112 | "nbconvert_exporter": "python", 113 | "pygments_lexer": "ipython3", 114 | "version": "3.8.12" 115 | } 116 | }, 117 | "nbformat": 4, 118 | "nbformat_minor": 5 119 | } 120 | -------------------------------------------------------------------------------- /exams/final_2022_2_B.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/final_2022_2_B.ipynb)" 8 | ] 9 | }, 10 | { 11 | "cell_type": "markdown", 12 | "metadata": {}, 13 | "source": [ 14 | "## Engineering Mathematics II (161007-21002) - Final Exam.\n", 15 | "December 12, 2022" 16 | ] 17 | }, 18 | { 19 | "cell_type": "markdown", 20 | "metadata": {}, 21 | "source": [ 22 | "**1[20].** $~$ Solve the problem\n", 23 | "\n", 24 | "> $\n", 25 | "\\begin{align*}\n", 26 | " u_t &= u_{xx}, \\;\\;0 $\n", 41 | "\\begin{align*}\n", 42 | " u_{tt} &= u_{xx}, \\;\\;-\\infty $\n", 56 | "u(x,0)=\\begin{cases}\n", 57 | " 2hx & 0 \\leq x \\leq 0.5 \\\\ \n", 58 | " 2h(1-x)& 0.5 \\leq x \\leq 1 \n", 59 | "\\end{cases}\n", 60 | "$\n", 61 | "\n", 62 | "> What is the subsequent motion of the string if it is suddenly released?" 63 | ] 64 | }, 65 | { 66 | "cell_type": "markdown", 67 | "metadata": {}, 68 | "source": [ 69 | "**4[20].** $~$ Solve the problem\n", 70 | "\n", 71 | "> $\n", 72 | "\\begin{align*}\n", 73 | " u_t &= u_{xx} +\\sin 3\\pi x, \\;\\; 0$\n", 88 | "\\begin{align*}\n", 89 | " \\nabla^2 u &= 0, \\;\\; 0 This is the problem where the boundary of the northern hemisphere is hot (+1), while the southern hemisphere is cold (-1)" 98 | ] 99 | } 100 | ], 101 | "metadata": { 102 | "kernelspec": { 103 | "display_name": "Python 3 (ipykernel)", 104 | "language": "python", 105 | "name": "python3" 106 | }, 107 | "language_info": { 108 | "codemirror_mode": { 109 | "name": "ipython", 110 | "version": 3 111 | }, 112 | "file_extension": ".py", 113 | "mimetype": "text/x-python", 114 | "name": "python", 115 | "nbconvert_exporter": "python", 116 | "pygments_lexer": "ipython3", 117 | "version": "3.8.12" 118 | } 119 | }, 120 | "nbformat": 4, 121 | "nbformat_minor": 4 122 | } 123 | -------------------------------------------------------------------------------- /exams/final_2023_1_A.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "id": "a26af8d5", 6 | "metadata": {}, 7 | "source": [ 8 | "### Engineering Mathematics I (161006-21003) - Final Exam.\n", 9 | "May 31, 2023" 10 | ] 11 | }, 12 | { 13 | "cell_type": "markdown", 14 | "id": "f492a544", 15 | "metadata": {}, 16 | "source": [ 17 | "**1[25].** The equation\n", 18 | "\n", 19 | "> $\\displaystyle (1 - x^2)y'' -xy' +\\alpha^2 y=0,$\n", 20 | "\n", 21 | " where $\\alpha$ is a constant, is called the *Chebyshev* equation.\n", 22 | " \n", 23 | " (a)$~$ Compute two linearly independent series solutions for $|x|<1$\n", 24 | " \n", 25 | " (b)$~$ Show that for every non-negative integer $\\alpha=n$ there is a polynomial solution of degree $n$. When appropriately normalized these are called the *Chebyshev polynomials*\n", 26 | "$~$\n", 27 | "\n", 28 | "$~$" 29 | ] 30 | }, 31 | { 32 | "cell_type": "markdown", 33 | "id": "67560fc1", 34 | "metadata": {}, 35 | "source": [ 36 | "**2[15].** $~$ Find the general solution of the differential equation on the interval $(0, \\infty)$\n", 37 | "\n", 38 | "> $\\displaystyle x y'' +y=0$ \n", 39 | "\n", 40 | "Hint) $~$Use the substitution $\\displaystyle y(x)=\\sqrt{x}u(x)$, $~2\\sqrt{x}=z$\n", 41 | "$~$\n", 42 | "\n", 43 | "$~$" 44 | ] 45 | }, 46 | { 47 | "cell_type": "markdown", 48 | "id": "717470bb", 49 | "metadata": {}, 50 | "source": [ 51 | "**3[15].** $~$ The given vectors span a subspace $W$ of $\\mathbb{R}^4$. Use the Gram-Schmidt orthogonalization process to construct an orthonormal basis for the subspace.\n", 52 | "\n", 53 | "> $\\mathbf{u}_1=\\left \\langle 4, 0, 2, -1 \\right \\rangle, \\; \\mathbf{u}_2=\\left \\langle 2, 1,-1,1 \\right \\rangle, \\;\\text{and}\\; \\mathbf{u}_3=\\left \\langle 1,1,-1,0 \\right \\rangle$\n", 54 | "$~$\n", 55 | "\n", 56 | "$~$" 57 | ] 58 | }, 59 | { 60 | "cell_type": "markdown", 61 | "id": "8d4ac247", 62 | "metadata": {}, 63 | "source": [ 64 | "**4[15].** $~$ Find the inverse of the given matrix or show that no inverse exists\n", 65 | "\n", 66 | "> $\\begin{pmatrix}\n", 67 | " \\phantom {-} 1 & \\phantom {-}2 & \\phantom {-}3 & \\phantom {-}1\\\\ \n", 68 | " -1& \\phantom {-}0 & \\phantom {-}2 &\\phantom {-}1 \\\\ \n", 69 | " \\phantom {-}2& \\phantom {-}1& -3& \\phantom {-}0\\\\ \n", 70 | " \\phantom {-}1& \\phantom {-}1 & \\phantom {-}2 & \\phantom {-}1\n", 71 | "\\end{pmatrix}$\n", 72 | "$~$\n", 73 | "\n", 74 | "$~$" 75 | ] 76 | }, 77 | { 78 | "cell_type": "markdown", 79 | "id": "f3cb2812", 80 | "metadata": {}, 81 | "source": [ 82 | "**5[30].** $~$ Simplify the following equations of quadric surfaces in three-dimensional space, and give the corresponding coordinate tranformations:\n", 83 | "\n", 84 | "> $x^2 -2y^2 +z^2 +6xy -2yz +8x +2y -12z -3 = 0$\n", 85 | "$~$\n", 86 | "\n", 87 | "$~$" 88 | ] 89 | } 90 | ], 91 | "metadata": { 92 | "kernelspec": { 93 | "display_name": "Python 3 (ipykernel)", 94 | "language": "python", 95 | "name": "python3" 96 | }, 97 | "language_info": { 98 | "codemirror_mode": { 99 | "name": "ipython", 100 | "version": 3 101 | }, 102 | "file_extension": ".py", 103 | "mimetype": "text/x-python", 104 | "name": "python", 105 | "nbconvert_exporter": "python", 106 | "pygments_lexer": "ipython3", 107 | "version": "3.9.13" 108 | } 109 | }, 110 | "nbformat": 4, 111 | "nbformat_minor": 5 112 | } 113 | -------------------------------------------------------------------------------- /exams/final_2023_1_B.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "id": "a26af8d5", 6 | "metadata": {}, 7 | "source": [ 8 | "### Engineering Mathematics I (161006-21004) - Final Exam.\n", 9 | "May 31, 2023" 10 | ] 11 | }, 12 | { 13 | "cell_type": "markdown", 14 | "id": "f492a544", 15 | "metadata": {}, 16 | "source": [ 17 | "**1[25].** The equation\n", 18 | "\n", 19 | "> $\\displaystyle y'' -2xy' +2\\alpha y=0,$\n", 20 | "\n", 21 | " where $\\alpha$ is a constant, is called the *Hermite* equation.\n", 22 | " \n", 23 | " (a)$~$ Find two linearly independent solutions for $-\\infty < x <\\infty$\n", 24 | " \n", 25 | " (b)$~$ Show that there is a polynomial solution of degree $n$, in case $\\alpha=n$ is a non-negative integer\n", 26 | "$~$\n", 27 | "\n", 28 | "$~$" 29 | ] 30 | }, 31 | { 32 | "cell_type": "markdown", 33 | "id": "3b9d888f", 34 | "metadata": {}, 35 | "source": [ 36 | "**2[15].** $~$ Find the general solution of the differential equation on the interval $(0, \\infty)$\n", 37 | "\n", 38 | "> $\\displaystyle x^2 y'' +2x y'+\\left(\\alpha^2x^2 -n(n+1)\\right)y=0, \\;\\;$ $n$ is an integer \n", 39 | "\n", 40 | "Hint) $~$Use the substitution $\\displaystyle y(x)=\\frac{1}{\\sqrt{x}}z(x)$\n", 41 | "$~$\n", 42 | "\n", 43 | "$~$" 44 | ] 45 | }, 46 | { 47 | "cell_type": "markdown", 48 | "id": "7f84baf3", 49 | "metadata": {}, 50 | "source": [ 51 | "**3[15].** $~$ The set of vectors $\\{ \\mathbf{u}_1, \\mathbf{u}_2, \\mathbf{u}_3 \\}$, where\n", 52 | "\n", 53 | "> $\\mathbf{u}_1=\\left \\langle 1, 1, 3 \\right \\rangle, \\; \\mathbf{u}_2=\\left \\langle 1,4,1 \\right \\rangle, \\;\\text{and}\\; \\mathbf{u}_3=\\left \\langle 1,10,-3 \\right \\rangle$\n", 54 | "\n", 55 | "is linearly dependent in $\\mathbb{R}^3$. Carry out the Gram-Schmidt orthogonalization process and then discuss that result.\n", 56 | "$~$\n", 57 | "\n", 58 | "$~$" 59 | ] 60 | }, 61 | { 62 | "cell_type": "markdown", 63 | "id": "0662a54d", 64 | "metadata": {}, 65 | "source": [ 66 | "**4[15].** $~$ Use Cramer's rule to determine the solution of the system\n", 67 | "\n", 68 | "> $\\begin{matrix}\n", 69 | " (2-k) x_1& + & kx_2 & = &4 \\\\ \n", 70 | " k x_1 & + & (3-k) x_2 & = &3 \n", 71 | "\\end{matrix}$\n", 72 | "\n", 73 | "For what value(s) of $k$ is the system inconsistent?\n", 74 | "\n", 75 | "$~$" 76 | ] 77 | }, 78 | { 79 | "cell_type": "markdown", 80 | "id": "717470bb", 81 | "metadata": { 82 | "slideshow": { 83 | "slide_type": "slide" 84 | } 85 | }, 86 | "source": [ 87 | "**5[30].** $~$ Simplify the following equations of quadric surfaces in three-dimensional space, and give the corresponding coordinate tranformations:\n", 88 | "\n", 89 | "> $5x^2 + 6y^2 +7z^2 -4xy +4yz -10x +8y +14z -6=0$\n", 90 | "$~$\n", 91 | "\n", 92 | "$~$" 93 | ] 94 | } 95 | ], 96 | "metadata": { 97 | "kernelspec": { 98 | "display_name": "Python 3 (ipykernel)", 99 | "language": "python", 100 | "name": "python3" 101 | }, 102 | "language_info": { 103 | "codemirror_mode": { 104 | "name": "ipython", 105 | "version": 3 106 | }, 107 | "file_extension": ".py", 108 | "mimetype": "text/x-python", 109 | "name": "python", 110 | "nbconvert_exporter": "python", 111 | "pygments_lexer": "ipython3", 112 | "version": "3.9.13" 113 | } 114 | }, 115 | "nbformat": 4, 116 | "nbformat_minor": 5 117 | } 118 | -------------------------------------------------------------------------------- /exams/final_2023_2_A.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "## Engineering Mathematics II (161007-21003) - Final Exam.\n", 8 | "December 13, 2023" 9 | ] 10 | }, 11 | { 12 | "cell_type": "markdown", 13 | "metadata": {}, 14 | "source": [ 15 | "**1[25].** $~$ Solve the problem\n", 16 | "\n", 17 | "$$\n", 18 | "\\begin{align*}\n", 19 | " u_t &= u_{xx}, \\;\\;0 by\n", 27 | "\n", 28 | "> (a) transforming it to one with zero BCs.\n", 29 | "\n", 30 | "> (b) solving the resulting problem." 31 | ] 32 | }, 33 | { 34 | "cell_type": "markdown", 35 | "metadata": {}, 36 | "source": [ 37 | "**2[25].** $~$ Find $u\\left( \\frac{3}{8}, 3 \\right)$\n", 38 | "\n", 39 | "$$\n", 40 | "\\begin{align*}\n", 41 | " u_{tt} &= u_{xx}, \\;\\;0 by\n", 27 | "\n", 28 | "> (a) changing the nonhomogeneous BCs to homogeneous ones.\n", 29 | "\n", 30 | "> (b) transforming into a new equation without the term $-u$.\n", 31 | "\n", 32 | "> (c) solving the resulting problem." 33 | ] 34 | }, 35 | { 36 | "cell_type": "markdown", 37 | "metadata": {}, 38 | "source": [ 39 | "**2[25].** $~$ Suppose the vibration of a string is described by $u_{tt}=u_{xx}$ and has an initial displacement as given by the following diagram:\n", 40 | "\n", 41 | "\n", 42 | "![fig](./figs/2023_final_2_B_2.jpg)\n", 43 | "\n", 44 | "> Assuming the initial velocity $u_t(x,0)=0$, describe the solution of this problem in the $xt$-plane." 45 | ] 46 | }, 47 | { 48 | "cell_type": "markdown", 49 | "metadata": {}, 50 | "source": [ 51 | "**3[25].** $~$ Solve the Dirichlet problem in a circle:\n", 52 | "$$\n", 53 | "\\begin{align*}\n", 54 | " \\nabla^2 u &= 0, \\;\\;1 < r < 2 \\\\ \n", 55 | " u(1,\\theta) &= \\frac{1}{2} + \\cos\\theta\\\\ \n", 56 | " u(2,\\theta) &= 1 + \\sin\\theta \n", 57 | "\\end{align*}\n", 58 | "$$" 59 | ] 60 | }, 61 | { 62 | "cell_type": "markdown", 63 | "metadata": {}, 64 | "source": [ 65 | "**4[25].