# Fourier coefficients of a trigonometric polynomial

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?

• Show that $\int_0^{2\pi} \sin (k_1 x) \cos (k_2 x)\, dx = 0$, and that $\int_0^{2\pi} \sin (k_1 x) \sin (k_2 x)\, dx = \int_0^{2\pi} \cos (k_1 x) \cos (k_2 x)\, dx = 0$ unless $k_1 = k_2$. – Connor Harris Sep 19 '19 at 14:05
• I tried to show this, but I did not manage it. Are you able to help me? – Mathlover Sep 24 '19 at 10:23
• Here's a hint: try to find values of $x$ about which the integrand is symmetrical. Plotting the integrand for a few values of $k_1, k_2$ may help. – Connor Harris Sep 24 '19 at 13:34
• For the first equation I found $k_1$ = 0 and $k_2$ $ℝ$. For the second equation I only found that its true for $k_1$ = $k_2$. I doubt that this is correct. – Mathlover Sep 25 '19 at 13:17
• 'Tis a standard result – Tojrah Sep 26 '19 at 12:57

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.
• That we recover the Fourier coefficients from $\int_0^{2\pi} ..$ is in his original publication gallica.bnf.fr/ark:/12148/bpt6k33707/… – reuns Sep 30 '19 at 4:09
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.