forked from Sasserisop/MATH201
revised fourier and added lec 34
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#fourier
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Remember the heat flow equation? We obtained that it's solution could be expressed in the form:
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$$\sum_{n=1}^\infty c_{n}\sin\left( \frac{n\pi x}{L} \right)\quad\text{for}\quad0\leq x\leq L$$
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But what is $c_{n}$? They are the coefficients of a fourier transform. We want to develop a way to compute them.
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Let's derive how to compute the coefficients of a fourier transform. (feel free to skip to the end)
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$f(x)=\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$ where L is length of the rod
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But what is $c_{n}$? They are the coefficients of a Fourier transform. We want to develop a way to compute them.
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Let's derive how to compute the coefficients of a Fourier transform. (feel free to skip to the end)
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$f(x)=\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$ where $L$ is length of the rod
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This is a Fourier series, it's a more general form of what we have above:
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This is a Fourier series: it's a more general form of what we have above.
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$f(x)=\frac{a_{0}}{2}+\sum_{n=1}^\infty\left( a_{n}\cos\left( \frac{n\pi x}{L} \right) + b_{n}\sin\left( \frac{n\pi x}{L}\right) \right)$
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$x \in [-L,L]$
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almost everywhere piecewise continuous (?)
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has a lot of benefits over taylor series. $f(x)$ doesn't have to be infinitely differentiable (analytic)
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f(x) can even have jump discontinuities
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lets assume the equation is true when $x \in [-L,L]$
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It converges to $f(x)$ almost everywhere (convergence will be discussed below)
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Has a lot of benefits over Taylor series. $f(x)$ doesn't have to be infinitely differentiable (analytic)
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$f(x)$ can even have jump discontinuities
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Let's assume the equation is true when $x \in [-L,L]$
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Integrate both sides, it will tell us the DC offset:
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$\int _{-L} ^L f(x) \, dx=\int _{-L}^L \frac{a_{0}}{2} \, dx+\int _{-L}^L (\text{put summation here}) \, dx$
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$\int _{-L}^L \cos\left( \frac{n\pi x}{L} \right) \, dx=\frac{L}{n\pi}\sin\left( \frac{n\pi x}{L} \right)|_{-L}^L=0$
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same for $\int _{-L}^L \sin\left( \frac{n\pi x}{L} \right)\, dx=0$ (it equals 0)
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so
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$\int _{-L} ^L f(x) \, dx=\int _{-L}^L \frac{a_{0}}{2} \, dx+\int _{-L}^L0 \, dx$
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$\int _{-L} ^L f(x) \, dx=\int _{-L}^L \frac{a_{0}}{2} \, dx+\int _{-L}^L0 \, dx$
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$a_{0}=\frac{1}{L}\int _{L}^{2?a_{0}L} f(x) \, dx$
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$\int _{-L}^L f(x)\cos\left( \frac{m\pi x}{L} \right)\, dx=\frac{a_{0}}{2}\cancelto{ 0 }{ \int _{-L}^L \cos\left( \frac{m\pi x}{L} \right) \, dx }+\sum_{n=1}^\infty\left( a_{n}\int _{-L}^L\cos\left( \frac{n\pi x}{L} \right)\cos\left( \frac{m\pi x}{L} \right) \, dx \right)+b_{n}\int _{-L}^L \sin\left( \frac{n\pi x}{L} \right)\cos\left( \frac{m\pi x}{L} \right) \, dx$
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$\int _{-L} ^L f(x) \, dx=a_{0}L$
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$a_{0}=\frac{1}{L}\int _{-L}^{L} f(x) \, dx$
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Now let's multiply both sides by $\cos\left( \frac{m\pi x}{L} \right)$ and integrate both sides, this will tell us the $\cos$ components:
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$\int _{-L}^L f(x)\cos\left( \frac{m\pi x}{L} \right)\, dx=\frac{a_{0}}{2}\cancelto{ 0 }{ \int _{-L}^L \cos\left( \frac{m\pi x}{L} \right) \, dx }+\sum_{n=1}^\infty\left( a_{n}\int _{-L}^L\cos\left( \frac{n\pi x}{L} \right)\cos\left( \frac{m\pi x}{L} \right) \, dx +b_{n}\int _{-L}^L \sin\left( \frac{n\pi x}{L} \right)\cos\left( \frac{m\pi x}{L} \right)\right) \, dx$
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use trig identities (will be provided on exam):
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$\cos(\alpha)\cos(\beta)=\frac{1}{2}(\cos(\alpha-\beta)+\cos(\alpha+\beta))$
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$\sin(\alpha)\cos(\beta)=\frac{1}{2}(\sin(\alpha+\beta)+\sin(\alpha-\beta))$
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$\sin(\alpha)\sin(\beta)=\frac{1}{2}(\cos(\alpha-\beta)-\cos(\alpha+\beta))$
