Testwiki:Reference desk/Archives/Mathematics/2022 January 21

{| width = "100%"

|- ! colspan="3" align="center" | Mathematics desk |- ! width="20%" align="left" | < January 20 ! width="25%" align="center"|<< Dec | January | Feb >> ! width="20%" align="right" |Current desk > |}

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

January 21

Where do Template:Math in Fourier coefficients come from?

Equantions for Template:Plain link are Template:NumBlk Template:NumBlk I cannot figure out the equations above because of the occurrence of $\frac{2}{P}$. I guess $\frac{2}{P}$ comes from the Template:Plain link being used for normalizing basis functions $\cos \left(\frac{2\pi }{P}nx\right)$ and $\sin \left(\frac{2\pi }{P}nx\right)$. Accroding to this, if I understand correctly, the inner product of $\cos \left(\frac{2\pi }{P}nx\right)$ with itself is Template:NumBlk where $P_0\in ℝ$. Therefore Template:NumBlk The normalized basis functions are Template:NumBlk So the Fourier coefficients I get should be like the scalar projection of $s(x)$ onto orthonormal basis in Template:EquationNote Template:NumBlk Unluckily, Template:EquationNote is different from Template:EquationNote. Any idea? - Justin545 (talk) 17:02, 21 January 2022 (UTC)

It is not defined symmetrically. So

$s(x)=\frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}(a_n\cos \frac{2\pi }{p}x+b_n\sin \frac{2\pi }{p}x)$

Ruslik_Zero 20:17, 21 January 2022 (UTC)

I still don't get it. Could you provide further explanation? - Justin545 (talk) 21:22, 21 January 2022 (UTC)

The basis for Fourier analysis of a periodic function is given by the following orthogonality properties of the sine and cosine functions. Let $m$ and $n$ be positive integers. Then

$\underset{0}{\overset{2\pi }{\int }}\sin mx\sin nxdx=\underset{0}{\overset{2\pi }{\int }}\cos mx\cos nxdx=\{\begin{array}{cc}0& \text{if }m\ne n,\\ \pi & \text{if }m=n,\end{array}$

$\underset{0}{\overset{2\pi }{\int }}\sin mx\cos nxdx=0.$

For the sake of simplicity, let us fix the period as $2\pi$. Let function $s$ be given by

$s(x)=\frac{1}{2}{a}_0+\sum _{m=1}^{\mathrm{\infty }}(a_m\cos mx+b_m\sin mx).$

Let us also assume the infinite summations converge. Now consider what happens if we multiply $s(x)$ by $\cos nx$, $n>0$, and integrate over the period:

$\underset{0}{\overset{2\pi }{\int }}s(x)\cos nxdx$

$=\underset{0}{\overset{2\pi }{\int }}(\frac{1}{2}{a}_0+\sum _{m=1}^{\mathrm{\infty }}(a_m\cos mx+b_m\sin mx))\cos nxdx$

$=\underset{0}{\overset{2\pi }{\int }}(\frac{1}{2}{a}_0\cos nx+\sum _{m=1}^{\mathrm{\infty }}(a_m\cos mx\cos nx+b_m\sin mx\cos nx))dx$

$=\underset{0}{\overset{2\pi }{\int }}\frac{1}{2}{a}_0\cos nxdx+\sum _{m=1}^{\mathrm{\infty }}(a_m\underset{0}{\overset{2\pi }{\int }}\cos mx\cos nxdx+b_m\underset{0}{\overset{2\pi }{\int }}\sin mx\cos nxdx)$

$=\pi a_n.$

(For the last step, split the summation into the cases $m=n$ and $m\ne n$ and apply the orthogonality formulas.) So, to find the value of $a_n$, we need to divide to result of the integral by $\pi$, that is, half the period. For $b_n$ we have the same story, except that we multiply $s(x)$ by $\cos nx.$  --Lambiam 07:29, 22 January 2022 (UTC)

Thanks guys, in particular Lambiam. The answer for $P=2\pi$ is understandable and crystal clear. I may try to figure out the term $\frac{a_0}{2}$ of $s(x)$ and ask for help if I'm stuck again. - Justin545 (talk) 10:23, 22 January 2022 (UTC)

For $a_0$, just integrate $s(x)$ without multiplier – which is equivalent to multiplying it by $\cos 0x.$  --Lambiam 14:26, 22 January 2022 (UTC)