Solution
The first step in saddle point method is to write the integral as \(\int _{0}^{\infty }e^{f\left ( E\right ) }dE\). Hence\begin{align} R & =N\int _{0}^{\infty }e^{\left ( -\beta E-\alpha E^{\frac{-1}{2}}+\ln E\right ) }dE\nonumber \\ & =N\int _{0}^{\infty }e^{f\left ( E\right ) }dE \tag{A} \end{align}
Where\begin{equation} f\left ( E\right ) =-\beta E-\alpha E^{\frac{-1}{2}}+\ln E \tag{1} \end{equation} The next step is to determine where \(f\left ( E\right ) \) is maximum. Therefore we need to solve \(f^{\prime }\left ( E\right ) =0\) in order to determine \(E_{0}\), where \(f\left ( E_{0}\right ) \) is maximum.\begin{align*} f^{\prime }\left ( E\right ) & =-\beta +\frac{1}{2}\alpha E^{\frac{-3}{2}}+\frac{1}{E}\\ & =0 \end{align*}
We need to make this dimensionless. Multiplying both sides of the above by \(\alpha ^{2}\) gives\[ -\alpha ^{2}\beta +\frac{1}{2}\alpha ^{3}E^{\frac{-3}{2}}+\frac{\alpha ^{2}}{E}=0 \] Let \(E=x\alpha ^{2}\), then the above becomes\begin{align} -\alpha ^{2}\beta +\frac{1}{2}\alpha ^{3}\left ( x\alpha ^{2}\right ) ^{\frac{-3}{2}}+\frac{\alpha ^{2}}{\left ( x\alpha ^{2}\right ) } & =0\nonumber \\ -\alpha ^{2}\beta +\frac{1}{2}\frac{1}{x^{\frac{3}{2}}}+\frac{1}{x} & =0 \tag{2} \end{align}
Case 1 Ignoring the term \(\frac{1}{x^{\frac{3}{2}}}\) in (2) results in\begin{align*} -\alpha ^{2}\beta +\frac{1}{x} & =0\\ \frac{1}{x} & =\alpha ^{2}\beta \\ x & =\frac{1}{\alpha ^{2}\beta } \end{align*}
Using this value for \(x\) we check if this is larger than or smaller than the term we ignored which is \(\frac{1}{x^{\frac{3}{2}}}\).\[ \left [ \frac{1}{x^{\frac{3}{2}}}\right ] _{x=\frac{1}{\alpha ^{2}\beta }}=\frac{1}{\left ( \frac{1}{\alpha ^{2}\beta }\right ) ^{\frac{3}{2}}}=\frac{1}{\left ( \frac{1}{\alpha \beta ^{2}}\right ) ^{3}}=\left ( \beta ^{2}\alpha \right ) ^{3}\] Since \(\left ( \alpha ^{2}\beta \right ) ^{\frac{1}{3}}\gg 1\), then \(\alpha ^{2}\beta \gg 1\) and hence \(x=\frac{1}{\alpha ^{2}\beta }\) is much smaller than \(\left ( \beta ^{2}\alpha \right ) ^{3}\). So our choice of ignoring \(\frac{1}{x^{\frac{3}{2}}}\) was wrong. Hence we need to ignore the term \(\frac{1}{x}\) from (2)
Case 2 Ignoring the term \(\frac{1}{x}\) results in\begin{align*} -\alpha ^{2}\beta +\frac{1}{2}\frac{1}{x^{\frac{3}{2}}} & =0\\ \frac{-2x^{\frac{3}{2}}\alpha ^{2}\beta +1}{2x^{\frac{3}{2}}} & =0\\ -2x^{\frac{3}{2}}\alpha ^{2}\beta +1 & =0\\ x^{\frac{3}{2}} & =\frac{-1}{-2\alpha ^{2}\beta } \end{align*}