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$\int _{-L}^L \cos \frac{n\pi x}{L}\cos \frac{m\pi x}{L}\, dx=\frac{1}{2}(\int _{-L}^L \left( \cos \frac{(n-m)\pi x}{L} \right) \, dx+\cancelto{ 0 }{ \frac{\cos(n+m)\pi x}{L} }dx$
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$\int _{-L}^L \cos \frac{n\pi x}{L}\cos \frac{m\pi x}{L}\, dx=\frac{1}{2}(\int _{-L}^L \left( \cos(\frac{(n-m)\pi x}{L} )+\cancelto{ 0 }{ \cos(\frac{(n+m)\pi x}{L} })\right)dx$
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$= \begin{cases}0 & n\ne m \\L & n=m\end{cases}$
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$\int _{-L}^L \sin \frac{n\pi x}{L}\cos \frac{m\pi}{L} \, dx=0;$
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$\int _{-L}^L \sin \frac{n\pi x}{L}\cos \frac{m\pi}{L} \, dx=\begin{cases}0 & n\ne m \\L & n=m\end{cases}$
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going back,
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$\int _{-L}^L \sin \frac{n\pi x}{L}\cos \frac{m\pi x}{L} \, dx=\int _{-L}^L \text{odd}\, dx=0$
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so:
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$\int _{-L} ^L f(x)\cos\left( \frac{m\pi x}{L} \right)\, dx=a_{m}L$
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Similarly can be done for when multiplying both sides by $\sin\left( \frac{m\pi x}{L} \right)$ and integrating both sides to find the $\sin$ coefficients:
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$\int _{-L}^L f(x)\sin\left( \frac{m\pi x}{L} \right)\, dx=\frac{a_{0}}{2}\cancelto{ 0 }{ \int _{-L}^L \sin\left( \frac{m\pi x}{L} \right) \, dx }+\sum_{n=1}^\infty\left( a_{n}\cancelto{ \text{odd} }{ \int _{-L}^L\cos\left( \frac{n\pi x}{L} \right)\sin\left( \frac{m\pi x}{L} \right) } \, dx +b_{n}\int _{-L}^L \sin\left( \frac{n\pi x}{L} \right)\sin\left( \frac{m\pi x}{L} \right)\right) \, dx$
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$\int _{-L} ^L \sin\left( \frac{n\pi x}{L} \right)\sin\left( \frac{m\pi x}{L} \right) \, dx=\frac{1}{2}\int_{-L}^L \cos\left(\frac{(n-m)\pi x}{L} \right)-\cos\left( \frac{(n+m)\pi x}{L} \right)dx$
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$=\begin{cases}0, & n\ne m \\L, & n=m\end{cases}$
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so:
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$\int _{-L} ^L f(x)\sin\left( \frac{m\pi x}{L} \right)\, dx=b_{m}L$
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In conclusion:
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$$a_{m}=\frac{1}{L}\int _{-L}^L f(x)\cos \frac{m\pi x}{L} \, dx \quad\text{valid for all }m=0,1,2,\dots$$
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$$b_{m}=\frac{1}{L}\int _{-L}^L f(x)\sin \frac{m\pi x}{L} \, dx=b_{m} \quad m=1,2,\dots$$
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now we know how to compute the coefficients for Fourier series!
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Now we know how to compute the coefficients for Fourier series!
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properties:
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for functions $f$, $g$, If $\int _{-L}^Lf(x)g(x) \, dx=\begin{cases}0 & f\ne g \\L & f=g \end{cases}$
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then $f, g$ are orthogonal
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the forier expantion is called an ortho normal expansion, taylor is not ortho normal.
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the Fourier expansion is called an ortho normal expansion, Taylor is not orthonormal.
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#end of lec 28
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#start of lec 29
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last lecture we derived how to find the coefficients in a fourier series.
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Last lecture we derived how to find the coefficients in a Fourier series.
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$f(x)=\frac{a_{0}}{2}+\sum_{n=1}^\infty\left( a_{n}\cos\left( \frac{n\pi x}{L} \right) + b_{n}\sin\left( \frac{n\pi x}{L}\right) \right)$
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$x \in [-L,L]$
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### Theorem:
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If $f$ and $f'$ are piecewise continuous on $[-L,L]$, then the fourier series converges to:
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### 1st convergence theorem:
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If $f$ and $f'$ are piecewise continuous on $[-L,L]$, then the Fourier series converges to:
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$\frac{1}{2}(f(x^-)+f(x^+))$ for all $x \in (-L,L)$
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Basically meaning, the fourier series converges.
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At $x=\pm L$ the fourier series converges to $\frac{1}{2}(f(-L^+)+f(L^-))$
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and on $x=\pm L$ the Fourier series converges to $\frac{1}{2}(f(-L^+)+f(L^-))$
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### Theorem:
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If f(x) is continuous on $(-\infty,\infty)$ and $2L$ periodic and if $f'$ is continuous, then the taylor series converges to $f(x)$ everywhere
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Recall the definition of piecewise continuous: $f(t)$ is piecewise continuous on an interval $I$ if $f(t)$ is continuous on $I$, except possibly at a <u>finite</u> number of points of <u>jump</u> discontinuity (horizontal asymptotes not allowed).