** $~$ 아래에 주어진 3차원 구에서의 Laplace 방정식 해를 설명하시오.\n", 66 | "\n", 67 | "$$\n", 68 | "\\begin{align*}\n", 69 | " \\nabla^2 u &= 0, \\;r<1 \\\\ \n", 70 | " u(1,\\phi,\\theta) &= g(\\phi,\\theta) \n", 71 | "\\end{align*}\n", 72 | "$$" 73 | ] 74 | } 75 | ], 76 | "metadata": { 77 | "kernelspec": { 78 | "display_name": "Python 3 (ipykernel)", 79 | "language": "python", 80 | "name": "python3" 81 | }, 82 | "language_info": { 83 | "codemirror_mode": { 84 | "name": "ipython", 85 | "version": 3 86 | }, 87 | "file_extension": ".py", 88 | "mimetype": "text/x-python", 89 | "name": "python", 90 | "nbconvert_exporter": "python", 91 | "pygments_lexer": "ipython3", 92 | "version": "3.9.18" 93 | } 94 | }, 95 | "nbformat": 4, 96 | "nbformat_minor": 4 97 | } 98 | -------------------------------------------------------------------------------- /exams/mid_2019_2_A.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2019_2_A.ipynb)" 8 | ] 9 | }, 10 | { 11 | "cell_type": "markdown", 12 | "metadata": {}, 13 | "source": [ 14 | "## Engineering Mathematics II – Mid. Exam." 15 | ] 16 | }, 17 | { 18 | "cell_type": "markdown", 19 | "metadata": {}, 20 | "source": [ 21 | "**1.** $\\text{ }$ Find an equation of the plane that satisfies the given conditions:\n", 22 | "\n", 23 | "> Contains $(2,3,-5)$ and is parallel to $x + y -4z =1$\n", 24 | "\n", 25 | ">**solution**\n", 26 | "\n", 27 | ">$\\begin{align*}\n", 28 | " \\underbrace{x +y -4z}_{\\text{parallel plane}} &= d\\\\ \n", 29 | " &\\Downarrow {\\scriptstyle\\text{contains } (2,3,-5)}\\\\ \n", 30 | " x +y -4z &= 25\n", 31 | "\\end{align*}$" 32 | ] 33 | }, 34 | { 35 | "cell_type": "markdown", 36 | "metadata": {}, 37 | "source": [ 38 | "**2.** $\\text{ }$ Determine whether the given set is a subspace of the vector space $C(-\\infty,\\infty)$\n", 39 | "\n", 40 | "> All nonnegative functions $f(x)$\n", 41 | "\n", 42 | "> **solution** $\\text{ }$ \n", 43 | "\n", 44 | "> The set of all nonnegative functions is not closed under scalar multiplication $\\Rightarrow$ **Not a subpace**" 45 | ] 46 | }, 47 | { 48 | "cell_type": "markdown", 49 | "metadata": {}, 50 | "source": [ 51 | "**3.** $\\text{ }$ Find the curvature at $t=\\pi$ of the cycloid that is described by\n", 52 | "\n", 53 | ">$\\mathbf{r}(t) =a(t -\\sin t)\\mathbf{i} +a(1 -\\cos t)\\mathbf{j}, \\;\\; a > 0$\n", 54 | "\n", 55 | ">**solution**\n", 56 | "\n", 57 | ">$\\begin{align*}\n", 58 | " \\mathbf{r}'(t) &= a(1-\\cos t)\\, \\mathbf{i} + a\\sin t\\, \\mathbf{j}, \\;\\; \\left \\| \\mathbf{r}'(t) \\right \\| =a\\sqrt{2(1 -\\cos t)}\\\\\n", 59 | " &\\Downarrow \\\\\n", 60 | " \\mathbf{T}(t) &= \\frac{\\mathbf{r}'(t)}{\\left \\| \\mathbf{r}'(t) \\right \\|} \n", 61 | " =\\frac{1}{\\sqrt{2}} \\left( \\sqrt{1-\\cos t}\\,\\mathbf{i} +\\frac{\\sin t}{\\sqrt{1 -\\cos t}}\\,\\mathbf{j} \\right)\\\\ \n", 62 | " \\mathbf{T}'(t) &= \\frac{1}{2\\sqrt{2}}\\left(\\frac{\\sin t}{\\sqrt{1 -\\cos t}}\\,\\mathbf{i}-\\sqrt{1-\\cos t}\\,\\mathbf{j} \\right), \\\\ \n", 63 | " &\\Downarrow \\\\ \n", 64 | " \\kappa(\\pi) &=\\frac{\\left\\| \\mathbf{T}'(\\pi) \\right \\|}{\\left \\| \\mathbf{r}'(\\pi) \\right \\|} = \\frac{1}{4a}\n", 65 | "\\end{align*}$" 66 | ] 67 | }, 68 | { 69 | "cell_type": "markdown", 70 | "metadata": {}, 71 | "source": [ 72 | "**4.** $\\text{ }$ Evaluate $\\displaystyle\\oint_C \\mathbf{F} \\cdot d\\mathbf{r}$, where $\\mathbf{F}=z^2 y \\cos xy\\, \\mathbf{i} +z^2 x(1+\\cos xy) \\,\\mathbf{j} +2z \\sin xy\\, \\mathbf{k}$; $C$ the boundary of the plane $z = 1 -y$ shown in\n", 73 | "\n", 74 | "![figure1](./figs/mid_2019_2_A_1.png)\n", 75 | "\n", 76 | ">**solution**\n", 77 | "\n", 78 | "> Using Stokes' Theorem\n", 79 | "\n", 80 | ">$\\begin{align*}\n", 81 | " \\oint_C \\mathbf{F} \\cdot d\\mathbf{r} &= \\iint_S \\left(\\nabla \\times \\mathbf{F}\\right) \\, \\cdot \\mathbf{n} \\,dS\\\\ \n", 82 | " \\text{ where }& \\\\ \n", 83 | " \\nabla \\times \\mathbf{F} &= \\left| \\begin{matrix}\n", 84 | " \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ \n", 85 | " \\frac{\\partial }{\\partial x} & \\frac{\\partial }{\\partial y} & \\frac{\\partial }{\\partial z}\\\\ \n", 86 | " z^2y\\cos xy & z^2x(1 +\\cos xy) & 2z\\sin xy\n", 87 | "\\end{matrix}\\right| = -2xz\\,\\mathbf{i} +z^2\\,\\mathbf{k}\\\\ \n", 88 | " \\mathbf{n}&= \\frac{\\nabla g}{\\left\\| \\nabla g\\right\\|}=\\frac{1}{\\sqrt{2}}(\\mathbf{j} +\\mathbf{k})\\\\ \n", 89 | " &\\Downarrow \\\\\n", 90 | " &= \\frac{1}{\\sqrt{2}} \\iint_S z^2 \\,dS = \\frac{1}{\\sqrt{2}} \\int_0^2 \\int_0^1 (1 -y)^2 \\,\\sqrt{2}\\,dy dx = \\frac{2}{3}\n", 91 | "\\end{align*}$" 92 | ] 93 | }, 94 | { 95 | "cell_type": "markdown", 96 | "metadata": {}, 97 | "source": [ 98 | "**5.** $\\text{ }$ (a) Find the Fourier series of $f$ on the given interval\n", 99 | "\n", 100 | ">$f(x) = \\begin{cases}\n", 101 | " \\;\\;0 & -\\pi < x< 0 \\\\ \n", 102 | " \\sin x & \\;\\;0 \\leq x <\\pi \n", 103 | "\\end{cases}$\n", 104 | "\n", 105 | ">**solution**\n", 106 | "\n", 107 | ">$\\begin{align*}\n", 108 | " f(x)&= \\frac{a_0}{2} + \\sum_{n=1}^\\infty \\left(a_n \\cos nx +b_n \\sin nx \\right ) \\\\ \n", 109 | " \\text{wher}&\\text{e} \\\\ \n", 110 | " a_0 &=\\frac{1}{\\pi} \\int_{-\\pi}^\\pi f(x) \\,dx =\\frac{1}{\\pi} \\int_0^\\pi \\sin x \\,dx = \\frac{2}{\\pi} \\\\ \n", 111 | " a_n &=\\frac{1}{\\pi}\\int_{-\\pi}^\\pi f(x) \\cos nx \\,dx =\\frac{1}{\\pi}\\int_{0}^\\pi \\sin x \\cos nx \\,dx \\\\ \n", 112 | " &= \\frac{1}{2\\pi} \\int_0^\\pi \\sin (n +1)x -\\sin(n -1)x \\,dx \\\\ \n", 113 | " &= \\frac{1}{2\\pi} \\left[ -\\frac{1}{n +1} \\cos (n +1)x +\\frac{1}{n -1} \\cos (n-1)x \\right]_{0}^{\\pi} \\\\\n", 114 | " &=-\\frac{2}{\\pi}\\frac{1}{(2m -1)(2m +1)}\\;\\;\\text{if } n = 2m \\text{ else } a_n = 0 \\\\\n", 115 | " b_n &=\\frac{1}{\\pi} \\int_{-\\pi}^\\pi f(x) \\sin nx\\, dx = \\frac{1}{\\pi} \\int_{0}^\\pi \\sin x \\sin nx\\, dx \n", 116 | " = \\frac{1}{2} \\text{ if } n = 1 \\text{ else } b_n = 0 \\\\\n", 117 | " &\\Downarrow \\\\\n", 118 | " f(x) &= \\frac{1}{\\pi} +\\frac{1}{2} \\sin x -\\frac{2}{\\pi} \\sum_{m=1}^\\infty \\frac{1}{(2m -1)(2m +1)}\\cos 2mx\n", 119 | "\\end{align*}$\n", 120 | "\n", 121 | "$\\text{ }\\;\\;\\;\\;$(b) Use the result of (a) to evaluate the following series:\n", 122 | "\n", 123 | ">$\\displaystyle \\frac{1}{2} + \\frac{1}{1 \\cdot 3} - \\frac{1}{3 \\cdot 5} +\\frac{1}{5 \\cdot 7} -\\frac{1}{7\\cdot 9} + \\cdots$\n", 124 | "\n", 125 | ">**solution**\n", 126 | "\n", 127 | ">$\\begin{align*}\n", 128 | " f(x) &= \\frac{1}{\\pi} +\\frac{1}{2} \\sin x -\\frac{2}{\\pi} \\sum_{m=1}^\\infty \\frac{1}{(2m -1)(2m +1)}\\cos 2mx\\\\ \n", 129 | " &\\Downarrow \\;\\; x = \\frac{\\pi}{2}\\\\ \n", 130 | " 1 &= \\frac{1}{\\pi} +\\frac{1}{2} -\\frac{2}{\\pi} \\sum_{m=1}^\\infty \\frac{(-1)^m}{(2m -1)(2m +1)} \\\\\n", 131 | " &\\Downarrow \\\\\n", 132 | " \\frac{\\pi}{4} &= \\frac{1}{2} +\\sum_{m=1}^\\infty \\frac{(-1)^{m +1}}{(2m -1)(2m +1)}\n", 133 | "\\end{align*}$" 134 | ] 135 | }, 136 | { 137 | "cell_type": "markdown", 138 | "metadata": {}, 139 | "source": [ 140 | "**6.** $\\text{ }$ Chebyshev’s differential equation $(1 -x^2)y'' -xy' +n^2y = 0$ has a polynomial solution $T_n(x)$ for $n = 0,1,2,\\cdots$. \n", 141 | "\n", 142 | "$\\text{ }\\;\\;\\;$ Specify the weight function $p(x)$ and the interval over which the set of Chebyshev polynomials $\\{ T_n(x) \\}$ is orthogonal.\n", 143 | "\n", 144 | ">**solution**\n", 145 | "\n", 146 | ">$\\begin{align*}\n", 147 | " (1 -x^2)y'' &-xy' +n^2y = 0 \\\\ \n", 148 | " &\\Downarrow\\; x \\in (-1, 1) \\\\ \n", 149 | " y'' -\\frac{x}{1 -x^2}& y' +\\frac{n^2}{1 -x^2} y = 0\\\\ \n", 150 | " &\\Downarrow \\;\\; \\times \\sqrt{1 -x^2} \\\\ \n", 151 | " \\frac{d}{dx} \\left[ \\sqrt{1 -x^2} \\frac{dy}{dx} \\right] &+ \\frac{n^2}{\\sqrt{1 -x^2}} y=0 \\;\\;\\text{Self-Adjoint form}\\\\ \n", 152 | " &\\Downarrow \\\\\n", 153 | " r(x) = \\sqrt{1 -x^2}, \\;\\; &p(x) = \\frac{1}{\\sqrt{1 -x^2}} \\\\\n", 154 | " \\int_{-1}^1 \\frac{1}{\\sqrt{1 -x^2}}\\, T_m(x)& T_n(x)\\,dx = 0,\\;\\; m \\neq n \n", 155 | "\\end{align*}$" 156 | ] 157 | } 158 | ], 159 | "metadata": { 160 | "kernelspec": { 161 | "display_name": "Python 3 (ipykernel)", 162 | "language": "python", 163 | "name": "python3" 164 | }, 165 | "language_info": { 166 | "codemirror_mode": { 167 | "name": "ipython", 168 | "version": 3 169 | }, 170 | "file_extension": ".py", 171 | "mimetype": "text/x-python", 172 | "name": "python", 173 | "nbconvert_exporter": "python", 174 | "pygments_lexer": "ipython3", 175 | "version": "3.8.12" 176 | } 177 | }, 178 | "nbformat": 4, 179 | "nbformat_minor": 4 180 | } 181 | -------------------------------------------------------------------------------- /exams/mid_2021_2_A.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2019_2_A.ipynb)" 8 | ] 9 | }, 10 | { 11 | "cell_type": "markdown", 12 | "metadata": {}, 13 | "source": [ 14 | "## Engineering Mathematics II – Mid. Exam." 15 | ] 16 | }, 17 | { 18 | "cell_type": "markdown", 19 | "metadata": {}, 20 | "source": [ 21 | "**1.** $\\text{ }$ Find an equation of the plane that satisfies the given conditions:\n", 22 | "\n", 23 | "> Contains $(2,3,-5)$ and is parallel to $x + y -4z =1$\n", 24 | "\n", 25 | ">**solution**\n", 26 | "\n", 27 | ">$\\begin{align*}\n", 28 | " \\underbrace{x +y -4z}_{\\text{parallel plane}} &= d\\\\ \n", 29 | " &\\Downarrow {\\scriptstyle\\text{contains } (2,3,-5)}\\\\ \n", 30 | " x +y -4z &= 25\n", 31 | "\\end{align*}$" 32 | ] 33 | }, 34 | { 35 | "cell_type": "markdown", 36 | "metadata": {}, 37 | "source": [ 38 | "**2.** $\\text{ }$ Determine whether the given set is a subspace of the vector space $C(-\\infty,\\infty)$\n", 39 | "\n", 40 | "> All nonnegative functions $f(x)$\n", 41 | "\n", 42 | "> **solution** $\\text{ }$ \n", 43 | "\n", 44 | "> The set of all nonnegative functions is not closed under scalar multiplication $\\Rightarrow$ **Not a subpace**" 45 | ] 46 | }, 47 | { 48 | "cell_type": "markdown", 49 | "metadata": {}, 50 | "source": [ 51 | "**3.** $\\text{ }$ Find the curvature at $t=\\pi$ of the cycloid that is described by\n", 52 | "\n", 53 | ">$\\mathbf{r}(t) =a(t -\\sin t)\\mathbf{i} +a(1 -\\cos t)\\mathbf{j}, \\;\\; a > 0$\n", 54 | "\n", 55 | ">**solution**\n", 56 | "\n", 57 | ">$\\begin{align*}\n", 58 | " \\mathbf{r}'(t) &= a(1-\\cos t)\\, \\mathbf{i} + a\\sin t\\, \\mathbf{j}, \\;\\; \\left \\| \\mathbf{r}'(t) \\right \\| =a\\sqrt{2(1 -\\cos t)}\\\\\n", 59 | " &\\Downarrow \\\\\n", 60 | " \\mathbf{T}(t) &= \\frac{\\mathbf{r}'(t)}{\\left \\| \\mathbf{r}'(t) \\right \\|} \n", 61 | " =\\frac{1}{\\sqrt{2}} \\left( \\sqrt{1-\\cos t}\\,\\mathbf{i} +\\frac{\\sin t}{\\sqrt{1 -\\cos t}}\\,\\mathbf{j} \\right)\\\\ \n", 62 | " \\mathbf{T}'(t) &= \\frac{1}{2\\sqrt{2}}\\left(\\frac{\\sin t}{\\sqrt{1 -\\cos t}}\\,\\mathbf{i}-\\sqrt{1-\\cos t}\\,\\mathbf{j} \\right), \\\\ \n", 63 | " &\\Downarrow \\\\ \n", 64 | " \\kappa(\\pi) &=\\frac{\\left\\| \\mathbf{T}'(\\pi) \\right \\|}{\\left \\| \\mathbf{r}'(\\pi) \\right \\|} = \\frac{1}{4a}\n", 65 | "\\end{align*}$" 66 | ] 67 | }, 68 | { 69 | "cell_type": "markdown", 70 | "metadata": {}, 71 | "source": [ 72 | "**3.