/sci/ - Science & Math

[Anonymous No.9720351]

How do I calculate this please?

[Anonymous No.9720357]

Expand the exponential using Euler's formula.

[Anonymous No.9720365]

>>9720351
interchange integral and sum sign, as long as the sum is finite that works

[Anonymous No.9720371]

First of all, write better so that we can understand what the fuck you're asking.

[Anonymous No.9720375]

>>9720351
>sum from n to N

[Anonymous No.9720378]

>>9720351
Switch the integral and sum, ALWAYS

[Anonymous No.9720411]

>>9720351
[eqn] e^{2 i \pi n x} = e^{2 i \pi n x} \frac{e^{2 i \pi x} - 1}{e^{2 i \pi x} - 1} \\
= \frac{e^{2 i \pi (n+1) x} - e^{2 i \pi n x}}{e^{2 i \pi x} - 1}
[/eqn]

[eqn] \sum_{n=-N}^N e^{2 i \pi n x} = \frac{e^{2 i \pi (N+1) x} - e^{-2 i \pi N x}}{e^{2 i \pi x} - 1}[/eqn]

[eqn]\int_{-\frac{1}{2}}^{\frac{1}{2}} \sum_{n=-N}^N e^{2 i \pi n x} dx = 1 [/eqn]

[Anonymous No.9721036]

>>9720351
The sum is a geometric series find the closed form and integrate easy peasy

[Anonymous No.9721067]

[math] \left| \int_{-1/2}^{1/2}\sum\limits_{n=-N}^Ne^{2\pi i n x} dx\right| \leq \int_{-1/2}^{1/2}\sum\limits_{n=-N}^N|e^{2\pi i n x}| dx \leq \int_{-1/2}^{1/2}\sum\limits_{n=-N}^N 1dx = 2N-1 [/math]

good enough

[Anonymous No.9721077]

[Anonymous No.9721083]

>>9721077
Slap FullSimplify[ ] on that shit

[Anonymous No.9721152]

>>9721083
Nothing happened.