\begin{equation} f\left ( x\right ) =e^{x}={\displaystyle \sum \limits _{k=-\infty }^{\infty }} c_{k}e^{ikx}\tag{1} \end{equation} Where\begin{align*} c_{k} & =\frac{1}{2\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} e^{x}e^{-ikx}dx\\ & =\frac{1}{2\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} e^{\left ( 1-ik\right ) x}dx\\ & =\frac{1}{2\pi }\left [ \frac{e^{\left ( 1-ik\right ) x}}{1-ik}\right ] _{-\pi }^{\pi }\\ & =\frac{1}{\pi \left ( 1-ik\right ) }\left [ \frac{e^{\pi \left ( 1-ik\right ) }-e^{-\pi \left ( 1-ik\right ) }}{2}\right ] \end{align*}
But \(\frac{e^{z}}{2}-\frac{e^{-z}}{2}=\sinh \left ( z\right ) \), hence the above reduces to\begin{equation} c_{k}=\frac{1}{\pi \left ( 1-ik\right ) }\sinh \left ( \pi \left ( 1-ik\right ) \right ) \tag{2} \end{equation} Substituting (2) into (1) gives\[ e^{x}={\displaystyle \sum \limits _{k=-\infty }^{\infty }} \frac{1}{\pi \left ( 1-ik\right ) }\sinh \left ( \pi \left ( 1-ik\right ) \right ) e^{ikx}\] Here are few terms in the series generated using symbolic software:
\[\begin{array} [c]{cc}k & C_{k}\\ -5 & \frac{(1-5i)e^{-\pi }-(1-5i)e^{\pi }}{52\pi }\\ -4 & \frac{\left ( \frac{1}{34}-\frac{2i}{17}\right ) e^{-\pi }\left ( e^{2\pi }-1\right ) }{\pi }\\ -3 & \frac{(1-3i)e^{-\pi }-(1-3i)e^{\pi }}{20\pi }\\ -2 & \frac{\left ( \frac{1}{10}-\frac{i}{5}\right ) e^{-\pi }\left ( e^{2\pi }-1\right ) }{\pi }\\ -1 & -\frac{\left ( \frac{1}{4}-\frac{i}{4}\right ) e^{-\pi }\left ( e^{2\pi }-1\right ) }{\pi }\\ 0 & -\frac{e^{-\pi }-e^{\pi }}{2\pi }\\ 1 & \frac{(1+i)e^{-\pi }-(1+i)e^{\pi }}{4\pi }\\ 2 & \frac{\left ( \frac{1}{10}+\frac{i}{5}\right ) e^{-\pi }\left ( e^{2\pi }-1\right ) }{\pi }\\ 3 & \frac{(1+3i)e^{-\pi }-(1+3i)e^{\pi }}{20\pi }\\ 4 & \frac{\left ( \frac{1}{34}+\frac{2i}{17}\right ) e^{-\pi }\left ( e^{2\pi }-1\right ) }{\pi }\\ 5 & \frac{(1+5i)e^{-\pi }-(1+5i)e^{\pi }}{52\pi }\end{array} \]
Here is a plot of Fourier series of \(e^{x}\) for \(k\) increasing range to compare with \(e^{x}\). To generate this plot the terms with \(c_{-k}+c_{k}\) were added in order together to obtain a real valued function before plotting. Plotting was done from \(x=-\pi \cdots \pi \). We see as more terms are added, the approximation improves. At 20 terms, the approximations became very good. Here is the plot
The even part of \(e^{x}\) are given by \(\frac{e^{x}+e^{-x}}{2}=\cosh x\) and the odd part is \(\frac{e^{x}-e^{-x}}{2}=\sinh x\). For \(e^{ix}\), the even part is \(\frac{e^{ix}+e^{-ix}}{2}=\cos x\) and the odd part is \(\frac{e^{ix}-e^{-ix}}{2}=i\sin x\)
Since \(f\left ( -\pi \right ) =-f\left ( -\pi \right ) \) then \(f\left ( x\right ) \) is an odd function. For an odd function all the \(a_{k}=0\) since these go with the even part.
\begin{align*} b_{k} & =\frac{1}{\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} f\left ( x\right ) \sin \left ( kx\right ) dx\\ & =\frac{1}{\pi }\left ({\displaystyle \int \limits _{-\pi }^{0}} f\left ( x\right ) \sin \left ( kx\right ) dx+{\displaystyle \int \limits _{0}^{\pi }} f\left ( x\right ) \sin \left ( kx\right ) dx\right ) \\ & =\frac{1}{\pi }\left ({\displaystyle \int \limits _{-\pi }^{0}} -\sin \left ( kx\right ) dx+{\displaystyle \int \limits _{0}^{\pi }} \sin \left ( kx\right ) dx\right ) \end{align*}
Changing the limits of integration changes the sign, hence the above can be written as\begin{align*} b_{k} & =\frac{1}{\pi }\left ({\displaystyle \int \limits _{0}^{\pi }} \sin \left ( kx\right ) dx+{\displaystyle \int \limits _{0}^{\pi }} \sin \left ( kx\right ) dx\right ) \\ & =\frac{2}{\pi }{\displaystyle \int \limits _{0}^{\pi }} \sin \left ( kx\right ) dx\\ & =\frac{2}{\pi }\left [ \frac{-\cos kx}{k}\right ] _{0}^{\pi }\\ & =\frac{-2}{\pi k}\left [ \cos kx\right ] _{0}^{\pi }\\ & =\frac{-2}{\pi k}\left [ \cos k\pi -\cos 0\right ] \\ & =\frac{2}{\pi k}\left ( 1-\cos k\pi \right ) \ \ \ \ \ k=1,2,3,\cdots \end{align*}
Hence\[ b_{k}=\left \{ \begin{array} [c]{ccc}\frac{4}{\pi k} & & k=1,3,5,\cdots \\ 0 & & \ \ \ k=2,4,6,\cdots \end{array} \right . \] Hence using \(f\left ( x\right ) ={\displaystyle \sum \limits _{k=1}^{\infty }} b_{k}\sin kx\), we can write the Fourier series of \(f\left ( x\right ) \) as\begin{align*} f\left ( x\right ) & ={\displaystyle \sum \limits _{k=1,3,\cdots }^{\infty }} \frac{4}{\pi k}\sin kx\\ & =\frac{4}{\pi }\sin x+\frac{4}{3\pi }\sin 3x+\frac{4}{5\pi }\sin 5x+\cdots \\ & =\frac{4}{\pi }\left ( \sin x+\frac{1}{3}\sin 3x+\frac{1}{5}\sin 5x+\cdots \right ) \end{align*}
Here is a plot showing the Fourier series approximation to the square wave from \(x=-\pi \cdots \pi \) as more terms are added