** $\\text{ }$ Evaluate $\\displaystyle \\oint_C \\mathbf{F} \\cdot d\\mathbf{r}$, $~$where $\\mathbf{F}=x^2y\\,\\mathbf{i} +(x +y^2)\\,\\mathbf{j} +xy^2z \\,\\mathbf{k}$ and $C$ the boundary of the surface shown in\n", 73 | "\n", 74 | "![figure1](./figs/mid_2019_2_B_1.png)\n", 75 | "\n", 76 | ">**solution**\n", 77 | "\n", 78 | "> Stokes' law: $~\\displaystyle\\oint_C \\mathbf{F} \\cdot d\\mathbf{r} = \\iint_S (\\nabla \\times \\mathbf{F}) \\cdot \\mathbf{n}\\,dS$\n", 79 | "\n", 80 | ">$\\begin{align*}\n", 81 | " \\mathbf{n}&= \\frac{\\nabla g}{\\left\\| \\nabla g\\right\\|}=\\frac{1}{\\sqrt{1 +4y^2}}(2y\\,\\mathbf{j} +\\mathbf{k})\\\\ \n", 82 | " &\\Downarrow \\\\\n", 83 | " \\iint_S (\\nabla \\times \\mathbf{F}) \\cdot \\mathbf{n}\\,dS &= \\iint_S \\frac{1}{\\sqrt{1 +4y^2}}\\left( -2y^3 z + 1 -x^2 \\right) \\,dS = \\int_0^3 \\int_0^{\\frac{y}{2}} \\frac{1}{\\sqrt{1 +4y^2}}\\left(-18y^3 +2y^5 + 1 -x^2\\right) \\,\\sqrt{1 +4y^2}\\,dx dy = - \\frac{138393}{1120}\n", 84 | "\\end{align*}$" 85 | ] 86 | }, 87 | { 88 | "cell_type": "markdown", 89 | "metadata": {}, 90 | "source": [ 91 | "#### **4.** $\\text{ }$ (a) Find the Fourier series of $f$ on the given interval\n", 92 | "\n", 93 | ">$f(x) = \\begin{cases}\n", 94 | " \\;\\;0 & -\\pi < x< 0 \\\\ \n", 95 | " \\sin x & \\;\\;0 \\leq x <\\pi \n", 96 | "\\end{cases}$\n", 97 | "\n", 98 | ">**solution**\n", 99 | "\n", 100 | ">$\\begin{align*}\n", 101 | " f(x)&= \\frac{a_0}{2} + \\sum_{n=1}^\\infty \\left(a_n \\cos nx +b_n \\sin nx \\right ) \\\\ \n", 102 | " \\text{wher}&\\text{e} \\\\ \n", 103 | " a_0 &=\\frac{1}{\\pi} \\int_{-\\pi}^\\pi f(x) \\,dx =\\frac{1}{\\pi} \\int_0^\\pi \\sin x \\,dx = \\frac{2}{\\pi} \\\\ \n", 104 | " a_n &=\\frac{1}{\\pi}\\int_{-\\pi}^\\pi f(x) \\cos nx \\,dx =\\frac{1}{\\pi}\\int_{0}^\\pi \\sin x \\cos nx \\,dx \\\\ \n", 105 | " &= \\frac{1}{2\\pi} \\int_0^\\pi \\sin (n +1)x -\\sin(n -1)x \\,dx \\\\ \n", 106 | " &= \\frac{1}{2\\pi} \\left[ -\\frac{1}{n +1} \\cos (n +1)x +\\frac{1}{n -1} \\cos (n-1)x \\right]_{0}^{\\pi} \\\\\n", 107 | " &=-\\frac{2}{\\pi}\\frac{1}{(2m -1)(2m +1)}\\;\\;\\text{if } n = 2m \\text{ else } a_n = 0 \\\\\n", 108 | " b_n &=\\frac{1}{\\pi} \\int_{-\\pi}^\\pi f(x) \\sin nx\\, dx = \\frac{1}{\\pi} \\int_{0}^\\pi \\sin x \\sin nx\\, dx \n", 109 | " = \\frac{1}{2} \\text{ if } n = 1 \\text{ else } b_n = 0 \\\\\n", 110 | " &\\Downarrow \\\\\n", 111 | " f(x) &= \\frac{1}{\\pi} +\\frac{1}{2} \\sin x -\\frac{2}{\\pi} \\sum_{m=1}^\\infty \\frac{1}{(2m -1)(2m +1)}\\cos 2mx\n", 112 | "\\end{align*}$\n", 113 | "\n", 114 | "$\\text{ }\\;\\;\\;\\;$(b) Use the result of (a) to evaluate the following series:\n", 115 | "\n", 116 | ">$\\displaystyle \\frac{1}{2} + \\frac{1}{1 \\cdot 3} - \\frac{1}{3 \\cdot 5} +\\frac{1}{5 \\cdot 7} -\\frac{1}{7\\cdot 9} + \\cdots$\n", 117 | "\n", 118 | ">**solution**\n", 119 | "\n", 120 | ">$\\begin{align*}\n", 121 | " f(x) &= \\frac{1}{\\pi} +\\frac{1}{2} \\sin x -\\frac{2}{\\pi} \\sum_{m=1}^\\infty \\frac{1}{(2m -1)(2m +1)}\\cos 2mx\\\\ \n", 122 | " &\\Downarrow \\;\\; x = \\frac{\\pi}{2}\\\\ \n", 123 | " 1 &= \\frac{1}{\\pi} +\\frac{1}{2} -\\frac{2}{\\pi} \\sum_{m=1}^\\infty \\frac{(-1)^m}{(2m -1)(2m +1)} \\\\\n", 124 | " &\\Downarrow \\\\\n", 125 | " \\frac{\\pi}{4} &= \\frac{1}{2} +\\sum_{m=1}^\\infty \\frac{(-1)^{m +1}}{(2m -1)(2m +1)}\n", 126 | "\\end{align*}$" 127 | ] 128 | }, 129 | { 130 | "cell_type": "markdown", 131 | "metadata": {}, 132 | "source": [ 133 | "**6.** $\\text{ }$ Chebyshev’s differential equation $(1 -x^2)y'' -xy' +n^2y = 0$ has a polynomial solution $T_n(x)$ for $n = 0,1,2,\\cdots$. \n", 134 | "\n", 135 | "$\\text{ }\\;\\;\\;$ Specify the weight function $p(x)$ and the interval over which the set of Chebyshev polynomials $\\{ T_n(x) \\}$ is orthogonal.\n", 136 | "\n", 137 | ">**solution**\n", 138 | "\n", 139 | ">$\\begin{align*}\n", 140 | " (1 -x^2)y'' &-xy' +n^2y = 0 \\\\ \n", 141 | " &\\Downarrow\\; x \\in (-1, 1) \\\\ \n", 142 | " y'' -\\frac{x}{1 -x^2}& y' +\\frac{n^2}{1 -x^2} y = 0\\\\ \n", 143 | " &\\Downarrow \\;\\; \\times \\sqrt{1 -x^2} \\\\ \n", 144 | " \\frac{d}{dx} \\left[ \\sqrt{1 -x^2} \\frac{dy}{dx} \\right] &+ \\frac{n^2}{\\sqrt{1 -x^2}} y=0 \\;\\;\\text{Self-Adjoint form}\\\\ \n", 145 | " &\\Downarrow \\\\\n", 146 | " r(x) = \\sqrt{1 -x^2}, \\;\\; &p(x) = \\frac{1}{\\sqrt{1 -x^2}} \\\\\n", 147 | " \\int_{-1}^1 \\frac{1}{\\sqrt{1 -x^2}}\\, T_m(x)& T_n(x)\\,dx = 0,\\;\\; m \\neq n \n", 148 | "\\end{align*}$" 149 | ] 150 | } 151 | ], 152 | "metadata": { 153 | "kernelspec": { 154 | "display_name": "Python 3", 155 | "language": "python", 156 | "name": "python3" 157 | }, 158 | "language_info": { 159 | "codemirror_mode": { 160 | "name": "ipython", 161 | "version": 3 162 | }, 163 | "file_extension": ".py", 164 | "mimetype": "text/x-python", 165 | "name": "python", 166 | "nbconvert_exporter": "python", 167 | "pygments_lexer": "ipython3", 168 | "version": "3.9.4" 169 | } 170 | }, 171 | "nbformat": 4, 172 | "nbformat_minor": 4 173 | } 174 | -------------------------------------------------------------------------------- /exams/mid_2022_1_A.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "id": "a26af8d5", 6 | "metadata": {}, 7 | "source": [ 8 | "## Engineering Mathematics I (161006-21001) - Mid. Exam.\n", 9 | "April 11, 2022" 10 | ] 11 | }, 12 | { 13 | "cell_type": "markdown", 14 | "id": "f492a544", 15 | "metadata": {}, 16 | "source": [ 17 | "**1[10].** $~$Solve\n", 18 | "\n", 19 | "> $\\displaystyle \\frac{dy}{dx}=\\frac{xy+3x-y-3}{xy -2x +4y-8}$\n", 20 | "\n", 21 | "$~$\n", 22 | "\n", 23 | "$~$" 24 | ] 25 | }, 26 | { 27 | "cell_type": "markdown", 28 | "id": "67560fc1", 29 | "metadata": {}, 30 | "source": [ 31 | "**2[10].** $~$ Solve the given initial-value problem\n", 32 | "\n", 33 | "> $\\displaystyle y\\frac{dx}{dy} -x =2y^2, \\;\\;y(1)=5$\n", 34 | "\n", 35 | "$~$\n", 36 | "\n", 37 | "$~$" 38 | ] 39 | }, 40 | { 41 | "cell_type": "markdown", 42 | "id": "717470bb", 43 | "metadata": {}, 44 | "source": [ 45 | "**3[10].** $~$ Solve $\\cos x dx + \\left( 1+\\frac{2}{y} \\right)\\sin x dy=0$\n", 46 | "\n", 47 | "$~$\n", 48 | "\n", 49 | "$~$" 50 | ] 51 | }, 52 | { 53 | "cell_type": "markdown", 54 | "id": "8d4ac247", 55 | "metadata": {}, 56 | "source": [ 57 | "**4[10].** $~$Solve $~\\displaystyle \\frac{dy}{dx} = \\frac{x}{y} + \\frac{y}{x} +1$\n", 58 | "\n", 59 | "$~$\n", 60 | "\n", 61 | "$~$" 62 | ] 63 | }, 64 | { 65 | "cell_type": "markdown", 66 | "id": "f3cb2812", 67 | "metadata": {}, 68 | "source": [ 69 | "**5[15].** $~$Solve $~y'' - 2y' + y = 4x^2 - 3 +x^{-1} e^x$\n", 70 | "\n", 71 | "$~$\n", 72 | "\n", 73 | "$~$" 74 | ] 75 | }, 76 | { 77 | "cell_type": "markdown", 78 | "id": "ad89515e", 79 | "metadata": {}, 80 | "source": [ 81 | "**6[15].** Solve $x^2y'' -4xy' + 6y = 2x^4 + x^2$\n", 82 | "\n", 83 | "$~$\n", 84 | "\n", 85 | "$~$" 86 | ] 87 | }, 88 | { 89 | "cell_type": "markdown", 90 | "id": "da779d82", 91 | "metadata": {}, 92 | "source": [ 93 | "**7[15].** $~$Solve\n", 94 | "\n", 95 | "> $\\begin{align*}\n", 96 | " x' + y' &= t\\\\ \n", 97 | " 4x + y' &= 0\n", 98 | "\\end{align*} ~\\text{ at } x(0)=1, y(0)=2$\n", 99 | "\n", 100 | "$~$\n", 101 | "\n", 102 | "$~$" 103 | ] 104 | }, 105 | { 106 | "cell_type": "markdown", 107 | "id": "5c81d569", 108 | "metadata": {}, 109 | "source": [ 110 | "**8[15].** $~$ One definition of the gamma function $\\Gamma(\\alpha)$ is given by the improper integral\n", 111 | "\n", 112 | " >$\\displaystyle \\Gamma(\\alpha) =\\int_0^\\infty t^{\\alpha-1} e^{-t}\\,dt, \\;\\;\\alpha > 0$\n", 113 | "\n", 114 | " > Use this definition to show that $\\Gamma(\\alpha + 1)=\\alpha \\Gamma(\\alpha)$ and show that\n", 115 | " \n", 116 | " >$\\displaystyle \\mathcal{L} \\{ t^\\alpha \\} = \\frac{\\Gamma (\\alpha+1)}{s^{\\alpha+1}}, \\;\\; \\alpha>-1$\n", 117 | " " 118 | ] 119 | } 120 | ], 121 | "metadata": { 122 | "kernelspec": { 123 | "display_name": "Python 3 (ipykernel)", 124 | "language": "python", 125 | "name": "python3" 126 | }, 127 | "language_info": { 128 | "codemirror_mode": { 129 | "name": "ipython", 130 | "version": 3 131 | }, 132 | "file_extension": ".py", 133 | "mimetype": "text/x-python", 134 | "name": "python", 135 | "nbconvert_exporter": "python", 136 | "pygments_lexer": "ipython3", 137 | "version": "3.8.12" 138 | } 139 | }, 140 | "nbformat": 4, 141 | "nbformat_minor": 5 142 | } 143 | -------------------------------------------------------------------------------- /exams/mid_2022_1_B.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "id": "c8a295c0", 6 | "metadata": {}, 7 | "source": [ 8 | "## Engineering Mathematics I (161006-21002) - Mid. Exam.\n", 9 | "April 13, 2022" 10 | ] 11 | }, 12 | { 13 | "cell_type": "markdown", 14 | "id": "c1f28567", 15 | "metadata": {}, 16 | "source": [ 17 | "**1[10].** $~$Find a solution of $x\\frac{dy}{dx}=y^2 - y$ that passes through the indicated point $(2, \\frac{1}{4})$\n", 18 | "\n", 19 | "$~$\n", 20 | "\n", 21 | "$~$" 22 | ] 23 | }, 24 | { 25 | "cell_type": "markdown", 26 | "id": "937ecce5", 27 | "metadata": {}, 28 | "source": [ 29 | "**2[10].** $~$Find a general solution of the equation $dx=(x + y^2)dy$\n", 30 | "\n", 31 | "$~$\n", 32 | "\n", 33 | "$~$" 34 | ] 35 | }, 36 | { 37 | "cell_type": "markdown", 38 | "id": "4b74f6de", 39 | "metadata": {}, 40 | "source": [ 41 | "**3[10].** $~$Solve $(4t^3y-15t^2 -y)dt +(t^4+3y^2-t)dy=0$\n", 42 | "\n", 43 | "$~$\n", 44 | "\n", 45 | "$~$" 46 | ] 47 | }, 48 | { 49 | "cell_type": "markdown", 50 | "id": "f2b9a9e7", 51 | "metadata": {}, 52 | "source": [ 53 | "**4[10].** $~$Solve $~\\displaystyle 2xyy' +y^2 =2x^2$\n", 54 | "\n", 55 | "$~$\n", 56 | "\n", 57 | "$~$" 58 | ] 59 | }, 60 | { 61 | "cell_type": "markdown", 62 | "id": "f0edd1ea", 63 | "metadata": {}, 64 | "source": [ 65 | "**5[15].** $~$Solve $~2y'' + 2y' + y = 4\\sqrt{x} +\\sin x$\n", 66 | "\n", 67 | "$~$\n", 68 | "\n", 69 | "$~$" 70 | ] 71 | }, 72 | { 73 | "cell_type": "markdown", 74 | "id": "4733cea6", 75 | "metadata": {}, 76 | "source": [ 77 | "**6[15].** Solve $x^2y'' -xy' + y = x^3$\n", 78 | "\n", 79 | "$~$\n", 80 | "\n", 81 | "$~$" 82 | ] 83 | }, 84 | { 85 | "cell_type": "markdown", 86 | "id": "960169ab", 87 | "metadata": {}, 88 | "source": [ 89 | "**7[15].** $~$Solve\n", 90 | "\n", 91 | "> $\\begin{align*}\n", 92 | " x'' + y'' &= e^{2t}\\\\ \n", 93 | " x' + y'' &= -e^{2t}\n", 94 | "\\end{align*} ~\\text{ at } x(0)=0, x'(0)=0, y(0)=0, y'(0)=0$\n", 95 | "\n", 96 | "$~$\n", 97 | "\n", 98 | "$~$" 99 | ] 100 | }, 101 | { 102 | "cell_type": "markdown", 103 | "id": "e17492b4", 104 | "metadata": {}, 105 | "source": [ 106 | "**8[15].