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### 2nd Convergence theorem (uniform convergence):
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If $f(x)$ is continuous on $(-\infty,\infty)$ and $2L$ periodic and if $f'$ is piecewise continuous on $[-L,L]$, then its Fourier series converges to $f(x)$ everywhere (i.e., the Fourier series converges uniformly).
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#ex lets compute the fourier transform of:
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$f(x)=\begin{cases}1, & -\pi\leq x\leq 0 \\x, & 0<x\leq \pi\end{cases}$
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#ex #fourier
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Let's compute the Fourier transform of:
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$$f(x)=\begin{cases}1, & -\pi\leq x\leq 0 \\x, & 0<x\leq \pi\end{cases}$$
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$L$ here is $\pi$ clearly.
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lets find the coefficients $a_{n}$ and $b_{n}$
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$a_{n}=\frac{1}{\pi}\left( \int _{-\pi}^0 \cos\left( \frac{n\pi x}{\pi} \right)\, dx +\int _{0}^\pi x\cos(nx)\, dx\right)$
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using integration by parts for second term:
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$=\frac{1}{\pi}\left( \frac{1}{n}\cancelto{ 0 }{ \sin(nx) }|_{-\pi}^0 +\frac{1}{n}\left( x\cancelto{ 0 }{ \sin(nx) }|_{0}^\pi-\int _{0}^\pi \sin(nx) \, dx \right)\right)$
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$a_{n}=\frac{1}{n^2\pi}(\cos(nx)|_{0}^\pi)=\frac{1}{n^2\pi}((-1)^n-1) \quad n=0,1,2,\dots$
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now lets find $b_{n}$
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Let's find the coefficients $a_{n}$ and $b_{n}$
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Use the formula we derived earlier:
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$a_{n}=\frac{1}{\pi}\left( \int _{-\pi}^0 1\cos\left( \frac{n\pi x}{\pi} \right)\, dx +\int _{0}^\pi x\cos(nx)\, dx\right)$
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Using integration by parts (for the second integral):
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$=\frac{1}{\pi}\left( \frac{1}{n}\cancelto{ 0 }{ \sin(nx) |_{-\pi}^0} + \frac{1}{n}x\cancelto{ 0 }{ \sin(nx) |_{0}^\pi}-\frac{1}{n}\int _{0}^\pi \sin(nx) \, dx \right)$
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$a_{n}=\frac{1}{n^2\pi}(\cos(nx)|_{0}^\pi)=\frac{1}{n^2\pi}(\underbrace{ \cos(n\pi) }_{ (-1)^n }-1)$
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$a_{n}=\frac{1}{n^2\pi}((-1)^n-1)\quad n=1,2,\dots$
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Now let's find $b_{n}$
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$b_{n}=\frac{1}{\pi}\left( \int _{-\pi}^0\sin(nx) \, dx+\int _{0}^\pi x\sin(nx) \, dx \right)$
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$=\frac{1}{\pi}\left( \frac{-1}{n}\cos(nx)|_{-\pi}^0-\frac{1}{n}\left( x\cos(nx)|_{0}^\pi-\int _{0}^\pi \cos(nx)\, dx \right) \right)$
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$=\frac{1}{\pi}[ \frac{-1}{n}\underbrace{ \cos(nx)|_{-\pi}^0 }_{ 1-(-1)^n }-\frac{1}{n}( \underbrace{ x\cos(nx)|_{0}^\pi }_{ \pi(-1)^n-0 }-\underbrace{ \int _{0}^\pi \cos(nx)\, dx }_{ 0 } ) ]$
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$b_{n}=\frac{1}{n\pi}((-1)^n-1-\pi(-1)^n)$
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$b_{n}=\frac{1}{n\pi}((-1)^n(1-\pi)-1) \quad n=1,2,\dots$
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we find that
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$a_{2n}=0$
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$a_{2k-1}=-\frac{2}{n^2\pi}$ for $k=1,2,\dots$
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we find that
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$$a_{2n}=0 \qquad n=1,2,3,\dots$$
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$$a_{2k-1}=-\frac{2}{n^2\pi} \qquad k=1,2,3\dots$$
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what about when $n=0$?
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$a_{0}=\frac{1}{\pi}\left( \int _{-\pi}^0 \, dx+\int _{0}^\pi x \, dx \right)$
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$=\frac{1}{\pi}\left( x|_{-\pi}^0+\frac{x^2}{2}|_{0}^\pi \right)$
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$=\frac{1}{\pi}\left( 0+\pi+\frac{\pi^2}{2} \right)=\frac{\pi}{2}+1$
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$a_{0}=\frac{1}{\pi}\left( x|_{-\pi}^0+\frac{x^2}{2}|_{0}^\pi \right)$
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$a_{0}=\frac{1}{\pi}\left( 0+\pi+\frac{\pi^2}{2} \right)$
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$$a_{0}=\frac{\pi}{2}+1$$
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#ex lets compute the fourier transform of $f(x)=x$ from $-\pi\leq x\leq \pi$
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we have to take a windowed form of $f$ to make this possible, $L=\pi$
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at the left and right edge of the interval, the fourier series is equal to 0. (from the previous theorem)
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find the coefficients:
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#ex Let's compute the Fourier transform of:
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$$f(x)=x \qquad -\pi\leq x\leq \pi$$
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We have to take a windowed form of $f$ to make this possible, $L=\pi$
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At the left and right edge of the interval, the Fourier series is equal to 0. (1st convergence theorem.)
