Find complex Fourier coefficients 
  
*
  
*let $f(x) = \sum^{10}_{m=1}(-1)^m \sin(2^m x)$.
  
  
  denote complex Fourier coefficients of $f(x)$ over $[-\pi, \pi]$ as  $c_n =  \frac{1}{2\pi} \int _{-\pi}^\pi f(x) e^{-inx}\,dx.$
Calculate $c_n$ for any $n$.

I don't have any idea how to even start.
Can you please explain me? thanks in advance.
 A: You can compute the integrals explicitly. Or use Euler's formulae to write
$$
\sin(2^mx)=\frac{e^{2^mxi}-e^{-2^mxi}}{2\,i}
$$
and substitute in the definition of $f$.
A: The coefficient $c_{0}=0$, because $f(x)$ is odd
\begin{equation*}
f(-x)=\sum_{m=1}^{10}(-1)^{m}\sin (-2^{m}x)=-\sum_{m=1}^{10}(-1)^{m}\sin
(2^{m}x)=-f(x).
\end{equation*}
For $n<0$, $n=-\left\vert n\right\vert $. Then 
\begin{eqnarray*}
c_{n} &=&\frac{1}{2\pi }\int_{-\pi }^{\pi }f(x)e^{-inx}\,dx=\frac{1}{2\pi }
\int_{-\pi }^{\pi }f(x)e^{i\left\vert n\right\vert x}\,dx \\
&=&\frac{1}{2\pi }\int_{-\pi }^{\pi }f(x)\overline{e^{-i\left\vert
n\right\vert x}}\,dx=\frac{1}{2\pi }\int_{-\pi }^{\pi }\overline{
f(x)e^{-i\left\vert n\right\vert x}}\,dx=\overline{c}_{\left\vert
n\right\vert }=\overline{c}_{-n},
\end{eqnarray*}
which means that the negative coefficients are easily computed from the
positive ones. For $n>0$, interchanging the order of the integration and the
summation in 
\begin{eqnarray*}
c_{n} &=&\frac{1}{2\pi }\int_{-\pi }^{\pi }f(x)e^{-inx}\,dx \\
&=&\frac{1}{2\pi }\int_{-\pi }^{\pi }\sum_{m=1}^{10}(-1)^{m}\sin
(2^{m}x)e^{-inx}\,dx,
\end{eqnarray*}
we conclude that
\begin{equation*}
c_{n}=\frac{1}{2\pi }\sum_{m=1}^{10}(-1)^{m}I(m,n),
\end{equation*}
where 
\begin{eqnarray*}
I(m,n) &=&\int_{-\pi }^{\pi }\sin (2^{m}x)e^{-inx}\,dx,\qquad e^{-inx}=\cos
(nx)-i\sin (nx) \\
&=&\int_{-\pi }^{\pi }\sin (2^{m}x)\cos (nx)\,dx-i\int_{-\pi }^{\pi }\sin
(2^{m}x)\sin (nx)\,dx.
\end{eqnarray*}
The first integral
\begin{equation*}
\int_{-\pi }^{\pi }\sin (2^{m}x)\cos (nx)\,dx=0,
\end{equation*}
because $g(x)=\sin (2^{m}x)\cos (nx)$ is odd ($g(-x)=-g(x)$). Hence $I(m,n)$
reduces to
\begin{equation*}
I(m,n)=-iJ(m,n)=-i\int_{-\pi }^{\pi }\sin (2^{m}x)\sin (nx)\,dx.
\end{equation*}
To evaluate this integral we can use the following trigonometric identities
\begin{eqnarray*}
\sin (2^{m}x)\sin (nx) &=&\frac{1}{2}\left[ \cos \left( (2^{m}-n)x\right)
-\cos \left( (2^{m}+n)x\right) \right]  \\
\sin ^{2}(2^{m}x) &=&\frac{1-\cos \left( 2\cdot 2^{m}x\right) }{2},
\end{eqnarray*}
to obtain the well known orthogonal relation (or in Orthonormality/Fourier Series)
\begin{equation*}
J(m,n)=\int_{-\pi }^{\pi }\sin (2^{m}x)\sin (nx)\,dx=\left\{ 
\begin{array}{c}
\pi  \\ 
0
\end{array}
\begin{array}{c}
\text{if} \\ 
\text{if}
\end{array}
\begin{array}{c}
2^{m}=n \\ 
2^{m}\neq n
\end{array}
\right. 
\end{equation*}
Consequently, 
\begin{equation*}
I(m,n)=\left\{ 
\begin{array}{c}
-i\pi  \\ 
0
\end{array}
\begin{array}{c}
\text{if} \\ 
\text{if}
\end{array}
\begin{array}{c}
n=2^{m} \\ 
n\neq 2^{m}.
\end{array}
\right. 
\end{equation*}


*

*For $1\leq n\neq 2^{m}$ as well as for $n=2^{m}$, $m>10$, $c_{n}=0=
\overline{c}_{\left\vert n\right\vert }=\overline{c}_{-n}$. For $n=2^{m}$, $
1\leq m\leq 10,$ I've computed the following values :
\begin{eqnarray*}
c_{2} &=&\frac{1}{2\pi }(-1)^{1}I(1,2)=-\frac{1}{2\pi }\left( -i\pi \right) =
\frac{1}{2}i=\overline{c}_{-2}, \\
c_{4} &=&\frac{1}{2\pi }(-1)^{2}I(2,4)=\frac{1}{2\pi }\left( -i\pi \right) =-
\frac{1}{2}i=\overline{c}_{-4} \\
c_{8} &=&\frac{1}{2\pi }(-1)^{3}I(3,8)=-\frac{1}{2\pi }\left( -i\pi \right) =
\frac{1}{2}i=\overline{c}_{-8} \\
&&\cdots  \\
c_{2^{2k-1}} &=&\frac{1}{2}i=\overline{c}_{-2^{2k-1}},\quad c_{2^{2k}}=-
\frac{1}{2}i=\overline{c}_{-2^{2k}},\quad 1\leq k\leq 5.
\end{eqnarray*}

