# how to reduce $f(x)$ to Fourier series?

Let $$f(t)=e^{iwt}$$ on $$(-\pi,\pi).$$Expand $$f(t)$$ in a complex exponential Fourier series of period $$2\pi$$.($$w$$ is not an integer.)

Fourier series of the function in complex form is $$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos nx +b_n\sin nx$$

$$\implies f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\frac{e^{inx}+e^{-inx}}{2}+\sum_{n=1}^{\infty}b_n\frac{e^{inx}-e^{-inx}}{2i}$$

$$\implies \frac{a_0}{2}+\sum_{n=1}^{\infty}(\frac{a_n+ib_n}{2})e^{-inx}+\sum_{n=1}^{\infty}(\frac{a_n-ib_n}{2})e^{inx}$$

$$\implies f(x)=\sum_{-\infty}^{\infty}C_n e^{inx}$$,

Here,$$C_0=\frac{a_0}{2}$$,$$C_n=\frac{a_n-ib_n}{2},C_{-n}=\frac{a_n+ib_n}{2}$$

The coefficients $$C_n$$ are called Complex Fourier coefficients.They are defined by the formulas

$$C_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{inx}dx$$

$$C_0=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{iwt}dt=\frac{1}{\pi w}\sin(w\pi)$$

$$C_n=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{iwt}e^{-int}dt=\frac{1}{\pi (w-n)}\sin(w-n)\pi$$

I'm not getting how to reduce $$f(x)$$ to Fourier series.

I've invested a great amount of time on solving this...Please give any hint

• I don't understand the question. You wrote what the $C_n$-s are (according to you, at least: I did not check). – Gae. S. Nov 25 '19 at 10:07
• @Gae.S.:See the edit... – Styles Nov 25 '19 at 10:20

There is a small typo in your coefficient, it should be $$c_n = \frac{1}{2\pi}\int_{-\pi}^\pi \mathrm{e}^{\mathrm{i}wt}\mathrm{e}^{-\mathrm{i}nt}\mathrm{d}t$$ for all $$n\in\mathbb{Z}$$ (there is a minus sign in the argument of the second exponential since you defined the serie with $$\sum_n c_n\mathrm{e}^{+\mathrm{i}nt}$$).
Your expression is correct. It is useless to compute $$c_0$$ separately here : $$c_n = \frac{1}{(w-n)\pi}\sin (w-n)\pi$$ for all $$n\in\mathbb{Z}$$.
Introducing the (normalized) sinc function you can write : $$c_n = \operatorname{sinc}(w-n)$$ for all $$n\in\mathbb{Z}$$. Thus, the notation is more compact and you can now extend the result to the general case where $$w$$ can be an integer (indeed $$\operatorname{sinc}(0)=1$$).
Thus, $$f(t) = \sum_{n=-\infty}^\infty \operatorname{sinc}(w-n)\mathrm{e}^{+\mathrm{i}nt}$$ for all $$t\in(-\pi,\pi)$$.
Now, if we introduce $$g$$ the "periodised"-$$f$$ function (a $$2\pi$$ periodic function such that $$g(t) = f(t)$$ for all $$t\in(-\pi,\pi)$$), we can see that $$g$$ can jump if $$w$$ is not an integer. Then, according to Dirichlet theorem
$$\tilde{g}(t) = \sum_{n=-\infty}^\infty \operatorname{sinc}(w-n)\mathrm{e}^{+\mathrm{i}nt}$$ for all $$t\in\mathbb{R}$$ where $$g(t) = \frac{g(t^+)+g(t^-)}{2}$$.