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

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The discussion revolves around evaluating the integral of u(t)^2, where u(t) is a combination of trigonometric functions. Participants suggest using Parseval's theorem and the orthogonality of sine and cosine functions to simplify the calculation. It is noted that the integral can be computed directly, leading to the result of 12π. The conversation emphasizes that the Fourier coefficients can be derived directly from u(t) without needing extensive calculations. Overall, the integration process and the application of Fourier analysis techniques are central to the discussion.
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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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