Show that Dirichlet's theorem can be written with one term

Question

I'm having trouble with a question in my Introduction to Quantum Mechanics book.

My attempt:

$$f(x) = \Sigma_{n=0}^\infty a_n \sin(\frac{in \pi x}{a}) + b_n \cos(\frac{in \pi x}{a})$$ $$=b_0 + \Sigma_{n=1}^\infty a_n(\frac{e^{\frac{in\pi x}{a}} - e^{\frac{-in \pi x}{a}}}{2i})+b_n(\frac{e^{\frac{in\pi x}{a}}+e^{\frac{-in \pi x}{a}}}{2})$$ $$=b_0 + \Sigma_{n=1}^\infty e^{\frac{in \pi x}{a}}(\frac{-ia_n +b_n}{2})+e^{\frac{-in \pi x}{a}}(\frac{b_n+ia_n}{2})$$

...but this is where I get stuck. Where should I go from here?

Answer 1

It's a bit easier if you do this starting from the latter: \begin{align*} \sum^\infty_{n=-\infty} c_n e^{in\pi x /a} &= c_0 + \sum^\infty_{n=1} c_n e^{in\pi x /a} + \sum^{-1}_{n=-\infty} c_n e^{in\pi x /a} \\ &= c_0 +\sum^\infty_{n=1} c_n e^{in\pi x /a} + \sum^\infty_{n=1} c_{-n} e^{-in\pi x /a} \\ &= c_0 + \sum^\infty_{n=1} c_n(\cos(n\pi x/a)) + i \sin(n\pi x/ a) + c_{-n}(\cos(-n\pi x/a) + i \sin(-n\pi x/ a)) \\ &=c_0 +\sum^\infty_{n=1} (c_n + c_{-n})\cos(n\pi x /a) + i(c_n - c_{-n}) \sin(n\pi x /a). \end{align*} This shows that the representations $$\sum_{n=0}^\infty a_n \cos(n\pi x /a) + b_n \sin(n\pi x /a) \,\,\,\,\,\,\,\, \text{ and } \,\,\,\,\,\,\,\,\sum^\infty_{n=-\infty} c_n e^{in\pi x /a}$$ are equivalent with the relationships $$a_0 = c_0, \,\,\,\,\,\,\, a_n = c_n + c_{-n}, \,\,\,\,\,\,\, b_{n} = i(c_n - c_{-n}).$$ Given the former representation, you can uniquely solve for $c_n$ in terms of $a_n$ and $b_n$.

Answer 2

Observe \begin{align} a_n \sin(n\pi x/a) + b_n \cos(n\pi x/a) =&\ \sqrt{a_n^2+b_n^2}\left( \frac{a_n}{\sqrt{a_n^2+b_n^2}} \sin(n\pi x/a) + \frac{b_n}{\sqrt{a_n^2+b_n^2}} \cos(n\pi x/a) \right)\\ =&\ \sqrt{a_n^2+b_n^2} \sin\left(\frac{n \pi x}{a}+\theta \right) \end{align} where $\theta$ has the properties \begin{align} \cos\theta = \frac{a_n}{\sqrt{a_n^2+b_n^2}} \ \ \text{ and } \ \ \sin\theta = \frac{b_n}{\sqrt{a_n^2+b_n^2}}. \end{align}

Now, it's easy to see \begin{align} \alpha \sin\left(\frac{n \pi x}{a}+\theta \right) =&\ \alpha\left( \frac{\exp(i\frac{n \pi x}{a}+i\theta)-\exp(-i\frac{n \pi x}{a}-i\theta)}{2i} \right)\\ =&\ \frac{\alpha e^{i\theta}}{2i}\exp\left(i\frac{n\pi x}{a}\right)- \frac{\alpha e^{-i\theta}}{2i} \exp\left(-i\frac{n\pi x}{a}\right). \end{align}