We first need to determine the Fourier series for \(\delta \left ( x\right ) \) and \(\delta \left ( x+\pi \right ) \). For \(\delta \left ( x\right ) \) we find\begin{align*} a_{0} & =\frac{1}{2\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} \delta \left ( x\right ) dx=\frac{1}{2\pi }\\ a_{k} & =\frac{1}{\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} \delta \left ( x\right ) \cos kxdx=\frac{1}{\pi }\text{ \ \ \ \ }(\text{since }\cos 0=1)\\ b_{k} & =\frac{1}{\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} \delta \left ( x\right ) \sin kxdx=0\text{ \ \ \ \ }(\text{since }\sin 0=0) \end{align*}
Hence \begin{align*} \delta \left ( x\right ) & =\frac{1}{2\pi }+{\displaystyle \sum \limits _{k=1}^{\infty }} a_{k}\cos kx\\ & =\frac{1}{2\pi }+\frac{1}{\pi }{\displaystyle \sum \limits _{k=1}^{\infty }} \cos kx\\ & =\frac{1}{2\pi }+\frac{1}{\pi }\left ( \cos x+\cos 2x+\cos 3x+\cdots \right ) \end{align*}
Now to determine Fourier series for \(\delta \left ( x+\pi \right ) \)\begin{align*} a_{0} & =\frac{1}{2\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} \delta \left ( x+\pi \right ) dx=\frac{1}{2\pi }\\ a_{k} & =\frac{1}{\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} \delta \left ( x+\pi \right ) \cos kxdx=\frac{\left ( -1\right ) ^{k}}{\pi }\text{ \ \ \ \ }(\text{since }\cos \left ( -k\pi \right ) =\cos k\pi =\left ( -1\right ) ^{k})\\ b_{k} & =\frac{1}{\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} \delta \left ( x\right ) \sin kxdx=0\text{ \ \ \ \ }(\text{since }\sin \left ( -k\pi \right ) =0) \end{align*}
Hence \begin{align*} \delta \left ( x+\pi \right ) & =\frac{1}{2\pi }+{\displaystyle \sum \limits _{k=1}^{\infty }} a_{k}\cos kx\\ & =\frac{1}{2\pi }+\frac{1}{\pi }{\displaystyle \sum \limits _{k=1}^{\infty }} \left ( -1\right ) ^{k}\cos kx\\ & =\frac{1}{2\pi }+\frac{1}{\pi }\left ( -\cos x+\cos 2x-\cos 3x+\cdots \right ) \end{align*}
Therefore\begin{align*} 2\delta \left ( x\right ) -2\delta \left ( x+\pi \right ) & =2\left [ \frac{1}{2\pi }+\frac{1}{\pi }\left ( \cos x+\cos 2x+\cos 3x+\cdots \right ) \right ] -2\left [ \frac{1}{2\pi }+\frac{1}{\pi }\left ( -\cos x+\cos 2x-\cos 3x+\cdots \right ) \right ] \\ & =\frac{1}{\pi }+\frac{2}{\pi }\left ( \cos x+\cos 2x+\cos 3x+\cdots \right ) -\frac{1}{\pi }+\frac{2}{\pi }\left ( \cos x-\cos 2x+\cos 3x-\cos 5x+\cdots \right ) \\ & =\frac{2}{\pi }\left ( 2\cos x+2\cos 3x+2\cos 5x+\cdots \right ) \\ & =\frac{4}{\pi }\left ( \cos x+\cos 3x+\cos 5x+\cdots \right ) \end{align*}
Hence \[ \frac{df}{dx}=\frac{4}{\pi }\left ( \cos x+\cos 3x+\cos 5x+\cdots \right ) \] Hence\[ f\left ( x\right ) =\frac{4}{\pi }\left ( \sin x+\frac{1}{3}\sin 3x+\frac{1}{5}\sin 5x+\cdots \right ) \]
We first need to determine the Fourier series for \(\delta \left ( x\right ) \) and \(\delta \left ( x+\pi \right ) \). For \(\delta \left ( x\right ) \) we find\[ c_{k}=\frac{1}{2\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} \delta \left ( x\right ) e^{-ikx}dx=\frac{1}{2\pi }\] Hence \begin{align*} \delta \left ( x\right ) & ={\displaystyle \sum \limits _{k=-\infty }^{\infty }} c_{k}e^{ikx}\\ & ={\displaystyle \sum \limits _{k=-\infty }^{\infty }} \frac{1}{2\pi }e^{ikx}\\ & =\frac{1}{2\pi }\left ( 1+e^{-ikx}+e^{ikx}+e^{-2ik}+e^{2ik}+\cdots \right ) \\ & =\frac{1}{2\pi }\left ( 1+2\cos kx+2\cos 2kx+2\cos 3kx+\cdots \right ) \end{align*}
Now to determine Fourier series for \(\delta \left ( x+\pi \right ) \)\[ c_{k}=\frac{1}{2\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} \delta \left ( x+\pi \right ) e^{-ikx}dx=\frac{1}{2\pi }e^{ik\pi }=\frac{1}{2\pi }\cos k\pi =\frac{\left ( -1\right ) ^{k}}{2\pi }\] Hence \begin{align*} \delta \left ( x+\pi \right ) & ={\displaystyle \sum \limits _{k=-\infty }^{\infty }} \frac{\left ( -1\right ) ^{k}}{2\pi }e^{ikx}\\ & =\frac{1}{2\pi }\left ( 1-e^{-ix}-e^{ix}+e^{-2ix}+e^{2ix}-e^{-3ix}-e^{3ix}+\cdots \right ) \\ & =\frac{1}{2\pi }\left ( 1-\left ( e^{-ix}+e^{ix}\right ) +e^{-2ix}+e^{2ix}-\left ( e^{-3ix}+e^{3ix}\right ) +\cdots \right ) \\ & =\frac{1}{2\pi }\left ( 1-2\cos x+2\cos 2x-2\cos 3x+\cdots \right ) \end{align*}
Therefore\begin{align*} 2\delta \left ( x\right ) -2\delta \left ( x+\pi \right ) & =2\left [ \frac{1}{2\pi }\left ( 1+2\cos x+2\cos 2x+2\cos 3x+\cdots \right ) \right ] -2\left [ \frac{1}{2\pi }\left ( 1-2\cos x+2\cos 2x-2\cos 3x+\cdots \right ) \right ] \\ & =\frac{1}{\pi }\left ( 1+2\cos x+2\cos 2x+2\cos 3x+\cdots \right ) -\frac{1}{\pi }\left ( 1-2\cos x+2\cos 2x-2\cos 3x+\cdots \right ) \\ & =\frac{1}{\pi }\left ( 4\cos x+4\cos 3x+4\cos 5x+\cdots \right ) \\ & =\frac{4}{\pi }\left ( \cos x+\cos 3x+\cos 5x+\cdots \right ) \end{align*}
Hence \[ \frac{df}{dx}=\frac{4}{\pi }\left ( \cos x+\cos 3x+\cos 5x+\cdots \right ) \] Therefore \[ f\left ( x\right ) =\frac{4}{\pi }\left ( \sin x+\frac{1}{3}\sin 3x+\frac{1}{5}\sin 5x+\cdots \right ) \] \(\,\)
Which is the same as above using the \(a_{k},b_{k}\) method.
