[University] Calculus/Fourier Series
Good morning,
I'm going over my fourier series class notes and can't understand where the 6/nPi comes from in step 3 of the a^n calculation comes from.
Can someone point me towards how to calculate this and what thought process I should be using to deal with this in the future?
Thanks.
Posted Nov 21, 2017, 4:19 AM UTC
25
It's a u-substitution.
2cos(nπt/3) = 2(3nπ)/(3nπ) cos(nπt/3)
= 6/(nπ) cos(nπt/3) (nπ/3).
= 6/(nπ) cos(u) (du/dt).
Then
∫2cos(nπt/3) dt = ∫6/(nπ) cos(u) du/dt dt
= 6/(nπ) ∫cos(u) du
= 6/(nπ) sin(u)
= 6/(nπ) sin(nπt/3)
So if we just take the cos(npi/3) it goes:
Cos(u) --> i/u sin(u)
Which I understand but i'm struggling with getting from nπ/3 to 3/nπ
Actually I think I understand now:
u = nπt/3 du = nπ/3
this gets pulled out as the reciprocal to give 3/nπ ∫cos(u)du
To give 3/nπ sin(u)
Substitute back u back in: 3/nπ sin(nπt/3)
Which times the two gives: 6/(nπ) sin(nπt/3)
1 / (nπ/3) = 3/(nπ)