Problem
Fourier Series
Solution
From the solution manual, they said:
Note that $f(1)=L=2$. Therefore: $$ f(1) = 2 = \frac{2}{2} + \frac{4}{\pi}\sum_{n=0}^{\infty} \frac{sin((2n+1)\frac{\pi}{2})}{2n+1}$$
Question
The solution seems very arbitrary- you just take a random guess that $f(1)=L=2$. How do you know that $f(1)=L=2$ and how do they know to set $n=0$ in the summation? Is there a more intuitive way to get the answer above?
$$ \mbox{Maybe, ( a "non Fourier" one )}\quad{\pi \over 4} = \int_{0}^{1}{\mathrm{d}x \over x^{2} + 1} = \sum_{n = 0}^{\infty}\left(\,-1\right)^{n}\int_{0}^{1}x^{2n}\,\mathrm{d}x = \sum_{n = 0}^{\infty}{\left(\,-1\right)^{n} \over 2n + 1} $$