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?

  • $\begingroup$ 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 $k_1 = k_2$. $\endgroup$ – Connor Harris Sep 19 '19 at 14:05
  • $\begingroup$ I tried to show this, but I did not manage it. Are you able to help me? $\endgroup$ – Mathlover Sep 24 '19 at 10:23
  • $\begingroup$ 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. $\endgroup$ – Connor Harris Sep 24 '19 at 13:34
  • $\begingroup$ 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. $\endgroup$ – Mathlover Sep 25 '19 at 13:17
  • $\begingroup$ 'Tis a standard result $\endgroup$ – 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.

  • $\begingroup$ That we recover the Fourier coefficients from $\int_0^{2\pi} ..$ is in his original publication gallica.bnf.fr/ark:/12148/bpt6k33707/… $\endgroup$ – reuns Sep 30 '19 at 4:09
  • $\begingroup$ @reuns Thanks for the reference. Hausdorff is one of the giant mathematicians of the XXth century, with multi-faceted talents. $\endgroup$ – Jean Marie Sep 30 '19 at 4:14
  • $\begingroup$ Hello sir. Could you please help me with this question: math.stackexchange.com/q/3380406/394202 $\endgroup$ – Soumee Oct 4 '19 at 17:55

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.

  • $\begingroup$ I tried to prove them, but I think I am doing something wrong because I do not get the right answer. I tried it several times now... $\endgroup$ – Mathlover Sep 26 '19 at 14:35
  • $\begingroup$ Hello sir. Could you please help me with this question: math.stackexchange.com/q/3380406/394202 $\endgroup$ – Soumee Oct 4 '19 at 17:55

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