Anyone have a clue what function f(x) (however weird it needs to be) will satisfy the following integral for any integer n>0:
$\displaystyle \int^{\pi}_{0} f(x)sin(nx)dx=(-1)^{n+1}$
In retrospect I don't think this is going to be much use, but here it isOriginally 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
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.
Notes on numerical experiments: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
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
RonL