-2

The question is to compute the Fourier Series of the function

$$h(x) = \sin\left(\frac{x}{4}\right)$$ on $[-\pi, \pi]$

My question is I know normally, we can use the following formula to compute $a_0, a_k, b_k$

$$a_0 = \frac{1}{L} \int_{-L}^{L}f(x)dx$$ $$a_k = \frac{1}{L} \int_{-L}^{L}f(x)\cos\left(\frac{k\pi x}{L}\right) dx$$ $$b_k = \frac{1}{L} \int_{-L}^{L}f(x)\sin\left(\frac{k\pi x}{L}\right) dx$$

But I noticed that for $\sin\left(\frac{x}{4}\right)$, the period $T = 2L = 8\pi$. If I use $L = 4\pi$ to plug into the formula, what is the meaning of $[-\pi, \pi]$? Could anyone explain for me?

Alain Remillard
  • 3,625
chen zhang
  • 17
  • 2
    It would appear that the problem is telling you to restrict $t \mapsto \sin(t/4)$ to the interval $[-\pi, \pi]$, so that it is a $2\pi$-periodic function, rather than an $8\pi$-periodic function. – Xander Henderson Mar 26 '24 at 18:21
  • Right, you are likely expected to treat it as a $2\pi$-periodic function with the given values for $t\in(-\pi,\pi]

    .$

     – Thomas Andrews Mar 26 '24 at 18:24
  • Certainly, as a function with period $8\pi,$ the function is its Fourier series. $$\sin(t/4)=\sum_n(a_n\cos(nt/4)+b_n\sin(nt/4))$$ easily gives all $a_n=0,$ $b_1=1,$ and $b_n=0$ for $n\neq1.$ – Thomas Andrews Mar 26 '24 at 18:29
  • So based on my understanding, I can just use $L = \pi$ to compute the coefficients – chen zhang Mar 26 '24 at 18:38

0 Answers0