** Reduce the given differential equation to a linear first-order DE in the transformed function $Y(s)=\\mathcal{L}\\{y(t)\\}$. Solve the first-order DE and then find $y(t)$. Note: Your solution may contain an arbitrary constant:\n", 107 | "\n", 108 | "> $ty'' -y' = 2t^2, \\;y(0)=y'(0)=0$" 109 | ] 110 | } 111 | ], 112 | "metadata": { 113 | "kernelspec": { 114 | "display_name": "Python 3 (ipykernel)", 115 | "language": "python", 116 | "name": "python3" 117 | }, 118 | "language_info": { 119 | "codemirror_mode": { 120 | "name": "ipython", 121 | "version": 3 122 | }, 123 | "file_extension": ".py", 124 | "mimetype": "text/x-python", 125 | "name": "python", 126 | "nbconvert_exporter": "python", 127 | "pygments_lexer": "ipython3", 128 | "version": "3.8.12" 129 | } 130 | }, 131 | "nbformat": 4, 132 | "nbformat_minor": 5 133 | } 134 | -------------------------------------------------------------------------------- /exams/mid_2022_2_A.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "## Engineering Mathematics II (161007-21001) - Mid. Exam.\n", 8 | "October 26, 2022" 9 | ] 10 | }, 11 | { 12 | "cell_type": "markdown", 13 | "metadata": {}, 14 | "source": [ 15 | "**1[20].** $~$ Evaluate the surface integral $\\displaystyle \\iint_S \\mathbf{F}\\cdot\\mathbf{n}\\, dS:$\n", 16 | "\n", 17 | "> $\\displaystyle \\mathbf{F} = y \\mathbf{j} +\\mathbf{k}$\n", 18 | "\n", 19 | "> where $S$ is the portion of the paraboloid $z=1-x^2-y^2$ above the $xy$-plane" 20 | ] 21 | }, 22 | { 23 | "cell_type": "markdown", 24 | "metadata": {}, 25 | "source": [ 26 | "**2[20].** $~$ Is it possible to write $\\mathbf{G}=\\nabla \\times \\mathbf{H}$? In the case where it is possible, find a possible $\\mathbf{H}$\n", 27 | "\n", 28 | "> $\\mathbf{G}=y\\mathbf{i} + z\\mathbf{j} +x\\mathbf{k}$" 29 | ] 30 | }, 31 | { 32 | "cell_type": "markdown", 33 | "metadata": {}, 34 | "source": [ 35 | "**3[20].** $~$ Each of the following functions is smooth in a simply connected region. Determine which of them may be written as the gradient of a scalar function, and for those which can, find that scalar function:\n", 36 | "\n", 37 | "> (a) $~\\mathbf{F}=y\\mathbf{i}$\n", 38 | "\n", 39 | "> (b) $~\\mathbf{F}=c\\mathbf{k}$, $c$ a constant\n", 40 | "\n", 41 | "> (c) $~\\mathbf{F}=yz\\mathbf{i} +xz\\mathbf{j} +xy\\mathbf{k}$\n" 42 | ] 43 | }, 44 | { 45 | "cell_type": "markdown", 46 | "metadata": {}, 47 | "source": [ 48 | "**4[20].** $~$ (a) $~$Find the Fourier series of $f$ on the given interval\n", 49 | "\n", 50 | "> $f(x) = x^2, \\;\\;-\\pi \\le x \\le \\pi$\n", 51 | "\n", 52 | "$~\\;\\;\\;\\;$(b) $~$Use the result of (a) and Parseval's Theorem to evaluate the following series:\n", 53 | "\n", 54 | "> $\\displaystyle 1 + \\frac{1}{2^4} + \\frac{1}{3^4} + \\frac{1}{4^4} + \\cdots$\n", 55 | "\n", 56 | "\n", 57 | "> **Parseval's Theorem** $~$If $f$ is piecewise continuous, of period $2\\pi$, then\n", 58 | "\n", 59 | "> $\\displaystyle\\frac{1}{\\pi} \\int_0^{2\\pi} \\left[ f(x) \\right]^2 \\,dx = \\frac{1}{2}a_0^2 + \\sum_{n=1}^\\infty \\left( a_n^2 + b_n^2 \\right)$\n", 60 | "\n", 61 | "> where $a_0$, $a_n$, and $b_n$ are Fourier coefficients" 62 | ] 63 | }, 64 | { 65 | "cell_type": "markdown", 66 | "metadata": {}, 67 | "source": [ 68 | "**5[20].** $\\text{ }$ Hermite differential equation $y'' -2xy' +2ny = 0$ has a polynomial solution $H_n(x)$ for $n = 0, 1,2,\\cdots$:\n", 69 | "\n", 70 | "\\begin{align*}\n", 71 | " H_0(x) &= 1\\\\ \n", 72 | " H_1(x) &= 2x\\\\ \n", 73 | " H_2(x) &= 4x^2 - 2 \\\\ \n", 74 | " H_3(x) &= 8x^3-12x \\\\\n", 75 | " &\\;\\,\\vdots \n", 76 | "\\end{align*}\n", 77 | "\n", 78 | "$\\text{ }\\;\\;\\;$ Specify the weight function $p(x)$ and the interval over which the set of Hemite polynomials $\\{ H_n(x): \\;n=0,1,2,\\cdots \\}$ is orthogonal" 79 | ] 80 | }, 81 | { 82 | "cell_type": "markdown", 83 | "metadata": {}, 84 | "source": [ 85 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2022_2_A.ipynb)" 86 | ] 87 | } 88 | ], 89 | "metadata": { 90 | "kernelspec": { 91 | "display_name": "Python 3 (ipykernel)", 92 | "language": "python", 93 | "name": "python3" 94 | }, 95 | "language_info": { 96 | "codemirror_mode": { 97 | "name": "ipython", 98 | "version": 3 99 | }, 100 | "file_extension": ".py", 101 | "mimetype": "text/x-python", 102 | "name": "python", 103 | "nbconvert_exporter": "python", 104 | "pygments_lexer": "ipython3", 105 | "version": "3.8.12" 106 | } 107 | }, 108 | "nbformat": 4, 109 | "nbformat_minor": 4 110 | } 111 | -------------------------------------------------------------------------------- /exams/mid_2022_2_B.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exams/mid_2022_2_B.ipynb)" 8 | ] 9 | }, 10 | { 11 | "cell_type": "markdown", 12 | "metadata": {}, 13 | "source": [ 14 | "## Engineering Mathematics II (161007-21002) - Mid. Exam.\n", 15 | "October 24, 2022" 16 | ] 17 | }, 18 | { 19 | "cell_type": "markdown", 20 | "metadata": {}, 21 | "source": [ 22 | "**1[20].** $~$ Evaluate the surface integral $\\displaystyle\\iint_S G(x, y, z) \\,dS$:\n", 23 | "\n", 24 | "> $G(x,y,z) = \\left(1 - x^2 -y^2 \\right )^{3/2}$\n", 25 | "\n", 26 | "> where $S$ is the hemisphere $z=\\left( 1 - x^2 -y^2 \\right)^{1/2}$" 27 | ] 28 | }, 29 | { 30 | "cell_type": "markdown", 31 | "metadata": {}, 32 | "source": [ 33 | "**2[10].** $~$ Is it possible to write $\\mathbf{G}=\\nabla \\times \\mathbf{H}$? In the case where it is possible, find $\\mathbf{H}$\n", 34 | "\n", 35 | "> $\\mathbf{G}=2x\\mathbf{i} -y\\mathbf{j} +z\\mathbf{k}$" 36 | ] 37 | }, 38 | { 39 | "cell_type": "markdown", 40 | "metadata": {}, 41 | "source": [ 42 | "**3[20].** $~$ Neihter of the following functions is smooth everywhere. Nonetheless each can be written as the gradient of a scalar function. Find that scalar function:\n", 43 | "\n", 44 | "> (a) $~\\displaystyle\\mathbf{F}=\\frac{\\mathbf{r}}{r^2}$, $~\\mathbf{r}=x\\mathbf{i} + y\\mathbf{j}$\n", 45 | "\n", 46 | "> (b) $~\\displaystyle\\mathbf{F}=\\frac{\\mathbf{r}}{r^{1/2}}$, $~\\mathbf{r}=x\\mathbf{i} + y\\mathbf{j} + z\\mathbf{k}$" 47 | ] 48 | }, 49 | { 50 | "cell_type": "markdown", 51 | "metadata": {}, 52 | "source": [ 53 | "**4[30].** $~$ (a) $~$Find the Fourier series of $f$ on the given interval\n", 54 | "\n", 55 | "> $f(x) = \\left| x \\right|, \\;\\;-\\pi \\le x \\le \\pi$\n", 56 | "\n", 57 | "$~\\;\\;\\;\\;$(b) $~$Use the result of (a) and Parseval's Theorem to evaluate the following series:\n", 58 | "\n", 59 | "> $\\displaystyle 1 + \\frac{1}{3^4} + \\frac{1}{5^4} + \\frac{1}{7^4} + \\cdots$\n", 60 | "\n", 61 | "\n", 62 | "> **Parseval's Theorem** $~$If $f$ is piecewise continuous, of period $2\\pi$, then\n", 63 | "\n", 64 | "> $\\displaystyle\\frac{1}{\\pi} \\int_0^{2\\pi} \\left[ f(x) \\right]^2 \\,dx = \\frac{1}{2}a_0^2 + \\sum_{n=1}^\\infty \\left( a_n^2 + b_n^2 \\right)$\n", 65 | "\n", 66 | "> where $a_0$, $a_n$, and $b_n$ are Fourier coefficients" 67 | ] 68 | }, 69 | { 70 | "cell_type": "markdown", 71 | "metadata": {}, 72 | "source": [ 73 | "**5[20].** $~$ Find all solutions of the problem\n", 74 | "\n", 75 | "> $y'' + \\lambda y = 0, \\;\\;y(0)+y'(0)=0, \\;y(2\\pi) + y'(2\\pi) = 0$" 76 | ] 77 | } 78 | ], 79 | "metadata": { 80 | "kernelspec": { 81 | "display_name": "Python 3 (ipykernel)", 82 | "language": "python", 83 | "name": "python3" 84 | }, 85 | "language_info": { 86 | "codemirror_mode": { 87 | "name": "ipython", 88 | "version": 3 89 | }, 90 | "file_extension": ".py", 91 | "mimetype": "text/x-python", 92 | "name": "python", 93 | "nbconvert_exporter": "python", 94 | "pygments_lexer": "ipython3", 95 | "version": "3.8.12" 96 | } 97 | }, 98 | "nbformat": 4, 99 | "nbformat_minor": 4 100 | } 101 | -------------------------------------------------------------------------------- /exams/mid_2023_1_A.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "id": "c8a295c0", 6 | "metadata": {}, 7 | "source": [ 8 | "### Engineering Mathematics I (161006-21003) - Mid. Exam.\n", 9 | "\n", 10 | "April 12, 2023" 11 | ] 12 | }, 13 | { 14 | "cell_type": "markdown", 15 | "id": "c1f28567", 16 | "metadata": {}, 17 | "source": [ 18 | "**1[20].** $~$Find a general solution of the equation $~\\displaystyle ydx+(2x+xy-3)dy=0$\n", 19 | "\n", 20 | "$~$\n" 21 | ] 22 | }, 23 | { 24 | "cell_type": "markdown", 25 | "id": "937ecce5", 26 | "metadata": {}, 27 | "source": [ 28 | "**2[15].** $~$Solve the differential equation $~\\displaystyle xv\\frac{dv}{dx} +v^2=32x$\n", 29 | "\n", 30 | "$~$\n", 31 | "\n" 32 | ] 33 | }, 34 | { 35 | "cell_type": "markdown", 36 | "id": "4b74f6de", 37 | "metadata": {}, 38 | "source": [ 39 | "**3[15].** $~$Solve $~y'' - 2y' +y = 4x^2 - 3 +x^{-1} e^x$\n", 40 | "\n", 41 | "$~$\n", 42 | "\n" 43 | ] 44 | }, 45 | { 46 | "cell_type": "markdown", 47 | "id": "f2b9a9e7", 48 | "metadata": {}, 49 | "source": [ 50 | "**4[15].** $~$Solve the given initial-value problem on the interval $(-\\infty, 0)$.\n", 51 | "\n", 52 | "$\\;\\;\\;\\;\\;\\;\\;\\;\\;4x^2y'' + y = 0, \\;y(-1)=2, \\;y'(-1)=4$\n", 53 | "\n", 54 | "$~$\n", 55 | "\n" 56 | ] 57 | }, 58 | { 59 | "cell_type": "markdown", 60 | "id": "f0edd1ea", 61 | "metadata": {}, 62 | "source": [ 63 | "**5[20].** $~$ $~y=e^{-t^2}$ is a solution of the initial value problem \n", 64 | "$~\\displaystyle\\frac{dy}{dt} +2ty=0, \\;y(0)=1$. $~$Find $~Y(s)=\\mathcal{L}\\left[ e^{-t^2}\\right]$ by applying\n", 65 | "the Laplace transform to the given differential equation\n", 66 | "\n", 67 | "\n", 68 | "$~$\n", 69 | "\n", 70 | "\n", 71 | "$\\,\\;\\;\\;\\;\\;\\;\\;\\;$Hint) $\\displaystyle\\int_0^\\infty e^{-t^2}\\, dt = \\frac{\\sqrt{\\pi}}{2}, \\;\\;\\mathrm{erf}(t) = \\frac{2}{\\sqrt{\\pi}} \\int_0^t e^{-u^2}\\,du, \\;\\;\\mathrm{erf}c(t) = 1 - \\mathrm{erf}(t)$\n", 72 | "\n", 73 | "\n", 74 | "$~$" 75 | ] 76 | }, 77 | { 78 | "cell_type": "markdown", 79 | "id": "4733cea6", 80 | "metadata": {}, 81 | "source": [ 82 | "**6[15].** Solve the given differential equation $\\displaystyle ~ y''+2y'+y=(t-1) u(t-1) +\\delta(t-5)$, $y(0)=0$, $y'(0)=0$\n", 83 | "\n", 84 | "$~$" 85 | ] 86 | } 87 | ], 88 | "metadata": { 89 | "kernelspec": { 90 | "display_name": "Python 3 (ipykernel)", 91 | "language": "python", 92 | "name": "python3" 93 | }, 94 | "language_info": { 95 | "codemirror_mode": { 96 | "name": "ipython", 97 | "version": 3 98 | }, 99 | "file_extension": ".py", 100 | "mimetype": "text/x-python", 101 | "name": "python", 102 | "nbconvert_exporter": "python", 103 | "pygments_lexer": "ipython3", 104 | "version": "3.9.13" 105 | } 106 | }, 107 | "nbformat": 4, 108 | "nbformat_minor": 5 109 | } 110 | -------------------------------------------------------------------------------- /exams/mid_2023_1_B.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "id": "c8a295c0", 6 | "metadata": {}, 7 | "source": [ 8 | "### Engineering Mathematics I (161006-21004) - Mid. Exam.