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

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January 21

Where do 2/P in Fourier coefficients come from?

 Resolved

Equantions for Fourier coefficients are

$a_{n}={\frac {2}{P}}\int _{P}s(x)\cdot \cos \left({\tfrac {2\pi }{P}}nx\right)\,dx$ (1)

$b_{n}={\frac {2}{P}}\int _{P}s(x)\cdot \sin \left({\tfrac {2\pi }{P}}nx\right)\,dx$ (2)

I cannot figure out the equations above because of the occurrence of ${\frac {2}{P}}$. I guess ${\frac {2}{P}}$ comes from the norm being used for normalizing basis functions $\cos \left({\tfrac {2\pi }{P}}nx\right)$ and $\sin \left({\tfrac {2\pi }{P}}nx\right)$. Accroding to this, if I understand correctly, the inner product of $\cos \left({\tfrac {2\pi }{P}}nx\right)$ with itself is

$C=\int _{P_{0}}^{P_{0}+P}\cos \left({\frac {2\pi }{P}}nx\right)\cos \left({\frac {2\pi }{P}}nx\right)\,dx={\frac {P}{2}}$ (3)

where $P_{0}\in \mathbb {R}$. Therefore

$a={\frac {1}{\sqrt {C}}}={\sqrt {\frac {2}{P}}}$ (4)

The normalized basis functions are

$a\cdot \cos \left({\frac {2\pi }{P}}nx\right)={\sqrt {\frac {2}{P}}}\cos \left({\frac {2\pi }{P}}nx\right)$ (5)

So the Fourier coefficients I get should be like the scalar projection of $s(x)$ onto orthonormal basis in (5)

$a_{n}^{\prime }=\int _{P_{0}}^{P_{0}+P}s(x)\cdot {\sqrt {\frac {2}{P}}}\cos \left({\frac {2\pi }{P}}nx\right)\,dx={\sqrt {\frac {2}{P}}}\int _{P}s(x)\cdot \cos \left({\frac {2\pi }{P}}nx\right)\,dx$ (6)

Unluckily, (6) is different from (1). Any idea? - Justin545 (talk) 17:02, 21 January 2022 (UTC)

It is not defined symmetrically. So

$s(x)={\frac {a_{0}}{2}}+\sum _{n=1}^{\infty }\left(a_{n}\cos {\frac {2\pi }{p}}x+b_{n}\sin {\frac {2\pi }{p}}x\right)$

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

$\int _{0}^{2\pi }\sin mx\sin nx~dx=\int _{0}^{2\pi }\cos mx\cos nx~dx={\begin{cases}0&{\mbox{if }}m\neq n,\\\pi &{\mbox{if }}m=n,\end{cases}}$

$\int _{0}^{2\pi }\sin mx\cos nx~dx=0.$

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

$s(x)={\frac {1}{2}}a_{0}+\sum _{m=1}^{\infty }(a_{m}\cos mx+b_{m}\sin mx).$

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:

$\int _{0}^{2\pi }s(x)\cos nx~dx$

$=\int _{0}^{2\pi }\left({\frac {1}{2}}a_{0}+\sum _{m=1}^{\infty }(a_{m}\cos mx+b_{m}\sin mx)\right)\cos nx~dx$

$=\int _{0}^{2\pi }\left({\frac {1}{2}}a_{0}\cos nx+\sum _{m=1}^{\infty }(a_{m}\cos mx\cos nx+b_{m}\sin mx\cos nx)\right)dx$

$=\int _{0}^{2\pi }{\frac {1}{2}}a_{0}\cos nx~dx+\sum _{m=1}^{\infty }\left(a_{m}\int _{0}^{2\pi }\cos mx\cos nx~dx+b_{m}\int _{0}^{2\pi }\sin mx\cos nx~dx\right)$

$=\pi a_{n}.$

(For the last step, split the summation into the cases $m=n$ and $m\neq n$ 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)