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

[Mathitalian]

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)...

 

[MisterX]

You might consider using Parseval's theorem.

 

[HallsofIvy]

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).

 

[stevenb]

HallsOfIvy,

I think you just squared each term, rather than squaring the whole expression?

 

[LCKurtz]

You might consider using Parseval's theorem.

..along with the fact that $u(t)$ is its own Fourier Series.

 

[Mathitalian]

Ok, thanks for your hints!

So

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

right? :)

 

[Dick]

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.

 

[LCKurtz]

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).

 

[Dick]

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.