Question
Show that if a real $f(x)$ is expanded in a complex exponential Fourier series $\sum_{-\infty}^{\infty} c_{n} e^{i n x},$ then $c_{-n}=\bar{c}_{n},$ where $\bar{c}_{n}$ means the complex conjugate of $c_{n}$.
Step 1
The coefficient $c_n$ is given by: \[ c_n = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) e^{-inx} \, dx \] Show more…
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Show that if a real $f(x)$ is expanded in a complex exponential Fourier series $\sum_{-\infty}^{\infty} c_{n} e^{i n x}$ then $\epsilon_{-n}=\bar{c}_{n}$, where $\bar{c}_{n}$ means the colnplex conjugate of $c_{n}$
FOURIER SERIES
Complex form of Fourier series
Complex Form of the Fourier Series. (a) Using the Euler formula $$e^{i \theta}=\cos \theta+i \sin \theta$$ $$\begin{array}{l}{i=\sqrt{-1, \text { prove that }}} \\ {\cos n x=\frac{e^{i n x}+e^{-i n x}}{2} \quad \text { and } \quad \sin n x=\frac{e^{i n x}-e^{-i n x}}{2 i}}\end{array}$$ (b) Show that the Fourier series $$\begin{aligned} f(x) & \sim \frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left\{a_{n} \cos n x+b_{n} \sin n x\right\} \\ &=c_{0}+\sum_{n=1}^{\infty}\left\{c_{n} e^{i n x}+c_{-n} e^{-i n x}\right\} \end{aligned}$$ where $$c_{0}=\frac{a_{0}}{2}, \quad c_{n}=\frac{a_{n}-i b_{n}}{2}, \quad c_{-n}=\frac{a_{n}+i b_{n}}{2}$$ (c) Finally, use the results of part (b) to show that $$f(x) \sim \sum_{n=-\infty}^{\infty} c_{n} e^{i n x}, \quad-\pi < x <\pi$$ where $$c_{n}=\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(x) e^{-i n x} d x$$
Partial Differential Equations
Fourier Series
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