A function f has the Fourier series
\displaystyle f(x) = \frac{a_0}{2} + \sum_{k=1}^\infty a_k \cos(kx) + b_k \sin(kx) with
\displaystyle a_k = \frac{(-1)^k}{A k} und
\displaystyle b_k = \frac{(-1)^{k+1}}{B k^{L}}.
Determine the Fourier coefficients
\displaystyle {\color{blue}A_{M}} und
\displaystyle {\color{red}B_{N}}
in the Fourier series of the derivative function
\displaystyle
f'(x) = \sum_{k=1}^\infty {\color{blue}A_k} \cos(kx) + {\color{red}B_k} \sin(kx).
{\color{blue}A_{M}}
=
pow(-1,M+1)/(B*pow(M,L-1))
{\color{red}B_{N}}
=
pow(-1,M+1)/A
We start with
\displaystyle f(x) = \frac{a_0}{2} + \sum_{k=1}^\infty a_k \cos(kx) + b_k \sin(kx)
and differentiate both sides.
Thus
\displaystyle f'(x) = \left( \frac{a_0}{2} + \sum_{k=1}^\infty a_k \cos(kx) + b_k \sin(kx)\right)' =
0+ \sum_{k=1}^\infty (a_k \cos(kx))' + (b_k \sin(kx))' =
\sum_{k=1}^\infty - k a_k \sin(kx) + kb_k \cos(kx).
Therefore by comparison:
\displaystyle {\color{blue}A_{k} = kb_k} and
\displaystyle {\color{red}B_{k} = -ka_k}.
Substituting the given indices gives the solutions
\displaystyle \color{blue}A_{M} =
fractionReduce(pow(-1,M+1),B*pow(M,L-1))
and
\displaystyle \color{red}B_{N} =
fractionReduce(pow(-1,M+1),A).