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I was interested in answering this question while I was waddling through the Internet. It goes like this:

Evaluate $ \sum_{n=1}^\infty (a_n \cos \left( \frac{n\pi x}{L}\right) + b_n \sin \left(\frac{n\pi x}{L}\right))$

My work

To evaluate $ \sum_{n=1}^\infty (a_n \cos \left( \frac{n\pi x}{L}\right) + b_n \sin \left(\frac{n\pi x}{L}\right))$, we'll do it manually. If we let $n = 1, 2, 3, 4,....$, we see that:

If $n = 1$: $$a_1 \cos \left( \frac{(1)\pi x}{L}\right) + b_1 \sin \left( \frac{(1)\pi x}{L}\right)$$

If $n = 2$: $$a_2 \cos \left( \frac{(2)\pi x}{L}\right) + b_2 \sin \left( \frac{(2)\pi x}{L}\right)$$

If $n = 3$: $$a_3 \cos \left( \frac{(3)\pi x}{L}\right) + b_3 \sin \left( \frac{(3)\pi x}{L}\right)$$

If $n = 4$: $$a_4 \cos \left( \frac{(4)\pi x}{L}\right) + b_4 \sin \left( \frac{(4)\pi x}{L}\right)$$

If $n = 5$: $$a_5 \cos \left( \frac{(5)\pi x}{L}\right) + b_5 \sin \left( \frac{(5)\pi x}{L}\right)$$

We recall that odd multiples of $n$ $(n = 1, 3, 5, 7, 9,....)$ in $\sin \pi n$ equals to $0$ while odd multiples of $n$ $(n = 1, 3, 5, 7, 9,....)$ in $\cos \pi n$ equals to $-1$.

We also remember that even multiples of $n$ $(n = 0, 2, 4, 6, 8,....)$ in $\sin \pi n$ equals to $0$ while even multiples of $n$ $(n = 0, 2, 4, 6, 8,....)$ in $\cos \pi n$ equals to $1$. With that in mind, it becomes:

$$\sum_{n=1}^\infty (a_n \cos \left( \frac{n\pi x}{L}\right) + b_n \sin \left( \frac{n\pi x}{L}\right))$$ $$ = (-a_1 + 0b_1)+(a_2 + 0b_2)+(-a_3 + 0b_3)+(a_4 + 0b_4)+(-a_5 + 0b_5)+......$$

If we let $a_1$ $=$ $b_1$ $= 1$ so we can analyze it easily, it becomes:

$$-a_1 + a_2 + -a_3 + a_4 + -a_5 + .....$$ $$-1 + 1 + -1 + 1 + -1.....=0$$

The sum of the oscillating series $(-1) + 1 + (-1) + 1 + (-1).....$ or $-1 + 1 - 1 + 1 - 1.....$ is zero. Therefore:

$$\color{green}{\sum_{n=1}^\infty \left(a_n \cos \left( \frac{n\pi x}{L}\right) + b_n \sin \left( \frac{n\pi x}{L}\right)\right) = 0}$$

Is my answer correct, ladies and gentlemen?

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    $\begingroup$ Each term is a function of $x$ so you can't say that odd multiples are $-1$ and so on. The series you have is an example of fourier series. It is used to represent periodic function as a sum of signs and cosines. The exact value of the curve depends on the coefficients $a_i, b_i$. You can't solve this to get s simple function. $\endgroup$ – Sonal_sqrt Feb 19 '18 at 4:06
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    $\begingroup$ For any $a_n,b_n$ and $x$? No, definitely not true. $\endgroup$ – Thomas Andrews Feb 19 '18 at 4:07
  • $\begingroup$ @AndrewLi You're right:-) I look it up on a book. It diverges.... $\endgroup$ – Palautot Ka Feb 19 '18 at 4:07
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There is a lot of confusion in this question, but the first thing I'll note is that just because $\sin \pi n=0$ doesn't tell us anything about the values of $\sin\frac{\pi n x}{L}$ for any $x$.

If $a_0=1,b_0=0,a_n=b_n=0$ for $n>0$, then your sum is $\sin \frac{\pi x}{L},$ which is not zero.

Also, while $1-1+1=1+\cdots$ doesn't converge, that says nothing about $a_0-a_1+a_2-a_3+\cdots.$

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Nope, it is not correct.

While $\sin(n \pi )=0$ when $n$ is odd, we can't conclude that $\sin\left( \frac{n\pi x}{L}\right)=0$

You might like to study about Fourier series.

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    $\begingroup$ $\sin nx$ is not zero when $n$ is odd, exept for certain values of $x.$ $\endgroup$ – Thomas Andrews Feb 19 '18 at 4:08
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Wrong.

What you have is a general Fourier series.

If $x=L$ your results might be correct. Since, in general, $x \ne L$, you cannot compute the values as you have done.

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  • $\begingroup$ It's still not true for $x$ a multiple of $L.$ $\endgroup$ – Thomas Andrews Feb 19 '18 at 4:09

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