Find exact value regarding a Fourier Series

Question

Let $f(x)$ be a function defined by $$f(x) = \sin|x| + 5\sin(2013x)$$ if $-\pi < x < \pi$ and $f(x + 2\pi) = f(x)$ for all $x \subseteq R$.
Let $$a_0 + \sum_{n=1}^{\infty}(a_n\cos(nx) + b_n\sin(nx))$$ be the Fourier series for $f(x)$.

Find the exact value of $$\sum_{n=1}^{\infty}b_n$$

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I understand that $\sin|x|$ is an even function and $5\sin(2013x)$ is an odd function, thus we only need to take the latter into consideration when finding values relevant to $b_n$. However, how should I go about finding the summation from $n=1$ to infinity?

Update: I understand that since $5\sin(2013x)5\sin(2013x)$ is an odd function, both $a_0$ and the summation involving $a_n$ will be zero since our period is from $-\pi < x < \pi$. Thus we'd be left with the summation of $b_n\sin(n2x)bn\sin(\frac{n}{2}x)$ from $n = 1$ to infinity. Is the aforementioned correct?

Thank you!

Answer 1

Hint:

By inspection, the infinite system of equations in $a_n,b_n$ $$a_0 + \sum_{n=1}^{\infty}(a_n\cos(nx) + b_n\sin(nx))=5\sin(2013x)$$

has an obvious solution.

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If you prefer the "hard way", use

$$\int_0^{2\pi}\cos(mx)\cos(nx)\,dx=\int_0^{2\pi}\sin(mx)\sin(nx)dx=\pi\delta_{mn},$$ $$\int_0^{2\pi}\cos(mx)\sin(nx)\,dx=0.$$

Answer 2

To find the sum over $b_n$ I would suggest first calculating them. Use that \begin{equation} b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) dx \end{equation} As already pointed out in the comments the part with $\sin |x|$ should not contribute to this, but it is always a good idea to check this explicitly. As soon as you have calculated $b_n$ you can perhaps find the value of the sum, by using what you know about e.g. geometrical series. However, in this case, you will find that $b_n \neq 0$ only for a single $n$, can you guess which one? Hence the summation will be trivial.

A useful relation is the following \begin{equation} \frac{1}{\pi} \int_{-\pi}^\pi \sin(mx) \sin(nx) dx = \delta_{nm} \end{equation} To prove this integral, we can use the trigonometric relation $\cos(a \pm b) = \sin(a) \sin(b) \mp \cos(a) \cos(b)$. Taking the difference between these two equations with the different signs we get \begin{align} \cos(a + b) - \cos(a - b) &= \cos(a) \cos(b) - \sin(a) \sin(b) - \cos(a) \cos(b) - \sin(a) \sin(b) \\ &= - 2 \sin(a)\sin(b) \end{align} Hence we get that \begin{equation} \sin(mx) \sin(nx) = \frac{1}{2} \left\{ \cos[(m-n) x] - \cos[(m+n) x] \right\}. \end{equation} Now we need to consider the two different cases where $n = m$ and $n \neq m$ separately. However, the resulting integrals are very easy. E.g. if $n \neq m$ we get \begin{align} \int_{-\pi}^\pi \sin(mx)\sin(nx)dx = \frac{1}{2} \int_{-\pi}^\pi \left\{ \cos[(m - n ) x] - \cos[(m+ n)x] \right\} dx = \\ \left. \frac{1}{2} \left( \frac{\sin[(m-n)x]}{m - n} - \frac{\sin[(m+n)x]}{m+n} \right) \right|_{-\pi}^\pi = 0 \end{align} To check the case where $n=m$, insert this into the integrand and you will see that the integral above is $\pi$ in that case.