2
$\begingroup$

This is one of my homework problems:

Let $a,b \in \mathbb{R}$ and $a>0$, show that: $$\sum_{m=0}^\infty \exp(-ma) \cos(mb)= \frac{1-\exp(-a)\cos b}{1- 2 \exp(-a) \cos b + \exp(-2a)} $$

I believe to have an attempt on how to solve this, but most likely diverge during the process.

My approach:

Note that $ \exp(imb) = \cos (mb) + i \sin (mb) $ such that $\cos(mb)=\Re( \exp (imb))$ where $\Re$ represents the real part of the expression. This would lead to that

$$ \exp(-ma)\cos(mb)=\exp(-ma) \cdot \Re ( \exp(imb))=\Re(\exp(-ma+imb)=\Re ( e^{-ma+imb})$$ my idea was to use this method to get it into the form of a geometric Series. $$\Re(e^{ma+imb})=\Re\left(\frac{1}{e^{ma-imb}} \right)=\Re \left(\left(\frac{1}{e^{a-ib}} \right)^m \right)$$ If that is all correct I could make use of the geometric Series $$\Re \left(\sum_{m=0}^\infty \left(\frac{1}{e^{-a+ib}} \right)^m \right)=\Re \left( \frac{e^{a-ib}}{e^{a-ib}-1}\right)=\Re \left(\frac{e^ae^{-ib}}{e^ae^{-ib}-1} \right)= \Re \left( \frac{e^a(\cos b - i \sin (b))}{e^a ( \cos b - i \sin (b))-1} \right)\\=\underbrace{\frac{e^a\cos b}{e^a \cos b-1}}_!$$ Which at least looks to a small degree like the answer, i.e. the $e^a \cos b$ part, however for the rest it's not even near, would someone please highlight my mistakes?

Mistakes:

$ \Re(z_1z_2) \neq \Re (z_1) \cdot \Re (z_2)$

$\endgroup$
2
  • $\begingroup$ I think the problem is in your last equality, $\Re(z/w)\neq \Re(z)/\Re(w)$. $\endgroup$
    – Ian Mateus
    Nov 16, 2013 at 17:27
  • $\begingroup$ There's a minus sign missing after my idea was to use this method to get it into the form of a geometric Series., in the exponent of the LHS of the chain of equalities. But you fixed it along the way. Also the geometric series only works under certain conditions which aren't necessarily met. Finally how do you get the very last equality? My first try would you be to note that $\cos (mb)=\dfrac{e^{imb}+e^{-imb}}{2}$. $\endgroup$
    – Git Gud
    Nov 16, 2013 at 17:28

2 Answers 2

3
$\begingroup$

$$\sum_{m=0}^{\infty} \exp(-ma)\cos(mb)=\Re\left(\sum_{m=0}^{\infty} \exp(-ma)\exp(imb)\right)\\=\Re\left(\sum_{m=0}^{\infty}\exp(imb-ma)\right)=\Re\left(\sum_{m=0}^{\infty}\exp(m(ib-a))\right)\\=\Re\left(\frac{1}{1-\exp(ib-a)}\right)$$

Simplfying, you will get the desired answer.

$\endgroup$
3
  • $\begingroup$ Note that I obtained this 'answer' already (after if that is all correct..., the 3rd equation from the left) just missing a proper division. It seems to be that I don't quite understand how to split things up now with the Real Operator. However I will look into this now :) $\endgroup$
    – Spaced
    Nov 16, 2013 at 17:43
  • $\begingroup$ Just manipulate the expression into the form $a+ib$ before taking the real part. @Spaced $\endgroup$ Nov 16, 2013 at 17:55
  • $\begingroup$ Thanks a lot for that advice @freak_warrior, I think I might be able to solve it with this. $\endgroup$
    – Spaced
    Nov 16, 2013 at 17:59
1
$\begingroup$

$$\text{Real part of}(z_1z_2) \neq \text{Real part of}z_1 \cdot \text{Real part of}z_2$$ $$\text{Real part of}\left(\dfrac{z_1}{z_2} \right) \neq \dfrac{\text{Real part of}z_1}{\text{Real part of}z_2}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .