Fourier coefficients of a trigonometric polynomial

Question

Let $g(x)$ $=$ $1/2$ $p_0$ + $\sum_{k=1}^{n}$ $(p_k \cos(kx)+ q_k \sin(kx))$ be a trigonometric polynomial.

How can I explain why its Fourier coefficients are $a_k$ $=$ $p_k$ and $b_k$ $=$ $q_k$ for $k$ $≤$ $n$,

while $a_k$ $=$ $b_k$ $=$ $0$ for $k$ $>$ $n$ $?$

Someone gave me the hint to solve the equations I have shown down here

1) $$\int_{0}^{2\pi}\sin(k_1x)\cos(k_2x)dx = 0$$

and that

2) $$\int_{0}^{2\pi}\sin(k_1x)\sin(k_2x)dx=\int_{0}^{2\pi}\cos(k_1x)\cos(k_2x)dx=0$$ unless $k_1 = k_2$

I found for 1) $k_1 = 0$ and then $k_2$ $\in$ ${R}$ and for 2) I found $k_1 = k_2$ but I doubt that this is correct.

Someone who’s able to help me?

Answer 1

You should not "solve" your equations 1) and 2), but prove them. This is most easily done by using formula $$\sin\alpha\cos\beta={1\over2}\bigl(\sin(\alpha+\beta)+\sin(\alpha-\beta)\bigr)\ $$ to "linearize" the integrand in 1): $$\sin(jx)\cos(kx)={1\over2}\bigl(\sin((j+k)x)+\sin((j-k)x)\bigr)\ .$$ Here $\int_0^{2\pi}$ of the RHS is obviously $=0$ when $j$, $k\in{\mathbb N}$. Similarly for the integrals 2), but there you have a special case to consider.

The principles 1) and 2) then allow you to compute the Fourier coefficients of your function $g$ (using the official formulas for the $a_k$, $b_k$), and you will see that exactly the expected happens.

Answer 2

There is a very simple argument : the uniqueness of Fourier coefficients, valid in a very general framework that we do not need (Here we deal with a $C^{\infty}$ function). Take a look at the interesting answer https://math.stackexchange.com/q/1939575 recalling the work of Hausdorff on this subject.