ENG 5300 (Fall 2020)
Advanced Engineering Mathematics

Final
Mingzuoyang Chen	Chris Collins	Jon-Michael X Grabowski	Chenghao Ji
Ferris Kimmil	Chaowei Li	Yichun Li	Yufan Lu
Brendan McClanahan	Deep Patel	Devendar Reddy Thallapureddy	Kevin Weltzin
Jin Xue	Ruihao Yang	Yanjun Yao	Dong Ye
Zachary Satawa
Upload to Black Board. A single b/w pdf in a folder with Matlab codes.

fplot

clear all; clc;
syms x v t 

for t=0:0.01:2
    fplot(1/(2*sqrt(pi*t))*(int((-2-v)*exp(-(x-v)^2/(4*t)),v,-2,-1)+int(v*exp(-(x-v)^2/(4*t)),v,-1,1)+int((2-v)*exp(-(x-v)^2/(4*t)),v,1,2)),[-3 3]);
    axis image; axis([-3 3 -1 1]); title(t); xlabel('x-axis'); ylabel('temperature');
    drawnow;
end

Quiz 4.
Mingzuoyang Chen	Chris Collins	Jon-Michael X Grabowski	Chenghao Ji
Ferris Kimmil	Chaowei Li	Yichun Li	Yufan Lu
Brendan McClanahan	Deep Patel	Devendar Reddy Thallapureddy	Kevin Weltzin
Jin Xue	Ruihao Yang	Yanjun Yao	Dong Ye
Zachary Satawa
Upload to Black Board. A single b/w pdf

Test 2
Mingzuoyang Chen	Chris Collins	Jon-Michael X Grabowski	Chenghao Ji
Ferris Kimmil	Chaowei Li	Yichun Li	Yufan Lu
Brendan McClanahan	Deep Patel	Devendar Reddy Thallapureddy	Kevin Weltzin
Jin Xue	Ruihao Yang	Yanjun Yao	Dong Ye
Zachary Satawa
Upload to Black Board. A single b/w pdf in a folder with Matlab codes.

