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

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SUMMARY

The discussion centers on evaluating the integral of the function u(t) = 2 - cos(t) + sin(2t) - cos(3t) + sin(4t) over the interval from 0 to 2π. Participants suggest using Parseval's theorem and orthogonality relations to simplify the calculation. The final result of the integral is confirmed to be 12π, with emphasis on the fact that the Fourier coefficients are directly available from the function u(t) without needing extensive calculations.

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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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