Complex Fourier series has a singular term

Ragnord
Messages
4
Reaction score
0

Homework Statement



Find the complex Fourier series for f(t)=t(1-t), 0<t<1

Homework Equations



\sum_{n=-\infty}^{\infty}c_{n}e^{2in\pi t}

where c_{n}=\int_{0}^{1}f(t)e^{-2in\pi t}dt


The Attempt at a Solution



I've worked out that c_{n}=-1/(2n^2 \pi^2). The problem is that for n=0, it is singular. Is there some way around this or does it mean that the complex Fourier series doesn't exist?
I tried using maple to graph the series with the n=0 term omitted and it comes out to the right shape, but is shifted vertically down some, leading me to believe that the singular term should be replaced by a constant or something.
 
Physics news on Phys.org
c_0=\int_0^1 f(t)\,dt
 
Well that was easy, just like a real Fourier series. Thanks.
I'm interested in knowing why that's the case though, I haven't seen anything about doing anything special for c_{0} in anything I've seen about complex Fourier series.
 
Just plug in n=0 before integrating instead of after!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top