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Find the coefficients:
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$a_{n}=\frac{1}{\pi}\int _{-\pi}^\pi x\cos(nx) \, dx=0$
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Why is it zero? because the integrand is an odd function. (odd times even is odd.) and because we are integrating from $-\pi$ to $\pi$ (symmetric interval)
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$$a_{n}=0$$
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Why is it zero? because the integrand is an odd function. (odd times even is odd.) and because we are integrating from $-\pi$ to $\pi$ (a symmetric interval)
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definition of odd: $f(x)=-f(-x)$
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definition of even: $f(x)=f(-x)$
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odd times even is odd.
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odd times odd is even.
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even times even is even.
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huge exam time saving technique.
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find $b_{n}$
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$b_{n}=\frac{1}{\pi}\int _{-\pi}^\pi x\sin(nx) \, dx=\frac{2}{\pi}\int _{0}^\pi x\sin(nx) \, dx$
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using integration by parts:
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$b_{n}=\frac{2}{n}(-1)^{n+1}$
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another take away: if $f$ is odd, the cos terms are 0
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if $f$ is even, the sin terms are 0.
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Find $b_{n}$:
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$b_{n}=\frac{1}{\pi}\int _{-\pi}^\pi x\sin(nx) \, dx=\frac{2}{\pi}\int _{0}^\pi x\sin(nx) \, dx$ <- that's even, don't be a silly goose and say it's $0$
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Using integration by parts:
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$b_{n}=\frac{2}{\pi}\left( x\left( -\frac{1}{n}\cos(nx)|_{0}^\pi \right)-\int_{0}^\pi -\frac{1}{n}\cos(nx) \, dx \right)$
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$b_{n}=\frac{2}{\pi}\left( -\frac{\pi}{n}(-1)^n+\frac{1}{n^2}\cancelto{ 0 }{ \sin(nx)|_{0}^\pi } \right)$
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$$b_{n}=\frac{2}{n}(-1)^{n+1}$$
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another take away: if $f$ is odd, the $\cos$ terms are $0$
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if $f$ is even, the $\sin$ terms are $0$.
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if f is only defined between 0 and L:
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you can create an odd extension: $\bar{f}=\begin{cases}f(x) & 0\leq x\leq L \\-f(-x), & -L\leq x<0 & & \end{cases}$
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this will contain only sin terms
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you also have a choise to extend it as an even function, symmetrically across the y axis.
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$\bar{f}=\begin{cases}f(x) & 0\leq x\leq L \\f(-x), & -L\leq x<0 & & \end{cases}$
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this will contain only cos terms.
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#end of lec 29
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if $f$ is only defined between $0$ and $L$:
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you can create an odd extension: $\bar{f}(x)=\begin{cases}f(x) & 0\leq x\leq L \\-f(-x), & -L\leq x<0 & & \end{cases}$
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this will contain only $\sin$ terms.
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You also have a choice to extend it as an even function, symmetrically across the $y$ axis.
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$\bar{f}(x)=\begin{cases}f(x) & 0\leq x\leq L \\f(-x), & -L\leq x<0 & & \end{cases}$
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This will contain only $\cos$ terms.
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#end of lec 29
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#start of lec 30
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from last lecture:
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From last lecture:
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$f(x)$ is defined on $[0,L]$
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odd extention:
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odd extension:
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$\bar{f}(x)=\begin{cases}f(x), & 0\leq x\leq L \\-f(-x,) & -L\leq x<0\end{cases}$
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and the $a$ coeffecients (cos terms) are zero.
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not only that, but the b terms are:
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$\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right); \qquad b_{n}=\frac{1}{L}\int _{-L}^L\bar{f}(x) \sin\left( \frac{n\pi x}{L} \right) \, dx$
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and the $a$ coefficients ($\cos$ terms) are zero.
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not only that, but the $b$ coefficients are:
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$b_{n}=\frac{1}{L}\int _{-L}^L\bar{f}(x) \sin\left( \frac{n\pi x}{L} \right) \, dx$
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$$b_{n}=\frac{2}{L}\int _{0}^L f(x)\sin\left( \frac{n\pi x}{L} \right)\, dx$$
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"How about that, this is called a foureir sin series."
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and $\bar{f}(x)$ is:
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$$\bar{f}(x)=\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$$
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"How about that, this is called a Fourier sine series."
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For even extension:
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$\bar{f}(x)=\begin{cases}f(x), & 0\leq x\leq L \\f(-x,) & -L\leq x<0\end{cases}$
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and the $b$ coeffecients (sin terms) are zero.
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not only that but the a terms are:
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$\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n}\cos\left( \frac{n\pi x}{L} \right)$
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$a_{n}=\frac{2}{L}\int _{0}^L f(x)\cos\left( \frac{n\pi x}{L} \right)\, dx$
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and the $b$ coefficients ($\sin$ terms) are zero.
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not only that but the $a$ coefficients are:
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$a_{n}=\frac{1}{L}\int _{-L}^L\bar{f}(x) \cos\left( \frac{n\pi x}{L} \right) \, dx$
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$$a_{n}=\frac{2}{L}\int _{0}^L f(x)\cos\left( \frac{n\pi x}{L} \right)\, dx$$
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and $\bar{f}$ is:
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$$\bar{f}(x)=\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n}\cos\left( \frac{n\pi x}{L} \right)$$
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Remember that $\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$ was the expansion of the eigen value function from the heat equation?
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then $\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n}\cos\left( \frac{n\pi x}{L} \right)$ is also an expansion of some related eigen value function. It's interesting to note.
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remember that $\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$ was the expansion of the eigen value function from the heat equation?
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then $\frac{a_{0}}{2}+\sum_{n=1}^\infty a_{n}\cos\left( \frac{n\pi x}{L} \right)$ is also an expansion of some related eigen value function.
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#ex Fourier sine series for $f(x)=x^2$ from $0\leq x\leq \pi$
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well that means we want the odd extension:
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#ex #fourier Fourier sine series for:
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$$f(x)=x^2 \qquad 0\leq x\leq \pi$$
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Well that means we want the odd extension:
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the $\cos()$ ($a_{n}$) terms are zero.
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the b terms are:
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the $a_{n}$ (cosine) terms are zero.
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the $b_{n}$ terms are:
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$b_{n}=\frac{2}{\pi}\int _{0}^\pi x^2\sin(nx) \, dx$
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$=-\frac{2}{\pi n}\left[ x^2\cos(nx)|_{0}^\pi-2\int _{0}^\pi x\cos(nx)\, dx \right]$
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$=-\frac{2}{n\pi}\left[ x^2\cos(nx)|_{0}^\pi-2\int _{0}^\pi x\cos(nx)\, dx \right]$
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$=-\frac{2}{n\pi}\left[ \pi^2(-1)^n-\frac{2}{n}\left( x\cancelto{ 0 }{ \sin(nx) }|_{0}^\pi-\int _{0}^\pi \sin(nx)\, dx \right) \right]$
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$b_{n}=-\frac{2}{n\pi}\left[ \pi^2(-1)^n-\frac{2}{n^2}\cos(nx)|_{0}^\pi \right]=\frac{2\pi}{n}(-1)^{n+1}+\frac{4}{n^3\pi}((-1)^n-1)$
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for $n=1,2,3,\dots$ note no $n=0$ so no divison by zero problems here.
|
||||
$b_{n}=-\frac{2}{n\pi}[ \pi^2(-1)^n-\frac{2}{n^2}\underbrace{ \cos(nx)|_{0}^\pi }_{ (-1)^n-1 }]=\frac{2\pi}{n}(-1)^{n+1}+\frac{4}{n^3\pi}((-1)^n-1)$ for $n=1,2,3,\dots$
|
||||
Note no $n=0$ so no division by zero problems here.
|
||||
|
||||
#ex fourier cosine series of $f(x)=\sin(x)$ for $0\leq x\leq \pi$
|
||||
#ex #fourier Fourier cosine series of $f(x)=\sin(x)$ for $0\leq x\leq \pi$
|
||||
|
||||
$a_{n}=\frac{2}{\pi}\int _{0}^\pi \sin(x)\cos(nx)\, dx$ for $n=0,1,2,\dots$
|
||||
use trig identity: (by the way the identites will be provided in the exam.)
|
||||
$b_{n}$ (sine) terms are all zero obviously, as it's asking for a Fourier cosine series, i.e., we are doing an even extension.
|
||||
$a_{n}=\frac{2}{\pi}\int _{0}^\pi \sin(x)\cos(nx)\, dx$ for $n=0,1,2,\dots$ <-Odd, but not symmetrical bounds. We can't rule out that it's zero.
|
||||
>Don't be a silly goose and try changing the bounds by removing that 2 in the front. If you did, you'd also have to change $\sin(x)$ to $\bar{f}$ which is $abs(\sin(x))$ and then you're integrating $a_{n}=\frac{1}{\pi}\int _{-\pi}^\pi |\sin(x)|\cos(nx)\, dx$ which is even$\times$even.
|
||||
|
||||
$=\frac{2}{\pi} \frac{1}{2}\int _{0}^\pi (\sin(1-n)x+\sin(n+1)x)\, dx$
|
||||
integrating gives you:
|
||||
Use trig identity: (by the way the identities will be provided in the final exam.)