\n", 9 | "\n", 10 | "April 10, 2023" 11 | ] 12 | }, 13 | { 14 | "cell_type": "markdown", 15 | "id": "c1f28567", 16 | "metadata": {}, 17 | "source": [ 18 | "**1[20].** $~$Solve the initial-value problem $\\displaystyle\\frac{dy}{dt} + 2(t+1) y^2=0, \\;\\;y(0)=-\\frac{1}{8}$ and give the largest interval $I$ on which the solution is defined:\n", 19 | "\n", 20 | "\n", 21 | "$~$\n", 22 | "\n", 23 | "$~$" 24 | ] 25 | }, 26 | { 27 | "cell_type": "markdown", 28 | "id": "937ecce5", 29 | "metadata": {}, 30 | "source": [ 31 | "**2[15].** $~$Solve the differential equation $~\\displaystyle \\frac{dy}{dx}=\\sin (x+y)$\n", 32 | "\n", 33 | "$~$\n", 34 | "\n", 35 | "$~$" 36 | ] 37 | }, 38 | { 39 | "cell_type": "markdown", 40 | "id": "4b74f6de", 41 | "metadata": {}, 42 | "source": [ 43 | "**3[15].** $~$Solve $~2y'' + 2y' +y = 4\\sqrt{x}$\n", 44 | "\n", 45 | "$~$\n", 46 | "\n", 47 | "$~$" 48 | ] 49 | }, 50 | { 51 | "cell_type": "markdown", 52 | "id": "f2b9a9e7", 53 | "metadata": {}, 54 | "source": [ 55 | "**4[15].** $~$Find a homogeneous Cauchy-Euler differential equation whose general solution is given: $y = c_1 + c_2 x +c_3 x \\ln x$\n", 56 | "\n", 57 | "$~$\n", 58 | "\n", 59 | "$~$" 60 | ] 61 | }, 62 | { 63 | "cell_type": "markdown", 64 | "id": "f0edd1ea", 65 | "metadata": {}, 66 | "source": [ 67 | "**5[15].** $~$Solve $~y'' + 4y' + 3y =e^t \\delta (t-1), \\;\\;y(0)=0, \\;y'(0)=2$\n", 68 | "\n", 69 | "$~$\n", 70 | "\n", 71 | "$~$" 72 | ] 73 | }, 74 | { 75 | "cell_type": "markdown", 76 | "id": "4733cea6", 77 | "metadata": {}, 78 | "source": [ 79 | "**6[20].** Use the Laplace transform to find the numerical value of the improper integral $~\\displaystyle \\int_0^\\infty t e^{-2t} \\sin 4t \\,dt$\n", 80 | "\n", 81 | "$~$\n", 82 | "\n", 83 | "$~$" 84 | ] 85 | } 86 | ], 87 | "metadata": { 88 | "kernelspec": { 89 | "display_name": "Python 3 (ipykernel)", 90 | "language": "python", 91 | "name": "python3" 92 | }, 93 | "language_info": { 94 | "codemirror_mode": { 95 | "name": "ipython", 96 | "version": 3 97 | }, 98 | "file_extension": ".py", 99 | "mimetype": "text/x-python", 100 | "name": "python", 101 | "nbconvert_exporter": "python", 102 | "pygments_lexer": "ipython3", 103 | "version": "3.9.13" 104 | } 105 | }, 106 | "nbformat": 4, 107 | "nbformat_minor": 5 108 | } 109 | -------------------------------------------------------------------------------- /exams/mid_2023_1_B_solution.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "id": "c8a295c0", 6 | "metadata": {}, 7 | "source": [ 8 | "### Engineering Mathematics I (161006-21004) - Mid. Exam.\n", 9 | "\n", 10 | "April 10, 2023" 11 | ] 12 | }, 13 | { 14 | "cell_type": "markdown", 15 | "id": "c1f28567", 16 | "metadata": {}, 17 | "source": [ 18 | "**1[20].** $~$Solve the initial-value problem $\\displaystyle\\frac{dy}{dt} + 2(t+1) y^2=0, \\;\\;y(0)=-\\frac{1}{8}$ and give the largest interval $I$ on which the solution is defined:\n", 19 | "\n", 20 | "\n", 21 | "$~$\n", 22 | "\n", 23 | "* **Solution**\n", 24 | "\n", 25 | "\\begin{align*}\n", 26 | " \\frac{dy}{dt} &+2(t+1)y^2 = 0 \\\\ \n", 27 | " &\\Downarrow \\\\\n", 28 | " \\frac{dy}{y^2} &= -2(t+1) \\, dt \\\\\n", 29 | " &\\Downarrow \\\\ \n", 30 | " y &= \\frac{1}{t^2 + 2t + c}, \\;\\; y(0)=-\\frac{1}{8} \\\\\n", 31 | " &\\Downarrow \\\\ \n", 32 | " \\color{red}{y} &\\color{red}{= \\frac{1}{(t-2)(t+4)}, \\;\\;-4 $\\displaystyle \\mathbf{x}(t) = \\left(t, t, t^2 \\right)$\n" 18 | ] 19 | }, 20 | { 21 | "cell_type": "markdown", 22 | "metadata": {}, 23 | "source": [ 24 | "**2[15].** $\\phantom{1}$ Let $\\mathbf{r} = x \\mathbf{i} + y \\mathbf{j} + z \\mathbf{k}$ and $r$ denote $\\| \\mathbf{r} \\|$. Evaluate the following:\n", 25 | "\n", 26 | "> $\\nabla \\cdot (r^n \\mathbf{r})$" 27 | ] 28 | }, 29 | { 30 | "cell_type": "markdown", 31 | "metadata": {}, 32 | "source": [ 33 | "**3[20].** $\\phantom{1}$ Let $\\mathbf{v} = u(x,y) \\mathbf{i} - v(x,y) \\mathbf{j}$ be an **incompressible**, **irrotational** velocity field\n", 34 | "\n", 35 | "> (a) Show that the functions $u$ and $v$ (which determine the component functions of $\\mathbf{v}$) satisfy the **Cauchy-Riemann equations**\n", 36 | "\n", 37 | " $$\\frac{\\partial u}{\\partial x} = \\frac{\\partial v}{\\partial y}\\,\\; \\text{and}\\; \\,\n", 38 | " \\frac{\\partial u}{\\partial y} = -\\frac{\\partial v}{\\partial x}$$\n", 39 | "\n", 40 | "> (b) Show that $u$ and $v$ are **harmonic**, that is, that\n", 41 | "\n", 42 | " $$\\frac{\\partial^2 u}{\\partial x^2} + \\frac{\\partial^2 u}{\\partial y^2}=0\\, \\;\\text{and}\\; \\,\n", 43 | " \\frac{\\partial^2 v}{\\partial x^2} + \\frac{\\partial^2 v}{\\partial y^2} = 0$$\n" 44 | ] 45 | }, 46 | { 47 | "cell_type": "markdown", 48 | "metadata": {}, 49 | "source": [ 50 | "**4[15].** $\\phantom{1}$ (a) $\\phantom{1}$Find the Fourier series of $f$ on the given interval\n", 51 | "\n", 52 | "$$f(x) = |x|, \\;\\;-\\pi \\le x \\le \\pi$$\n", 53 | "\n", 54 | "> (b) $\\phantom{1}$Use the result of (a) and Parseval's Theorem to evaluate the following series:\n", 55 | "\n", 56 | "$$ 1 + \\frac{1}{3^4} + \\frac{1}{5^4} + \\frac{1}{7^4} + \\cdots$$\n", 57 | "\n", 58 | "\n", 59 | "> **Parseval's Theorem** $\\phantom{1}$If $f$ is piecewise continuous, of period $2\\pi$, then\n", 60 | "\n", 61 | "$$\\frac{1}{\\pi} \\int_0^{2\\pi} \\left[ f(x) \\right]^2 \\,dx = \\frac{1}{2}a_0^2 + \\sum_{n=1}^\\infty \\left( a_n^2 + b_n^2 \\right)$$\n", 62 | "\n", 63 | "> where $a_0$, $a_n$, and $b_n$ are Fourier coefficients" 64 | ] 65 | }, 66 | { 67 | "cell_type": "markdown", 68 | "metadata": {}, 69 | "source": [ 70 | "**5[15].** $\\phantom{1}$ (a) Find the eigenvalues and eigenfunctions of the boundary-value problem\n", 71 | "\n", 72 | "> $y''+y'+\\lambda y=0, \\;y(0)=0, \\; y(2)=0$\n", 73 | "\n", 74 | "> (b) Put the differential equation in self-adjoint form\n", 75 | "\n", 76 | "> (c) Give an orthogonality condition" 77 | ] 78 | }, 79 | { 80 | "cell_type": "markdown", 81 | "metadata": {}, 82 | "source": [ 83 | "**6[20].** $\\phantom{1}$ Expand $f(x)=x^4$, $-1 $\\displaystyle \\mathbf{x}(t) = \\left(2t, e^{2t} \\right)$\n" 18 | ] 19 | }, 20 | { 21 | "cell_type": "markdown", 22 | "metadata": {}, 23 | "source": [ 24 | "**2[15].** $\\phantom{1}$ Let $\\mathbf{r} = x \\mathbf{i} + y \\mathbf{j} + z \\mathbf{k}$ and $r$ denote $\\| \\mathbf{r} \\|$. Evaluate the following:\n", 25 | "\n", 26 | "> $\\nabla \\times (r^n \\mathbf{r})$" 27 | ] 28 | }, 29 | { 30 | "cell_type": "markdown", 31 | "metadata": {}, 32 | "source": [ 33 | "**3[20].** $\\phantom{1}$ Let a particle of mass $m$ travel along a differentiable path $\\mathbf{x}$ in a Newtonian vector field $\\mathbf{F}$ (i.e., one that satisfies Newton's second law $\\mathbf{F} = m\\mathbf{a}$, where $\\mathbf{a}$ is the acceleration of $\\mathbf{x}$). We define the **angular momentum** $\\mathbf{l}(t)$ of the particle to be the cross product of the position vector and the linear momentum $m\\mathbf{v}$, i.e.,\n", 34 | "\n", 35 | "$$\\mathbf{l}(t) = \\mathbf{x}(t) \\times m \\mathbf{v}(t)$$\n", 36 | "\n", 37 | "> (Here $\\mathbf{v}$ denotes the velocity of $\\mathbf{x}$.) The **torque** about the origin of the coordinate system due to the force $\\mathbf{F}$ is the cross product of position and force:\n", 38 | "\n", 39 | "$$\\mathbf{M}(t) = \\mathbf{x}(t) \\times \\mathbf{F}(t)$$" 40 | ] 41 | }, 42 | { 43 | "cell_type": "markdown", 44 | "metadata": {}, 45 | "source": [ 46 | "> (a) Show that \n", 47 | "\n", 48 | " $$\\frac{d\\mathbf{l}}{dt} = \\mathbf{M}$$\n", 49 | " \n", 50 | "> Thus, we see that the rate of change of angular momentum is equal to the torque imparted to the particle by the vector field $\\mathbf{F}$\n", 51 | "\n", 52 | "> (b) Suppose that $\\mathbf{F}$ is a central force (i.e., a force that always points directly towards or away from the origin). Show that in this case the angular momentum is **conserved**, that is, that it must remain constant" 53 | ] 54 | }, 55 | { 56 | "cell_type": "markdown", 57 | "metadata": {}, 58 | "source": [ 59 | "**4[15].** $\\phantom{1}$ The **root-mean-square value** of a function $f(x)$ defined over an interval $(a, b)$ is given by\n", 60 | "\n", 61 | "$$\\text{RMS}(f) = \\sqrt{\\frac{\\int_a^b \\,f^2(x)\\,dx}{b-a}}$$" 62 | ] 63 | }, 64 | { 65 | "cell_type": "markdown", 66 | "metadata": {}, 67 | "source": [ 68 | "> Show that the RMS value of $f$ over the interval $(-p, p)$ is given by\n", 69 | "\n", 70 | "$$\\text{RMS}(f) = \\sqrt{\\frac{1}{4} a_0^2 +\\frac{1}{2} \\sum_{n=1}^\\infty \\left(a_n^2 + b_n^2 \\right)}$$\n", 71 | "\n", 72 | "> where $a_0$, $a_n$, and $b_n$ are the Fourier coefficients, respectively" 73 | ] 74 | }, 75 | { 76 | "cell_type": "markdown", 77 | "metadata": {}, 78 | "source": [ 79 | "**5[15].** $\\phantom{1}$ Find the eigenfuctions and the equation that defines the eigenvalues for the given boundary-value problem:\n", 80 | "\n", 81 | "> $\\displaystyle y''+\\lambda y = 0, \\;y(0) +y'(0) = 0, \\; y(1)=0$" 82 | ] 83 | }, 84 | { 85 | "cell_type": "markdown", 86 | "metadata": {}, 87 | "source": [ 88 | "**6[20].** $\\phantom{1}$ Expand the given function in a Fourier-Bessel series using Bessel functions of the same order as in the indicated boundary condition: \n", 89 | "\n", 90 | "> $\\displaystyle f(x)=x^2, \\;\\;0$\\displaystyle \\frac{dT}{dt}=k(T -T_m), \\;\\;T(0)=T_0$\n", 25 | "\n", 26 | ">" 27 | ] 28 | }, 29 | { 30 | "cell_type": "markdown", 31 | "metadata": { 32 | "slideshow": { 33 | "slide_type": "fragment" 34 | } 35 | }, 36 | "source": [ 37 | "$\\text{ }$ **solution**\n", 38 | "\n", 39 | ">$\\begin{align*}\n", 40 | " \\frac{dT}{dt} &=k(T -T_m), \\;\\;T(0)=T_0 \\\\ \n", 41 | " &\\Downarrow \\\\ \n", 42 | " T(0) &= T_0 \\\\\n", 43 | " T(\\infty) &= T_m \\\\\n", 44 | " \\text{plot }\\ln \\left( \\frac{T -T_m}{T_0 -T_m} \\right) &\\text{ vs. } t \\;\\rightarrow \\; k \\\\\n", 45 | "\\end{align*}$" 46 | ] 47 | }, 48 | { 49 | "cell_type": "markdown", 50 | "metadata": { 51 | "slideshow": { 52 | "slide_type": "slide" 53 | } 54 | }, 55 | "source": [ 56 | "**19.** After a mass $m$ is attached to a spring, it stretches $s$ units and then hangs at rest in the equilibrium position as shown in Figure (b). After the spring/mass system has been set in motion, let $x(t)$ denote the directed distance of the mass beyond the equilibrium position. As indicated in Figure (c), assume that the downward direction is positive, that the motion takes place in a vertical straight line through the center of gravitiy of the mass, and that the only forces acting on the system are the weight of the mass and the restoring force of the stretched spring. Use Hooke's law: The restoring force of a spring is proportional to its total elongation. Determine a differential equation for the displacement $x(t)$ a time $t$\n", 57 | "\n", 58 | ">" 59 | ] 60 | }, 61 | { 62 | "cell_type": "markdown", 63 | "metadata": { 64 | "slideshow": { 65 | "slide_type": "fragment" 66 | } 67 | }, 68 | "source": [ 69 | "$\\text{ }$ **solution**\n", 70 | "\n", 71 | ">" 72 | ] 73 | } 74 | ], 75 | "metadata": { 76 | "celltoolbar": "Slideshow", 77 | "kernelspec": { 78 | "display_name": "Python 3 (ipykernel)", 79 | "language": "python", 80 | "name": "python3" 81 | }, 82 | "language_info": { 83 | "codemirror_mode": { 84 | "name": "ipython", 85 | "version": 3 86 | }, 87 | "file_extension": ".py", 88 | "mimetype": "text/x-python", 89 | "name": "python", 90 | "nbconvert_exporter": "python", 91 | "pygments_lexer": "ipython3", 92 | "version": "3.9.18" 93 | } 94 | }, 95 | "nbformat": 4, 96 | "nbformat_minor": 4 97 | } 98 | -------------------------------------------------------------------------------- /exercises/ch02_exer02.