Hello Folks,
I am trying to find out if there is a closed form solution to the following infinite summation. Could anyone give me a few pointers about how to proceed?
S = sum(cos((n+1)*x)/n/(n+1)) where n = 1 to infinity.
Thanks,
- VL.
Hello Folks,
I am trying to find out if there is a closed form solution to the following infinite summation. Could anyone give me a few pointers about how to proceed?
S = sum(cos((n+1)*x)/n/(n+1)) where n = 1 to infinity.
Thanks,
- VL.
Hi,
I've no clue whether my approach leads to a correct result, but I'd first try with some very dirty tricks (beware!!):
$\displaystyle f(x) = \sum_n \frac{\cos((n+1)x)}{n(n+1)} = \mathfrak{Re}\left{ \sum_n \frac{e^{j(n+1)x}}{n(n+1)} \right}$
Differentiate the complex series with respect to $\displaystyle x$ and you get
$\displaystyle \frac{\partial f}{\partial x} = j e^{jx} \sum_n \frac{(e^{jx})^n}{n}$
Now I'll just take the identity $\displaystyle \ln(1-x) = - \sum_n \frac{x^n}{n}$ valid for $\displaystyle -1 \leq x < 1$ and hope it converges for most complex numbers on the border of the closed disk $\displaystyle |x| \leq 1$ (needs checking!). I get
$\displaystyle \frac{\partial f}{\partial x} = - j e^{jx} \ln(1-e^{jx})$
Now what's the logarithm of a complex number? For a complex number $\displaystyle z = a+jb$, I have
$\displaystyle e^{\ln(a+jb)} = a+jb = \sqrt{a^2+b^2} e^{j \arctan(\tfrac{b}{a})} = e^{\ln(\sqrt{a^2+b^2}) + j \arctan(\tfrac{b}{a})}$
whence $\displaystyle \ln(a+jb) = \ln(\sqrt{a^2+b^2}) + j \arctan(\tfrac{b}{a})$. Hence
$\displaystyle \frac{\partial f}{\partial x} = - j e^{jx} \left[ \ln\left(\sqrt{(1-\cos(x))^2+\sin^2(x)}\right) + j \arctan\left(\frac{\sin(x)}{1-\cos(x)}\right) \right]$
$\displaystyle \frac{\partial f}{\partial x} = - j e^{jx} \left[ \ln\left(\sqrt{2(1-\cos(x))}\right) + j \arctan\left(\frac{\sin(x)}{1-\cos(x)}\right) \right]$
You can work out the rest. Take the real part, and try to find the antiderivative. Check by numerical simulation, before trying to rigorize all the above.
I'd also be interested in a rigorous (and maybe more elegant?) derivation. Thanks to anybody who can provide it!
Jens
Hi,
Ok, let me know.
I've just noticed that I defined my function $\displaystyle f$ as the real part of a complex series, but then I went on as if it were the complex series itself. Just a notational mistake!
I hope that didn't confuse anybody…