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Find the Fourier cosine series of $f(x) = x^2(1-x)^2$ over the interval $[0,1/2]$. Use it to to evaluate
$$\sum_{n=1}^{\infty} \frac{1}{n^4}.$$
You may use a computer algebra system to perform the integration for you.
The Fourier series for $f(x)$ over the interval $[0,\frac{1}{2}]$ is
$$\frac{a_0}{2} + \sum_{n=1}^{\infty}a_n\cos(2\pi nx),$$
where
$$a_n = 4\int_{0}^{\frac{1}{2}}x^2(1-x)^2\cos(2\pi nx)dx.$$
By using Mathematica to evaluate this integral for us, we find that $a_0 = \frac{1}{15}$ and $a_n = -\frac{3}{n^4\pi^4}$. Hence,
$$f(x) = \frac{1}{30} - \frac{3}{\pi^4}\sum_{n=1}^{\infty}\frac{1}{n^4}.$$
So
$$f(0) = 0 = \frac{1}{30} - \frac{3}{\pi^4}\sum_{n=1}^{\infty}\frac{1}{n^4}.$$
$$=> \frac{1}{30} = \frac{3}{\pi^4}\sum_{n=1}^{\infty}\frac{1}{n^4}.$$
Therefore,
$$\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{\pi^4}{90}$$
Comments
The Fourier series for $f(x)$ over the interval $[0,\frac{1}{2}]$ is
$$\frac{a_0}{2} + \sum_{n=1}^{\infty}a_n\cos(2\pi nx),$$
where
$$a_n = 4\int_{0}^{\frac{1}{2}}x^2(1-x)^2\cos(2\pi nx)dx.$$
By using Mathematica to evaluate this integral for us, we find that $a_0 = \frac{1}{15}$ and $a_n = -\frac{3}{n^4\pi^4}$. Hence,
$$f(x) = \frac{1}{30} - \frac{3}{\pi^4}\sum_{n=1}^{\infty}\frac{1}{n^4}.$$
So
$$f(0) = 0 = \frac{1}{30} - \frac{3}{\pi^4}\sum_{n=1}^{\infty}\frac{1}{n^4}.$$
$$=> \frac{1}{30} = \frac{3}{\pi^4}\sum_{n=1}^{\infty}\frac{1}{n^4}.$$
Therefore,
$$\sum_{n=1}^{\infty}\frac{1}{n^4} = \frac{\pi^4}{90}$$