Question

Series and Sequences:

What are other ways to find the sequence that describes a series of numbers besides trial and error of educated guesses?

This will be for finding the coefficients of a Fourier Series, which are often sequences.

Adding an example:

Answer 1

The textbook jumps straight from:

\[\frac{ 1 }{ \pi } \int\limits_{0}^{\pi/2} \cos(2n - 1)x + \cos(2n + 1)x dx\]

to

\[\frac{ 2(-1)^{n+1} }{ \pi(4n^2 - 1) }\]

I understand it and the trial and error method of finding it after integrating and plugging in a few ns.

Answer 2

1. regression
2. recurrence

Answer 3

or just simple decomposition?

Answer 4

well the integral itself isn't that bad:$$\int\cos((2n-1)x)\,dx=\frac1{2n-1}\sin ((2n-1)x
)\\\int\cos((2n+1)x)\,dx=\frac1{2n+1}\sin((2n+1)x)\\$$Evaluating at \(x=0\) just yields \(0\), so only consider \(x=\pi\):$$\sin\left(\pi n+\frac\pi2\right)=\cos\pi n\\\sin\left(\pi n-\frac\pi2\right)=-\cos\pi n$$... and we get:$$\frac{\cos\pi n}{2n-1}-\frac{\cos\pi n}{2n+1}=\frac{(2n+1)\cos \pi n-(2n -1)\cos \pi n}{4n^2-1}=\frac{2\cos\pi n}{4n^2-1}$$For even \(n\) we find \(\cos \pi n=1\) yet for odd \(n\) we have \(\cos \pi n=-1\) so \(\cos\pi n=(-1)^{n+1}\) and rewrite:$$
\frac{2(-1)^{n+1}}{4n^2-1}$$and don't forget our original \(1/\pi\):$$\frac{2(-1)^{n+1}}{\pi(4n^2-1)}$$