Fourier series convergence test

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SUMMARY

The discussion focuses on the convergence of the Fourier series defined for the piecewise function f(x), specifically at the point x = π/2. The Fourier series is given by F(x) = π/16 + (1/π)Σ[(1/n²)(cos(nπ) - cos(nπ/2))cos(nx) - (1/(2n²))(nπ cos(nπ) + 2sin(nπ/2))sin(nx)], where n ranges from 1 to infinity. The key insight is that cos(nπ) simplifies to (-1)ⁿ, and the evaluation of F(π/2) requires analyzing the behavior of cos(nx) and sin(nx) based on the parity of n. This leads to a clearer understanding of the series' convergence properties.

PREREQUISITES
  • Understanding of Fourier series and their convergence criteria
  • Knowledge of piecewise functions and their representations
  • Familiarity with trigonometric identities, particularly cosine and sine functions
  • Basic calculus concepts, including limits and infinite series
NEXT STEPS
  • Study the convergence tests for Fourier series, focusing on Dirichlet's conditions
  • Learn about the implications of piecewise continuity on Fourier series convergence
  • Explore the properties of sine and cosine functions in Fourier series analysis
  • Investigate the impact of discontinuities on the convergence of Fourier series
USEFUL FOR

Mathematicians, engineering students, and anyone studying Fourier analysis or signal processing who seeks to understand the convergence behavior of Fourier series for piecewise functions.

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Homework Statement



A function f(x) is given as follows

f(x) = 0, , -pi <= x <= pi/2
f(x) = x -pi/2 , pi/2 < x <= pi

determine if it's Fourier series (given below)

[itex]F(x)=\pi/16 + (1/\pi)\sum=[ (1/n^{2})(cos(n\pi) - cos(n\pi/2))cos(nx) <br /> - (1/(2n^{2}))( n\pi cos(n\pi) + 2sin(n\pi/2) )sin(nx) ][/itex]

for n= 1 to infinity


converges to it for the case of x = pi/2

Homework Equations



F(x) = 0.5*[f(x+) + f(x-) ]


The Attempt at a Solution



see pdf attachment
 

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Your work is kind of hard to follow, but it looks like you incorrectly got rid of the cosine terms. You have

[tex]F(x)=\frac{\pi}{16} + \frac{1}{\pi}\sum_{n=1}^\infty \frac{\cos n\pi -\cos \frac{n\pi}{2}}{n^2} \cos nx - \frac{1}{\pi}\sum_{n=1}^\infty \frac{n\pi \cos n\pi + 2\sin \frac{n\pi}{2}}{2n^2}\sin nx[/tex]

You can simplify it a bit by noting that [itex]\cos n\pi = (-1)^n[/itex]. Next, when [itex]x=\pi/2[/itex], the values of cos nx and sin nx depend on whether n is odd or even. Use those facts to get the series for [itex]F(\pi/2)[/itex].
 
Last edited:

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