Solving gives\[ x=\left ( \frac{1}{2\alpha ^{2}\beta }\right ) ^{\frac{2}{3}}\] But \(E=x\alpha ^{2}\), and from the above we the energy \(E_{0}\) which makes \(f\left ( E\right ) \) maximum as\begin{align*} E_{0} & =\alpha ^{2}\left ( \frac{1}{2\alpha ^{2}\beta }\right ) ^{\frac{2}{3}}\\ & =\frac{\alpha ^{2-\frac{4}{3}}}{2^{\frac{2}{3}}\beta ^{\frac{2}{3}}}\\ & =\frac{\alpha ^{\frac{2}{3}}}{2^{\frac{2}{3}}\beta ^{\frac{2}{3}}} \end{align*}
Hence\[ \fbox{$E_0=\left ( \frac{\alpha }{2\beta }\right ) ^\frac{2}{3}$}\] Now that we found which value of \(E\) makes \(f\left ( E\right ) \) maximum, we can expand \(f\left ( E\right ) \) in Taylor series around \(E_{0}\)\[ f\left ( E\right ) =f\left ( E_{0}\right ) +f^{\prime }\left ( E_{0}\right ) \left ( E-E_{0}\right ) +\frac{f^{\prime \prime }\left ( E_{0}\right ) }{2!}\left ( E-E_{0}\right ) ^{2}+H.O.T \] But \(f^{\prime }\left ( E_{0}\right ) =0\) then the above becomes, after ignoring H.O.T.\begin{equation} f\left ( E\right ) =f\left ( E_{0}\right ) +\frac{f^{\prime \prime }\left ( E_{0}\right ) }{2!}\left ( E-E_{0}\right ) ^{2}\tag{3} \end{equation} Since \(f^{\prime }\left ( E\right ) =-\beta +\frac{1}{2}\alpha E^{\frac{-3}{2}}+\frac{1}{E}\) then\[ f^{\prime \prime }\left ( E_{0}\right ) =-\frac{3}{4}\alpha E_{0}^{\frac{-5}{2}}-E_{0}^{-2}\]
Since \(E_{0}^{\frac{-5}{2}}\gg E_{0}^{-2}\) the above becomes
\begin{align} f^{\prime \prime }\left ( E_{0}\right ) & =-\frac{3}{4}\alpha E_{0}^{\frac{-5}{2}}\nonumber \\ & \simeq -\frac{3}{2}\frac{\beta ^{2}}{E_{0}}\tag{4} \end{align}
Equation (A) now becomes\begin{align*} R & =N\int _{0}^{\infty }e^{f\left ( E\right ) }dE\\ & =N\int _{0}^{\infty }e^{f\left ( E_{0}\right ) +\frac{f^{\prime \prime }\left ( E_{0}\right ) }{2!}\left ( E-E_{0}\right ) ^{2}}dE\\ & =Ne^{f\left ( E_{0}\right ) }\int _{0}^{\infty }e^{\frac{f^{\prime \prime }\left ( E_{0}\right ) }{2!}\left ( E-E_{0}\right ) ^{2}}dE \end{align*}
We would like to write the above as \(\int _{0}^{\infty }e^{-ax^{2}}dx=\sqrt{\frac{\pi }{a}}\). Therefore, assuming \(u=E-E_{0}\), hence \(\frac{du}{dE}=1\). When \(E=0\) then \(u=-E_{0}\) and when \(E=\infty \) then \(u=\infty \). Hence the above becomes\begin{align*} R & =Ne^{f\left ( E_{0}\right ) }\int _{-E_{0}}^{\infty }e^{\frac{f^{\prime \prime }\left ( E_{0}\right ) }{2!}u^{2}}du\\ & =Ne^{f\left ( E_{0}\right ) }\int _{-E_{0}}^{\infty }e^{-\frac{3}{4}\frac{\beta ^{2}}{E_{0}}u^{2}}du \end{align*}
Since \(E_{0}\) is positive, then contribution from lower limit \(u=-E_{0}\) to the value of the integral is Negligible. We can then let lower limit go to \(-\infty \) without affecting the overall result of the integral. The above becomes\[ R=Ne^{f\left ( E_{0}\right ) }\int _{-\infty }^{\infty }e^{-\frac{3}{4}\frac{\beta ^{2}}{E_{0}}u^{2}}du \] This is now in the form of Gaussian \(\int _{-\infty }^{\infty }e^{-ax^{2}}dx=\sqrt{\frac{\pi }{a}}\). Hence we can write the above, using \(a=\frac{3}{4}\frac{\beta ^{2}}{E_{0}}\)\begin{align*} R & =Ne^{f\left ( E_{0}\right ) }\sqrt{\frac{\pi }{\frac{3}{4}\frac{\beta ^{2}}{E_{0}}}}\\ & =Ne^{f\left ( E_{0}\right ) }\sqrt{\frac{4\pi E_{0}}{3\beta ^{2}}} \end{align*}