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": { 6 | "slideshow": { 7 | "slide_type": "skip" 8 | } 9 | }, 10 | "source": [ 11 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exercises/ch02_exer02.ipynb)" 12 | ] 13 | }, 14 | { 15 | "cell_type": "markdown", 16 | "metadata": { 17 | "slideshow": { 18 | "slide_type": "slide" 19 | } 20 | }, 21 | "source": [ 22 | "$\\text{ }$ Solve the given differential equation by separation of variables\n", 23 | "\n", 24 | "**1.** $\\text{ }\\displaystyle \\frac{dy}{dx}=\\sin 5 x$\n", 25 | "\n", 26 | "$\\text{ }$ **solution**\n", 27 | "\n", 28 | ">$\\begin{align*}\n", 29 | " \\frac{dy}{dx} &=\\sin 5x\\\\ \n", 30 | " &\\Downarrow \\\\ \n", 31 | " dy &= \\sin 5x \\,dx \\\\\n", 32 | " &\\Downarrow \\\\ \n", 33 | " y &=-\\frac{1}{5} \\cos 5x +c \n", 34 | "\\end{align*}$" 35 | ] 36 | }, 37 | { 38 | "cell_type": "markdown", 39 | "metadata": { 40 | "slideshow": { 41 | "slide_type": "slide" 42 | } 43 | }, 44 | "source": [ 45 | "**3.** $\\text{ }\\displaystyle dx +e^{3x} dy = 0$\n", 46 | "\n", 47 | "$\\text{ }$ **solution**\n", 48 | "\n", 49 | ">$\\begin{align*}\n", 50 | " dy &=-e^{-3x} \\,dx\\\\ \n", 51 | " &\\Downarrow \\\\ \n", 52 | " y &= \\frac{1}{3}e^{-3x} +c \n", 53 | "\\end{align*}$" 54 | ] 55 | }, 56 | { 57 | "cell_type": "markdown", 58 | "metadata": { 59 | "slideshow": { 60 | "slide_type": "slide" 61 | } 62 | }, 63 | "source": [ 64 | "**21.** $\\text{ }\\displaystyle \\frac{dy}{dx} = x\\sqrt{1 -y^2}$\n", 65 | "\n", 66 | "$\\text{ }$ **solution**\n", 67 | "\n", 68 | ">$\\begin{align*}\n", 69 | " \\frac{1}{\\sqrt{1 -y^2}}dy &= x \\,dx\\\\ \n", 70 | " &\\Downarrow \\\\ \n", 71 | " \\arcsin y &= \\frac{1}{2}x^2 +c \\\\\n", 72 | " &\\Downarrow \\\\\n", 73 | " y &= \\sin \\left(\\frac{1}{2}x^2 +c \\right)\n", 74 | "\\end{align*}$" 75 | ] 76 | }, 77 | { 78 | "cell_type": "markdown", 79 | "metadata": { 80 | "slideshow": { 81 | "slide_type": "slide" 82 | } 83 | }, 84 | "source": [ 85 | "Find an implicit and an explicit solution of the given initial-value problem\n", 86 | "\n", 87 | "**25.** $\\text{ }\\displaystyle x^2 \\frac{dy}{dx} = y -xy, \\;\\;y(-1)=-1$\n", 88 | "\n", 89 | "$\\text{ }$ **solution**\n", 90 | "\n", 91 | ">$\\begin{align*}\n", 92 | " \\frac{1}{y}dy &= \\frac{1-x}{x^2} \\,dx\\\\ \n", 93 | " &\\Downarrow \\\\ \n", 94 | " \\ln |y| &= -\\left(\\frac{1}{x} +\\ln |x| \\right) +c \\\\\n", 95 | " &\\Downarrow y(-1)=-1 \\; \\Rightarrow \\; c=-1\\\\\n", 96 | " \\ln |y| &= -\\left(1 +\\frac{1}{x} \\right) -\\ln |x| \\\\\n", 97 | " &\\Downarrow \\\\ \n", 98 | " y &= \\frac{1}{x} e^{-\\left( 1 +\\frac{1}{x}\\right)}\n", 99 | "\\end{align*}$" 100 | ] 101 | } 102 | ], 103 | "metadata": { 104 | "celltoolbar": "Slideshow", 105 | "kernelspec": { 106 | "display_name": "Python 3 (ipykernel)", 107 | "language": "python", 108 | "name": "python3" 109 | }, 110 | "language_info": { 111 | "codemirror_mode": { 112 | "name": "ipython", 113 | "version": 3 114 | }, 115 | "file_extension": ".py", 116 | "mimetype": "text/x-python", 117 | "name": "python", 118 | "nbconvert_exporter": "python", 119 | "pygments_lexer": "ipython3", 120 | "version": "3.9.13" 121 | } 122 | }, 123 | "nbformat": 4, 124 | "nbformat_minor": 4 125 | } 126 | -------------------------------------------------------------------------------- /exercises/ch02_exer07.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exercises/ch02_exer07.ipynb)" 8 | ] 9 | }, 10 | { 11 | "cell_type": "markdown", 12 | "metadata": {}, 13 | "source": [ 14 | "**39.** $\\text{ }$ **Evaporating Raindrop** $\\text{ }$ As a raindrop falls, it evaporates while retaining its spherical shape. If we make the further assumptions that the rate at which the raindrop evaporates is proportional to its surface area and that air resistance is negligible, then a model for the velocity $v(t)$ of the raindrop is\n", 15 | "\n", 16 | "$$\n", 17 | "\\frac{dv}{dt} + \\frac{3(k/\\rho)}{(k/\\rho) t +r_0} v = g\n", 18 | "$$\n", 19 | "\n", 20 | "Here $\\rho$ is the density of water, $r_0$ is the radius of the raindrop at $t=0$, $k < 0$ is the constant of proportionality $\\displaystyle \\frac{dm}{dt}=4\\pi r^2 k$, and the downward direction is taken to be the positive direction\n", 21 | "\n", 22 | "(a)$\\text{ }$ Solve for $v(t)$ if the raindrop falls from the rest\n", 23 | "\n", 24 | "(b)$\\text{ }$ Show that the radius of the raindrop at time $t$ is $r(t)=(k/\\rho)t +r_0$\n", 25 | "\n", 26 | "(c)$\\text{ }$ If $r_0=0.01$ ft and $r =0.007$ ft at time $t=10$ sec after the raindrop falls from a cloud, determine the time at which the raindrop has evaporated completely " 27 | ] 28 | }, 29 | { 30 | "cell_type": "markdown", 31 | "metadata": {}, 32 | "source": [ 33 | "**solution**\n", 34 | "\n", 35 | "(a)$\\text{ }$\n", 36 | "\n", 37 | "\\begin{align*}\n", 38 | " \\frac{dv}{dt} &+ \\frac{3(k/\\rho)}{(k/\\rho) t +r_0} v = g \\\\ \n", 39 | " &\\Downarrow \\text{ multiply by the integral factor } \\\\ \n", 40 | " \\frac{d}{dt} \\left[ \\left(\\frac{k}{\\rho}t +r_0 \\right)^3 v \\right] &= \\left(\\frac{k}{\\rho}t +r_0 \\right)^3 g\\\\ \n", 41 | " &\\Downarrow \\\\\n", 42 | " v(t)=\\frac{g\\rho}{4k} \\left( \\frac{k}{\\rho} t +r_0 \\right) &+c\\left(\\frac{k}{\\rho}t +r_0 \\right)^{-3} \\\\\n", 43 | " &\\Downarrow \\text{ } v(0)=0 \\; \\rightarrow \\;c = -\\frac{g\\rho}{4k} r_0^4 \\\\\n", 44 | " v(t) = \\frac{g\\rho}{4k} \\left(\\frac{k}{\\rho}t +r_0 \\right) & \\left[ 1 - \\left( \\frac{r_0}{ \\frac{k}{\\rho}t +r_0 } \n", 45 | " \\right)^4 \\right] \\\\ \n", 46 | "\\end{align*}\n", 47 | "\n", 48 | "(b)$\\text{ }$\n", 49 | "\n", 50 | "\\begin{align*}\n", 51 | " m &= \\frac{4}{3}\\pi r^3 \\rho\\\\ \n", 52 | " &\\Downarrow \\\\ \n", 53 | " \\frac{dm}{dt} &= 4\\pi r^2 \\rho \\frac{dr}{dt}\\\\ \n", 54 | " &\\Downarrow \\text{ } {\\scriptstyle \\frac{dm}{dt}=4\\pi r^2 k}\\\\\n", 55 | " \\frac{dr}{dt} &= \\frac{k}{\\rho} \\\\\n", 56 | " &\\Downarrow \\text{ } {\\scriptstyle r=r_0 \\text{ at } t=0} \\\\\n", 57 | " r(t) &= \\frac{k}{\\rho} t +r_0 \n", 58 | "\\end{align*}\n", 59 | "\n", 60 | "(c)$\\text{ }$\n", 61 | "\n", 62 | "\\begin{align*}\n", 63 | " r(10) &= \\frac{k}{\\rho} \\times 10 +0.01 =0.007 \\; \\Rightarrow \\; \\frac{k}{\\rho} = -0.0003\\\\ \n", 64 | " &\\Downarrow \\\\ \n", 65 | " r(t) &= -0.0003 t + 0.01 = 0 \\; \\Rightarrow \\; t =\\frac{0.01}{0.0003} \\approx 33.34 \\;\\text{sec}\n", 66 | "\\end{align*}" 67 | ] 68 | } 69 | ], 70 | "metadata": { 71 | "kernelspec": { 72 | "display_name": "Python 3", 73 | "language": "python", 74 | "name": "python3" 75 | }, 76 | "language_info": { 77 | "codemirror_mode": { 78 | "name": "ipython", 79 | "version": 3 80 | }, 81 | "file_extension": ".py", 82 | "mimetype": "text/x-python", 83 | "name": "python", 84 | "nbconvert_exporter": "python", 85 | "pygments_lexer": "ipython3", 86 | "version": "3.7.5" 87 | } 88 | }, 89 | "nbformat": 4, 90 | "nbformat_minor": 4 91 | } 92 | -------------------------------------------------------------------------------- /exercises/ch04_exer06.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exercises/ch04_exer06.ipynb)" 8 | ] 9 | }, 10 | { 11 | "cell_type": "markdown", 12 | "metadata": {}, 13 | "source": [ 14 | "**14.** $\\text{ }$ Derive the system of differential equations describing the straight-line vertical motion of the coupled springs shown in equilibrium in the following figure. Use the Laplace transform to solve the system when $k_1=1$, $k_2=1$, $k_3=1$, $m_1=1$, $m_2=1$ and $x_1(0)=0$, $\\dot{x}_1(0)=-1$, $x_2(0)=0$, $\\dot{x}_2(0)=1$.\n", 15 | "\n", 16 | "$\\text{ }$\n", 17 | "\n", 18 | "
\"coupled
" 19 | ] 20 | }, 21 | { 22 | "cell_type": "markdown", 23 | "metadata": {}, 24 | "source": [ 25 | "**solution**\n" 26 | ] 27 | } 28 | ], 29 | "metadata": { 30 | "kernelspec": { 31 | "display_name": "Python 3", 32 | "language": "python", 33 | "name": "python3" 34 | }, 35 | "language_info": { 36 | "codemirror_mode": { 37 | "name": "ipython", 38 | "version": 3 39 | }, 40 | "file_extension": ".py", 41 | "mimetype": "text/x-python", 42 | "name": "python", 43 | "nbconvert_exporter": "python", 44 | "pygments_lexer": "ipython3", 45 | "version": "3.7.5" 46 | } 47 | }, 48 | "nbformat": 4, 49 | "nbformat_minor": 4 50 | } 51 | -------------------------------------------------------------------------------- /exercises/ch10_exer02.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exercises/ch10_exer02.ipynb)" 8 | ] 9 | }, 10 | { 11 | "cell_type": "markdown", 12 | "metadata": {}, 13 | "source": [ 14 | "**1.** $\\text{ }$ Find the general solution of the given system\n", 15 | "\n", 16 | "$~$\n", 17 | "\n", 18 | ">$\\begin{align*}\n", 19 | " \\frac{dx}{dt}&= x +2y\\\\ \n", 20 | " \\frac{dy}{dt}&= 4x +3y \n", 21 | "\\end{align*}$" 22 | ] 23 | }, 24 | { 25 | "cell_type": "markdown", 26 | "metadata": {}, 27 | "source": [ 28 | "**solution**\n" 29 | ] 30 | }, 31 | { 32 | "cell_type": "markdown", 33 | "metadata": {}, 34 | "source": [ 35 | "**13.** $\\text{ }$ Solve the given initial-value problem\n", 36 | "\n", 37 | ">$\n", 38 | "\\mathbf{x}' = \\begin{pmatrix}\n", 39 | " \\frac{1}{2}& 0\\\\ \n", 40 | " 1& -\\frac{1}{2} \n", 41 | "\\end{pmatrix} \\mathbf{x}, \\;\\;\\mathbf{x}(0)=\\begin{pmatrix}\n", 42 | "3\\\\5\n", 43 | "\\end{pmatrix}\n", 44 | "$" 45 | ] 46 | }, 47 | { 48 | "cell_type": "markdown", 49 | "metadata": {}, 50 | "source": [ 51 | "**solution**" 52 | ] 53 | }, 54 | { 55 | "cell_type": "markdown", 56 | "metadata": {}, 57 | "source": [ 58 | "**19.** $\\text{ }$ Find the general solution of the given system\n", 59 | "\n", 60 | "$~$\n", 61 | "\n", 62 | ">$\\begin{align*}\n", 63 | " \\frac{dx}{dt}&= 3x -y\\\\ \n", 64 | " \\frac{dy}{dt}&= 9x -3y \n", 65 | "\\end{align*}$" 66 | ] 67 | }, 68 | { 69 | "cell_type": "markdown", 70 | "metadata": {}, 71 | "source": [ 72 | "**solution**" 73 | ] 74 | }, 75 | { 76 | "cell_type": "markdown", 77 | "metadata": {}, 78 | "source": [ 79 | "**35.