Answer to Question #132025 in Real Analysis for Joel George

Question

Expand f(x) = x^2, 0 < x < 2π, in a Fourier cosine series;

Answer 1

$$Solution$$

$f(x) = x^2, 0 < x < 2π,$ in a Fourier cosine series;

$f(x)=a_0+\sum_{n=1}^\infin a_n\ cos\ nx + \sum_{n=1}^\infin b_n\ sin\ nx\\ \therefore\ a_0=\frac{1}{2\pi}\intop_0^{2\pi} f(x)\ dx= \frac{1}{2\pi}\intop_0^{2\pi} x^2\ dx\\ =\frac{1}{2\pi}|\frac{x^3}{3}|_0^{2\pi} = \frac{1}{2\pi}[\frac{(2\pi)^3}{3}-\frac{0}{3}]\\ \implies a_0=\frac{1}{2\pi}[\frac{8\pi^3}{3}]=\frac{4\pi^3}{3}----------i\\ \therefore a_n=\frac{1}{\pi}\intop_0^{2\pi} F(x)\ cos\ nx\ dx =\frac{1}{\pi}\intop_0^{2\pi} x^2\ cos\ nx\ dx\\ =\frac{1}{\pi}|x^2(\frac{sin\ nx}{n})-2x(\frac{-cos\ nx}{n^2})+2(\frac{-sin\ nx}{n^3})|_0^{2\pi}\\ =\frac{1}{\pi}[4\pi^2(\frac{sin\ 2\pi n}{n})-4\pi(\frac{-cos\ 2\pi n}{n^2})+2(\frac{sin\ 2\pi n}{n^3})]\\ =\frac{1}{\pi}[4\pi(\frac{cos\ 2\pi n}{n^2})]=\frac{1}{\pi}(\frac{4\pi}{n^2})=\frac{4}{n^2}\\ \implies a_n=\frac{4}{n^2}--------------ii\\ \therefore b_n=\frac{1}{\pi}\intop_0^{2\pi} f(x)\ sin\ nx\ dx =\frac{1}{\pi}\intop_0^{2\pi} x^2\ sin\ nx\ dx \\ =\frac{1}{\pi}|x^2(\frac{-cos\ nx}{n})-2x(\frac{-sin\ nx}{n^2})+2(\frac{cos\ nx}{n^3})|_0^{2\pi}\\ =\frac{1}{\pi}[4\pi^2(\frac{-cos2\pi n}{n})-4\pi(\frac{-sin2\pi n}{n^2})+2(\frac{cos2\pi n}{n^3})]_0^{2\pi}\\ =\frac{1}{\pi}(-\frac{4\pi^2}{n})\ (\because\ cos2\pi n =cos\ 0 =1 )\\ =b_n= -\frac{4\pi}{n}---------------iii\\ \therefore F(x)=a_0+\sum_{n=1}^\infin a_n\ cos\ nx + \sum_{n=1}^\infin b_n\ sin\ nx\\ \therefore\ x^2=\frac{4\pi^2}{3}+\frac{4}{n^2} \sum_{n=1}^{\infin}\ cos\ nx- 4\pi \sum_{n=1}^{\infin} \frac{1}{n}\ sin\ nx\\ =\frac{4\pi^2}{3}+4(\frac{1}{1^2}cos\ x+\frac{1}{2^2}cos\ 2x+\frac{1}{3^2}cos\ 3x+----)-4\pi(\frac{1}{1}sin\ x+\frac{1}{2}sin\ 2x+\frac{1}{3}sin\ 3x+----)\\$

$\therefore$ Put value of $x=\pi$

$\therefore\ \pi^2=\frac{4\pi^2}{3}+4(\frac{1}{1^2}cos\ \pi+\frac{1}{2^2}cos\ 2\pi+\frac{1}{3^2}cos\ 3\pi+----)-4\pi(\frac{1}{1}sin\ n\pi+\frac{1}{2}sin\ 2\pi+\frac{1}{3}sin\ 3\pi+----)\\ \therefore\ \pi^2=\frac{4\pi^2}{3}+4(-\frac{1}{1^2}+\frac{1}{2^2}-\frac{1}{3^2}+-----)\\ \therefore\ \pi^2-\frac{4\pi^2}{3}=4(-\frac{1}{1^2}+\frac{1}{2^2}-\frac{1}{3^2}+-----)\\ \therefore\ \frac{\pi^2}{3}=4(-\frac{1}{1^2}+\frac{1}{2^2}-\frac{1}{3^2}+-----)\\ \therefore\ -\frac{\pi^2}{12}=(-\frac{1}{1^2}+\frac{1}{2^2}-\frac{1}{3^2}+-----)\\ \implies\ [\frac{\pi^2}{12}=(\frac{1}{1^2}-\frac{1}{2^2}+\frac{1}{3^2}--)]---->Answer\\$