From above we found that the Fourier series for square wave is\[ f\left ( x\right ) =\frac{4}{\pi }\left ( \sin x+\frac{1}{3}\sin 3x+\frac{1}{5}\sin 5x+\cdots \right ) \] Therefore at \(x=\frac{\pi }{2}\), the above becomes\[ 1=\frac{4}{\pi }\left ( \sin \frac{\pi }{2}+\frac{1}{3}\sin 3\frac{\pi }{2}+\frac{1}{5}\sin 5\frac{\pi }{2}+\cdots \right ) \] Hence \begin{align*} \pi & =4\left ( \sin \frac{\pi }{2}+\frac{1}{3}\sin 3\frac{\pi }{2}+\frac{1}{5}\sin 5\frac{\pi }{2}+\cdots \right ) \\ & =4\left ( 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots \right ) \end{align*}
We found that only the \(b_{k}\) survive for the Fourier series of the wave function. They are\[ b_{k}=\left \{ \begin{array} [c]{ccc}\frac{4}{\pi k} & & k=1,3,5,\cdots \\ 0 & & \ \ \ k=2,4,6,\cdots \end{array} \right . \] Applying Parseval’s formula leads to\[ \pi \left ( b_{1}^{2}+b_{3}^{2}+b_{5}^{2}+\cdots \right ) ={\displaystyle \int \limits _{-\pi }^{\pi }} \left \vert f\left ( x\right ) \right \vert ^{2}dx=2\pi \] Where we used only the odd \(b_{k}\) terms since all others are zero. The above becomes\begin{align*} \pi \left ( \left ( \frac{4}{\pi }\right ) ^{2}+\left ( \frac{4}{3\pi }\right ) ^{2}+\left ( \frac{4}{5\pi }\right ) ^{2}+\cdots \right ) & =2\pi \\ \pi \left ( \frac{1}{\pi ^{2}}4^{2}+\frac{1}{\pi ^{2}}\left ( \frac{4}{3}\right ) ^{2}+\frac{1}{\pi ^{2}}\left ( \frac{4}{5}\right ) ^{2}+\cdots \right ) & =2\pi \\ \left ( 4^{2}+\left ( \frac{4}{3}\right ) ^{2}+\left ( \frac{4}{5}\right ) ^{2}+\cdots \right ) & =2\pi ^{2}\\ \pi ^{2} & =8\left ( 1+\left ( \frac{1}{3}\right ) ^{2}+\left ( \frac{1}{5}\right ) ^{2}+\cdots \right ) \end{align*}
Hence\[ \pi ^{2}=8\left ( 1+\frac{1}{9}+\frac{1}{25}+\cdots \right ) \]
ps. In the solution below, I was using \(T\) when I should be using \(\frac{T}{2}\) in all the limits. Need to correct later. Or just let period be \(2T\) then the math works ok.
In this problem, the basic idea is to observe that when the period was \(2\pi \) then\begin{align*} f\left ( x\right ) & ={\displaystyle \sum \limits _{k=0}^{\infty }} a_{k}\cos kx+{\displaystyle \sum \limits _{k=1}^{\infty }} b_{k}\sin kx\\ f\left ( x\right ) & ={\displaystyle \sum \limits _{k=-\infty }^{\infty }} c_{k}e^{ikx} \end{align*}
Now when the period is a general value \(T\) we use \(\left ( \frac{2\pi }{T}k\right ) \) in place of just \(k\). So the above becomes\begin{align} f\left ( x\right ) & ={\displaystyle \sum \limits _{k=0}^{\infty }} a_{k}\cos \left ( k\frac{2\pi }{T}x\right ) +{\displaystyle \sum \limits _{k=1}^{\infty }} b_{k}\sin \left ( k\frac{2\pi }{T}x\right ) \tag{1}\\ f\left ( x\right ) & ={\displaystyle \sum \limits _{k=-\infty }^{\infty }} c_{k}e^{i\left ( \frac{2\pi }{T}k\right ) x} \tag{2} \end{align}
We now need to determine \(a_{k},b_{k},c_{k}\) using (1) and (2) in similar way we did when the period was \(2\pi \).
To find \(a_{k}\) we multiply (1) by \(\cos \left ( m\frac{2\pi }{T}x\right ) \) where \(m\) is some integer between \(1\cdots \infty \), and integrating from \(-T\) to \(T\) gives\begin{align*}{\displaystyle \int \limits _{-T}^{T}} f\left ( x\right ) \cos \left ( m\frac{2\pi }{T}x\right ) dx & ={\displaystyle \int \limits _{-T}^{T}}{\displaystyle \sum \limits _{k=0}^{\infty }} a_{k}\cos \left ( k\frac{2\pi }{T}x\right ) \cos \left ( m\frac{2\pi }{T}x\right ) dx+{\displaystyle \int \limits _{-T}^{T}}{\displaystyle \sum \limits _{k=1}^{\infty }} b_{k}\sin \left ( k\frac{2\pi }{T}x\right ) \cos \left ( m\frac{2\pi }{T}x\right ) dx\\ & ={\displaystyle \sum \limits _{k=0}^{\infty }}{\displaystyle \int \limits _{-T}^{T}} a_{k}\cos \left ( k\frac{2\pi }{T}x\right ) \cos \left ( m\frac{2\pi }{T}x\right ) dx+{\displaystyle \sum \limits _{k=1}^{\infty }}{\displaystyle \int \limits _{-T}^{T}} b_{k}\sin \left ( k\frac{2\pi }{T}x\right ) \cos \left ( m\frac{2\pi }{T}x\right ) dx \end{align*}
Due to orthogonality between the \(\sin \) and \(\cos \), all the product of \(\sin \cos \) vanish, and only one term in the product of \(\cos \cos \) remain which is the one when \(k=m\), hence the above reduces to\[{\displaystyle \int \limits _{-T}^{T}} f\left ( x\right ) \cos \left ( m\frac{2\pi }{T}x\right ) dx={\displaystyle \int \limits _{-T}^{T}} a_{m}\cos \left ( m\frac{2\pi }{T}x\right ) \cos \left ( m\frac{2\pi }{T}x\right ) dx \] Since \(m\) is arbitrary, we can rename it back to \(k\) to keep the same naming as before. \begin{equation}{\displaystyle \int \limits _{-T}^{T}} f\left ( x\right ) \cos \left ( k\frac{2\pi }{T}x\right ) dx={\displaystyle \int \limits _{-T}^{T}} a_{k}\cos ^{2}\left ( k\frac{2\pi }{T}x\right ) dx\tag{3} \end{equation} When \(k=0\) we find\begin{align*}{\displaystyle \int \limits _{-T}^{T}} f\left ( x\right ) dx & ={\displaystyle \int \limits _{-T}^{T}} a_{0}dx\\ & =2a_{0}T \end{align*}
Hence \[ a_{0}=\frac{1}{2T}{\displaystyle \int \limits _{-T}^{T}} f\left ( x\right ) dx \] Notice, when \(T=\pi \), the above reduces to \(a_{0}=\frac{1}{2\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} f\left ( x\right ) dx\). Now to find \(a_{k}\) for \(k\geq 1\), then from (3)\begin{align*}{\displaystyle \int \limits _{-T}^{T}} f\left ( x\right ) \cos \left ( k\frac{2\pi }{T}x\right ) dx & ={\displaystyle \int \limits _{-T}^{T}} a_{k}\cos ^{2}\left ( k\frac{2\pi }{T}x\right ) dx\\ & =a_{k}T \end{align*}
Hence \[ a_{k}=\frac{1}{T}\int _{-T}^{T}f(x)\cos \left ( k\frac{2\pi }{T}x\right ) \,dx \] Notice that when \(T=\pi \) the above reduces to \(a_{k}=\frac{1}{\pi }{\displaystyle \int \limits _{-\pi }^{\pi }}f(x)\cos (kx)\,dx\) as before.