But \(f\left ( E_{0}\right ) \) from (1) is \(f\left ( E_{0}\right ) =-\beta E_{0}-\alpha E_{0}^{\frac{-1}{2}}+\ln E_{0}\), hence the above becomes\begin{align*} R & =NE_{0}e^{-\beta E_{0}-\alpha E_{0}^{\frac{-1}{2}}}\sqrt{\frac{4\pi E_{0}}{3\beta ^{2}}}\\ & =NE_{0}e^{-\beta E_{0}-\alpha E_{0}^{\frac{-1}{2}}}\sqrt{\frac{4\pi }{3\alpha E_{0}^{\frac{-5}{2}}}} \end{align*}
But \(E_{0}=\left ( \frac{\alpha }{2\beta }\right ) ^{2/3}\), therefore the above becomes, after some more simplifications\[ R=N\left ( \frac{\alpha }{2\beta }\right ) ^{2/3}\exp \left ( -\beta \left ( \frac{\alpha }{2\beta }\right ) ^{2/3}-\alpha \left ( \frac{\alpha }{2\beta }\right ) ^{-2/6}\right ) \sqrt{\frac{4\pi }{3\alpha \left ( \frac{\alpha }{2\beta }\right ) ^{-10/6}}}\] Simplifies to
\[ R=\sqrt{\frac{\pi }{3}}N\left ( k_{\beta }T\right ) ^{\frac{3}{2}}\alpha e^{-\left ( \frac{\alpha ^{2}}{4}k_{\beta }T\right ) ^{\frac{1}{3}}}\]
This was a hard problem. See key solution.
Solution
Let \(u=g\left ( x\right ) \), hence \begin{equation} \frac{du}{dx}=g^{\prime }\left ( x\right ) \tag{1} \end{equation} But \begin{align*} x & =g^{-1}\left ( g\left ( x\right ) \right ) \\ & =g^{-1}\left ( u\right ) \end{align*}
Replacing \(x\) in (1) by the above results (so everything is in terms of \(u\)) gives\[ \frac{du}{dx}=g^{\prime }\left ( g^{-1}\left ( u\right ) \right ) \] Now we take care of the limits of integration. When \(x=a\) then \(u=g\left ( a\right ) \) and when \(x=b\) then \(u=g\left ( b\right ) \). Now the integral \(I\) becomes in terms of \(u\) the following\begin{align} I & =\int _{g\left ( a\right ) }^{g\left ( b\right ) }f\left ( g^{-1}\left ( u\right ) \right ) \delta \left ( u\right ) \frac{du}{g^{\prime }\left ( g^{-1}\left ( u\right ) \right ) }\nonumber \\ & =\int _{g\left ( a\right ) }^{g\left ( b\right ) }\delta \left ( u\right ) \left [ \frac{f\left ( g^{-1}\left ( u\right ) \right ) }{g^{\prime }\left ( g^{-1}\left ( u\right ) \right ) }\right ] du \tag{2} \end{align}