|
||||
$=\frac{2}{\pi} \frac{1}{2}\int _{0}^\pi \left[\sin((1-n)x)+\sin((n+1)x)\right]\, dx$
|
||||
Integrating gives you:
|
||||
$\frac{1}{\pi}( \frac{-1}{1-n}\underbrace{ \cos((1-n)x)|_{0}^\pi }_{ (-1)^{n+1}-1 } +\frac{-1}{n+1}\underbrace{ \cos((n+1)x)|_{0}^\pi }_{ (-1)^{n+1}-1 })$
|
||||
$a_{n}=-\frac{1}{\pi} \frac{1}{n+1}(-1)^{n+1}+\frac{1}{\pi} \frac{1}{n+1}+\frac{1}{\pi} \frac{1}{-(1-n)}(-1)^{n+1}+\frac{1}{\pi} \frac{1}{1-n}$
|
||||
$a_{n}=-\frac{1}{\pi} \frac{1}{n+1} (-1)^{n+1}+\frac{1}{\pi} \frac{1}{n+1}+\frac{1}{\pi} \frac{1}{n-1}(-1)^{n-1}-\frac{1}{\pi} \frac{1}{n-1}$
|
||||
what about when n=0 or n=1?
|
||||
$a_{0}=\frac{4}{\pi}=\frac{2}{\pi}\int _{0}^\pi \sin(x) \, dx$
|
||||
Assuming that $n\ne0,1$. (note: $n=-1$ is a non-issue since negative coefficients are never considered when taking a Fourier transform.)
|
||||
So what is $a_{0}, a_{1}$?
|
||||
$a_{0}=\frac{2}{\pi}\int _{0}^\pi \sin(x) \, dx=\frac{4}{\pi}$
|
||||
$a_{1}=\frac{2}{\pi}\int _{0}^\pi \sin(x)\cos(x) \, dx=\frac{1}{\pi}\int _{0}^\pi \sin(2x)\, dx=0$
|
||||
"0 is a very very special number it took humanity many numbers of years to invent 0" referring to when dividing by 0.
|
||||
additionally we know that the terms cancel when:
|
||||
<i>"zero is a very very special number it took humanity many numbers of years to invent zero"</i> referring to when dividing by 0.
|
||||
Additionally we know that the terms cancel when:
|
||||
$a_{2k-1}=0$ for $k=1,2,\dots$
|
||||
$a_{2k}=\frac{2}{\pi} \frac{1}{2k+1}+\frac{2}{\pi} \frac{1}{2k-1}$ for $k=1,2,\dots$
|
||||
$a_{2k}=\frac{2}{\pi} \frac{1}{2k+1}-\frac{2}{\pi} \frac{1}{2k-1}$ for $k=1,2,\dots$
|
||||
then:
|
||||
$$\bar{f}(x)=\frac{2}{\pi}+\frac{2}{\pi}\sum_{k=1}^\infty\left( \frac{1}{2k+1}+\frac{1}{2k-1} \right)\cos(2kx)$$
|
||||
|
||||
$$\bar{f}(x)=\frac{2}{\pi}+\frac{2}{\pi}\sum_{k=1}^\infty\left( \frac{1}{2k+1}-\frac{1}{2k-1} \right)\cos(2k\pi x)$$
|
||||
Even with 10 terms, we get a pretty good approximation:
|
||||
|
||||
We have prepared ourselves now, now we start solving PDE's. He's encouraging us to attend the lectures in these last two weeks. He's making it sound like PDE's are hard.
|
||||
|
|
@ -231,9 +231,74 @@ you complaining that your exams are hard theyre not hard. I'm talking about 40 y
|
|||
now we consider a guitar string:
|
||||
|
||||
assuming the thickness of the string is much smaller than the length of the string, which is true.
|
||||
$\frac{ \partial u^2 }{ \partial t^2 }=\alpha^2 \frac{ \partial^2 u }{ \partial x^2 } \quad 0\leq x\leq L, t>0$
|
||||
^ Reminds me of the wave equation from phys 130.
|
||||
$\frac{ \partial u^2 }{ \partial t^2 }=\alpha^2 \frac{ \partial^2 u }{ \partial x^2 } \quad 0\leq x\leq L,\quad t>0$
|
||||
^ Reminds me of the wave equation from Phys 130.
|
||||
Since the string is tied down on the ends we have the following initial conditions:
|
||||
$u(t,0)=u(t,L)=0 \qquad t>0$
|
||||
$u(0,x)=f(x)$ $0\leq x\leq L$
|
||||
$\frac{ \partial u }{ \partial t }(0,x)=g(x)$ $0\leq x\leq L$
|
||||
#end of lec 33
|
||||
^IBVP of the system.
|
||||
#end of lec 33
|
||||
#start of lec 34
|
||||
The wave equation follows many phenomena in electrical engineering.
|
||||
separation of variables:
|
||||
$u(t,x)=X(x)T(t)$
|
||||
plug in to equation:
|
||||
$T''X=\alpha^2TX''$
|
||||
$\frac{T''}{\alpha^2T}=\frac{X''}{X}=-\lambda$
|
||||
^ #evp !
