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Chapter 5: Problem 17

Consider the function \(f(x)=x,-\pi

Short Answer

Expert verified
The given expression in part (a) is a Fourier sine series. After integrating the series, as required in part (b), the result matches with the Fourier cosine series of \(f(x)=x^{2}\) on \((0, \pi)\), computed in part (c); thus validating the computations.

Step by step solution

01

Show the Fourier series

Start by writing the left-hand side of the equation and carefully compute the given sum. The right-hand side involves a sine series. Each sine term is multiplied by \(-1^{n+1}\)/n. It's a classic Fourier sine series.
02

Integrate the series

First, integrate the left-hand side to get \(x^2 \)/2. Then, integrate each term of the series on then right-hand side. Use the property that the integral of \( \sin(nx)\) is \(-\cos(nx)/n\). Do not forget to apply the constant of integration which will be found by setting \(x = 0\). Compare the two expressions and simplify to find the constant.
03

Find the Fourier cosine series

To find the Fourier cosine series of \(f(x)=x^{2}\) on \((0, \pi)\), recall the formula for the Fourier cosine series: \(f(x)=a_{0}/2 + \sum_{n=1}^\infty[a_{n}\cos(nx)]\) where \(a_{n}=(2/\pi)\int_{0}^{\pi}f(x)\cos(nx)dx\). Substitute \(f(x)=x^{2}\) into the formula and calculate the integrals.
04

Compare the results

Compare the series obtained in parts (b) and (c). They should match, confirming that the calculations in previous steps are correct.

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