3

I have calculated the Fourier Series of $\cos(\frac{x}{2})$ as:

$$f(x) \sim \frac{2}{\pi} + \frac{4}{\pi} \sum_{n=0}^\infty \frac{(-1)^n}{1-4n^2} \cos(nx)$$

and the Fourier Series of $\cos(x)$ as:

$f(x) \sim \sum_{n=0}^\infty a_n \cos nx$ where $a_n = 1$ for $n= 1$ and $0$ otherwise.

My question is: Can you obtain the Fourier Series of $\cos(x)$ from the Fourier Series of $\cos(\frac{x}{2})$ by changing the variable from $x$ to $2x$?

I can't see how you can go from having all the $a_n$ with values to just the one with a value, just by changing $x$?

amWhy
  • 209,954
Jessie
  • 109

1 Answers1

0

Firstly, the first series should start with $n = 1$, not $n = 0$ (the constant term is already stated explicitly).

Secondly, the sum of the first series is $$ \frac{2}{\pi} + \frac{4}{\pi} \sum_{n = 1}^\infty \frac{(-1)^n}{1-4n^2} \cos n x = \left| \cos \frac{x}{2} \right| $$ whereas the second series is just $\cos x$ with the only nonzero coefficient being $a_1 = 1$.

These functions are not similar to each other in a way you suggest, i.e. you don't get the latter from the former merely by substituting $x \to 2x$, because that won't get you rid of the absolute value (also note, that the function $\left| \cos (x/2) \right|$ is not differentiable at $\pm \pi$, $\pm 3 \pi$, $\pm 5 \pi$, ..., whereas $\cos x$ is differentiable everywhere). Instead, substituting $x \to 2x$ would yield function $\left| \cos x \right|$ which is different from $\cos x$. They are only equal to each other on intervals $\left\langle 4 n \pi -\pi, 4 n \pi + \pi \right\rangle, n \in \mathbb{N}$.

user16320
  • 696