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I am trying to calculate the fourier series of $f(x)=\sin x$ for $0<x<\pi$ and $f(x)=0$ for $-\pi<x<0$ and make the plots. I tried

f[x_] = If[x > 0, Sin[x], 0];
a[n_] := (2/L)*Integrate[f[x]*Cos[2 n*Pi*x/L], {x, -L/2, L/2}]
a[0] = (1/L)*Integrate[f[x], {x, -L/2, L/2}]
b[n_] := (2/L)*Integrate[f[x]*Sin[2 n*Pi*x/L], {x, -L/2, L/2}]
F[x_, N_] := 
a[o] + Sum[a[n]*Cos[2 n*Pi*x/L] + b[n]*Sin[2 n*Pi*x/L], {n, 1, N}]
p[N_, a_] := 
 Plot[Evaluate[F[x, N]], {x, -a, a}, PlotRange -> All, 
 PlotPoints -> 200]
L = 2Pi;
a[n]
a[0]
b[n]
p[20, 1]    

`

but the Plot doesn't seems correct. Then, for $f(x)=0?.$ What about the converge of the series?

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    $\begingroup$ Move L = 2Pi; from the bottom of your code to the top and in your definition of F[x_,N_] change a[o] to a[0] and that will fix two of your problems. $\endgroup$
    – Bill
    Commented Jan 16, 2020 at 17:13
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    $\begingroup$ Have you seen FourierSeries and FourierCoefficient? $\endgroup$ Commented Jan 16, 2020 at 17:52

1 Answer 1

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You can just use FourierSeries

ClearAll[x]
nTerms = 10;
f[x_] := Piecewise[{{Sin[x], 0 < x < Pi}, {0, -Pi < x < 0}}]
series = ExpToTrig[FourierSeries[f[x], x, nTerms]]

$$ \frac{\sin (x)}{2}-\frac{2 \cos (2 x)}{3 \pi }-\frac{2 \cos (4 x)}{15 \pi }-\frac{2 \cos (6 x)}{35 \pi }-\frac{2 \cos (8 x)}{63 \pi }-\frac{2 \cos (10 x)}{99 \pi }+\frac{1}{\pi } $$

enter image description here

Manipulate[
 Module[{series},
  series = ExpToTrig[FourierSeries[f[x], x, nTerms]];
  Grid[{
    {series},
    {Plot[series, {x, -a, a}, ImageSize -> 400]}
    }, Frame -> All]
  ]
 ,
 {{nTerms, 4, "How many terms?"}, 1, 10, 1, Appearance -> "Labeled"},
 {{a, Pi, "Range?"}, Pi/10, 5 Pi, Pi/10, Appearance -> "Labeled"},
 ContinuousAction -> False,
 TrackedSymbols :> {a, nTerms},
 Initialization :>
  (f[x_] := Piecewise[{{Sin[x], 0 < x < Pi}, {0, -Pi < x < 0}}])
 ]

There is also lots of Fourier series animations here, all done using Mathematica.

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