1. Functions, the Fourier series expansion of functions, the relationship between Fourier series and functions
1.1 Derivation of coefficient formulas in Fourier series
Let’s first deduce the coefficient formula in Fourier series. In fact, the author has already written a related note. For details, see: Why should a function be decomposed into trigonometric functions? (Fourier series)
f ( x ) ∼ a 0 2 + ∑ n = 1 ∞ ( a n cos n π x l + b n sin n π x l ) ( x ∈ R ) f ( x ) = a 0 2 + ∑ n = 1 ∞ ( a n cos n π x l + b n sin n π x l ) ( − l < x < l ) a 0 = 1 l ∫ − l l f ( x ) d x = 2 l ∫ 0 l f ( x ) d x a n = 1 l ∫ − l l f ( x ) cos n π x l d x = 2 l ∫ 0 l f ( x ) cos n π x l d x b n = 1 l ∫ − l l f ( x ) sin n π x l d x = 2 l ∫ 0 l f ( x ) sin n π x l d x f(x)\sim \frac{a_0}{2}+\sum\limits_{n=1}^{\infty}(a_n\cos\frac{n\pi x}{l}+b_n\sin\frac{n\pi x}{l})(x\in \boldsymbol{R})\\ ~\\ f(x)= \frac{a_0}{2}+\sum\limits_{n=1}^{\infty}(a_n\cos\frac{n\pi x}{l}+b_n\sin\frac{n\pi x}{l})(-l\lt x\lt l)\\ ~\\ a_0=\frac{1}{l}\int_{-l}^{l}f(x)dx=\frac{2}{l}\int_{0}^{l}f(x)dx\\ ~\\ a_n=\frac{1}{l}\int_{-l}^{l}f(x)\cos\frac{n\pi x}{l}dx=\frac{2}{l}\int_{0}^{l}f(x)\cos\frac{n\pi x}{l}dx\\ ~\\ b_n=\frac{1}{l}\int_{-l}^{l}f(x)\sin\frac{n\pi x}{l}dx=\frac{2}{l}\int_{0}^{l}f(x)\sin\frac{n\pi x}{l}dx f(x)∼2a0+n=1∑∞(ancoslnπx+bnsinlnπx)(x∈R) f(x)=2a0+n=1∑∞(ancoslnπx+bnsinlnπx)(−l<x<l) a0=l1∫−llf(x)dx=l2∫0lf(x)dx an=l1∫−llf(x)coslnπxdx=l2∫0lf(x)coslnπxdx bn=l1∫−llf(x)sinlnπxdx=l2∫0lf(x)sinlnπxd xIf l = π l=\pi
in the above formulal=π
f ( x ) ∼ a 0 2 + ∑ n = 1 ∞ ( an cos nx + bn sin nx ) ( x ∈ R ) f ( x ) = a 0 2 + ∑ n = 1 ∞ ( an cos nx + bn sin nx ) ( − π < x < π ) a 0 = 1 π ∫ − π π f ( x ) dx = 2 π ∫ 0 π f ( x ) dx an = 1 π ∫ − π π f ( x ) cos nxdx = 2 π ∫ 0 π f ( x ) cos nxdx bn = 1 π ∫ − π π f ( x ) sin nxdx = 2 π ∫ 0 π f ( x ) sin nxdxf(x)\sim \frac{a_0}{2}+\sum\limits_{n=1}^{\infty}(a_n\cos nx+b_n\sin nx)(x\in \boldsymbol{R})\\ ~\\ f (x)= \frac{a_0}{2}+\sum\limits_{n=1}^{\infty}(a_n\cos nx+b_n\sin nx)(-\pi\lt x\lt \pi) \\ ~\\ a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)dx=\frac{2}{\pi}\int_{0}^{ \pi}f(x)dx\\ ~\\ a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos nxdx=\frac{2}{ \pi}\int_{0}^{\pi}f(x)\cos nxdx\\ ~\\ b_n=\frac{1}{\pi}\int_{-\pi}^{\pi}f( x)\sin nxdx=\frac{2}{\pi}\int_{0}^{\pi}f(x)\sin nxdxf(x)∼2a0+n=1∑∞(ancosnx+bnsinn x ) ( x∈R) f(x)=2a0+n=1∑∞(ancosnx+bnsinn x ) ( − π<x<p ) a0=Pi1∫− ppf(x)dx=Pi2∫0pf(x)dx an=Pi1∫− ppf(x)cosn x d x=Pi2∫0pf(x)cosn x d x bn=Pi1∫− ppf(x)sinn x d x=Pi2∫0pf(x)sinn x d x
1.2 Expand the function into a Fourier series
Number 1 in 2008
Set the function f ( x ) = 1 − x 2 ( 0 ≤ x ≤ π ) f(x)=1-x^2(0\leq x\leq \pi)f(x)=1−x2(0≤x≤π ) is expanded into a Fourier series in the form of cosine, and the series∑ n = 1 ∞ ( − 1 ) n − 1 n 2 \sum\limits_{n=1}^{\infty}\frac{(- 1)^{n-1}}{n^2}n=1∑∞n2(−1)n−1Since it
expands into a Fourier series in the form of cosine, it does not contain a sine series and bn = 0 b_n=0bn=0
