Find the series of the function $f(x) = \cos(x)$ on the interval $[0,\pi]$
Find the sine series of the function $f(x) = \cos(x)$ on the interval $0\leq x \leq \pi$. For each $x$ in this interval, to what value does the sine series for $f(x)$ converge?
Attempted solution - The Fourier since series on $0\leq x \leq \pi$ is given by $$f(x) = \sum_{n=1}^\infty B_n \sin(nx)$$ where $$B_n = \frac{2}{\pi}\int_{0}^{\pi}f(x)\sin(nx)\,dx$$ Thus for $f(x) = \cos(x)$ on $[0,\pi]$ we have \begin{align*} B_n &= \frac{2}{\pi}\cos(x)\sin(nx) \,dx\\ &= \frac{1}{\pi}\int_{0}^{\pi}(\sin((n-1)x) + \sin((n+1)x))\,dx \ \ \ (\text{Using trig identity } \sin(\alpha)\cos(\beta) = \frac{1}{2}(\sin(\alpha - \beta) + \sin(\alpha + \beta))\\ &= \frac{1}{\pi}\int_0^\pi \sin((n+1)x)\,dx + \frac{1}{\pi}\int_0^\pi \sin((n-1)x)\,dx\\ &= \frac{1}{\pi(n+1)}\int_0^{\pi(n+1)}\sin(u)du + \frac{1}{\pi}\sin((n-1)x)\,dx \ \ (\text{substituting} \ u = (n+1)x \ du = (n+1)\,dx)\\ &= \frac{\cos(n\pi) + 1}{n\pi + \pi} + \frac{1}{\pi}\int_{0}^{\pi}\sin((n-1)x)dx\\ &= \frac{\cos(n\pi) + 1}{n\pi + \pi} + \frac{1}{\pi(n-1)}\int_0^{\pi(n-1)}\sin(s)ds \ \ (\text{substituting} \ s = (n-1)x \ ds = (n-1)\,dx)\\ &= \frac{\cos(n\pi) + 1}{n\pi + \pi} + \frac{\cos(n\pi) + 1}{\pi(n-1)}\\ &= \frac{2n(\cos(n\pi) + 1)}{\pi(n^2 - 1)}\\ &= \frac{2n((-1)^n + 1)}{\pi(n^2 - 1)} \end{align*} Therefore the Fourier sine series is $$\cos(x) = \sum_{n=1}^\infty \frac{2n((-1)^n + 1)}{\pi(n^2 - 1)} \sin(nx)$$
By the question's statement I believe we find the Fourier sine series for $f(x) = \cos(x)$. What I do not understand is the next question, "to what value does the sine series for $f(x)$ converge? Doesn't that depend on $n$ and $x$? I am not sure if what I did was correct any suggestions are greatly appreciated.
$\endgroup$ 31 Answer
$\begingroup$You are in fact expanding the "odd-extension" of $\cos x$.
The series converges to $\frac{1}{2}(f(x^+)+f(x^-))$.
So it converges for all $x\in [0,\pi]$.
But since there is discontinuity at $0$, and $\pi$. So it converges to $\cos x$ only for $x\in (0,\pi)$.
At $0$, and $\pi$, it converges to the mean value of the discontinuity, viz., zero.
