First of all, I would like to apologise for my English. It is not my native language. I hope that you can understand everything.
I am currently having difficulties with Fourier series. Most of the problems I could already solve, but with one I am unsure and with another problem, I am unable to find the right approach. I would be very grateful if you could give me feedback on my solution for task 1 and hints on how to solve task 2.
I was given the following equations
\begin{align} \phi_{ n } &= \frac{ 1 }{ \sqrt{ L } } e^{ ik_{ n }x }, & k &= \frac{ 2 n \pi }{ L }, & \langle \phi_{ n } \lvert \phi_{ m } \rangle &= \int_0^L{ \phi_{ n }(x)^{ * } \phi_{ m }(x) dx = \delta_{ nm }} \end{align}
and asked to
- Show that the given basis is orthonormal.
Solution
For $m \neq n$
\begin{align} \int_0^L { \frac{ 1 }{ \sqrt{ L } } e^{ -ik_{ n }nx } \frac{ 1 }{ \sqrt{ L } } e^{ ik_{ n }mx } dx } &= \frac{ 1 }{ L } \int_0^L{ e^{ ik_{ n }(m-n)x } dx} \\ &= \frac{ 1 }{ L } \left[ \frac{ 1 }{ ik_{ n }(m-n) } e^{ ik_{ n }(m-n)x } \right]^{ L }_{ 0 } \end{align}
Next, using
$$e^{ ik_{ n }(m-n)x }=e^{ ik_{ n }ax }=e^{ in\frac{ 2\pi }{ L }aL }=e^{ ina2\pi }= \cos(2an\pi)+i \sin(2an\pi)=1$$
we have
$$\frac{ 1 }{ L } \left[ \frac{ 1 }{ ik_{ n }(m-n) } e^{ ik_{ n }(m-n)x } \right]^{ L }_{ 0 } = \frac{ 1 }{ L } \left( \frac{ 1 }{ ik_{ n }(m-n) } (1-e^{ 0 })\right)=0$$
For $m=n$
\begin{align} \int_0^L{ \frac{ 1 }{ \sqrt{ L } } e^{ -ik_{ n }nx } \frac{ 1 }{ \sqrt{ L } } e^{ ik_{ n }mx } dx } &= \frac{ 1 }{ L } \int_0^L{ e^{ ik_{ n }(m-n)x } dx} \\ &= \frac{ 1 }{ L }\int_0^L{ e^{ 0 }dx } \\ &= 1 \\ \end{align}
- Show that for any function $f(x)$ that can be represented as a linear combination of the basis functions $\phi_{ n }=\frac{ 1 }{ \sqrt{ L } } e^{ ik_{ n }x }$ i.e $$f(x) = \sum_{ n = -\infty }^{ \infty } { c_{ n } \phi_{ n } }$$ the complex Fourier coefficients $c_{ n }$ are given by $c_{ n } = \langle \phi_{ n } \lvert f \rangle$.
$$f(x) = \sum_{ n = -\infty }^{ \infty } { c_{ n } \phi_{ n } } | *\phi_{ m }^*$$ $$f(x)*\phi_{ m }^* = \sum_{ n = -\infty }^{ \infty } { c_{ n } \phi_{ n } }\phi_{ m }^*$$ $$<\phi_{ m } | f>=\int_0^L{ \sum_{ n = -\infty }^{ \infty } { c_{ n } \phi_{ n } }\phi_{ m }^* dx } \\ = \sum_{ n = -\infty }^{ \infty } { c_{ n }} \int_0^L{ \frac{ 1 }{ \sqrt{ L } } e^{ -ik_{ n }mx }* \frac{ 1 }{ \sqrt{ L } } e^{ ik_{ n }nx } dx}\\ =\sum_{ n = -\infty }^{ \infty } { c_{ n }} * \frac{ 1 }{ L } \int_0^L{ e^{ ik_{ n }(n-m)x } dx }$$
From task 1 I know that for $m \neq n$ the whole term will equal 0. For $m=n$ the terms on the right hand side of $c_{ n }$ equal 1, leading to $<\phi_{ m } | f>=\sum_{ n = -\infty }^{ \infty } { c_{ n }}$.
Thank you in advance and kind regards from Germany