# Find exact value regarding a Fourier Series

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 $$\sum_{n=1}^{\infty}b_n$$

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!

• Maybe focus on what the Fourier series for $\sin(2013 x)$ can be...
– user65203
Nov 17, 2016 at 8:20
• @YvesDaoust Hi, would I be correct to say that since $5sin(2013x)$ 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 $b_n sin(\frac{n}{2}x)$ from $n=1$ to infinity? Nov 17, 2016 at 8:40
• To continue with Yves's line of thought: There are some functions that have a very simple Fourier series. For example the Fourier series of $1$ is just $1,$ the Fourier series of $\cos 4x$ is just $\cos 4x$ etc.
– zhw.
Nov 17, 2016 at 9:10
• @zhw. Hi zhw, thanks for the prompt but I am struggling to derive the proof for the $cos4x$ case... I know it is probably something really simple but I'm just.. not getting it as of now.. Are you able to point me in a specific direction? Nov 17, 2016 at 9:33

## 2 Answers

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.

If you prefer the "hard way", use

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

• Wow.. that's an interesting way to approach the question. I can see that by comparison, when $n = 2013$, Summation of $b_n = 5$, is that right? However, how should I go about proofing that when $n \neq 2013$, $b_n = 0$? Would I have to evaluate the integral as written by @JezuzStardust above to proof that? Thanks for the help thus far. Nov 17, 2016 at 9:45
• @stephchia: I wrote all you need, but you still missed the obvious point.
– user65203
Nov 17, 2016 at 9:55
• Can you solve $a+b\cos x+c\sin x+d\cos 2x+e\sin 2x+f\cos3x+g\sin 3x=4\sin 2x$ ?
– user65203
Nov 17, 2016 at 9:58
• Hi... this is probably as frustrating for you as it is for me currently but I am indeed failing to see the obvious point you're stating to guide me to solving the question, and I'm not sure about the above either.. :/ (sigh) Nov 17, 2016 at 10:06
• @YvesDaoust Thanks so much for your help! I really appreciate it. Nov 17, 2016 at 10:55

To find the sum over $b_n$ I would suggest first calculating them. Use that $$b_n = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(nx) dx$$ 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 $$\frac{1}{\pi} \int_{-\pi}^\pi \sin(mx) \sin(nx) dx = \delta_{nm}$$ 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 $$\sin(mx) \sin(nx) = \frac{1}{2} \left\{ \cos[(m-n) x] - \cos[(m+n) x] \right\}.$$ 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.

• Hi, I am embarking on the arduous journey of finding $b_n$ currently. May I know if, since our $f(x)$ is an odd function and the multiple of odd functions, such as the case of that within the integral, yields an even function - that we could change the limit from $-\pi < x < \pi$ to $0 < x < \pi$ by multiplying the whole function by 2? Nov 17, 2016 at 9:48
• Hi, I've so for gotten $$\frac{5}{\pi} ( \frac{sin(2013-n)\pi}{2013-n} - \frac{sin(2013+n)\pi}{2013+n})$$ but I don't seem to find a way for $n = 2013$ to stand out as a unique case... Did I do something wrong? Nov 17, 2016 at 9:57
• Hi there, I've followed your integrals and reached $$\frac{5}{\pi} ( \frac{sin(2013-n)\pi}{2013-n} - \frac{sin(2013+n)\pi}{2013+n})$$ as written above, but I am failing to see how we could get an integral of $\pi$ from the above, I understand that we will get 0 on the left side since $sin0 = 0$, but for the other side how should I approach it? Nov 17, 2016 at 10:54
• Use that $\sin ( n \pi) = 0$ for any integer $n$. $2013 \pm n$ is an integer if and only if $n$ is an integer. Nov 17, 2016 at 11:53
• Regarding your previous question, you must go back to the integral expression when considering the case $m = n$, otherwise you get division by zero. Nov 17, 2016 at 12:52