I have a function $ f $ and it's derivative $f'$ that are both continuous on $ [0,2\pi] $ and for any $x$ in this range :
$$ f'(x) = a_0 + \sum_{n=1}^\infty (a_n\cos(nx) + b_n\sin(nx))$$
I've been asked to:
Determine a formula for $b_8$ in terms of $f$ and I'm completely stuck on what to do.
(Assuming we now know the function is even) Justify whether $a_{12} = 0$ or not? My initial idea here is that since we know the function is even, all $b_n = 0$ and $a_n$ are not equal to $0$ so this statement must be false and $a_{12}$ cannot equal $0$. I'm not sure if this is correct or if there is more I need to show for this to be sufficient.
Any help would be much appreciated, especially for the first part of the question.
Assume f and f' are continuous on [0,2π].
Assume further $a_0 = 0$ and
$f(-π) = f(π)$ and $f ' (- π) = f ' (π)$.
If Fourier Series for f' is
$$ f'(x) = a_0 + \sum_{n=1}^\infty (a_n\cos(nx) + b_n\sin(nx))$$ then its corresponding Fourier Series for f becomes $$ f (x) = b'_0 + \sum_{n=1}^\infty(b'_n\cos(nx) +a'_n\sin(nx))$$ Where ${a'_n} * n = a_n ~\text{and} -{b'_n} * n = b_n$
Define the inner product of $f$ and $g$ as $$\langle f,g\rangle = \int_{0}^{2π} {f(x) g(x)}\mathrm dx$$
$$ b'_n = -b_n/n = 1/π \cdot \langle f(x),\cos(nx)\rangle, a'_n = a_n/n = 1/π \cdot \langle f(x), \sin(nx)\rangle$$
or $$b'_0 = 1/π \cdot \langle f(x),1\rangle, b'_n = 1/π \cdot \langle f(x),\cos(nx)\rangle, a'_n = 1/π \cdot \langle f(x), \sin(nx)\rangle$$
Conclusion:
$b_8 = -8/π \cdot \langle f(x), \cos(nx)\rangle $
If $f$ is even, then $a_n = 0$ for all n, in particular $a_{12} = 0.$
Anyone interested to get a Fourier Series for $f$ or $f'$, let me know and please
indicate whether they are for $f(x)$ or $f'(x)$. We will try to get its Fourier
coefficients $a_n, b_n$ from the Internet!
Have fun and enjoy a nice day!