** $\\text{ }$ Find the general solution of the given system\n", 80 | "\n", 81 | "$~$\n", 82 | "\n", 83 | ">$\\begin{align*}\n", 84 | " \\frac{dx}{dt}&= 6x -y\\\\ \n", 85 | " \\frac{dy}{dt}&= 5x +2y \n", 86 | "\\end{align*}$" 87 | ] 88 | }, 89 | { 90 | "cell_type": "markdown", 91 | "metadata": {}, 92 | "source": [ 93 | "**solution**" 94 | ] 95 | } 96 | ], 97 | "metadata": { 98 | "kernelspec": { 99 | "display_name": "Python 3", 100 | "language": "python", 101 | "name": "python3" 102 | }, 103 | "language_info": { 104 | "codemirror_mode": { 105 | "name": "ipython", 106 | "version": 3 107 | }, 108 | "file_extension": ".py", 109 | "mimetype": "text/x-python", 110 | "name": "python", 111 | "nbconvert_exporter": "python", 112 | "pygments_lexer": "ipython3", 113 | "version": "3.7.5" 114 | } 115 | }, 116 | "nbformat": 4, 117 | "nbformat_minor": 4 118 | } 119 | -------------------------------------------------------------------------------- /exercises/ch10_exer03.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exercises/ch10_exer03.ipynb)" 8 | ] 9 | }, 10 | { 11 | "cell_type": "markdown", 12 | "metadata": {}, 13 | "source": [ 14 | "**1.** $\\text{ }$ Use the diagonalization to solve the given system\n", 15 | "\n", 16 | ">$\n", 17 | "\\mathbf{x}' = \\begin{pmatrix}\n", 18 | " 5& \\;\\;6\\\\ \n", 19 | " 3& -2 \n", 20 | "\\end{pmatrix} \\mathbf{x}\n", 21 | "$" 22 | ] 23 | }, 24 | { 25 | "cell_type": "markdown", 26 | "metadata": {}, 27 | "source": [ 28 | "**solution**" 29 | ] 30 | } 31 | ], 32 | "metadata": { 33 | "kernelspec": { 34 | "display_name": "Python 3", 35 | "language": "python", 36 | "name": "python3" 37 | }, 38 | "language_info": { 39 | "codemirror_mode": { 40 | "name": "ipython", 41 | "version": 3 42 | }, 43 | "file_extension": ".py", 44 | "mimetype": "text/x-python", 45 | "name": "python", 46 | "nbconvert_exporter": "python", 47 | "pygments_lexer": "ipython3", 48 | "version": "3.7.5" 49 | } 50 | }, 51 | "nbformat": 4, 52 | "nbformat_minor": 4 53 | } 54 | -------------------------------------------------------------------------------- /exercises/ch10_exer04.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exercises/ch10_exer04.ipynb)" 8 | ] 9 | }, 10 | { 11 | "cell_type": "markdown", 12 | "metadata": {}, 13 | "source": [ 14 | "**3.** $\\text{ }$ Use the method of undetermined coefficients to solve the given system\n", 15 | "\n", 16 | ">$\n", 17 | "\\mathbf{x}' = \\begin{pmatrix}\n", 18 | " 1& 3\\\\ \n", 19 | " 3& 1 \n", 20 | "\\end{pmatrix} \\mathbf{x}+\\begin{pmatrix}\n", 21 | "-2t^2\\\\t +5\n", 22 | "\\end{pmatrix}\n", 23 | "$" 24 | ] 25 | }, 26 | { 27 | "cell_type": "markdown", 28 | "metadata": {}, 29 | "source": [ 30 | "**solution**" 31 | ] 32 | }, 33 | { 34 | "cell_type": "markdown", 35 | "metadata": {}, 36 | "source": [ 37 | "**7.** $\\text{ }$ Use the method of undetermined coefficients to solve the given system\n", 38 | "\n", 39 | ">$\n", 40 | "\\mathbf{x}' = \\begin{pmatrix}\n", 41 | " 1& 1 & 1\\\\ \n", 42 | " 0 & 2 & 3 \\\\\n", 43 | " 0 & 0 & 5\n", 44 | "\\end{pmatrix} \\mathbf{x}+\\begin{pmatrix}\n", 45 | "\\;\\;1 \\\\ -1 \\\\ \\;\\;2\n", 46 | "\\end{pmatrix}e^{4t}\n", 47 | "$" 48 | ] 49 | }, 50 | { 51 | "cell_type": "markdown", 52 | "metadata": {}, 53 | "source": [ 54 | "**solution**\n" 55 | ] 56 | }, 57 | { 58 | "cell_type": "markdown", 59 | "metadata": {}, 60 | "source": [ 61 | "**33.** $\\text{ }$ Solve the given initial-value problem\n", 62 | "\n", 63 | "$~$\n", 64 | "\n", 65 | ">$\n", 66 | "\\mathbf{x}' = \\begin{pmatrix}\n", 67 | " \\;\\; 3& -1\\\\ \n", 68 | " -1& \\;\\;3 \n", 69 | "\\end{pmatrix} \\mathbf{x}+\\begin{pmatrix}\n", 70 | "4e^{2t} \\\\ 4e^{4t}\n", 71 | "\\end{pmatrix}, \\;\\; \\mathbf{x}(0) = \\begin{pmatrix} 1 \\\\ 1 \\end{pmatrix}\n", 72 | "$" 73 | ] 74 | }, 75 | { 76 | "cell_type": "markdown", 77 | "metadata": {}, 78 | "source": [ 79 | "**solution**" 80 | ] 81 | } 82 | ], 83 | "metadata": { 84 | "kernelspec": { 85 | "display_name": "Python 3", 86 | "language": "python", 87 | "name": "python3" 88 | }, 89 | "language_info": { 90 | "codemirror_mode": { 91 | "name": "ipython", 92 | "version": 3 93 | }, 94 | "file_extension": ".py", 95 | "mimetype": "text/x-python", 96 | "name": "python", 97 | "nbconvert_exporter": "python", 98 | "pygments_lexer": "ipython3", 99 | "version": "3.7.5" 100 | } 101 | }, 102 | "nbformat": 4, 103 | "nbformat_minor": 4 104 | } 105 | -------------------------------------------------------------------------------- /exercises/ch10_exer05.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": {}, 6 | "source": [ 7 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/exercises/ch10_exer05.ipynb)" 8 | ] 9 | }, 10 | { 11 | "cell_type": "markdown", 12 | "metadata": {}, 13 | "source": [ 14 | "**1.** $\\text{ }$ Compute $e^{\\mathbf{A}t}$ and $e^{-\\mathbf{A}t}$\n", 15 | ">$\n", 16 | "\\begin{pmatrix}\n", 17 | " 1& 0\\\\ \n", 18 | " 0& 2 \n", 19 | "\\end{pmatrix} \n", 20 | "$" 21 | ] 22 | }, 23 | { 24 | "cell_type": "markdown", 25 | "metadata": {}, 26 | "source": [ 27 | "**solution**" 28 | ] 29 | }, 30 | { 31 | "cell_type": "markdown", 32 | "metadata": {}, 33 | "source": [ 34 | "**5.** $\\text{ }$ Find the general solution of the given system\n", 35 | "\n", 36 | ">$\n", 37 | "\\mathbf{x}' = \\begin{pmatrix}\n", 38 | " 1& 0 \\\\ \n", 39 | " 0 & 2 \n", 40 | "\\end{pmatrix} \\mathbf{x}\n", 41 | "$" 42 | ] 43 | }, 44 | { 45 | "cell_type": "markdown", 46 | "metadata": {}, 47 | "source": [ 48 | "**solution**\n" 49 | ] 50 | }, 51 | { 52 | "cell_type": "markdown", 53 | "metadata": {}, 54 | "source": [ 55 | "**9.** $\\text{ }$ Find the general solution of the given system\n", 56 | "\n", 57 | ">$\n", 58 | "\\mathbf{x}' = \\begin{pmatrix}\n", 59 | " 1& 0 \\\\ \n", 60 | " 0 & 2 \n", 61 | "\\end{pmatrix} \\mathbf{x} +\\begin{pmatrix} \\;\\;3 \\\\-1\\end{pmatrix} \n", 62 | "$" 63 | ] 64 | }, 65 | { 66 | "cell_type": "markdown", 67 | "metadata": {}, 68 | "source": [ 69 | "**solution**\n" 70 | ] 71 | }, 72 | { 73 | "cell_type": "markdown", 74 | "metadata": {}, 75 | "source": [ 76 | "**23.** $\\text{ }$ If the matrix $\\matrix{A}$ can be diagonalized, then $\\mathbf{P}^{-1} \\mathbf{A} \\mathbf{P}=\\mathbf{D}$ or\n", 77 | "$\\text{ } \\mathbf{A} =\\mathbf{P}\\mathbf{D}\\mathbf{P}^{-1} $. Use this last result and show that $e^{\\mathbf{A} t}=\\mathbf{P}e^{\\mathbf{D} t}\\mathbf{P}^{-1} $" 78 | ] 79 | }, 80 | { 81 | "cell_type": "markdown", 82 | "metadata": {}, 83 | "source": [ 84 | "**solution**" 85 | ] 86 | } 87 | ], 88 | "metadata": { 89 | "kernelspec": { 90 | "display_name": "Python 3", 91 | "language": "python", 92 | "name": "python3" 93 | }, 94 | "language_info": { 95 | "codemirror_mode": { 96 | "name": "ipython", 97 | "version": 3 98 | }, 99 | "file_extension": ".py", 100 | "mimetype": "text/x-python", 101 | "name": "python", 102 | "nbconvert_exporter": "python", 103 | "pygments_lexer": "ipython3", 104 | "version": "3.7.5" 105 | } 106 | }, 107 | "nbformat": 4, 108 | "nbformat_minor": 4 109 | } 110 | -------------------------------------------------------------------------------- /exercises/figs/ch01_exer01.png: -------------------------------------------------------------------------------- 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Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/supplements/LegendrePolynomials_IntegralRelations.ipynb)" 12 | ] 13 | }, 14 | { 15 | "cell_type": "markdown", 16 | "metadata": { 17 | "slideshow": { 18 | "slide_type": "slide" 19 | } 20 | }, 21 | "source": [ 22 | "## **Legendre Polynomials Integral Relations** " 23 | ] 24 | }, 25 | { 26 | "cell_type": "markdown", 27 | "metadata": { 28 | "slideshow": { 29 | "slide_type": "fragment" 30 | } 31 | }, 32 | "source": [ 33 | " We consider the function $B(a, b)$ defined by\n", 34 | " \n", 35 | " >$\\displaystyle B(a, b)=\\int_0^1 t^a (1 -t)^b\\,dt$\n", 36 | " \n", 37 | " A partial integration shows\n", 38 | " \n", 39 | " >$\\displaystyle\n", 40 | " B(a, b)=\\int_0^1 t^a (1 -t)^b\\,dt=\\frac{b}{a +1} \\int_0^1 t^{a +1} (1 -t)^{b -1}\\,dt\n", 41 | " =\\frac{b}{a +1} B(a +1, b -1)\n", 42 | " $\n", 43 | " \n", 44 | " Since $\\displaystyle B(a +b, 0)=\\frac{1}{a +b +1}$, it then follows rather easily\n", 45 | " \n", 46 | " >$\\displaystyle \n", 47 | " \\begin{align*}\n", 48 | " B(a,b)&= \\frac{b}{a +1} B(a+1,b-1)\\\\ \n", 49 | " &= \\frac{b}{a +1} \\frac{b-1}{a +2} B(a+2,b-2)\\\\ \n", 50 | " &= \\frac{b}{a +1} \\frac{b-1}{a +2} \\frac{b-2}{a +3} B(a+3, b-3)\\\\ \n", 51 | " &\\;\\vdots \\\\ \n", 52 | " &= \\frac{b}{a +1} \\frac{b-1}{a +2} \\frac{b-2}{a +3} \\cdots \\frac{1}{a +b} B(a+b,0)\n", 53 | " = \\frac{a! b!}{(a +b +1)!}\n", 54 | " \\end{align*}$\n", 55 | " \n", 56 | " Then we deduce\n", 57 | " \n", 58 | " >$\n", 59 | " \\begin{align*}\n", 60 | " \\int_{-1}^1 (1 -x^2)^l\\, dx &= \\int_{-1}^1 (1 +x)^l (1 -x)^l\\, dx\\\\ \n", 61 | " &\\;\\big\\Downarrow \\; x= 2t-1 \\\\ \n", 62 | " &= 2^{2l +1} \\int_0^1 t^l (1 -t)^l\\, dt \n", 63 | " = 2^{2l +1} B(l,l)\n", 64 | " = 2^{2l +1} \\frac{l!l!}{(2l+1)!}\n", 65 | " \\end{align*}\n", 66 | " $\n", 67 | " \n", 68 | " **Let us now show that**\n", 69 | " \n", 70 | " >$\\displaystyle\n", 71 | " \\int_{-1}^1 P_k(x) P_l(x) \\,dx = \\delta_{k,l} \\frac{2}{2l +1},\\;\\;k \\leq l\n", 72 | " $\n", 73 | " \n", 74 | " Plugging Rodrigues' formula, we have\n", 75 | " \n", 76 | " >$\\displaystyle\n", 77 | " \\int_{-1}^1 P_k(x) P_l(x) \\,dx \n", 78 | " =\\frac{1}{2^{k +l} k! l!} \\int_{-1}^1 \\frac{d^k}{dx^k} \\left(x^2 -1\\right)^k\n", 79 | " \\frac{d^l}{dx^l} \\left(x^2 -1\\right)^l \\,dx\n", 80 | " $\n", 81 | " \n", 82 | " Since $\\displaystyle\\frac{d^a}{dx^a} \\left(x^2 -1\\right)^b$ vanishes at $x = \\pm 1$ if $a < b$, partial integration leads to\n", 83 | " \n", 84 | " >$\\displaystyle\n", 85 | " \\int_{-1}^1 P_k(x) P_l(x) \\,dx \n", 86 | " =\\frac{(-1)^l}{2^{k +l} k! l!} \\int_{-1}^1 \\frac{d^{k +1}}{dx^{k +l}} \\left(x^2 -1\\right)^k \n", 87 | " \\cdot\\left(x^2 -1\\right)^l \\,dx\n", 88 | " $\n", 89 | " \n", 90 | " If $k \\neq l$, we may assume $l > k$ and then $\\displaystyle \\frac{d^{k+l}}{dx^{k +l}}(x^2 − 1)^k = 0$.\n", 91 | " Hence the integral vanishes. When $k = l$, we use \n", 92 | " \n", 93 | " >$\\displaystyle \\frac{d^{2l}}{dx^{2l}}(x^2 -1)^l = \\frac{d^{2l}}{dx^{2l}}x^{2l}=(2l)!\\;$\n", 94 | " and $\\;\\left(x^2 -1\\right)^l=(-1)^l \\left(1 -x^2\\right)^l $\n", 95 | " \n", 96 | " to find\n", 97 | " \n", 98 | " >$\n", 99 | " \\displaystyle\n", 100 | " \\int_{-1}^1 P_l^2(x) \\,dx \n", 101 | " =\\frac{(2l)!}{2^{2l} l! l!} \\int_{-1}^1 \\left(1 -x^2\\right)^l \\,dx =\\frac{2}{2l +1}\n", 102 | " $\n", 103 | " \n", 104 | " **Let us now show that**\n", 105 | " \n", 106 | " >$\n", 107 | " \\displaystyle\n", 108 | " \\int_{-1}^1 x P_s(x) P_r(x) \\,dx = \\delta_{r,s +1} \\frac{2r}{(2r -1)(2r +1)}, \\; r \\geq s \n", 109 | " $\n", 110 | " \n", 111 | " First we consider the case that $r = s$. In this case the integral is easily found to be zero:\n", 112 | " $P_r$ has parity $(-1)^r$, and thus $x(P_r)^2$ has parity $-1$. Therefore the integral over \n", 113 | " the symmetric interval $[-1,\\; 1]$ vanishes.