Fourier Series Limit proof

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May 28th 2009, 09:51 AM
TiRune

Fourier Series Limit proof

I'm working on a problem at the moment and I just can't figure it out, here goes:

$\lim_{n \rightarrow \infty } \frac{1}{\pi} \int_{-l}^l f(x) \frac{sin(nx)}{x} dx = f(0)$

with l > 0 and f(x) differentiable on the interval [-l,l]

Things that come to mind are Riemann-Lebesgue Lemma -- from Wolfram MathWorld Riemann's lemma which states that any limit of n to infinity of an integral of a function on -l to l of the function times sin(nx) or cos(nx) tends to 0.
And ofcourse that $\int_0^{\infty}\frac{sin(x)}{x}dx = \frac{\pi}{2}$

Does anyone have any ideas on how I can proceed?
May 28th 2009, 12:23 PM
Media_Man

Clarity?

Substitute $y=\frac{L}{\pi}x$ , so $dx=\frac{\pi}{L}dy$ , and sub $g(x)=\frac{f(x)}{x}$

Then $ \lim_{n \rightarrow \infty } \frac{1}{\pi} \int_{-l}^l f(x) \frac{sin(nx)}{x} dx $ becomes $ \frac{L}{\pi^2} \lim_{n \rightarrow \infty }\int_{-\pi}^\pi g(y)\sin(ny) dy $

So by the Riemann-Lebesgue Lemma, the LHS is zero, so $f(0)=0$

*What exactly is it you have to prove? Are you trying to find a function $f(x)$ that satisfies the equation, or are you supposed to prove the equality holds for all $f(x)$ ? (Thinking)
May 28th 2009, 02:49 PM
TiRune

We have to prove that the limit over that integral is equal to f(0) (not zero, but the value f(0) for any f differentiable above -l,l).

Although in the case x is not 0 this proves that the LHS goes to 0, but not that it goes to f(0) :(
May 28th 2009, 03:13 PM
Sampras

Quote:

Originally Posted by Media_Man

Substitute $y=\frac{L}{\pi}x$ , so $dx=\frac{\pi}{L}dy$ , and sub $g(x)=\frac{f(x)}{x}$

Then $ \lim_{n \rightarrow \infty } \frac{1}{\pi} \int_{-l}^l f(x) \frac{sin(nx)}{x} dx $ becomes $ \frac{L}{\pi^2} \lim_{n \rightarrow \infty }\int_{-\pi}^\pi g(y)\sin(ny) dy $

So by the Riemann-Lebesgue Lemma, the LHS is zero, so $f(0)=0$

*What exactly is it you have to prove? Are you trying to find a function $f(x)$ that satisfies the equation, or are you supposed to prove the equality holds for all $f(x)$ ? (Thinking)

So you basically "transformed" the LHS so that you could use the Riemann-Lebesgue Lemma showing that it is $0$ . But this does not mean its $f(0)$ .
May 28th 2009, 03:45 PM
Sampras

Quote:

Originally Posted by Media_Man

Substitute $y=\frac{L}{\pi}x$ , so $dx=\frac{\pi}{L}dy$ , and sub $g(x)=\frac{f(x)}{x}$

Then $ \lim_{n \rightarrow \infty } \frac{1}{\pi} \int_{-l}^l f(x) \frac{sin(nx)}{x} dx $ becomes $ \frac{L}{\pi^2} \lim_{n \rightarrow \infty }\int_{-\pi}^\pi g(y)\sin(ny) dy $

So by the Riemann-Lebesgue Lemma, the LHS is zero, so $f(0)=0$

*What exactly is it you have to prove? Are you trying to find a function $f(x)$ that satisfies the equation, or are you supposed to prove the equality holds for all $f(x)$ ? (Thinking)

Shouldn't it be $\frac{L}{\pi^2} \lim_{n \to \infty} \int_{-L^2/\pi}^{L^2/\pi} g(y) \sin \left (\frac{n \pi y}{L} \right) \ dy$ ?
May 28th 2009, 05:04 PM
Media_Man

Blooper, and yes, f(0)=0

Quote:

Originally Posted by Media_Man

Substitute $y=\frac{L}{\pi}x$ , so $dx=\frac{\pi}{L}dy$ , and sub $g(x)=\frac{f(x)}{x}$

This was a typo. Sub $y=\frac{\pi}{L}x$ and $dx=\frac{\pi}{L}dy$ , in order to change the limits of integration from $[-L,L]$ to $[-\pi,\pi]$ in accordance with the R-L lemma. Thanks for catching that blooper, Sampras. (Itwasntme)

Quote:

But this does not mean its $f(0)=0$ .

