Integrating u(t)^2: A Shortcut to Finding Fourier Coefficients?

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Homework Help Overview

The problem involves evaluating the integral of the square of a function defined as u(t) = 2 - cos(t) + sin(2t) - cos(3t) + sin(4t) over the interval from 0 to 2π. The context relates to Fourier coefficients and potentially using Parseval's theorem.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss different methods for evaluating the integral, including direct integration and the application of Parseval's theorem. Some question the approach of squaring each term versus the whole expression.

Discussion Status

There are multiple lines of reasoning being explored, with some participants suggesting direct integration while others mention the use of Fourier series. Hints and guidance have been provided, but no consensus has been reached on a specific method or solution.

Contextual Notes

Participants note the importance of trigonometric identities and orthogonality relations in evaluating the integrals involved. There is also mention of the average values of sine and cosine functions in the context of Fourier analysis.

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Homework Statement



Let u(t)=2-\cos(t)+\sin(2t)- \cos(3t)+ \sin(4t)

Evaluate:

\int_0^{2\pi}u^2(t)\mbox{d}t

Homework Equations





The Attempt at a Solution



Sorry, I don't have any idea :(... As I can see

\int_0^{2\pi}u^2(t)\mbox{d}t

is similar to the first term of Fourier Serie of the function u^2(t)...
 
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You might consider using Parseval's theorem.
 
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I don't see any reason to think about "Fourier Series" at all- just integrate directly:
\int_0^{2\pi} 4+ cos^2(t)+ sin^2(2t)+ cos^2(3t)+ sin^2(4t) dt
is easy.
2\int_0^{2\pi}-2cos(t)+ 2sin(2t)- 2cos(3t)+ 2sin(4t)dt
2\int_0^{2\pi}- cos(t)sin(2t)- cos(t)cos(3t)+ cos(t)sin(4t)dt
2\int_0^{2\pi}-sin(2t)cos(3t)+ sin(2t)sin(4t)- cos(3t)sin(4t)dt

You may need the trig identities for cos(a)cos(b), sin(a)sin(b), and sin(a)cos(b).
 
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HallsOfIvy,

I think you just squared each term, rather than squaring the whole expression?
 
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MisterX said:
You might consider using Parseval's theorem.

..along with the fact that ##u(t)## is its own Fourier Series.
 
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Ok, thanks for your hints!

So

\int_{0}^{2\pi}u^2(t)\mbox{d}t= 12\pi

right? :)
 
Mathitalian said:
Ok, thanks for your hints!

So

\int_{0}^{2\pi}u^2(t)\mbox{d}t= 12\pi

right? :)

Right. But if you think about orthogonality relations, you only have to evaluate the first integral Halls gave in post 3. The rest are obviously 0. And that one doesn't take much calculation either if you remember that the 'average value' of sin(nt)^2 or cos(nt)^2 where n>1 is 1/2.
 
Last edited:
Dick said:
Right. But if you think about orthogonality relations, you only have to evaluate the first integral Halls gave in post 3. The rest are obviously 0. And that one doesn't take much calculation either if you remember that the 'average value' of sin(nt)^2 or cos(nt)^2 where n>1 is 1/2.

Actually, you don't have to evaluate any integrals. The Fourier coefficients are sitting right there in the function u(t).
 
LCKurtz said:
Actually, you don't have to evaluate any integrals. The Fourier coefficients are sitting right there in the function u(t).

Sure, if you use Parseval's. I wasn't sure which track the OP was following.
 

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