# Infinite series calculation of a potential

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• Vitani1
In summary, the potential can be expressed as a closed form solution using the half hyperbolic tangent and the principle branch of the logarithm.
Vitani1
TL;DR Summary
I'm having trouble beginning this...
I'm trying to get from the formula in the top to the formula in the bottom (See image: Series). My approach was to complexify the sine term and then use the fact that (see image: Series 1) for the infinite sum of 1/ne^-n. Then use the identity (see image: Series 2). Any other ideas?

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Let ##f(x)=\sum_{n=1}^\infty \frac{e^{-nx}}{n}##. Then ##f'(x)=\sum_{n=1}^\infty e^{-nx}=\frac{1}{e^x-1}##. Integrate to get ##f(x)=x-ln(e^x-1)##. When ##x=1##, we get ##f(1)=1-ln(e-1)##.

Vitani1 and BvU
Vitani1 said:
Any other ideas?
Since you haven't included a homework statement and it's been awhile since your original post, I offer my solution to the problem.
We seek a closed form solution to the potential,
$$V(x,y)=\frac{4V_0}{\pi}S_n$$
$$S_n=\sum_{n=0}^{\infty}\frac{e^{-(2n+1)\frac{\pi}{a}x}\sin((2n+1)\frac{\pi}{a}y)}{(2n+1)}$$
I make the substitutions ##z=\frac{\pi}{a}(x+iy)## and ##\bar z=\frac{\pi}{a}(x-iy)## which scales the complex plane by ##\frac{\pi}{a}##. We now have
$$S_n=\frac{1}{2i}(\sum_{n=0}\frac{e^{-(2n+1)\bar z}}{(2n+1)}-\sum_{n=0}\frac{e^{-(2n+1) z}}{(2n+1)})$$
I observe
$$\frac{e^{-(2n+1)z}}{(2n+1)}=\int_{z}^{\infty} e^{-(2n+1) z'}d z'$$
and thus
$$\sum_{n=0}\frac{e^{-(2n+1)z}}{(2n+1)}=\sum_{n=0}\int_{z}^{\infty} dz'e^{-(2n+1)z'}$$
and interchanging the order of summation and integration we have
$$S_n=\int_{\bar z}^{\infty} d\bar{z'}\sum_{n=0}e^{-(2n+1)\bar{z'}}-\int_{z}^{\infty} dz'\sum_{n=0}e^{-(2n+1)z'}\\$$
The sum ##\sum_{n=0}e^{-(2n+1)z'}## can be evaluated as a geometric series
$$\sum_{n=0}e^{-(2n+1)z'}=e^{-z'}(1+e^{-2z'} + e^{-4z'} + ...)$$
$$=\frac{e^{-z'}}{1-e^{-2z'}}=\frac{1}{2\sinh(z')}$$
and we have
$$S_n=\frac{1}{2i}(\int_{z}^{\infty}\frac{d\bar{z'}}{2\sinh(\bar{z'})}-\int_{ z}^{\infty}\frac{d z'}{2\sinh(z')})$$
The upper limit of integration evaluates to ##\log(\tanh(\infty))=0## and we have
$$S_n=\frac{1}{4i}\left [\log(\tanh(\frac{ z}{2}))-\log(\tanh(\frac{\bar z}{2}))\right ]\\$$
The half hyperbolic tangent can be expressed as
$$\tanh(\frac{ z}{2})=\frac{(e^z-1)}{(e^z+1)}$$
and so
$$S_n=\frac{1}{4i}\left [\log(\frac{(e^z-1)(e^{\bar z}+1)}{(e^{\bar z}-1)(e^z+1)}) \right ]$$
$$=\frac{1}{4i}\left [ \log(\frac{\sinh(\frac{\pi}{a}x)+i\sin(\frac{\pi}{a}y)}{\sinh(\frac{\pi}{a}x)-i\sin(\frac{\pi}{a}y)}) \right ]$$
$$=\frac{1}{4i}\left [\log(-1) -\log(\frac{i-\frac{\sinh(\frac{\pi}{a}x)}{\sin(\frac{\pi}{a}y)}}{i+\frac{\sinh(\frac{\pi}{a}x)}{\sin(\frac{\pi}{a}y)}})\right ]$$
The ##\arctan(x)## can be expressed as
$$\arctan(x)=\frac{1}{2i}(\frac{i-x}{i+x})$$
and taking the principle branch of the logarithm we have
$$S_n=\frac{1}{2}\left [ \frac{\pi}{2}-\arctan(\frac{\sinh(\frac{\pi}{a}x)}{\sin(\frac{\pi}{a}y)})\right ]$$
$$V(x,y)=\frac{2V_0}{\pi}\left [\frac{\pi}{2}-\arctan(\frac{\sinh(\frac{\pi}{a}x)}{\sin(\frac{\pi}{a}y)})\right ]$$

Vitani1 and Infrared

## 1. What is an infinite series?

An infinite series is a sum of infinitely many terms. It is written in the form of a_n + a_(n+1) + a_(n+2) + ..., where a_n represents the nth term in the series.

## 2. How is an infinite series used to calculate potential?

Infinite series can be used to calculate the potential of a system by representing the potential as a sum of infinitely many terms. By using a finite number of terms in the series, an approximation of the potential can be obtained.

## 3. What is the formula for calculating an infinite series?

The formula for calculating an infinite series is given by S = a/(1-r), where S is the sum of the series, a is the first term, and r is the common ratio between consecutive terms.

## 4. What is the significance of convergence in infinite series calculation?

Convergence is important in infinite series calculation as it determines whether the series will have a finite sum or not. If the series converges, it means that the sum of the terms is finite and the calculation can be accurately approximated. If the series diverges, the calculation will not be accurate.

## 5. How can we improve the accuracy of infinite series calculation for potential?

To improve the accuracy of infinite series calculation for potential, we can increase the number of terms used in the series or use more advanced techniques such as numerical methods. It is also important to check for convergence and make sure that the series is being used within its applicable range.

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