Now we find \(b_{k}\) similarly. We multiply (1) by \(\sin \left ( m\frac{2\pi }{T}x\right ) \) where \(m\) is some integer between \(1\cdots \infty \), and integrating from \(-T\) to \(T\) gives\[{\displaystyle \int \limits _{-T}^{T}} f\left ( x\right ) \sin \left ( m\frac{2\pi }{T}x\right ) dx={\displaystyle \sum \limits _{k=0}^{\infty }}{\displaystyle \int \limits _{-T}^{T}} a_{k}\cos \left ( k\frac{2\pi }{T}x\right ) \sin \left ( m\frac{2\pi }{T}x\right ) dx+{\displaystyle \sum \limits _{k=1}^{\infty }}{\displaystyle \int \limits _{-T}^{T}} b_{k}\sin \left ( k\frac{2\pi }{T}x\right ) \sin \left ( m\frac{2\pi }{T}x\right ) dx \] Due to orthogonality between the \(\sin \) and \(\cos \), all the products of \(\sin \cos \) vanish, and only one term in the product of \(\sin \sin \) remain which is the one when \(k=m\), hence the above reduces to\[{\displaystyle \int \limits _{-T}^{T}} f\left ( x\right ) \sin \left ( m\frac{2\pi }{T}x\right ) dx={\displaystyle \int \limits _{-T}^{T}} b_{m}\sin \left ( m\frac{2\pi }{T}x\right ) \sin \left ( m\frac{2\pi }{T}x\right ) dx \] Since \(m\) is arbitrary, we can rename it back to \(k\) to keep the same naming as before. \begin{align*}{\displaystyle \int \limits _{-T}^{T}} f\left ( x\right ) \sin \left ( k\frac{2\pi }{T}x\right ) dx & ={\displaystyle \int \limits _{-T}^{T}} b_{k}\sin ^{2}\left ( k\frac{2\pi }{T}x\right ) dx\\ & =b_{k}T \end{align*}
Hence \[ b_{k}=\frac{1}{T}{\displaystyle \int } T^{T}f\left ( x\right ) \sin \left ( k\frac{2\pi }{T}x\right ) dx \] Notice that when \(T=\pi \) the above reduces to \(b_{k}=\frac{1}{\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} f\left ( x\right ) \sin \left ( kx\right ) dx\) as before. We now find \(c_{k}\).\[ f\left ( x\right ) ={\displaystyle \sum \limits _{k=-\infty }^{\infty }} c_{k}e^{i\left ( k\frac{2\pi }{T}\right ) x}\] Multiplying both side by \(e^{-i\left ( m\frac{2\pi }{T}\right ) x}\) and integrating over the period\[{\displaystyle \int \limits _{-T}^{T}} f\left ( x\right ) e^{-i\left ( m\frac{2\pi }{T}\right ) x}dx={\displaystyle \sum \limits _{k=-\infty }^{\infty }}{\displaystyle \int \limits _{-T}^{T}} c_{k}e^{i\left ( k\frac{2\pi }{T}\right ) x}e^{-i\left ( m\frac{2\pi }{T}\right ) x}dx \] All terms other than ones which \(k=m\) remain. Hence the above becomes\begin{align*}{\displaystyle \int \limits _{-T}^{T}} f\left ( x\right ) e^{-i\left ( m\frac{2\pi }{T}\right ) x}dx & ={\displaystyle \int \limits _{-T}^{T}} c_{m}e^{i\left ( m\frac{2\pi }{T}\right ) x}e^{-i\left ( m\frac{2\pi }{T}\right ) x}dx\\ & ={\displaystyle \int \limits _{-T}^{T}} c_{m}dx \end{align*}
Therefore, since \(m\) is now arbitrary, we rename it back to \(k\) and simplifying\begin{align*}{\displaystyle \int \limits _{-T}^{T}} f\left ( x\right ) e^{-i\left ( k\frac{2\pi }{T}\right ) x}dx & =2Tc_{k}\\ c_{k} & =\frac{1}{2T}{\displaystyle \int \limits _{-T}^{T}} f\left ( x\right ) e^{-i\left ( k\frac{2\pi }{T}\right ) x}dx \end{align*}
The \(a_{0}\) term in the Fourier series of \(\cos ^{2}x\) is the constant term. Hence it is the constant that is closest to \(\cos ^{2}x\) in the square sense. Therefore\begin{align*} a_{0} & =\frac{1}{2\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} \cos ^{2}xdx\\ & =\frac{1}{2\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} \cos ^{2}xdx\\ & =\frac{1}{2} \end{align*}
To find the multiple of \(\cos x\) which is closest to \(\cos ^{3}x\), we find \(a_{1}\) term in the Fourier series of \(\cos ^{3}x\) since that is the term which has \(a_{1}\cos x\) in it. Hence\begin{align*} a_{1} & =\frac{1}{\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} \cos ^{3}x\cos xdx\\ & =\frac{1}{\pi }\left ( \frac{3\pi }{4}\right ) \\ & =\frac{3}{4} \end{align*}
The function we are approximating using Fourier series is
Since it is even, we only need to determine \(a_{k}\)\begin{align*} a_{k} & =\frac{1}{\pi }{\displaystyle \int \limits _{-\pi }^{\pi }} f\left ( x\right ) \cos kxdx=\frac{2}{\pi }{\displaystyle \int \limits _{0}^{\pi }} \left ( 1-\frac{x}{\pi }\right ) \cos kxdx\\ & =\frac{2}{\pi }\left ( \frac{1-\cos k\pi }{k^{2}\pi }\right ) \end{align*}
Hence\begin{align*} f\left ( x\right ) & =a_{0}+{\displaystyle \sum \limits _{k=1}^{\infty }} a_{k}\cos kx\\ & =\frac{1}{2}+\frac{2}{\pi }\left ( \frac{1-\cos \pi }{\pi }\right ) \cos x+\frac{2}{\pi }\left ( \frac{1-\cos 2\pi }{4\pi }\right ) \cos 2x+\frac{2}{\pi }\left ( \frac{1-\cos 3\pi }{9\pi }\right ) \cos 3x+\cdots \\ & =\frac{1}{2}+\frac{2}{\pi }\left ( \frac{2}{\pi }\right ) \cos x+\frac{2}{\pi }\left ( \frac{2}{9\pi }\right ) \cos 3x+\frac{2}{\pi }\left ( \frac{2}{25\pi }\right ) \cos 5x+\cdots \\ & =\frac{1}{2}+\frac{4}{\pi ^{2}}\cos x+\frac{4}{9\pi ^{2}}\cos 3x+\frac{4}{25\pi }\cos 5x+\cdots \end{align*}
Here is a plot showing the approximation as more terms are added. The label of each plot show the number of terms used. The more terms we use, the better the approximation