Since we do not know the sign of \(g^{\prime }\left ( x_{0}\right ) \), as it can be positive or negative, so we take its absolute value in the above, so that the limits of integration do not switch. Hence (2) becomes\begin{equation} I=\int _{g\left ( a\right ) }^{g\left ( b\right ) }\delta \left ( u\right ) \left [ \frac{f\left ( g^{-1}\left ( u\right ) \right ) }{\left \vert g^{\prime }\left ( g^{-1}\left ( u\right ) \right ) \right \vert }\right ] du \tag{3} \end{equation} We are given that there is one point \(x_{0}\) between \(g\left ( a\right ) \), and \(g\left ( b\right ) \) where \(g\left ( x_{0}\right ) =0\) which is the same as saying \(u=0\) at that point. Hence by applying the standard property of Dirac delta function, which says that \(\int _{a}^{b}\delta \left ( 0\right ) \phi \left ( z\right ) dz=\phi \left ( 0\right ) \) to equation (3) gives\[ I=\frac{f\left ( g^{-1}\left ( 0\right ) \right ) }{\left \vert g^{\prime }\left ( g^{-1}\left ( 0\right ) \right ) \right \vert }\] But \(g^{-1}\left ( 0\right ) =x_{0}\), therefore the above becomes\[ \int _{a}^{b}f\left ( x\right ) \delta \left ( g\left ( x\right ) \right ) dx=\frac{f\left ( x_{0}\right ) }{\left \vert g^{\prime }\left ( x_{0}\right ) \right \vert }\] Which is the result required to show.
Solution
A plot of the function to approximate is (using \(L=1\)) for illustration
The function period is \(T=L\). Hence the Fourier series is given by \[ f\left ( x\right ) \sim \frac{a_{0}}{2}+\sum _{n=1}^{\infty }a_{n}\cos \left ( \frac{2\pi }{L}nx\right ) +b_{n}\cos \left ( \frac{2\pi }{L}nx\right ) \] Since \(f\left ( x\right ) \) is an even function, then \(b_{n}=0\) and the above simplifies to\[ f\left ( x\right ) \sim \frac{a_{0}}{2}+\sum _{n=1}^{\infty }a_{n}\cos \left ( \frac{2\pi }{L}nx\right ) \] Where \[ a_{0}=\frac{2}{L}\int _{-\frac{L}{2}}^{\frac{L}{2}}f\left ( x\right ) dx \] We can calculate this integral, but it is easier to find \(a_{0}\) knowing that \(\frac{a_{0}}{2}\) represent the average of the area under the function \(f\left ( x\right ) \).
We see right away that the area is \(2\left ( \frac{1}{2}\frac{L}{2}\right ) =\frac{L}{2}\). Hence, solving \(\frac{a_{0}}{2}L=\frac{L}{2}\) for \(a_{0}\) gives \(a_{0}=1\). Now we find \(a_{n}\)\[ a_{n}=\frac{2}{L}\int _{-\frac{L}{2}}^{\frac{L}{2}}f\left ( x\right ) \cos \left ( \frac{2\pi }{L}nx\right ) dx \] Since \(f\left ( x\right ) \) is even and \(\cos \left ( \frac{2\pi }{L}nx\right ) \) is even, then the above simplifies to\begin{align} a_{n} & =\frac{4}{L}\int _{0}^{\frac{L}{2}}f\left ( x\right ) \cos \left ( \frac{2\pi }{L}nx\right ) dx\nonumber \\ & =\frac{4}{L}\int _{0}^{\frac{L}{2}}\left ( 1-\frac{2x}{L}\right ) \cos \left ( \frac{2\pi }{L}nx\right ) dx\nonumber \\ & =\frac{4}{L}\left ( \int _{0}^{\frac{L}{2}}\cos \left ( \frac{2\pi }{L}nx\right ) dx-\frac{2}{L}\int _{0}^{\frac{L}{2}}x\cos \left ( \frac{2\pi }{L}nx\right ) dx\right ) \tag{1} \end{align}