Yes it does. Completing the RHS of the substitution, $ \frac{L}{\pi^2} \lim_{n \rightarrow \infty }\int_{-\pi}^\pi g(y)\sin(ny) dy=f(0) $

Regardless of what f is, it needs to sit on the origin. Here is a proof: If $\int_{-a}^a h(x)dx=0$ for all $a\in\mathbb{R}$ , then h must be an even function, that is, symmetric about the y-axis. So, we can think of our integral in question as "tending towards" an even function as n grows large. Obviously, there does not have to be (probably won't be) an n large enough that $h(-x)=h(x)$ but $\lim_{n\rightarrow\infty} h(x)-h(-x)$ does need to be zero.

Consider this counterexample: $f(x)=x+1$ , so $f(0)\neq0$ Theorem: $\lim_{n \rightarrow \infty } \frac{1}{\pi} \int_{-l}^l \frac{x+1}x sin(nx) dx\neq0$

Proof: $\lim_{n\rightarrow\infty} \frac{x+1}x sin(nx) - \frac{-x+1}{-x} sin(-nx)$ = $\lim_{n\rightarrow\infty} \frac{\sin(nx)}x [(x+1)-(1-x)]$ = $\lim_{n\rightarrow\infty} 2\sin(nx)\neq 0$

*I'm not sure how rigorous this is, but because the $\frac{x+1}x sin(nx)$ is not tending to an even function as n gets large, the area bounded under it on either side of the y-axis will always be inequal for some choice of L.
May 28th 2009, 05:07 PM
Sampras

Quote:

Originally Posted by Media_Man

This was a typo. Sub $y=\frac{\pi}{L}x$ and $dx=\frac{\pi}{L}dy$ , in order to change the limits of integration from $[-L,L]$ to $[-\pi,\pi]$ in accordance with the R-L lemma. Thanks for catching that blooper, Sampras. (Itwasntme)

Yes it does. Completing the RHS of the substitution, $ \frac{L}{\pi^2} \lim_{n \rightarrow \infty }\int_{-\pi}^\pi g(y)\sin(ny) dy=f(0) $

Regardless of what f is, it needs to sit on the origin. Here is a proof: If $\int_{-a}^a h(x)dx=0$ for all $a\in\mathbb{R}$ , then h must be an even function, that is, symmetric about the y-axis. So, we can think of our integral in question as "tending towards" an even function as n grows large. Obviously, there does not have to be (probably won't be) an n large enough that $h(-x)=h(x)$ but $\lim_{n\rightarrow\infty} h(x)-h(-x)$ does need to be zero.

Consider this counterexample: $f(x)=x+1$ , so $f(0)\neq0$ Theorem: $\lim_{n \rightarrow \infty } \frac{1}{\pi} \int_{-l}^l \frac{x+1}x sin(nx) dx\neq0$

Proof: $\lim_{n\rightarrow\infty} \frac{x+1}x sin(nx) - \frac{-x+1}{-x} sin(-nx)$ = $\lim_{n\rightarrow\infty} \frac{\sin(nx)}x [(x+1)-(1-x)]$ = $\lim_{n\rightarrow\infty} 2\sin(nx)\neq 0$

*I'm not sure how rigorous this is, but because the $\frac{x+1}x sin(nx)$ is not tending to an even function as n gets large, the area bounded under it on either side of the y-axis will always be inequal for some choice of L.

$dx = \frac{L}{\pi} dy$ .
May 28th 2009, 05:12 PM
Media_Man

Gahh

(Headbang)
May 28th 2009, 06:38 PM
TiRune

I totally get it now, thanks a lot both of you! XD