The first step is to obtain the \(a_{k},b_{k}\) coefficients by expanding the boundary value of the solution using Fourier series. On the boundary \[ u_{0}=\left \{ \begin{array} [c]{cc}1 & 0<\theta <\pi \\ 0 & -\pi <\theta <0 \end{array} \right . \] Hence \[ a_{0}=\frac{1}{2\pi }{\displaystyle \int \limits _{0}^{\pi }} d\theta =\frac{1}{2}\] And \[ a_{k}=\frac{1}{\pi }{\displaystyle \int \limits _{0}^{\pi }} \cos k\theta d\theta =\frac{1}{k\pi }\left [ \sin k\theta \right ] _{0}^{\pi }=0 \] And\begin{align*} b_{k} & =\frac{1}{\pi }{\displaystyle \int \limits _{0}^{\pi }} \sin k\theta d\theta =\frac{1}{k\pi }\left [ -\cos k\theta \right ] _{0}^{\pi }=0=\frac{-1}{k\pi }\left [ \cos k\pi -\cos 0\right ] \\ & =\left \{ \frac{2}{\pi },\frac{2}{3\pi },\frac{2}{5\pi },\cdots \right \} \end{align*}
Only odd values of \(k\) survive. Now that we found the Fourier coefficient, we use them in the solution given in equation (22), page 276 on the book\begin{align*} u\left ( r,\theta \right ) & =a_{0}+b_{1}r\sin \theta +b_{3}r^{3}\sin 3\theta +b_{5}r^{5}\sin ^{5}\theta +\cdots \\ & =\frac{1}{2}+\frac{2}{\pi }\left ( r\sin \theta +\frac{1}{3}r^{3}\sin 3\theta +\frac{1}{5}r^{5}\sin ^{5}\theta +\cdots \right ) \end{align*}
At the origin, let \(r=0\)\[ u\left ( 0,\theta \right ) =\frac{1}{2}\]
A sketch of the function (string) is below.
Since \(f\left ( x\right ) \) is odd, we only need to determine \(b_{k}\)\begin{align*} b_{k} & =\frac{2}{\pi }{\displaystyle \int \limits _{0}^{\pi }} f\left ( x\right ) \sin kxdx\\ & =\frac{2}{\pi }\left ({\displaystyle \int \limits _{0}^{p}} \frac{x}{p}\sin kxdx+{\displaystyle \int \limits _{p}^{\pi }} \frac{x-\pi }{p-\pi }\sin kxdx\right ) \\ & =\frac{2}{\pi }\left ( \frac{\sin kp-kp\cos kp}{k^{2}p}+\frac{k\left ( \pi -p\right ) \cos kp+\sin kp-\sin k\pi }{k^{2}\left ( \pi -p\right ) }\right ) \\ & =\frac{2\left ( \pi \sin kp-p\sin k\pi \right ) }{k^{2}p\pi \left ( \pi -p\right ) } \end{align*}
For \(k=2\)\begin{align*} b_{2} & =\frac{\left ( \pi \sin 2p-p\sin 2\pi \right ) }{2p\pi \left ( \pi -p\right ) }\\ & =\frac{\pi \sin 2p}{2p\pi \left ( \pi -p\right ) } \end{align*}
For zero, we need \begin{align*} 0 & =\pi \sin 2p\\ \sin 2p & =0 \end{align*}
Hence \[ p=\frac{\pi }{2}\]
Two functions \(f,g\) are if the inner product is zero \({\displaystyle \int \limits _{-1}^{1}} f\left ( x\right ) g\left ( x\right ) dx=0\). Hence\[{\displaystyle \int \limits _{-1}^{1}} P_{2}P_{0}dx={\displaystyle \int \limits _{-1}^{1}} \left ( x^{2}-1\right ) dx=\left ( \frac{x^{3}}{3}-x\right ) _{-1}^{1}=0 \] And\[{\displaystyle \int \limits _{-1}^{1}} P_{2}P_{1}dx={\displaystyle \int \limits _{-1}^{1}} \left ( x^{2}-1\right ) xdx=\left ( \frac{x^{4}}{4}-\frac{x^{2}}{2}\right ) _{-1}^{1}=0 \] Now let \(P_{3}=x^{3}-cx\), we want this to be orthogonal to \(P_{0},P_{1},P_{2}\). Hence\begin{align*}{\displaystyle \int \limits _{-1}^{1}} P_{3}P_{0}dx & ={\displaystyle \int \limits _{-1}^{1}} x^{3}-cxdx=\left ( \frac{x^{4}}{4}-c\frac{x^{2}}{2}\right ) _{-1}^{1}=\left ( \frac{1}{4}-c\frac{1}{2}\right ) -\left ( \frac{1}{4}-c\frac{1}{2}\right ) \\ 0 & =0 \end{align*}
This equation did not help us find \(c\). We try the next one\begin{align*}{\displaystyle \int \limits _{-1}^{1}} P_{3}P_{1}dx & ={\displaystyle \int \limits _{-1}^{1}} \left ( x^{3}-cx\right ) xdx=\left ( \frac{x^{5}}{5}-c\frac{x^{3}}{3}\right ) _{-1}^{1}=\left ( \frac{1}{5}-c\frac{1}{3}\right ) -\left ( -\frac{1}{5}+c\frac{1}{3}\right ) =\frac{2}{5}-\frac{2}{3}c\\ \frac{2}{5}-\frac{2}{3}c & =0\\ c & =\frac{2}{5}\frac{3}{2}\\ & =\frac{3}{5} \end{align*}
Hence \[ P_{3}=x^{3}-\frac{3}{5}x \]
The proposed solution is \begin{equation} u\left ( x,y\right ) ={\displaystyle \sum }{\displaystyle \sum } \frac{b_{kl}\sin kx\sin ly}{\left ( k^{2}+l^{2}\right ) } \tag{1} \end{equation} To see if this solves \begin{equation} -u_{xx}-u_{yy}=f={\displaystyle \sum }{\displaystyle \sum } b_{kl}\sin kx\sin ly \tag{1A} \end{equation} we will take (1) and substitute in the LHS of Poisson equation (1A) and see if we get the RHS of (1A) which is \(f\).\begin{align} \frac{\partial u}{\partial x} & ={\displaystyle \sum }{\displaystyle \sum } \frac{b_{kl}k\cos kx\sin ly}{\left ( k^{2}+l^{2}\right ) }\nonumber \\ \frac{\partial ^{2}u}{\partial x^{2}} & ={\displaystyle \sum }{\displaystyle \sum } \frac{-b_{kl}k^{2}\sin kx\sin ly}{\left ( k^{2}+l^{2}\right ) } \tag{2} \end{align}
And\begin{align} \frac{\partial u}{\partial y} & ={\displaystyle \sum }{\displaystyle \sum } \frac{b_{kl}\sin \left ( kx\right ) l\cos ly}{\left ( k^{2}+l^{2}\right ) }\nonumber \\ \frac{\partial ^{2}u}{\partial y^{2}} & ={\displaystyle \sum }{\displaystyle \sum } \frac{-b_{kl}\sin \left ( kx\right ) l^{2}\sin ly}{\left ( k^{2}+l^{2}\right ) } \tag{3} \end{align}
Substituting (2) and (3) in the LHS of (1A) gives\begin{align*} -u_{xx}-u_{yy} & ={\displaystyle \sum }{\displaystyle \sum } \frac{b_{kl}k^{2}\sin kx\sin ly}{\left ( k^{2}+l^{2}\right ) }+{\displaystyle \sum }{\displaystyle \sum } \frac{b_{kl}\sin \left ( kx\right ) l^{2}\sin ly}{\left ( k^{2}+l^{2}\right ) }\\ & ={\displaystyle \sum }{\displaystyle \sum } \frac{b_{kl}k^{2}\sin kx\sin ly+b_{kl}\sin \left ( kx\right ) l^{2}\sin ly}{\left ( k^{2}+l^{2}\right ) }\\ & ={\displaystyle \sum }{\displaystyle \sum } \frac{\left ( b_{kl}\sin kx\sin ly\right ) \left ( k^{2}+l^{2}\right ) }{\left ( k^{2}+l^{2}\right ) }\\ & ={\displaystyle \sum }{\displaystyle \sum } b_{kl}\sin kx\sin ly \end{align*}
Which is \(f\). Hence \(u\left ( x,y\right ) ={\displaystyle \sum }{\displaystyle \sum } \frac{b_{kl}\sin kx\sin ly}{\left ( k^{2}+l^{2}\right ) }\) is the solution verified.
