Truncated form of a infinite series

  • #1

Main Question or Discussion Point

Hi,

In griffith's "Introduction to Electrodynamics" he indicates that a specific infinite series has a truncated form (the series and truncated form are given below)
And he says the reader can try to show that it indeed has that form.

[itex] \frac{4V_0}{\pi}\sum_{n=1,3,5,7,...}^{\infty}\frac{1}{n}e^{-n\pi x/a}sin(n\pi y/a)=\frac{2V_0}{\pi}arctg(\frac{sin(\pi y/a)}{sinh(\pi x/a)}) [/itex]

However I can't figure out how to get that result. Can anybody help me figure it out, without starting from the truncated form?
 
Last edited:

Answers and Replies

  • #2
Mute
Homework Helper
1,388
10
Hi,

In griffith's "Introduction to Electrodynamics" he indicates that a specific infinite series has a truncated form (the series and truncated form are given below)
And he says the reader can try to show that it indeed has that form.

[itex] \frac{4V_0}{\pi}\sum_{n=1,3,5,7,...}^{\infty}\frac{1}{n}e^{-n\pi x/a}sin(n\pi y/a)=\frac{2V_0}{\pi}arctg(\frac{sin(\pi y/a)}{sinh(\pi x/a)}) [/itex]

However I can't figure out how to get that result. Can anybody help me figure it out, without starting from the truncated form?
Hm. I would first start by writing this as

$$V(x,y) = \frac{4V_0}{\pi} \mbox{Im}\left[ \sum_{n~odd}^\infty \frac{\exp(-n\pi x/a + i n\pi y/a)}{n}\right],$$
where "Im" denotes the imaginary part of the expression and I have used Euler's identity.

If we write ##z = x - iy##, then we can write this as a complex function

$$\tilde{V}(z) = \frac{4V_0}{\pi} \sum_{k=0}^\infty \frac{e^{-(2k+1)\pi z/a}}{2k+1}.$$

If we differentiate once with respect to z, and assuming the sum uniformly(?) converges so that we can exchange derivatives and infinite sums, we have

$$\frac{d\tilde{V}}{dz} = \frac{-4V_0}{a} \sum_{k=0}^\infty e^{-(2k+1)\pi z/a} = -\frac{4V_0}{a}e^{-\pi z/a} \sum_{k=0}^\infty (e^{-2\pi z/a})^k.$$

This should now be in a form you can sum up explicitly. You can write the result in terms of a hyperbolic trig function that you can find the antiderivative of to get ##\tilde{V}(z)## back. The trick then is to find the imaginary part of the resulting expression.
 
Last edited:
  • #3
Hi,

Thanks for the help :). I managed to do it
 
  • #4
Mute
Homework Helper
1,388
10
Glad you got it. By the way, about your terminology, the phrase "truncated form of an infinite series" suggests to me that you have taken only a finite number of terms of the series (i.e., you truncated the series). Typically people call the result of summing up an infinite series into an expression in terms of finitely many elementary functions a "closed form".
 

Related Threads on Truncated form of a infinite series

Replies
2
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
6
Views
2K
Replies
9
Views
19K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
5
Views
22K
Replies
3
Views
6K
Replies
2
Views
3K
Replies
2
Views
632
  • Last Post
Replies
9
Views
2K
Top