Anyone have a clue what function f(x) (however weird it needs to be) will satisfy the following integral for any integer n>0:

$\displaystyle \int^{\pi}_{0} f(x)sin(nx)dx=(-1)^{n+1}$

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- Jul 4th 2006, 04:05 AMDMTFinding a function from Fourier coefficients
Anyone have a clue what function f(x) (however weird it needs to be) will satisfy the following integral for any integer n>0:

$\displaystyle \int^{\pi}_{0} f(x)sin(nx)dx=(-1)^{n+1}$ - Jul 4th 2006, 06:07 AMCaptainBlackQuote:

Originally Posted by**DMT**

anyway:

Consider the formal Fourier series:

$\displaystyle g(x)=k+\frac{1}{\pi}\sum_1^{\infty} \frac{(-1)^n}{n}\cos(nx)$

Now if I have done this right your $\displaystyle f$ is the formal derivative of $\displaystyle g$.

RonL - Jul 4th 2006, 06:43 AMmalaygoelQuote:

Originally Posted by**DMT**

$\displaystyle n - \frac{1}{\pi sin(nx)}$

Keep Smiling

Malay - Jul 4th 2006, 06:52 AMDMT
Well, the point was to find an f(x) that gave a specific Fourier series of which those were the coefficients. I did find it finally though, I think.

I'm pretty sure it's just a version of an alternating delta function, with peaks at pi and -pi and then made 2pi periodic. I think the peaks go to infinity with infinite terms, but I'm not sure. This became a lot clearer with the similar cosine series. - Jul 4th 2006, 07:23 AMCaptainBlackQuote:

Originally Posted by**DMT**

I'm not sure, .. how about the derivative of the delta function at pi.

(made periodic, factors of 2 or 1/2 inserted as needed etc etc ..)

RonL - Jul 4th 2006, 10:01 AMCaptainBlackQuote:

Originally Posted by**CaptainBlack**

the partial sums look like they may be heading for the derivative of

the delta function, its the term wise integrals that do not look like they

are heading for the delta or the unit step functions :confused:

However since the main property of the derivative of the delta is:

$\displaystyle

\int_{R} \delta'(x-a) f(x) dx = f'(a)

$

for any sufficiently well behaved function $\displaystyle f$ and with slight

abuse of notation.

It does look as though its what we need (if we slightly modify the

statment of the problem). On second thoughts no, it will introduce

an extra factor of 1/n where we don't want/need it :mad:

RonL - Jul 4th 2006, 03:28 PMDMT
Thansk for looking at this, but I have what I need now so you don't need to bother anymore.

- Jul 5th 2006, 02:50 AMCaptainBlackQuote:

Originally Posted by**DMT**

Anyway I think I know what it is (more or less) now, so I will now go

back to sleep.

RonL