\n", 114 | " \n", 115 | " Now we consider the case $r > s$. We find, using Rodrigues' formula, partial integration\n", 116 | " and $(xf)^{(n)} = xf^{(n)}+nf^{(n -1)}$,\n", 117 | " \n", 118 | " >$\n", 119 | " \\displaystyle\n", 120 | " \\begin{align*}\n", 121 | " \\int_{-1}^1 x P_s(x) P_r(x) \\,dx &= \\frac{1}{2^{s +r} s! r!} \\int_{-1}^1 x \\frac{d^s}{dx^s}\n", 122 | " \\left(x^2 -1\\right)^s \\frac{d^r}{dx^r} \\left(x^2 -1\\right)^r \\,dx\\\\\n", 123 | " &=\\frac{(-1)^r}{2^{s +r} s! r!} \\int_{-1}^1\n", 124 | " \\left[ x \\frac{d^{s +r}}{dx^{s +r}} \\left(x^2 -1\\right)^s \n", 125 | " +r\\frac{d^{s +r -1}}{dx^{s +r -1}} \\left(x^2 -1\\right)^s \\right] \\left(x^2 -1\\right)^r \\,dx\n", 126 | " \\end{align*}\n", 127 | " $\n", 128 | " \n", 129 | " We see that the first term vanishes since $r > s$; the second term only survives if $r -1 \\leq s$.\n", 130 | " Hence we need only consider the case $s = r -1$.\n", 131 | " \n", 132 | " >$\n", 133 | " \\displaystyle\n", 134 | " \\begin{align*}\n", 135 | " \\int_{-1}^1 x P_{r -1}(x) P_r(x) \\,dx \n", 136 | " =\\frac{(-1)^r}{2^{2r-1} (r -1)! r!} \\int_{-1}^1 \\left[ r \\frac{d^{2r -2}}{dx^{2r -2}} \n", 137 | " \\left(x^2 -1\\right)^{r -1} \\right ] \\left(x^2 -1\\right)^r \\,dx\n", 138 | " \\end{align*}\n", 139 | " $\n", 140 | " \n", 141 | " We find, again using Rodrigues' formula, partial integration and $(xf)^{(n)} = xf^{(n)}+nf^{(n -1)}$,\n", 142 | " but also that the $(2r -2)$-th derivative of $(x^2 -1)^{r -1}$ equals $(2r -2)!$, that\n", 143 | " \n", 144 | " >$\\displaystyle\n", 145 | " \\begin{align*}\n", 146 | " \\int_{-1}^1 x P_{r -1}(x) P_r(x) \\,dx \n", 147 | " =\\frac{r(2r -2)!}{2^{2r-1} (r -1)! r!} \\int_{-1}^1 \\left(1 -x^2\\right)^r \\,dx\n", 148 | " =\\frac{2r}{(2r -1)(2r +1)}\n", 149 | " \\end{align*}\n", 150 | " $ " 151 | ] 152 | } 153 | ], 154 | "metadata": { 155 | "celltoolbar": "Slideshow", 156 | "kernelspec": { 157 | "display_name": "Python 3 (ipykernel)", 158 | "language": "python", 159 | "name": "python3" 160 | }, 161 | "language_info": { 162 | "codemirror_mode": { 163 | "name": "ipython", 164 | "version": 3 165 | }, 166 | "file_extension": ".py", 167 | "mimetype": "text/x-python", 168 | "name": "python", 169 | "nbconvert_exporter": "python", 170 | "pygments_lexer": "ipython3", 171 | "version": "3.8.12" 172 | } 173 | }, 174 | "nbformat": 4, 175 | "nbformat_minor": 4 176 | } 177 | -------------------------------------------------------------------------------- /supplements/Poisson_Equation_in_Sphere.ipynb: -------------------------------------------------------------------------------- 1 | { 2 | "cells": [ 3 | { 4 | "cell_type": "markdown", 5 | "metadata": { 6 | "slideshow": { 7 | "slide_type": "skip" 8 | } 9 | }, 10 | "source": [ 11 | "[![Open In Colab](https://colab.research.google.com/assets/colab-badge.svg)](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/supplements/Poisson_Equation_in_Sphere.ipynb)" 12 | ] 13 | }, 14 | { 15 | "cell_type": "markdown", 16 | "metadata": { 17 | "slideshow": { 18 | "slide_type": "slide" 19 | } 20 | }, 21 | "source": [ 22 | "## Poisson's Equation for the Sphere" 23 | ] 24 | }, 25 | { 26 | "cell_type": "markdown", 27 | "metadata": { 28 | "slideshow": { 29 | "slide_type": "fragment" 30 | } 31 | }, 32 | "source": [ 33 | "We consider the problem\n", 34 | "\n", 35 | ">$\\begin{align*}\n", 36 | " u_{rr} +\\frac{2}{r} u_r &+\\frac{1}{r^2}u_{\\phi\\phi} +\\frac{\\cot\\phi}{r^2} u_{\\phi} +\\frac{1}{r^2\\sin^2\\phi}u_{\\theta\\theta} = -F(r,\\phi,\\theta), \\;\\; r < R \\\\ \n", 37 | " u(R,\\phi,\\theta) &= 0\\\\ \n", 38 | "\\end{align*}\\;\\;$ **(P2)**\n", 39 | "\n", 40 | "(a) $~$solve by taking the finite Fourier Transforms, $~$(b) $~$derive $\\,G(r,\\phi,\\theta; \\rho,\\varphi,\\vartheta)\\,$ in the equivalent integral formula:\n", 41 | "\n", 42 | ">$\\displaystyle u(r,\\phi,\\theta) = \\int_0^{2\\pi} \\int_0^\\pi \\int_0^R G(r,\\phi,\\theta; \\rho,\\varphi,\\vartheta)\\, F(\\rho, \\varphi,\\vartheta) \\,\\rho^2 \\sin\\varphi \\,d\\rho \\,d\\varphi \\, d\\vartheta$" 43 | ] 44 | }, 45 | { 46 | "cell_type": "markdown", 47 | "metadata": { 48 | "slideshow": { 49 | "slide_type": "slide" 50 | } 51 | }, 52 | "source": [ 53 | "**(a)** $\\text{ }$ We expand $u$ and $F$ in Fourier series in $P_n^m(\\cos\\phi)\\cos m\\theta$ and $P_n^m(\\cos\\phi)\\sin m\\theta$. $~$That is, $~$we take their finite Fourier transforms:\n", 54 | "\n", 55 | ">$\\begin{align*}\n", 56 | " a_{nm}(r) &= \\frac{2n +1}{2\\pi} \\frac{(n -m)!}{(n +m)!}\n", 57 | " \\int_{0}^{2\\pi} \\int_{0}^\\pi u(r,\\phi,\\theta) \\,P_n^m(\\cos\\phi) \\,\\cos m\\theta \\,\\sin\\phi \\,d\\phi \\,d\\theta\\\\ \n", 58 | " b_{nm}(r) &= \\frac{2n +1}{2\\pi} \\frac{(n -m)!}{(n +m)!}\n", 59 | " \\int_{0}^{2\\pi} \\int_{0}^\\pi u(r,\\phi,\\theta) \\,P_n^m(\\cos\\phi) \\,\\sin m\\theta \\,\\sin\\phi \\,d\\phi \\,d\\theta\\\\ \n", 60 | " A_{nm}(r) &= \\frac{2n +1}{2\\pi} \\frac{(n -m)!}{(n +m)!}\n", 61 | " \\int_{0}^{2\\pi} \\int_{0}^\\pi F(r,\\phi,\\theta) \\,P_n^m(\\cos\\phi) \\,\\cos m\\theta \\,\\sin\\phi \\,d\\phi \\,d\\theta\\\\ \n", 62 | " B_{nm}(r) &= \\frac{2n +1}{2\\pi} \\frac{(n -m)!}{(n +m)!}\n", 63 | " \\int_{0}^{2\\pi} \\int_{0}^\\pi F(r,\\phi,\\theta) \\,P_n^m(\\cos\\phi) \\,\\sin m\\theta \\,\\sin\\phi \\,d\\phi \\,d\\theta\n", 64 | "\\end{align*}$\n", 65 | "\n", 66 | "Taking finite Fourier transforms of the differential equation and integrating by parts leads to the system\n", 67 | "\n", 68 | ">$\\begin{align*}\n", 69 | " r^2 a_{nm}'' +2r a_{nm}' &-n(n +1) a_{nm} = -r^2 A_{nm}, \\; \n", 70 | " a_{nm}(R)= 0, \\;a_{nm} \\sim \\text{bounded}\\\\ \n", 71 | " r^2 b_{nm}'' +2r b_{nm}' &-n(n +1) b_{nm} = -r^2 B_{nm}, \\; \n", 72 | " b_{nm}(R)= 0, \\;b_{nm} \\sim \\text{bounded} \n", 73 | "\\end{align*}$\n" 74 | ] 75 | }, 76 | { 77 | "cell_type": "markdown", 78 | "metadata": { 79 | "slideshow": { 80 | "slide_type": "slide" 81 | } 82 | }, 83 | "source": [ 84 | "These problems have the solutions\n", 85 | "\n", 86 | ">$\\begin{align*}\n", 87 | " a_{nm}(r) &=\\int_0^R G_n(r,\\rho)\\,A_{nm}(\\rho)\\,\\rho^2 \\,d\\rho \\\\ \n", 88 | " b_{nm}(r) &=\\int_0^R G_n(r,\\rho)\\,B_{nm}(\\rho)\\,\\rho^2 \\,d\\rho\n", 89 | "\\end{align*}\\;\\;$ **(E5)**\n", 90 | "\n", 91 | "where [Colab: Green's function](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/supplements/Green's_Function.ipynb) or [local: Green's function](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/supplements/Green's_Function.ipynb)\n", 92 | "\n", 93 | ">$\\displaystyle G_n(r,\\rho) = \n", 94 | "\\begin{cases}\n", 95 | " \\displaystyle\\frac{1}{(2n +1)R}\\left(\\frac{r}{R} \\right)^n\\left[\\left(\\frac{\\rho}{R} \\right)^{-(n+1)} -\\left(\\frac{\\rho}{R} \\right)^{n} \\right] & \\text{ for } r \\leq \\rho \\\\ \n", 96 | " \\displaystyle\\frac{1}{(2n +1)R}\\left(\\frac{\\rho}{R} \\right)^n\\left[\\left(\\frac{r}{R} \\right)^{-(n+1)} -\\left(\\frac{r}{R} \\right)^{n} \\right]& \\text{ for } r \\geq \\rho \n", 97 | "\\end{cases}$\n", 98 | "\n", 99 | "If the problem (P2) has a solution, $~$it is\n", 100 | "\n", 101 | ">$\\displaystyle u(r,\\phi,\\theta) = \\sum_{n=0}^\\infty \\left[ \\frac{1}{2} a_{n0}(r) P_n(\\cos\\phi) +\\sum_{m=1}^n \\left(a_{nm}(r) \\cos m\\theta +b_{nm}(r) \\sin m\\theta \\right) P_n^m(\\cos\\phi) \\right]\\;\\;$ **(E6)**" 102 | ] 103 | }, 104 | { 105 | "cell_type": "markdown", 106 | "metadata": { 107 | "slideshow": { 108 | "slide_type": "slide" 109 | } 110 | }, 111 | "source": [ 112 | "**(b)** $\\text{ }$ We substitute the definitions of $A_{nm}(r)$ and $B_{nm}(r)$ in (E5), $~$and formally interchange integration and summation in (E6). $~$This process gives the formal solution \n", 113 | "\n", 114 | ">$\\displaystyle u(r,\\phi,\\theta) = \\int_0^{2\\pi} \\int_0^\\pi \\int_0^R G(r,\\phi,\\theta; \\rho,\\varphi,\\vartheta)\\, F(\\rho, \\varphi,\\vartheta) \\,\\rho^2 \\sin\\varphi \\,d\\rho \\,d\\varphi \\, d\\vartheta$\n", 115 | "\n", 116 | "where for $r > \\rho$\n", 117 | "\n", 118 | ">$\\begin{align*}\n", 119 | " G(r,\\phi,\\theta;\\rho,\\varphi,\\vartheta) &=\\sum_{n=0}^\\infty \\frac{1}{(2n +1)R}\n", 120 | " \\left(\\frac{\\rho}{R} \\right)^n \n", 121 | " \\left[ \\left(\\frac{r}{R} \\right)^{-(n +1)} -\\left(\\frac{r}{R} \\right)^{n} \\right] \n", 122 | " K_n(\\phi,\\theta;\\varphi,\\vartheta)\\\\ \n", 123 | " &= \\frac{1}{4\\pi R} \\sum_{n=0}^\\infty \\left(\\frac{\\rho}{R} \\right)^n \n", 124 | " \\left[ \\left(\\frac{r}{R} \\right)^{-(n +1)} -\\left(\\frac{r}{R} \\right)^{n} \\right] P_n(\\cos\\varphi') \n", 125 | "\\end{align*}$\n", 126 | "\n", 127 | "For $r < \\rho$, $~$we must only interchange $r$ and $\\rho$ in this formula. To evaluate the series in the above equation, $~$we use the generating function:\n", 128 | "\n", 129 | ">$\\begin{align*}\n", 130 | " G(r,\\phi,\\theta;\\rho,\\varphi,\\vartheta) &=\\frac{1}{4\\pi} \\frac{1}{\\sqrt{r^2+\\rho^2-2r\\rho\\cos\\varphi'}}\n", 131 | " -\\frac{1}{4\\pi} \\frac{1}{\\sqrt{R^2 +\\frac{r^2\\rho^2}{R^2} -2r\\rho\\cos\\varphi'}}\n", 132 | "\\end{align*}\n", 133 | "$\n", 134 | "\n", 135 | "This, then, is the Green's function for $r > \\rho$. $~$It is already symmetric in $r$ and $\\rho$, $~$so that the same formula holds for $r < \\rho$, and even for $r=\\rho$. $~$We note that $G=0$ for $r=R$" 136 | ] 137 | } 138 | ], 139 | "metadata": { 140 | "celltoolbar": "Slideshow", 141 | "kernelspec": { 142 | "display_name": "Python 3 (ipykernel)", 143 | "language": "python", 144 | "name": "python3" 145 | }, 146 | "language_info": { 147 | "codemirror_mode": { 148 | "name": "ipython", 149 | "version": 3 150 | }, 151 | "file_extension": ".py", 152 | "mimetype": "text/x-python", 153 | "name": "python", 154 | "nbconvert_exporter": "python", 155 | "pygments_lexer": "ipython3", 156 | "version": "3.9.13" 157 | } 158 | }, 159 | "nbformat": 4, 160 | "nbformat_minor": 4 161 | } 162 | -------------------------------------------------------------------------------- /supplements/README.md: -------------------------------------------------------------------------------- 1 | ## EngMath Suppliments 2 | 3 | * [Laplace Equation in Sphere](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/supplements/Laplace_Equation_in_Sphere.ipynb) 4 | * [Poisson Equation in Sphere](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/supplements/Poisson_Equation_in_Sphere.ipynb) 5 | * [Green's Function](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/supplements/Green's_Function.ipynb) 6 | * [The Hydrogen Atom](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/supplements/hydrogen.ipynb) 7 | 8 | * [Root Locus](https://colab.research.google.com/github/SeoulTechPSE/EngMath/blob/master/supplements/RootLocus.ipynb) 9 | --------------------------------------------------------------------------------