But \begin{align*} \int _{0}^{\frac{L}{2}}\cos \left ( \frac{2\pi }{L}nx\right ) dx & =\frac{1}{\frac{2n\pi }{L}}\left [ \sin \left ( \frac{2\pi }{L}nx\right ) \right ] _{0}^{\frac{L}{2}}\\ & =\frac{L}{2n\pi }\left ( \sin \left ( \frac{2\pi }{L}n\frac{L}{2}\right ) \right ) \\ & =\frac{L}{2n\pi }\sin \left ( \pi n\right ) \\ & =0 \end{align*}
And\(\ \int _{0}^{\frac{L}{2}}x\cos \left ( \frac{2\pi }{L}nx\right ) dx\) is integrated by parts. Let \(u=x,dv=\cos \left ( \frac{2\pi }{L}nx\right ) \), hence \(du=1\) and \(v=\frac{1}{\frac{2n\pi }{L}}\sin \left ( \frac{2\pi }{L}nx\right ) \). Therefore\begin{align*} \int _{0}^{\frac{L}{2}}x\cos \left ( \frac{2\pi }{L}nx\right ) dx & =uv-\int vdu\\ & =\frac{1}{\frac{2n\pi }{L}}\left [ x\sin \left ( \frac{2\pi }{L}nx\right ) \right ] _{0}^{\frac{L}{2}}-\frac{1}{\frac{2n\pi }{L}}\int \sin \left ( \frac{2\pi }{L}nx\right ) dx\\ & =-\frac{L}{2n\pi }\int \sin \left ( \frac{2\pi }{L}nx\right ) dx\\ & =\frac{L}{2n\pi }\left [ \frac{\cos \left ( \frac{2\pi }{L}nx\right ) }{\frac{2\pi }{L}n}\right ] _{0}^{\frac{L}{2}}\\ & =\left ( \frac{L}{2n\pi }\right ) ^{2}\left ( \cos \left ( \frac{2\pi }{L}n\frac{L}{2}\right ) -1\right ) \\ & =\left ( \frac{L}{2n\pi }\right ) ^{2}\left ( \cos \left ( n\pi \right ) -1\right ) \\ & =\left ( \frac{L}{2n\pi }\right ) ^{2}\left ( \left ( -1\right ) ^{n}-1\right ) \end{align*}
Substituting these results in (1) gives\begin{align*} a_{n} & =-\frac{4}{L}\left ( \frac{2}{L}\left ( \frac{L}{2n\pi }\right ) ^{2}\left ( \left ( -1\right ) ^{n}-1\right ) \right ) \\ & =-\frac{2}{n^{2}\pi ^{2}}\left ( \left ( -1\right ) ^{n}-1\right ) \end{align*}
When \(n\) is even we see that \(a_{n}=0\) and when \(n\) is odd, then \(a_{n}=\frac{4}{n^{2}\pi ^{2}}\). Therefore\begin{align*} f\left ( x\right ) & =\frac{a_{0}}{2}+\sum _{n=1}^{\infty }a_{n}\cos \left ( \frac{2\pi }{L}nx\right ) \\ & =\frac{1}{2}+\sum _{n=1,3,5,\cdots }^{\infty }\left ( \frac{4}{n^{2}\pi ^{2}}\right ) \cos \left ( \frac{2\pi }{L}nx\right ) \\ & =\frac{1}{2}+\frac{4}{\pi ^{2}}\sum _{n=1}^{\infty }\frac{1}{\left ( 2n-1\right ) ^{2}}\cos \left ( \frac{2\pi }{L}\left ( 2n-1\right ) x\right ) \end{align*}
Solution
A plot of the above function is
We first need to find the Fourier series of the function \(f\left ( x\right ) \). Since the function is odd, then we only need to determine \(b_{n}\)\[ f\left ( x\right ) \sim \sum _{n=1}^{\infty }b_{n}\sin \left ( nx\right ) \] Where\[ b_{n}=\frac{1}{\pi }\int _{0}^{2\pi }f\left ( x\right ) \sin \left ( nx\right ) dx \] Since \(f\left ( x\right ) \) is odd, and \(\sin \) is odd, then the product is even, and the above simplifies to\begin{align*} b_{n} & =\frac{2}{\pi }\int _{0}^{\pi }f\left ( x\right ) \sin \left ( nx\right ) dx\\ & =\frac{2}{\pi }\int _{0}^{\pi }\sin \left ( nx\right ) dx\\ & =\frac{2}{\pi }\left ( -\frac{\cos nx}{n}\right ) _{0}^{\pi }\\ & =\frac{-2}{n\pi }\left ( \cos nx\right ) _{0}^{\pi }\\ & =\frac{-2}{n\pi }\left ( \cos n\pi -1\right ) \\ & =\frac{-2}{n\pi }\left ( \left ( -1\right ) ^{n}-1\right ) \\ & =\frac{2}{n\pi }\left ( 1-\left ( -1\right ) ^{n}\right ) \end{align*}
When \(n\) is even, then \(b_{n}=0\) and when \(n\) is odd then \(b_{n}=\frac{4}{n\pi }\), therefore\[ f\left ( x\right ) \sim \frac{4}{\pi }\sum _{n=1,3,5,\cdots }^{\infty }\frac{1}{n}\sin \left ( nx\right ) \] Which can be written as\begin{equation} f\left ( x\right ) \sim \frac{4}{\pi }\sum _{n=1}^{\infty }\frac{1}{\left ( 2n-1\right ) }\sin \left ( \left ( 2n-1\right ) x\right ) \tag{1} \end{equation} Next, 4 plots were made to see the approximation for \(n=1,5,10,20\).
The source code used is
The partial sum of (1) is \begin{equation} f_{N}\left ( x\right ) =\frac{4}{\pi }\sum _{n=1}^{N}\frac{1}{\left ( 2n-1\right ) }\sin \left ( \left ( 2n-1\right ) x\right ) \tag{2} \end{equation} To determine the overshoot, we need to first find \(x_{0}\) where the local maximum near \(x=0\) is. This is an illustration, showing the Fourier series approximation to the right of \(x=0\). This plot uses \(n=100\).
Hence we need to determine \(f^{\prime }\left ( x\right ) \) and then solve for \(f^{\prime }\left ( x\right ) =0\) in order to find \(x_{0}\)\begin{align*} f_{N}^{\prime }\left ( x\right ) & =\frac{4}{\pi }\sum _{n=1}^{N}\cos \left ( \left ( 2n-1\right ) x\right ) \\ & =\frac{2}{\pi }\frac{\sin \left ( 2Nx\right ) }{\sin x} \end{align*}
Derivation that shows the above is included in the appendix of this problem. Therefore solving\(\frac{\sin \left ( 2Nx\right ) }{\sin x}=0\) implies \(\sin \left ( 2Nx\right ) =0\) or \(2Nx=\pi \) (since we want to be on the right side of \(x=0\), we do not pick \(0\), but the next zero, this means \(\pi \) is first value). This implies that local maximum to the right of \(x=0\) is located at\[ \fbox{$x_0=\frac{\pi }{2N}$}\] Therefore we need to determine \(f_{N}\left ( x_{0}\right ) \) to calculate the overshoot due to the Gibbs effect to the right of \(x=0\). From (2) and using \(x_{0}\) now instead of \(x\) gives\begin{align*} f_{N}\left ( \frac{\pi }{2N}\right ) & =\frac{4}{\pi }\sum _{n=1}^{N}\frac{1}{\left ( 2n-1\right ) }\sin \left ( \left ( 2n-1\right ) \frac{\pi }{2N}\right ) \\ & =\frac{4}{\pi }\left ( \frac{\sin \left ( \frac{\pi }{2N}\right ) }{1}+\frac{\sin \left ( 3\frac{\pi }{2N}\right ) }{3}+\frac{\sin \left ( 5\frac{\pi }{2N}\right ) }{5}+\cdots +\frac{\sin \left ( \left ( 2N-1\right ) \frac{\pi }{2N}\right ) }{2N-1}\right ) \end{align*}