f ( x ) ∼ a 0 2 + ∑ n = 1 ∞ an cos nx ( x ∈ R ) f ( x ) = a 0 2 + ∑ n = 1 ∞ an cos nx ( − π ≤ x ≤ π ) a 0 = 1 π ∫ − π π 1 − x 2 dx = 2 π ∫ 0 π 1 − x 2 dx = 2 ( 1 − π 2 3 ) an = 1 π ∫ − π π ( 1 − x 2 ) cos nxdx = 2 π ∫ 0 π ( 1 − x 2 ) cos nxdx = ( − 1 ) n + 1 ⋅ 4 n 2 f(x)\sim \frac{a_0}{2}+\sum\limits_{n= 1}^{\infty}a_n\cos nx(x\in \boldsymbol{R})\\ ~\\ f(x)= \frac{a_0}{2}+\sum\limits_{n=1}^ {\infty}a_n\cos nx(-\pi\leq x\leq \pi)\\ ~\\ a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}1- x^2dx=\frac{2}{\pi}\int_{0}^{\pi}1-x^2dx=2(1-\frac{\pi^2}{3})\\ ~\\ a_n=\frac{1}{\pi}\int_{-\pi}^{\pi}(1-x^2)\cos nxdx=\frac{2}{\pi}\int_{0}^{ \pi}(1-x^2)\cos nxdx=\frac{(-1)^{n+1}\cdot4}{n^2}f(x)∼2a0+n=1∑∞ancosn x ( x∈R) f(x)=2a0+n=1∑∞ancosn x ( − π≤x≤p ) a0=Pi1∫− pp1−x2dx=Pi2∫0p1−x2dx=2(1−3Pi2) an=Pi1∫− pp(1−x2)cosn x d x=Pi2∫0p(1−x2)cosn x d x=n2(−1)n+1⋅4
f ( x ) = 1 − π 2 3 + ∑ n = 1 ∞ ( − 1 ) n + 1 ⋅ 4 n 2 cos nx ( − π ≤ x ≤ π ) f ( 0 ) = 1 − π 2 3 + ∑ n = 1 ∞ ( − 1 ) n + 1 n 2 ( − π ≤ x ≤ π ) f ( 0 ) = 1 − 0 2 = 1 1 = 1 − π 2 3 + 4 ∑ n = 1 ∞ ( − ) n + 1 n 2 ( − π ≤ x ≤ π ) ∑ n = 1 ∞ ( − 1 ) n + 1 n 2 = π 2 12 f(x)= 1-\frac{\pi^2}{3} +\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}\cdot4}{n^2}\cos nx(-\pi\leq x\leq \pi )\\ ~\\ f(0)=1-\frac{\pi^2}{3}+4\sum\limits_{n=1}^{\infty}\frac{(-1)^{n +1}}{n^2}(-\pi\leq x\leq \pi)\\ ~\\ f(0)=1-0^2=1\\ ~\\ 1=1-\frac{ \pi^2}{3}+4\sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}(-\pi\leq x \leq \pi)\\ ~\\ \sum\limits_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}=\frac{\pi^ 2}{12}f(x)=1−3Pi2+n=1∑∞n2(−1)n+1⋅4cosn x ( − π≤x≤p ) f(0)=1−3Pi2+4n=1∑∞n2(−1)n+1( − p≤x≤p ) f(0)=1−02=1 1=1−3Pi2+4n=1∑∞n2(−1)n+1( − p≤x≤p ) n=1∑∞n2(−1)n+1=12Pi2
1.3 Analysis of the relationship between functions, Fourier series expansion of functions, Fourier series and functions
The graphing process is as follows:
function f ( x ) = 1 − x 2 ( 0 ≤ x ≤ π ) f(x)=1-x^2(0\leq x\leq \pi)f(x)=1−x2(0≤x≤π )
Partial cosine waves in cosine series
a n cos n x a_n\cos nx ancosn x nn
in this picturen only takes from 1 to 5, a total of 5 cosine waves.
Each cosine wave (green) is superimposed to form a cosine series (orange)
∑ n = 1 ∞ an cos nx \sum\limits_{n=1}^{\infty }a_n\cos nxn=1∑∞ancosn x
cosine series plusa 0 / 2 a_0/2a0Image of /2
(blue) a 0 2 + ∑ n = 1 ∞ an cos nx \frac{a_0}{2}+\sum\limits_{n=1}^{\infty}a_n\cos nx2a0+n=1∑∞ancosnx
Function f ( x ) f(x)Image of f ( x ) (red) and its Fourier series expanded into cosine form (blue)
In the above figure, the red color is the function, the blue color is the Fourier series expansion of the function, the overlapping part of the function and the Fourier series expansion of the function is the sum function, and the part of the function that exceeds the convergence region is the sum function of the Fourier series. no longer equal
When n is taken from 1 to 30, the sum function of the Fourier series basically matches the function exactly! !
Overview of the drawing process
 f(x)
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 a_ncos(nx)
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cosine series
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Functions and cosine series
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