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i need to solve the following question using parsevel's identity.

  1. $\displaystyle\int_{-\pi}^{\pi} \cos^{4} x\, dx = \frac{3\pi}{4}$.

  2. $\displaystyle\int_{-\pi}^{\pi} \cos^{6} x\, dx = \frac{5\pi}{8}$.

I tried using $f\left(x\right)$ as $\cos^{2}\left(x\right)$ but didn't get the answer after finding $a_{0}$, $a_n{ }$, and $b_{n}$. I got $a_{0}=1$, $a_{n}=0$, $b_{n}=0$.

Please help me.

I know that I have to use the Parseval's Indentity relation but I don't know whose function's expansion makes this possible.

Andrew D. Hwang
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rashmi
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2 Answers2

2

Okay, if $f(x)=\cos^2(x)$, then its Fourier series is $f(x)=\sum_{n \in \mathbb{Z}}c_ne^{inx}$, with $$c_n=\frac{1}{2\pi}\int_{-\pi}^\pi \cos^2(x)e^{-in x}dx=\frac{1}{4\pi}\int_{-\pi}^\pi \cos(2x)+1e^{-in x}dx=\frac{1}{4\pi}\begin{cases} \pi & \text{ if } n=\pm 2 \\ 0 & \text{ otherwise} \end{cases}+\frac{1}{4\pi} \begin{cases} 2\pi & \text{ if } n=0\\ 0 & \text{ otherwise} \end{cases}$$ Thus, applying Parseval's identity, $$\int_{-\pi}^\pi |\cos^2(x)|^2dx=\int_{-\pi}^\pi\cos^4(x)dx=2\pi\sum_{n=-\infty}^\infty \left|c_n\right|^2=2\pi \left( \frac{1}{16}+\frac{1}{16}+\frac{1}{4}\right)=\frac{3}{4}\pi.$$ The other one is tackled similarly ($c_n$ for $f(x)= \cos^3(x)$ is just a bit harder to work out), though this seems like overkill: just use double angle formulae to work out the original integrals!

Meow
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Using the identity $\displaystyle \cos(x)=\frac{e^{-2\pi i x}+e^{2\pi i x}}{2}$ we can expand $\cos^2x$ as

$$\cos^2x=\frac{(e^{-2\pi i x})^2+2(e^{-2\pi i x})(e^{2\pi i x})+(e^{2\pi i x})^2}{4}=\frac{1}{4}e^{-4\pi i x}+\frac{1}{2}+\frac{1}{4}e^{4\pi i x}.$$

Therefore Parseval's tells us

$$\frac{1}{2\pi}\int_{-\pi}^\pi |\cos^2 x|^2dx=\cdots+0^2+\frac{1}{4^2}+0^2+0^2+0^2+\frac{1}{2^2}+0^2+0^2+0^2+\frac{1}{4^2}+0^2+\cdots$$

Now can you do $f(x)=\cos^3x$ on your own?

anon
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