But \(\frac{\sin \left ( \pi z\right ) }{\pi z}=\operatorname{sinc}\left ( z\right ) \), therefore we rewrite the above as\begin{align*} f_{N}\left ( \frac{\pi }{2N}\right ) & =4\left ( \frac{\sin \left ( \frac{\pi }{2N}\right ) }{\pi }+\frac{\sin \left ( 3\frac{\pi }{2N}\right ) }{3\pi }+\frac{\sin \left ( 5\frac{\pi }{2N}\right ) }{5\pi }+\cdots +\frac{\sin \left ( \left ( 2N-1\right ) \frac{\pi }{2N}\right ) }{\left ( 2N-1\right ) \pi }\right ) \\ & =4\left ( \frac{1}{2N}\frac{\sin \left ( \pi \frac{1}{2N}\right ) }{\pi \frac{1}{2N}}+\frac{1}{2N}\frac{\sin \left ( \pi \frac{3}{2N}\right ) }{3\pi \frac{1}{2N}}+\frac{1}{2N}\frac{\sin \left ( \pi \frac{5}{2N}\right ) }{5\pi \frac{1}{2N}}+\cdots +\frac{1}{2N}\frac{\sin \left ( \pi \frac{\left ( 2N-1\right ) }{2N}\right ) }{\left ( 2N-1\right ) \pi \frac{1}{2N}}\right ) \\ & =4\left ( \frac{1}{2N}\operatorname{sinc}\left ( \frac{1}{2N}\right ) +\frac{1}{2N}\operatorname{sinc}\left ( \frac{3}{2N}\right ) +\frac{1}{2N}\operatorname{sinc}\left ( \frac{5}{2N}\right ) +\cdots +\frac{1}{2N}\operatorname{sinc}\left ( \frac{2N-1}{2N}\right ) \right ) \end{align*}
Therefore\begin{align*} f_{N}\left ( \frac{\pi }{2N}\right ) & =2\left ( \frac{1}{N}\operatorname{sinc}\left ( \frac{1}{2N}\right ) +\frac{1}{N}\operatorname{sinc}\left ( \frac{3}{2N}\right ) +\frac{1}{N}\operatorname{sinc}\left ( \frac{5}{2N}\right ) +\cdots +\frac{1}{N}\operatorname{sinc}\left ( \frac{2N-1}{2N}\right ) \right ) \\ & =2\left \{ \left [ \operatorname{sinc}\left ( \frac{1}{2N}\right ) +\operatorname{sinc}\left ( \frac{3}{2N}\right ) +\operatorname{sinc}\left ( \frac{5}{2N}\right ) +\cdots +\operatorname{sinc}\left ( \frac{2N-1}{2N}\right ) \right ] \frac{1}{N}\right \} \end{align*}
Therefore, if we consider a length of \(1\) and \(\frac{1}{N}\) is partition length, then the sum inside \(\{\}\) above is a Riemann sum and the above becomes In the limit, as \(N\rightarrow \infty \)\[ \lim _{N\rightarrow \infty }f_{N}\left ( \frac{\pi }{2N}\right ) =2\int _{0}^{1}\operatorname{sinc}\left ( x\right ) dx \]
Therefore\[ \lim _{N\rightarrow \infty }f_{N}\left ( \frac{\pi }{2N}\right ) =2\int _{0}^{1}\frac{\sin \left ( \pi x\right ) }{\pi x}dx \] The \(\int _{0}^{1}\frac{\sin \left ( \pi x\right ) }{\pi x}dx\) is known as \(\operatorname{Si}\). I could not solve it analytically. It has numerical value of \(0.5894898772\). Therefore\begin{align*} \lim _{N\rightarrow \infty }f_{N}\left ( \frac{\pi }{2N}\right ) & =2\left ( 0.5894898772\right ) \\ & =1.17897974 \end{align*}
