Finding Fourier Coefficient for f(x) = 2δ(x-1) + δ(x-2)

[Zaare]

I'm somewhat confused about how to find this particular Fourier coefficient. Let me explain:

I have these two formulas from the book:

$$(1) \qquad f\left(x\right) = \frac{a_0}{2}+\sum_{n=1}^\infty\left[a_n cos\left(n \Omega x\right)+b_n sin\left(n \Omega x \right)\right], \quad \Omega = \frac{2 \pi}{T}$$

$$(2) \qquad b_n = \frac{2}{T} \int_{a}^{a+T} f(x) sin(n \Omega x) dx$$
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where $$T$$ is the period of the function $$f(x)$$.

And I'm supposed to find the Fourier coeffecient in the following:

$$(3) \qquad f(x) = 2 \delta (x-1) + \delta (x-2) = \sum_{n=1}^\infty b_n sin\left(\frac{n \pi x}{3} \right), \quad 0 \leq x \leq 3.$$
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Now, if I compare (3) and (1):

$$sin\left(\frac{n \pi x}{3} \right) = sin(n \Omega x) \Rightarrow \Omega = \frac{\pi}{3} \Rightarrow T = 6.$$
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According to (2), I'm supposed to use 0 and 6 as the limits for the integral. But (3) limits x to between 0 and 3.
So what limits am I supposed to use for the integral, and why?
I'd appreciate any help.

 

[LeonhardEuler]

From what I can tell, you are not trying to find the coefficients of the Forier series, but of the sin series. By that I mean you are not trying to find coefficients for the cos terms. Is that correct? If it is, then you must extend the function in an odd way because the sin function is odd. You can only have a sin series if f(-x)=-f(x) because this will be true of each of the terms in the series. So, what you have to do is extent the function to be $f(x)=2 \delta (x-1) + \delta (x-2) - 2 \delta (x+1) - \delta (x+2)$. What they mean by the bounds is that the series should represent the function on the interval for x between 0 and 3. Since you have not changed the value of the function on this interval, the series you get will represent the function on that interval. Then you extend it again to be periodic. It should repeat every 6 units. Then just compute the coefficients. Of course, this is all wrong if you were supposed to use cos terms too.

 

[lurflurf]

so you need
a<=0<=3<=a+6 (since T=6)
thus any a in [-3,0] will work
If f was not zero out side of [0,3] the coefficient would depend upon the choice of a (and thus the periodic extensions would differ) but here they do not.

 

[Zaare]

I think you're correct in your assumption, LeonhardEuler, since the problem is: "Fourier series: Find the coefficient". So I assume they only want the coefficient of the sin terms.
Thank you for the help.

And, lurflurf, thanks for the explanation about the limits.