\begin{align*} f\left ( x\right ) & =\frac{1}{2\pi }{\displaystyle \int \limits _{k=-\infty }^{\infty }} \delta \left ( k\right ) e^{ikx}dk\\ & =\frac{1}{2\pi }\left [ e^{ikx}\right ] _{k=0}=\frac{1}{2\pi } \end{align*}
\begin{align} f\left ( x\right ) & =\frac{1}{2\pi }{\displaystyle \int \limits _{k=-\infty }^{\infty }} e^{-\left \vert k\right \vert }e^{ikx}dk\nonumber \\ & =\frac{1}{2\pi }\left ({\displaystyle \int \limits _{k=-\infty }^{0}} e^{k}e^{ikx}dk+{\displaystyle \int \limits _{0}^{\infty }} e^{-k}e^{ikx}dk\right ) \nonumber \\ & =\frac{1}{2\pi }\left ({\displaystyle \int \limits _{k=-\infty }^{0}} e^{k\left ( 1+ix\right ) }dk+{\displaystyle \int \limits _{0}^{\infty }} e^{k\left ( -1+ix\right ) }dk\right ) \nonumber \\ & =\frac{1}{2\pi }\left ( \left [ \frac{e^{k\left ( 1+ix\right ) }}{1+ix}\right ] _{-\infty }^{0}+\left [ \frac{e^{k\left ( -1+ix\right ) }}{-1+ix}\right ] _{0}^{\infty }\right ) \tag{1} \end{align}
Looking at the first integral result \[ \left [ \frac{e^{k\left ( 1+ix\right ) }}{1+ix}\right ] _{-\infty }^{0}=\frac{1}{1+ix}-\frac{e^{-\infty \left ( 1+ix\right ) }}{1+ix}=\frac{1}{1+ix}\] Where we looked at real part of \(e^{-\infty \left ( 1+ix\right ) }=0\) so that we can make \(e^{-\infty \left ( 1+ix\right ) }\) to be zero.
Looking at the second integral result\[ \left [ \frac{e^{k\left ( -1+ix\right ) }}{-1+ix}\right ] _{0}^{\infty }=\frac{e^{\infty \left ( -1+ix\right ) }}{-1+ix}-\frac{1}{-1+ix}=-\frac{1}{-1+ix}\] Where we looked at real part of \(e^{\infty \left ( -1+ix\right ) }=0\) so that we can make \(e^{\infty \left ( -1+ix\right ) }\) to be zero. Hence, using the above two results in (1) gives\begin{align*} f\left ( x\right ) & =\frac{1}{2\pi }\left ( \frac{1}{1+ix}-\frac{1}{-1+ix}\right ) \\ & =\frac{1}{2\pi }\left ( \frac{1}{1+ix}+\frac{1}{1-ix}\right ) \\ & =\frac{1}{2\pi }\left ( \frac{\left ( 1-ix\right ) +\left ( 1+ix\right ) }{\left ( 1+ix\right ) \left ( 1-ix\right ) }\right ) \\ & =\frac{1}{2\pi }\left ( \frac{2}{1+x^{2}}\right ) \\ & =\frac{1}{\pi }\frac{1}{1+x^{2}} \end{align*}
For \(f\left ( x\right ) =\delta \left ( x\right ) \)
\[ 2\pi{\displaystyle \int \limits _{-\infty }^{\infty }} \delta ^{2}\left ( x\right ) dx=2\pi \lim _{n\rightarrow \infty }{\displaystyle \int \limits _{-\infty }^{\infty }} \delta \left ( x\right ) g_{n}\left ( x\right ) dx \] Where \(g_{n}\left ( x\right ) \) is sequence of Gaussian functions. The RHS above becomes\[ 2\pi{\displaystyle \int \limits _{-\infty }^{\infty }} \delta ^{2}\left ( x\right ) dx=2\pi \lim _{n\rightarrow \infty }g_{n}\left ( 0\right ) \] But \(\lim _{n\rightarrow \infty }g_{n}\left ( 0\right ) =\infty \) hence\[ 2\pi{\displaystyle \int \limits _{-\infty }^{\infty }} \delta ^{2}\left ( x\right ) dx=\infty \] Now \(\hat{f}\left ( k\right ) =1\) \(\ \) for the Dirac delta. Hence \begin{align*} \hat{f}\left ( k\right ) & ={\displaystyle \int \limits _{-\infty }^{\infty }} \left ( 1\right ) e^{-ikx}dx\\ & ={\displaystyle \int \limits _{-\infty }^{\infty }} e^{-ikx}dx\\ & =\left [ \frac{e^{-ikx}}{-ik}\right ] _{-\infty }^{\infty }=\frac{1}{-ik}\left ( e^{-ik\infty }-e^{+ik\infty }\right ) =\frac{1}{-ik}\left ( 0-\infty \right ) =\infty \end{align*}
Hence verified for \(\delta \) OK.
