Science:Math Exam Resources/Courses/MATH307/December 2010/Question 06 (a)/Solution 1

Let

$\displaystyle {}e_{n}(x)=e^{2\pi {}inx}$

so that we want $c_{i}$ such that

$x=\sum _{i=-\infty }^{\infty }{}c_{i}e_{i}(x).$

From Hint 1 (and setting L=1) we have that $e_{n}$ are orthogonal i.e.,

$\int _{0}^{1}e_{n}(x){\overline {e_{m}}}(x){\textrm {d}}x={\begin{cases}1,\qquad {}n=m\\0,\qquad {}n\neq {}m\end{cases}}$

where the overbar indicates the complex conjugate

$\displaystyle {}{\overline {e_{n}}}(x)=e^{-2\pi {}inx}.$

We can get the coefficients of $x$ using this orthogonality. Take the inner product with each $e_{n}(x)$

$\langle e_{n},x\rangle =\sum _{i=-\infty }^{\infty }c_{i}\langle e_{n},e_{i}\rangle =c_{n}\langle e_{n},e_{n}\rangle =c_{n}$

$c_{n}=\int _{0}^{1}e^{-2\pi inx}xdx$.

If $n\neq 0,$ we use integration by parts: $\int udv=uv-\int vdu$ with $u=x$, $dv=e^{-2\pi inx}dx$ to obtain

${\begin{aligned}c_{n}&=\left[{\frac {x}{-2\pi in}}e^{-2\pi inx}\right]_{0}^{1}+\int _{0}^{1}{\frac {1}{2\pi in}}e^{-2\pi inx}dx\\&=\left[-{\frac {x}{2\pi in}}e^{-2\pi inx}-{\frac {1}{4\pi ^{2}i^{2}n^{2}}}e^{-2\pi inx}\right]_{0}^{1}\\&=\left[-{\frac {x}{2\pi in}}e^{-2\pi inx}+{\frac {1}{4\pi ^{2}n^{2}}}e^{-2\pi inx}\right]_{0}^{1}\\&=\left[\left({\frac {1}{4\pi ^{2}n^{2}}}-{\frac {x}{2\pi in}}\right)e^{-2\pi inx}\right]_{0}^{1}\\&=\left[\left({\frac {1}{4\pi ^{2}n^{2}}}-{\frac {1}{2\pi in}}\right)e^{-2\pi in(1)}\right]-\left[\left({\frac {1}{4\pi ^{2}n^{2}}}-0\right)e^{-2\pi in(0)}\right]\\&=\left({\frac {1}{4\pi ^{2}n^{2}}}-{\frac {1}{2\pi in}}\right)e^{-2\pi in}-{\frac {1}{4\pi ^{2}n^{2}}}\\&=\left({\frac {1}{4\pi ^{2}n^{2}}}-{\frac {1}{2\pi in}}\right)(1)-{\frac {1}{4\pi ^{2}n^{2}}}\\&=-{\frac {1}{2\pi in}}\\&={\frac {i}{2\pi {}n}}\end{aligned}}$

Here we see the importance of having $n\neq 0$, otherwise the coefficient diverges using this formula. If n = 0,

$c_{0}=\int _{0}^{1}f(x)dx=\left[{\frac {1}{2}}x^{2}\right]_{0}^{1}={\frac {1}{2}}$

Therefore, we have all the coefficients for the series.