Here we want use Euler's Identity \(\boxed{e^{i\theta}=\cos \theta +i\sin \theta }\) to repackage $$ f(x)=\color{blue}{\int\limits_0^\infty} \left[ A(\omega)\cdot \cos \omega x + B(\omega)\cdot \sin \omega x \right] \color{blue}{d\omega} $$ with $$ \boxed{A(\omega)=\frac{1}{\pi} \int\limits_{-\infty}^\infty f(v)\cdot \cos \omega v\, dv} \,\,\, \text{ and } \,\,\, \boxed{B(\omega)=\frac{1}{\pi} \int\limits_{-\infty}^\infty f(v)\cdot \sin \omega v\, dv} $$ into a more compact form.
First, substitute the integral expressions for \(A(\omega)\) and \(B(\omega)\) into the integral expression for \(f(x)\) $$ \begin{align} f(x) &= \color{blue}{\int\limits_0^\infty} \left[ A(\omega)\cdot \cos \omega x + B(\omega)\cdot \sin \omega x \right] \color{blue}{d\omega} \\ &= \frac{1}{\pi}\color{blue}{\int\limits_0^\infty} \int\limits_{-\infty}^\infty f(v) \underbrace{\left[ \cos \omega v \cdot \cos \omega x + \sin \omega v \cdot \sin \omega x \right]}_{\cos \alpha \cos \beta + \sin \alpha \sin \beta = \cos\left(\alpha -\beta \right)} \, dv\, \color{blue}{d\omega} \\ &= \frac{1}{\pi}\color{blue}{\int\limits_0^\infty} \int\limits_{-\infty}^\infty f(v) \underbrace{\cos\left( \omega x -\omega v \right)}_{\text{even function of }\omega} \, dv \, \color{blue}{d\omega} \\ &= \frac{1}{2\pi}\color{blue}{\int\limits_{-\infty}^\infty} \int\limits_{-\infty}^\infty f(v) \cos\left( \omega x -\omega v \right) \, dv \, \color{blue}{d\omega} + \underbrace{\frac{1}{2\pi}\color{blue}{\int\limits_{-\infty}^\infty} \int\limits_{-\infty}^\infty f(v) \overbrace{\sin\left( \omega x -\omega v \right)}^{\text{odd function of }\omega} \, dv \, \color{blue}{d\omega}}_{=0} \\ &= \frac{1}{2\pi}\color{blue}{\int\limits_{-\infty}^\infty} \int\limits_{-\infty}^\infty f(v) \cos\left( \omega x -\omega v \right) \, dv \, \color{blue}{d\omega}+ \color{red}{i}\frac{1}{2\pi}\color{blue}{\int\limits_{-\infty}^\infty} \int\limits_{-\infty}^\infty f(v) \sin\left( \omega x -\omega v \right) \, dv \, \color{blue}{d\omega} \\ &= \frac{1}{2\pi}\color{blue}{\int\limits_{-\infty}^\infty} \int\limits_{-\infty}^\infty f(v) \underbrace{\left[\cos\left( \omega x -\omega v \right)+\color{red}{i}\sin\left( \omega x -\omega v \right)\right]}_{e^{\color{red}{i}(\omega x -\omega v)}} \, dv \, \color{blue}{d\omega} \\ &= \frac{1}{2\pi}\color{blue}{\int\limits_{-\infty}^\infty} \int\limits_{-\infty}^\infty f(v) e^{\color{red}{i}(\omega x -\omega v)} \, dv \, \color{blue}{d\omega} \\ &= \frac{1}{\sqrt{2\pi}}\color{blue}{\int\limits_{-\infty}^\infty} \underbrace{\left[ \frac{1}{\sqrt{2\pi}}\int\limits_{-\infty}^\infty f(v)e^{-i\omega v}\, dv \right]}_{\hat{f}(\omega)}\, e^{i\omega x}\, \color{blue}{d\omega} \end{align} $$

The goal here is to extend the Fourier series representation of periodic functions \(f(x+p)=f(x)\) with period \(p=2L\) to nonperiodic functions by taking the limit \(\lim\limits_{L\rightarrow \infty} f(x)\) as illustrated in the figure below

