Let
![{\displaystyle \displaystyle {}e_{n}(x)=e^{2\pi {}inx}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/2d23dbb598cc6018ba9953ea6417765d58cb87c3)
so that we want
such that
![{\displaystyle x=\sum _{i=-\infty }^{\infty }{}c_{i}e_{i}(x).}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/469a5a8d92cf25cefdd245fa65fcabd72fee4ff4)
From Hint 1 (and setting L=1) we have that
are orthogonal i.e.,
![{\displaystyle \int _{0}^{1}e_{n}(x){\overline {e_{m}}}(x){\textrm {d}}x={\begin{cases}1,\qquad {}n=m\\0,\qquad {}n\neq {}m\end{cases}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/05412adc1f89c5ff05ec6e7695d41dec2a7c7fe5)
where the overbar indicates the complex conjugate
![{\displaystyle \displaystyle {}{\overline {e_{n}}}(x)=e^{-2\pi {}inx}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f5183a621ee01842087a13298a200ed6234f3806)
We can get the coefficients of
using this orthogonality. Take the inner product with each
![{\displaystyle \langle e_{n},x\rangle =\sum _{i=-\infty }^{\infty }c_{i}\langle e_{n},e_{i}\rangle =c_{n}\langle e_{n},e_{n}\rangle =c_{n}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/36248c1f60f3188d403bcc3d4be49b502c27f837)
.
If
we use integration by parts:
with
,
to obtain
![{\displaystyle {\begin{aligned}c_{n}&=\left[{\frac {x}{-2\pi in}}e^{-2\pi inx}\right]_{0}^{1}+\int _{0}^{1}{\frac {1}{2\pi in}}e^{-2\pi inx}dx\\&=\left[-{\frac {x}{2\pi in}}e^{-2\pi inx}-{\frac {1}{4\pi ^{2}i^{2}n^{2}}}e^{-2\pi inx}\right]_{0}^{1}\\&=\left[-{\frac {x}{2\pi in}}e^{-2\pi inx}+{\frac {1}{4\pi ^{2}n^{2}}}e^{-2\pi inx}\right]_{0}^{1}\\&=\left[\left({\frac {1}{4\pi ^{2}n^{2}}}-{\frac {x}{2\pi in}}\right)e^{-2\pi inx}\right]_{0}^{1}\\&=\left[\left({\frac {1}{4\pi ^{2}n^{2}}}-{\frac {1}{2\pi in}}\right)e^{-2\pi in(1)}\right]-\left[\left({\frac {1}{4\pi ^{2}n^{2}}}-0\right)e^{-2\pi in(0)}\right]\\&=\left({\frac {1}{4\pi ^{2}n^{2}}}-{\frac {1}{2\pi in}}\right)e^{-2\pi in}-{\frac {1}{4\pi ^{2}n^{2}}}\\&=\left({\frac {1}{4\pi ^{2}n^{2}}}-{\frac {1}{2\pi in}}\right)(1)-{\frac {1}{4\pi ^{2}n^{2}}}\\&=-{\frac {1}{2\pi in}}\\&={\frac {i}{2\pi {}n}}\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f47a14a29785b6dfdda2214b8d80efca8d50b2de)
Here we see the importance of having
, otherwise the coefficient diverges using this formula. If n = 0,
![{\displaystyle c_{0}=\int _{0}^{1}f(x)dx=\left[{\frac {1}{2}}x^{2}\right]_{0}^{1}={\frac {1}{2}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/71991cf2631300fc1f1200909e45bf03af24adb0)
Therefore, we have all the coefficients for the series.