|
||||
consider the $X$ side:
|
||||
$X''+\lambda X=0, \quad X(0)=X(L)=0$
|
||||
We've solved this before.
|
||||
the only non-trivial solutions for that Eigen value problem is:
|
||||
$\lambda_{n}=(\frac{n\pi}{L})^2$
|
||||
$X_{n}(x)=\sin\left( \frac{n\pi x}{L} \right)$ for $n=1,2,3,\dots$
|
||||
$\frac{T''}{\alpha^2T}=-\left( \frac{n\pi}{L} \right)^2$
|
||||
$T_{n}''+\left( \frac{\alpha n\pi}{L} \right)^2T_{n}=0$
|
||||
characteristic equation:
|
||||
$r^2+\left( \frac{\alpha n\pi}{L} \right)^2=0$
|
||||
$r_{1,2}=\pm i \frac{\alpha n\pi}{L}$
|
||||
"Don't memorize the steps. If you try to memorize you will mess up the final for sure. Ask yourself, why am I doing this here?"
|
||||
$T_{n}(t)=b_{n}\cos\left( \frac{\alpha n\pi}{L}t \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right)$
|
||||
$u_{n}(t,x)=\left( b_{n}\cos\left( \frac{\alpha n\pi}{L} \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right) \right)\sin\left( \frac{n\pi x}{L} \right)$
|
||||
if you sum all these <u>modes</u>, you get the solution:
|
||||
$$u_{n}(t,x)=\sum_{n=1}^\infty\left( b_{n}\cos\left( \frac{\alpha n\pi}{L} \right)+a_{n}\sin\left( \frac{\alpha n\pi}{L}t \right) \right)\sin\left( \frac{n\pi x}{L} \right)$$
|
||||
|
||||
"mathematics and reality do align very well, if the speed of the string is different the solution differs aswell."
|
||||
$u(0,x)=f(x)=\sum_{n=1}^\infty b_{n}\sin\left( \frac{n\pi x}{L} \right)$
|
||||
^ that's starting to look familiar.
|
||||
$\implies b_{n}=\frac{2}{L}\int _{0}^L f(x)\sin\left( \frac{n\pi x}{L} \right) \, dx$
|
||||
where does it converge? well $f(x)$ and $f'(x)$ are both continuous, so it converges everywhere.
|
||||
$\frac{ \partial u }{ \partial t }(0,x)=g(x)=\sum_{n=1}^\infty\underbrace{ a_{n} \frac{\alpha n\pi}{L} }_{ }\sin\left( \frac{\alpha n\pi}{L} \right)$
|
||||
$a_{n} \frac{\alpha n\pi}{L}$ are the Fourier $\sin$ coefficients of $g(x)$
|
||||
$a_{n}=\frac{2}{\alpha n\pi}\int _{0}^L g(x)\sin\left( \frac{n\pi x}{L} \right)\, dx$
|
||||
$\alpha^2$ is the Hooke modulus of the string btw.
|
||||
remember the heat equation, the amplitude is exponentionally decreasing,
|
||||
here the amplitude is oscillatory and doesnt increae in time. boi-oi-oi-oing
|
||||
to make it more releasitic we have to add a term for resistance, and we end up with a b in the characteristic equation for T.
|
||||
btw this equation models the electromagnetic feild, to some approximation.
|
||||
the lowest mode is called the fundemental mode, the following terms after are called harmonics.
|
||||
If two instruments play the same note (same fundemental frequency), they still sound different! and that's because of the difference in harmonics.
|
||||
The modes are standing waves in the string.
|
||||
"my claim is that any object, including social objects ,can be described by waves. Everything is a wave."
|
||||
you can model elementary particle behaviours with solitons (non linear waves.)
|
||||
in life in the real world, all waves have finite speed.
|
||||
So thats why its important to learn the wave equation. its the prototype to waves.
|
||||
"waves are the fundamental object. [...]. So that's why it's important, these are the fundamental objects of nature here."
|
||||
$f(x)=\begin{cases}x, & 0\leq x\leq \frac{\pi}{2}\\ \ \pi-x, & \frac{\pi}{2}<x\leq \pi\end{cases}$
|
||||
$\alpha^2=1$
|
||||
$g(x)=\sin(x)$
|
||||
$b_{n}=\frac{2}{\pi}\left( \int _{0} ^\frac{\pi}{2} x\sin(nx) \, dx +\int _{\frac{\pi}{2}} ^\pi (\pi-x)\sin(nx) \, dx\right)$
|
||||
$b_{n}=\frac{4}{n^2\pi}\sin\left( \frac{n\pi}{2} \right)$
|
||||
$b_{2k}=0$
|
||||
half of the b coefficients are 0.
|
||||
$b_{2k-1}=\frac{4}{(2k-1)^2\pi}(-1)^{k+1}$
|
||||
plug in g(x) to get a_n terms
|
||||
however theres a short cut here. from definition of g(x) earlier:
|
||||
$g(x)=a_{1}\sin(x)+2a_{2}\sin(2x)+3a_{3}\sin(3x)+\dots$
|
||||
but g(x) is sin x.