First let's recall the Fourier series of a periodic function \(f_L(x+p)=f_L(x)\) with period \(p=2L\) $$ \begin{align} f_L(x)&=a_0+\sum\limits_{n=1}^\infty a_n\cos \left( \frac{n\pi}{L}x \right)+b_n\sin \left( \frac{n\pi}{L}x \right) \\ a_0 &= \frac{1}{2L}\int\limits_{-L}^L f_L(v)dv \\ a_n &= \frac{1}{L}\int\limits_{-L}^L f_L(v)\cdot \cos\left( \frac{n\pi}{L}v \right)dv \\ b_n &= \frac{1}{L}\int\limits_{-L}^L f_L(v)\cdot \sin\left( \frac{n\pi}{L}v \right)dv \end{align} $$ Next, use $$ \omega_n=\frac{n\pi}{L}, \,\,\, \text{ and } \,\,\, \Delta \omega = \omega_{n+1}-\omega_n = \frac{(n+1)\pi}{L}-\frac{n\pi}{L}=\frac{\pi}{L} \,\,\, \text{ which leads to } \,\,\, \frac{1}{L}=\frac{\Delta \omega}{\pi} $$ and substitute the formulas for the coefficients \(\{a_0,\, a_n, \, b_n\}\) back into the Fourier series to get $$ \begin{align} f_L(x) &= \underbrace{\frac{1}{2L}\overbrace{\int\limits_{-L}^L f_L(v)dv}^{\text{finite } \because}}_{\int\limits_{[-L,L]}f_L(v)dv\, \leq \int\limits_{[-L,L]}|f_L(v)|dv \, < \infty}+\sum\limits_{n=1}^\infty \left[ \overbrace{\frac{1}{L}\int\limits_{-L}^L f_L(v)\cdot \cos\left( \frac{n\pi}{L}v \right)dv}^{a_n} \cdot \cos\omega_n x+ \overbrace{\frac{1}{L}\int\limits_{-L}^L f_L(v)\cdot \sin\left( \frac{n\pi}{L}v \right)dv}^{b_n} \cdot \sin\omega_n x\right]\\ &= \frac{1}{2L}\int\limits_{-\infty}^\infty f_L(v)dv+\frac{1}{\pi}\sum\limits_{n=1}^\infty \left[ \left(\cos \omega_n x\right) \Delta\omega \underbrace{\int\limits_{-L}^Lf_L(v)\cos \omega_n v \, dv}_{\text{later becomes part of }A(\omega)} + \left(\sin \omega_n x\right) \Delta\omega \underbrace{\int\limits_{-L}^Lf_L(v)\sin \omega_n v \, dv}_{\text{later becomes part of }B(\omega)} \right] \end{align} $$ Now, take the limit \( L\rightarrow \infty \), hence we write \( \lim\limits_{L\rightarrow \infty} f_L(x)=f(x) \), and note that \( \boxed{\lim\limits_{L\rightarrow \infty} \frac{1}{2L}\int\limits_{-\infty}^\infty f_L(v)dv=0} \) and \( \boxed{\lim\limits_{L\rightarrow \infty} \Delta \omega = d\omega} \). Moreover, since \(n\in \{1,2,3,\ldots\}\) and \(L\rightarrow \infty\), we have \(\boxed{\boxed{\frac{\pi}{L}\leq \omega_n < \infty} \implies \boxed{0\leq \omega < \infty}} \) $$ \begin{align} f(x)= \lim\limits_{L\rightarrow \infty} f_L(x) &= \color{blue}{\lim\limits_{L\rightarrow \infty}} \frac{1}{\pi}\color{blue}{\sum\limits_{n=1}^\infty} \left[ \left(\cos \omega_n x\right) \color{blue}{\Delta\omega} \underbrace{\int\limits_{-L}^Lf_L(v)\cos \omega_n v \, dv}_{\text{later becomes part of }A(\omega)} + \left(\sin \omega_n x\right) \color{blue}{\Delta\omega} \underbrace{\int\limits_{-L}^Lf_L(v)\sin \omega_n v \, dv}_{\text{later becomes part of }B(\omega)} \right] \\ &= \frac{1}{\pi} \color{blue}{\int\limits_0^\infty}\left[ \cos \omega x\cdot \underbrace{\int\limits_{-\infty}^\infty f(v)\cos \omega v\, dv}_{\text{function of } \omega \text{ only}} + \sin \omega x\cdot \underbrace{\int\limits_{-\infty}^\infty f(v)\sin \omega v\, dv}_{\text{function of }\omega\text{ only}} \right]\color{blue}{d\omega} \\ &= \color{blue}{\int\limits_0^\infty} \left[ A(\omega)\cdot \cos \omega x + B(\omega)\cdot \sin \omega x \right] \color{blue}{d\omega} \end{align} $$ Where $$ A(\omega)=\frac{1}{\pi} \int\limits_{-\infty}^\infty f(v)\cdot \cos \omega v\, dv \,\,\, \text{ and } \,\,\, B(\omega)=\frac{1}{\pi} \int\limits_{-\infty}^\infty f(v)\cdot \sin \omega v\, dv $$

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The blue sum became blue integral because by the high-school definition of the Riemann integral \(\boxed{ \lim\limits_{\Delta x_{\text{max}}\rightarrow 0}\sum\limits_{n=1}^\infty g(x^*_n)\Delta x_n =\int\limits_a^b g(x)\, dx}\) we get: $$ \color{blue}{ \lim\limits_{L\rightarrow \infty}\sum\limits_{n=1}^\infty g\left( \frac{n\pi}{L} \, x \right) \frac{\pi}{L}= \lim\limits_{L\rightarrow \infty}\sum\limits_{n=1}^\infty g(\omega_n \, x) \Delta \omega = \int\limits_0^\infty g(\omega x)\, d\omega } $$

