Basic Trig Integral Question.


by NewtonianAlch
Tags: basic, integral, trig
NewtonianAlch
NewtonianAlch is offline
#1
Apr9-12, 02:07 AM
P: 440
1. The problem statement, all variables and given/known data
This question is part of Fourier Series in Circuit Analysis. There were fairly straightforward integrals which I calculated and confirmed using MAPLE to be correct, however the book gives somewhat different answers. I would presume that what I did was correct and the solutions manual made an error, however since it's a fairly large question with answers being carried forward I want to make doubly sure. Sorry about the size of the images, I will remove them after the problem

This is the integral essentially, the definite integral from 2 to 4 is left out because it's zero,

f(t) = 5 for 0 < t < 1
f(t) = 10 for 1 < t < 2



2. Relevant equations

cos (Pi/2) = (-1)[itex]^{\frac{n-1}{2}}[/itex]

cos (Pi) = (-1)[itex]^{n}[/itex]

3. The attempt at a solution

My answer came to this:



EDIT: cos(nPi/2) goes to (-1)^n/2 - still doesn't reconcile my answers with the book though.

The MAPLE output was:

[itex]5\,{\frac {1+\cos \left( 1/2\,n\pi \right) -2\,\cos \left( n\pi
\right) }{n\pi }}[/itex]

The answer in the book was (last line before the table):



As you can imagine, because the answers are different, the values in the table are going to be different and hence whatever I have to plot afterwards will be different.
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gneill
gneill is online now
#2
Apr9-12, 09:45 AM
Mentor
P: 11,408
##(-1)^{integer}## cannot produce any zeros, it can only produce +1 or -1. So it doesn't replace ##cos(n \pi / 2)##.
NewtonianAlch
NewtonianAlch is offline
#3
Apr9-12, 10:29 AM
P: 440
I should have been clearer, cos (n*Pi/2) is replaced by (-1)^n/2

So if n = 3, I'm guessing that term is ignored because you can't compute that. At least that's the identity they gave in the book.


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