|
||||
so $a_{1}=1$ and every other term is 0
|
||||
plug this in to get solution:
|
||||
$$u(t,x)=\left( \frac{4}{\pi}\cos t+\sin t \right)\sin(x)+\sum_{k=1}^\infty \frac{4(-1)^{k+1}}{\pi(2k+1)^2}\cos((2k+1)t)\sin((2k+1)x)$$
|
||||
typo in his notes, not 2k-1 its 2k+1
|
||||
#end of lec 34
|
||||
|
|
@ -21,7 +21,8 @@ $\Delta u=u(t+\Delta t,x)-u(t,x)$
|
|||
Amount of heat to change temperature over $\Delta t$ with $\Delta u$ is $\underbrace{ C(x) }_{ \text{specific heat capacity } }\underbrace{ \rho(x) }_{ \text{density} }\Delta u\underbrace{ a\Delta x }_{ \text{volume} }$
|
||||
|
||||
$C(x)\rho(x)a\Delta x(u(t+\Delta t,x)-u(t,x))=a\Delta t(k(x+\Delta x) \frac{\partial u}{\partial x}(t,x+\Delta x)-k(x) \frac{\partial u}{\partial x}(t,x))+Q(t,x)\Delta xa\Delta t$
|
||||
>I'm gonna be honest, I'm lost in this derivation already. What is $Q(t,x)$?
|
||||
$Q(t,x)$ is a source term, a term to model internally produced heat.
|
||||
>I'm gonna be honest, I'm lost in this derivation already.
|
||||
|
||||
divide by $a\Delta x\Delta t$
|
||||
$C(x)\rho(x)\frac{(u(t+\Delta t,x)-u(t,x))}{\Delta t}=\frac{ k(x+\Delta x) \frac{\partial u}{\partial x}(t,x+\Delta x)-k(x) \frac{\partial u}{\partial x}(t,x)}{\nabla x}+Q(t,x)$
|
||||
|
|
@ -32,7 +33,7 @@ thermodynamics can be very important for electrical engineers, for instance the
|
|||
## Separation of variables & Eigen value problems
|
||||
#ex #SoV #evp
|
||||
(This is more of a case study than an example.)
|
||||
Rewrite the equation we derived by grouping the constant terms into one constant $D$
|
||||
Rewrite the equation we derived by grouping the constant terms into one constant $D$ and assume for this problem that the internally generated heat is $0$ (i.e., $Q(x,t)=0$)
|
||||
$\frac{ \partial u }{ \partial t }=D \frac{ \partial^2 u }{ \partial x^2 }, \quad 0\leq x\leq L, \quad t>0$
|
||||
boundary conditions:
|
||||
$u(t,0)=u(t,L)=0 , \quad t>0$ (simple case)
|
||||
|
|
@ -116,7 +117,7 @@ lets go back to the problem and focus on $T$:
|
|||
$\frac{T'}{DT}=\frac{X''}{X}=-\lambda$
|
||||
$\frac{T'}{T}=-\left( \frac{n\pi}{L} \right)^2D$
|
||||
this is a separable equation.
|
||||
We can treat the function T as a variable:
|
||||
We can treat the function $T$ as a variable:
|
||||
$\frac{dT}{dt} \frac{1}{T}=-\left( \frac{n\pi}{L} \right)^2D$
|
||||
$\int{dT} \frac{1}{T}=\int-\left( \frac{n\pi}{L} \right)^2Ddt$
|
||||
$\ln(T)=-\left( \frac{n\pi}{L} \right)Dt+c_{n}$
|
||||
|
|
|
|||
|
|
@ -24,8 +24,8 @@ I have written these notes for myself, I thought it would be cool to share them.
|
|||
[Systems of linear equations (lec 21-22)](systems-of-linear-equations-lec-21-22.html)
|
||||
[Power series (lec 22-25)](power-series-lec-22-25.html)
|
||||
[Separation of variables & Eigen value problems (lec 26-28)](separation-of-variables-eigen-value-problems-lec-26-28.html)
|
||||
[Fourier series (lec 28-29)](fourier-series-lec-28-29.html) (raw notes, not reviewed or revised yet.)
|
||||
[Partial differential equations (lec 30-33)](partial-differential-equations-lec-30-33.html) (raw notes, not reviewed or revised yet.)
|
||||
[Fourier series (lec 28-29)](fourier-series-lec-28-29.html)
|
||||
[Partial differential equations (lec 30-34)](partial-differential-equations-lec-30-34.html) (raw notes, not reviewed or revised yet.)
|
||||
</br>
|
||||
[How to solve any DE, a flow chart](Solve-any-DE.png) (Last updated Oct 1st, needs revision. But it gives a nice overview.)
|
||||
[Big LT table (.png)](drawings/bigLTtable.png)
|
||||
|
|
|
|||
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Reference in New Issue