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Example Fourier series and integral for $$ f_L(x)=\begin{cases} 0 & \text{ if } -L< x<-1 \\ |x| & \text{ if } -1\leq x \leq 1 \\ 0 & \text{ if } +1 < x < L \end{cases} \text{ and } f(x+2L)=f(x) \text{ for } L\in\{1,2,4,8,\infty \} $$ and the corresponding spectra \(a_n\) and \(A(\omega)\)

Test 1
Mingzuoyang Chen	Chris Collins	Jon-Michael X Grabowski	Chenghao Ji
Ferris Kimmil	Chaowei Li	Yichun Li	Yufan Lu
Brendan McClanahan	Deep Patel	Devendar Reddy Thallapureddy	Kevin Weltzin
Jin Xue	Ruihao Yang	Yanjun Yao	Dong Ye
Zachary Satawa
Don't use decimal approximations. Upload to Black Board. A single b/w pdf

Quiz 3.
Mingzuoyang Chen	Chris Collins	Jon-Michael X Grabowski	Chenghao Ji
Ferris Kimmil	Chaowei Li	Yichun Li	Yufan Lu
Brendan McClanahan	Deep Patel	Devendar Reddy Thallapureddy	Kevin Weltzin
Jin Xue	Ruihao Yang	Yanjun Yao	Dong Ye
Zachary Satawa
Don't use decimal approximations. Upload to Black Board. A single b/w pdf

Quiz 2.
Mingzuoyang Chen	Chris Collins	Jon-Michael X Grabowski	Chenghao Ji
Ferris Kimmil	Chaowei Li	Yichun Li	Yufan Lu
Brendan McClanahan	Deep Patel	Devendar Reddy Thallapureddy	Kevin Weltzin
Jin Xue	Ruihao Yang	Yanjun Yao	Dong Ye
Zachary Satawa
Don't use decimal approximations. Upload to Black Board. A single b/w pdf

Quiz 1.
Mingzuoyang Chen	Chris Collins	Jon-Michael X Grabowski	Chenghao Ji
Ferris Kimmil	Chaowei Li	Yichun Li	Yufan Lu
Brendan McClanahan	Deep Patel	Devendar Reddy Thallapureddy	Kevin Weltzin
Jin Xue	Ruihao Yang	Yanjun Yao	Dong Ye
Zachary Satawa
Don't use decimal approximations. Upload to Black Board. A single b/w pdf

clc; close all; clear;

U=[0:0.1:2*pi];
V=[-pi/2:0.1:pi/2];
[u,v]=meshgrid(U,V); % region in parameter space

x=v.*cos(u);
y=v.*sin(u);
z=v.^2;

surf(x,y,z);
axis image;

clc; close all; clear;

U=[0:0.1:2*pi];
V=[-pi/2:0.1:pi/2];
[u,v]=meshgrid(U,V); % region in parameter space

x=v.*cos(u);
y=v.*sin(u);
z=v;

xx=v.*cos(u);
yy=v.*sin(u);
zz=v.^2;

for t=0:0.01:1
    surf((1-t)*x+t*xx,(1-t)*y+t*yy,(1-t)*z+t*zz);
    axis([-2 2 -2 2 -2 2]);
    drawnow; 
end

Gradient Search computer assignment. Due September 9.
Mingzuoyang Chen	Chris Collins	Jon-Michael X Grabowski	Chenghao Ji
Ferris Kimmil	Chaowei Li	Yichun Li	Yufan Lu
Brendan McClanahan	Deep Patel	Devendar Reddy Thallapureddy	Kevin Weltzin
Jin Xue	Ruihao Yang	Yanjun Yao	Dong Ye
Zachary Satawa
single email with MatLab codes, and explanations. No m$Word or pdf's