For \(f\left ( x\right ) =e^{-\frac{x^{2}}{2}}\) then\begin{align*} 2\pi{\displaystyle \int \limits _{-\infty }^{\infty }} \left \vert f\left ( x\right ) \right \vert ^{2}dx & =2\pi{\displaystyle \int \limits _{-\infty }^{\infty }} \left \vert e^{-\frac{x^{2}}{2}}\right \vert ^{2}dx\\ & =2\pi{\displaystyle \int \limits _{0}^{\infty }} e^{-x^{2}}dx\\ & =2\pi \left ( \frac{\sqrt{\pi }}{2}\right ) \\ & =\pi ^{\frac{3}{2}} \end{align*}
Now \(\hat{f}\left ( k\right ) \) for the above function is\begin{align*} \hat{f}\left ( k\right ) & ={\displaystyle \int \limits _{-\infty }^{\infty }} f\left ( x\right ) e^{-ikx}dx\\ & ={\displaystyle \int \limits _{-\infty }^{\infty }} e^{-\frac{x^{2}}{2}}e^{-ikx}dx\\ & =e^{-\frac{k^{2}}{2}}\sqrt{2\pi } \end{align*}
Hence \begin{align*}{\displaystyle \int \limits _{-\infty }^{\infty }} \left \vert \hat{f}\left ( k\right ) \right \vert ^{2}dk & ={\displaystyle \int \limits _{-\infty }^{\infty }} \left \vert e^{-\frac{k^{2}}{2}}\sqrt{2\pi }\right \vert ^{2}dk\\ & =2\pi{\displaystyle \int \limits _{-\infty }^{\infty }} \left \vert e^{-\frac{k^{2}}{2}}\right \vert ^{2}dk\\ & =2\pi{\displaystyle \int \limits _{0}^{\infty }} e^{-k^{2}}dk\\ & =2\pi \left ( \sqrt{\frac{\pi }{2}}\right ) \\ & =\pi ^{\frac{3}{2}} \end{align*}
Which is the same as before. Hence verified.
For \(f\left ( x\right ) =e^{-\frac{x^{2}}{2}}\)\begin{align*} W_{x}^{2} & =\frac{\int _{-\infty }^{\infty }x^{2}\left \vert f\left ( x\right ) \right \vert ^{2}dx}{\int _{-\infty }^{\infty }\left \vert f\left ( x\right ) \right \vert ^{2}dx}\\ & =\frac{\int _{-\infty }^{\infty }x^{2}\left \vert e^{-\frac{x^{2}}{2}}\right \vert ^{2}dx}{\int _{-\infty }^{\infty }\left \vert e^{-\frac{x^{2}}{2}}\right \vert ^{2}dx}\\ & =\frac{\int _{0}^{\infty }x^{2}e^{-x^{2}}dx}{\int _{0}^{\infty }e^{-x^{2}}dx}\\ & =\frac{\frac{\sqrt{\pi }}{4}}{\frac{\sqrt{\pi }}{2}}=\frac{1}{2} \end{align*}
Now \(\hat{f}\left ( k\right ) =\int _{-\infty }^{\infty }f\left ( x\right ) e^{-ikx}dx=\int _{-\infty }^{\infty }e^{-\frac{x^{2}}{2}}e^{-ikx}dx=e^{-\frac{k^{2}}{2}}\sqrt{2\pi }\), hence\begin{align*} W_{k}^{2} & =\frac{\int _{-\infty }^{\infty }k^{2}\left \vert \hat{f}\left ( k\right ) \right \vert ^{2}dx}{\int _{-\infty }^{\infty }\left \vert \hat{f}\left ( k\right ) \right \vert ^{2}dx}\\ & =\frac{\int _{-\infty }^{\infty }k^{2}\left \vert e^{-\frac{k^{2}}{2}}\sqrt{2\pi }\right \vert ^{2}dx}{\int _{-\infty }^{\infty }\left \vert e^{-\frac{k^{2}}{2}}\sqrt{2\pi }\right \vert ^{2}dx}\\ & =\frac{2\pi \int _{0}^{\infty }k^{2}e^{-k^{2}}dx}{2\pi \int _{0}^{\infty }e^{-k^{2}}dx}\\ & =\frac{\frac{\sqrt{\pi }}{4}}{\frac{\sqrt{\pi }}{2}}=\frac{1}{2} \end{align*}
Hence \[ W_{x}W_{k}=\sqrt{\frac{1}{2}}\sqrt{\frac{1}{2}}=\frac{1}{2}\] But uncertainty principle says that \(W_{x}W_{k}\geq \frac{1}{2}\). Hence verified OK.
Using 4L(1), let \(f\left ( x\right ) =e^{-\frac{x^{2}}{2}}\), which has \(\hat{f}\left ( k\right ) =\int _{-\infty }^{\infty }e^{-\frac{x^{2}}{2}}e^{-ikx}dx=\sqrt{2\pi }e^{-\frac{k^{2}}{2}}\), hence \(\frac{d}{dx}f\left ( x\right ) \) will have the transform \(ik\hat{f}\left ( k\right ) \), therefore, \[\mathcal{F}\left ( \frac{d}{dx}f\left ( x\right ) \right ) =\mathcal{F}\left ( -xe^{-\frac{x^{2}}{2}}\right ) =ik\sqrt{2\pi }e^{-\frac{k^{2}}{2}}\] Therefore \(xe^{-\frac{x^{2}}{2}}\) has the transform \(-ik\sqrt{2\pi }e^{-\frac{k^{2}}{2}}\)
Let \(f\left ( x\right ) =xe^{-\frac{x^{2}}{2}}\), which has \(\hat{f}\left ( k\right ) =-ik\sqrt{2\pi }e^{-\frac{k^{2}}{2}}\) from part(a). But \(\frac{d}{dx}f\left ( x\right ) =e^{-\frac{x^{2}}{2}}-x^{2}e^{-\frac{x^{2}}{2}}\). Hence the transform of \(\frac{d}{dx}f\left ( x\right ) =ik\hat{f}\left ( k\right ) \). Therefore\begin{align*} \mathcal{F}\left ( e^{-\frac{x^{2}}{2}}-x^{2}e^{-\frac{x^{2}}{2}}\right ) & =ik\left ( -ik\sqrt{2\pi }e^{-\frac{k^{2}}{2}}\right ) \\\mathcal{F}\left ( e^{-\frac{x^{2}}{2}}\right ) -\mathcal{F}\left ( x^{2}e^{-\frac{x^{2}}{2}}\right ) & =k^{2}\sqrt{2\pi }e^{-\frac{k^{2}}{2}} \end{align*}
But \(\mathcal{F}\left ( e^{-\frac{x^{2}}{2}}\right ) =\sqrt{2\pi }e^{-\frac{k^{2}}{2}}\), hence\begin{align*} \mathcal{F}\left ( x^{2}e^{-\frac{x^{2}}{2}}\right ) & =\sqrt{2\pi }e^{-\frac{k^{2}}{2}}-k^{2}\sqrt{2\pi }e^{-\frac{k^{2}}{2}}\\\mathcal{F}\left ( x^{2}e^{-\frac{x^{2}}{2}}\right ) & =\sqrt{2\pi }e^{-\frac{k^{2}}{2}}\left ( 1-k^{2}\right ) \end{align*}
Therefore\[\mathcal{F}\left ( x^{2}e^{-\frac{x^{2}}{2}}\right ) =\sqrt{2\pi }e^{-\frac{k^{2}}{2}}\left ( 1-k^{2}\right ) \]
Let \(\hat{u}\left ( k\right ) \) be the Fourier transform of \(u\left ( x\right ) \). Using \(\mathcal{F}\left ( \frac{du}{dx}\right ) =ik\hat{u}\left ( k\right ) \) and \(\mathcal{F}\left ( \delta \right ) =1\), then applying Fourier transform on the ODE gives\[ ik\hat{u}\left ( k\right ) +a\hat{u}\left ( k\right ) =1 \] Solving for \(\hat{u}\left ( k\right ) \) \begin{align*} \hat{u}\left ( k\right ) \left ( a+ik\right ) & =1\\ \hat{u}\left ( k\right ) & =\frac{1}{a+ik} \end{align*}
