Find complex Fourier coefficients

Question

1. 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.

Answer 1

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*}

Answer 2

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 $f$.