Since \(f\left ( x\right ) =1\) between \(0\) and \(\pi \), then we see that the overshoot is the difference, which is\begin{align*} \lim _{N\rightarrow \infty }\delta _{N} & =1.17897974-1\\ & =0.1789 \end{align*}
For \(4\) decimal places. The above result gives good agreement with the plot showing that the overshoot is a little less than \(0.2\) when viewed on the computer screen. The only use for computation used by the computer for this part of the problem was the evaluation of \(\int _{0}^{1}\frac{\sin \left ( \pi x\right ) }{\pi x}dx\). The code is
Here we show the following result used in the above solution. \[ \frac{4}{\pi }\sum _{n=1}^{N}\cos \left ( \left ( 2n-1\right ) x\right ) =\frac{2}{\pi }\frac{\sin \left ( 2Nx\right ) }{\sin x}\] Since \(\cos z=\operatorname{Re}\left ( e^{iz}\right ) \), then \(\cos \left ( \left ( 2n-1\right ) x\right ) =\operatorname{Re}\left ( e^{i\left ( 2n-1\right ) x}\right ) \). Hence the above is the same as\begin{equation} \frac{4}{\pi }\sum _{n=1}^{N}\cos \left ( \left ( 2n-1\right ) x\right ) =\frac{4}{\pi }\operatorname{Re}\sum _{n=1}^{N}e^{i\left ( 2n-1\right ) x} \tag{1} \end{equation} But\begin{align*} \sum _{n=1}^{N}e^{i\left ( 2n-1\right ) x} & =\sum _{n=1}^{N}e^{2ixn-ix}\\ & =e^{-ix}\sum _{n=1}^{N}e^{2ixn}\\ & =e^{-ix}\sum _{n=1}^{N}\left ( e^{2ix}\right ) ^{n} \end{align*}
Using partial sum property \(\sum _{n=1}^{N}r^{n}=r\frac{1-r^{N}}{1-r}\), then we can write the above using \(r=e^{2ix}\) as\begin{align*} \sum _{n=1}^{N}e^{i\left ( 2n-1\right ) x} & =e^{-ix}\left ( e^{2ix}\frac{1-e^{2iNx}}{1-e^{2ix}}\right ) \\ & =e^{ix}\frac{1-e^{2iNx}}{1-e^{2ix}}\\ & =\frac{1-e^{2iNx}}{e^{-ix}-e^{ix}}\\ & =\frac{e^{2iNx}-1}{e^{ix}-e^{-ix}}\\ & =\frac{e^{2iNx}-1}{2i\sin \left ( x\right ) }\\ & =\frac{\cos \left ( 2Nx\right ) +i\sin (2Nx)-1}{2i\sin \left ( x\right ) } \end{align*}
Multiplying numerator and denominator by \(i\) gives\begin{align*} \sum _{n=1}^{N}e^{i\left ( 2n-1\right ) x} & =\frac{i\cos \left ( 2Nx\right ) -\sin (2Nx)-i}{-2\sin \left ( x\right ) }\\ & =i\frac{\left ( \cos \left ( 2Nx\right ) -1\right ) }{-2\sin x}+\frac{\sin \left ( 2Nx\right ) }{2\sin \left ( x\right ) } \end{align*}
The real part of the above is \(\frac{\sin \left ( 2Nx\right ) }{2\sin \left ( x\right ) }\), hence (1) becomes\begin{align*} \frac{4}{\pi }\sum _{n=1}^{N}\cos \left ( \left ( 2n-1\right ) x\right ) & =\frac{4}{\pi }\operatorname{Re}\sum _{n=1}^{N}e^{i\left ( 2n-1\right ) x}\\ & =\frac{4}{\pi }\left ( \frac{\sin \left ( 2Nx\right ) }{2\sin \left ( x\right ) }\right ) \\ & =\frac{2}{\pi }\frac{\sin \left ( 2Nx\right ) }{\sin \left ( x\right ) } \end{align*}
Which is the result was needed to show.