Hence, from page 310 in text book, it gives the inverse Fourier transform for the above as \[ u\left ( x\right ) =\left \{ \begin{array} [c]{cc}e^{-ax} & x>0\\ 0 & x<0 \end{array} \right . \]
Comparing the integral equation \begin{equation}{\displaystyle \int \limits _{-\infty }^{\infty }} e^{-\left \vert x-y\right \vert }u\left ( y\right ) dy-2u\left ( x\right ) =f\left ( x\right ) \tag{1} \end{equation} with the one in the textbook, page 322 in example one, where the Fourier transform of\[{\displaystyle \int \limits _{-\infty }^{\infty }} e^{-\left \vert x-y\right \vert }u\left ( y\right ) dy=f\left ( x\right ) \] Is given as \[ \frac{2}{1+\omega ^{2}}\hat{u}\left ( \omega \right ) =\hat{f}\left ( \omega \right ) \] The only difference is that in this problem we have an extra \(-2u\left ( x\right ) \) term, whose Fourier transform is \(-2\hat{u}\left ( \omega \right ) \). Hence the Fourier transform for (1) becomes\[ \frac{2}{1+\omega ^{2}}\hat{u}\left ( \omega \right ) -2\hat{u}\left ( \omega \right ) =\hat{f}\left ( \omega \right ) \] Solving for \(\hat{u}\left ( \omega \right ) \)\begin{align*} \hat{u}\left ( \omega \right ) \left ( \frac{2}{1+\omega ^{2}}-2\right ) & =\hat{f}\left ( \omega \right ) \\ \hat{u}\left ( \omega \right ) \left ( \frac{2-2\left ( 1+\omega ^{2}\right ) }{1+\omega ^{2}}\right ) & =\hat{f}\left ( \omega \right ) \\ \hat{u}\left ( \omega \right ) & =\frac{1+\omega ^{2}}{-2\omega ^{2}}\hat{f}\left ( \omega \right ) \end{align*}
We need to write the above as \(\hat{u}\left ( \omega \right ) =\frac{-1}{2}f+\frac{1}{2}g.\,\) Hence\begin{equation} \hat{u}\left ( \omega \right ) =\frac{-1}{2}\hat{f}\left ( \omega \right ) +\frac{1}{-2\omega ^{2}}\hat{f}\left ( \omega \right ) \tag{2} \end{equation} Let \(f\left ( x\right ) =e^{-\left \vert x\right \vert }\), then \begin{align*} \hat{f}\left ( \omega \right ) & ={\displaystyle \int \limits _{-\infty }^{\infty }} f\left ( x\right ) e^{-i\omega x}dx\\ & ={\displaystyle \int \limits _{-\infty }^{\infty }} e^{-\left \vert x\right \vert }e^{-i\omega x}dx\\ & ={\displaystyle \int \limits _{-\infty }^{0}} e^{x}e^{-i\omega x}dx+{\displaystyle \int \limits _{0}^{\infty }} e^{-x}e^{-i\omega x}dx\\ & =\left [ \frac{e^{x\left ( 1-i\omega \right ) }}{1-i\omega }\right ] _{-\infty }^{0}+\left [ \frac{e^{-x\left ( 1+i\omega \right ) }}{1+i\omega }\right ] _{0}^{\infty }\\ & =\frac{1}{1-i\omega }-\frac{1}{1+i\omega }\\ & =\frac{\left ( 1+i\omega \right ) -\left ( 1-i\omega \right ) }{\left ( 1-i\omega \right ) \left ( 1+i\omega \right ) }\\ & =\frac{2}{1+\omega ^{2}} \end{align*}
Hence using (2)\begin{align*} \hat{u}\left ( \omega \right ) & =\frac{-1}{2}\hat{f}\left ( \omega \right ) +\frac{1}{-2\omega ^{2}}\hat{f}\left ( \omega \right ) \\ & =\frac{-1}{2}\frac{2}{1+\omega ^{2}}+\frac{1}{-2\omega ^{2}}\frac{2}{1+\omega ^{2}}\\ & =-\frac{1}{\omega ^{2}} \end{align*}
Hence \[ u\left ( x\right ) =\frac{-1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} \frac{1}{\omega ^{2}}e^{i\omega x}d\omega \] Using tables \(u\left ( x\right ) =\frac{-1}{2}\left \vert x\right \vert \).
The equation is \[ \frac{d^{4}G\left ( x\right ) }{dx^{4}}-2a^{2}\frac{d^{2}G\left ( x\right ) }{dx^{2}}+a^{4}G\left ( x\right ) =\delta \] Taking Fourier transform, and using \(\frac{d^{n}G}{dx^{n}}\Longrightarrow \left ( ik\right ) ^{n}\hat{g}\left ( k\right ) ,\)hence \(G^{\prime }\left ( x\right ) \Longrightarrow ik\hat{g}\left ( k\right ) ,G^{\prime \prime }\left ( x\right ) \Longrightarrow -k^{2}\hat{g}\left ( k\right ) ,G^{\prime \prime \prime \prime }\left ( x\right ) \Longrightarrow \left ( ik\right ) ^{4}\hat{g}\left ( k\right ) =k^{4}\hat{g}\left ( k\right ) \). Therefore the Fourier transform of the above differential equation is\[ k^{4}\hat{g}\left ( k\right ) +2a^{2}k^{2}\hat{g}\left ( x\right ) +a^{4}\hat{g}\left ( k\right ) =1 \] Solving for \(\hat{g}\left ( k\right ) \)\begin{align*} \hat{g}\left ( k\right ) \left ( k^{4}+2a^{2}k^{2}+a^{4}\right ) & =1\\ \hat{g}\left ( k\right ) & =\frac{1}{k^{4}+2a^{2}k^{2}+a^{4}}\\ & =\frac{1}{\left ( k^{2}+a^{2}\right ) ^{2}} \end{align*}
To find \(G\left ( x\right ) \) we need to find the inverse Fourier transform. \[ G\left ( x\right ) =\frac{1}{2\pi }{\displaystyle \int \limits _{-\infty }^{\infty }} \frac{1}{\left ( k^{2}+a^{2}\right ) ^{2}}e^{ikx}dk \] With the help of computer, I obtained the following result\[ G\left ( x\right ) =\frac{\left ( 1+a\left \vert x\right \vert \right ) }{4a^{3}}e^{-}a\left \vert x\right \vert \]