This exercise once again makes use of the table of Fourier series.
We saw in class that
\[ \theta = 2 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \sin{n\theta} \]
for all \(\theta \in (-\pi, \pi)\).
Then, starting from entry 16 of the table, make use of the Integration Theorem to show that:
\(\theta^3 - \pi^2 \theta = 12 \sum_{n=1}^{\infty} \frac{(-1)^{n}}{n^3} \sin{n\theta}\)
\(\theta^4 - 2\pi^2 \theta^2 = 48 \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^4} \cos{n\theta}\)
\(\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}\)
Coming soon!
When is term-by-term differentiation of a Fourier series valid? This exercise explores this question in a specific case.
Consider the \(2\pi\)-periodic extension \(f\) of the function with \(f(\theta) = e^\theta\) for \(\theta \in (-\pi, \pi]\). This function is piecewise smooth, so we know that
\[ f(\theta) = \sum_{n=-\infty}^{\infty} c_n e^{i n \theta} \]
at all points of continuity, where the \(c_n\) are the Fourier coefficients of \(f\). If we formally differentiate both sides of this equation (the right side term-by-term), we obtain
\[ e^\theta = f'(\theta) = \sum_{n=-\infty}^{\infty} i n c_n e^{i n \theta}. \]
But by uniqueness of Fourier coefficients, this would say \(c_n = i n c_n\) for all \(n\), which only happens if \(c_n=0\). But this is clearly false! Discuss what went wrong.
The Differentiation Theorem only applies to the Fourier series of continuous functions. The function \(f\) is discontinuous when \(\theta\) is an odd multiple of \(\pi\), so the Differentiation Theorem does not apply, and we cannot differentiate the Fourier series of \(f\) term-by-term.