# questions on fourier series. answers are provided. but I dont understand the process

• November 2nd 2008, 06:51 AM
panda
questions on fourier series. answers are provided. but I dont understand the process
In the first picture, there are two circles i've indicated. My question is how to get the second one from the first circle. I can see that n changed from n≠ 0 to n=1. And I also don't understand why I have to change these n values.
Moreover, in the second picture, I wasn't sure why that circled one has disappeared in the next equation.

• November 2nd 2008, 07:04 AM
Peritus
$
\sum\limits_{n \ne 0} {\frac{{\left( { - 1} \right)^{n + 1} e^{inx} }}
{{in}}} = \sum\limits_{ - \infty }^{ - 1} {\frac{{\left( { - 1} \right)^{n + 1} e^{inx} }}
{{in}} + } \sum\limits_1^\infty {\frac{{\left( { - 1} \right)^{n + 1} e^{inx} }}
{{in}}} =$

$
= - \sum\limits_1^\infty {\frac{{\left( { - 1} \right)^{n + 1} e^{ - inx} }}
{{in}} + } \sum\limits_1^\infty {\frac{{\left( { - 1} \right)^{n + 1} e^{inx} }}
{{in}}} = \sum\limits_1^\infty {\frac{{\left( { - 1} \right)^{n + 1} \left( {e^{inx} - e^{ - inx} } \right)}}
{{in}}}$

regarding the second exercise:
$
e^{ - i2\pi n}$
equals one for any integral n:

$
\int\limits_0^1 {\frac{{e^{ - i2\pi nx} }}
{{i2\pi n}}} = \left. {\frac{{e^{ - i2\pi nx} }}
{{4\pi ^2 n^2 }}} \right|_0^1 = \frac{{e^{ - i2\pi n} - 1}}
{{4\pi ^2 n^2 }} = 0$