# Solving series with complex exponentials.

1. Dec 18, 2013

### yungman

1. The problem statement, all variables and given/known data
I don't know how to come up with this final solution:
$$1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}=\frac{a^2-r^2}{a^2-2at\cos(\theta-\phi)+r^2}$$
2. Relevant equations

3. The attempt at a solution
For $\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}$, multiply by $\frac{a-re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}$

$$\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}(a-re^{j(\theta-\phi)})$$
$$=\frac{ar^n e^{jn(\theta-\phi)}}{a^n}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}+\frac{ar^{n-1} e^{j(n-1)(\theta-\phi)}}{a^{n-1}}-\frac{r^{n}e^{j(n(\theta-\phi)}}{a^{n-1}}\cdot \cdot \cdot \cdot\cdot \cdot+\frac{are^{j(\theta-\phi)}}{a}-\frac{r^2e^{j2(\theta-\phi)}}{a}$$
$$=re^{j(\theta-\phi)}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}$$
$$\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}=\frac{re^{j(\theta-\phi)}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}}{a-re^{j(\theta-\phi)}}$$

The same steps are use:
$$\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=\frac{re^{-j(\theta-\phi)}-\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}}{a-re^{-j(\theta-\phi)}}$$

I expand the series out. You can see most cancel each other and leaving only two terms which factored out into 4 simpler terms. You can see two of the terms match the answer:
$$1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}-\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}} +\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}-\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}$$

Compare with $1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}$

$$\Rightarrow\;\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}}+\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}=0$$

Thanks

Last edited: Dec 18, 2013
2. Dec 18, 2013

### clamtrox

I can't follow at all what you have done, so I can't comment on it.

Perhaps the best approach would be to use the commonly known result for geometric series: $$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$ for $|x| < 1$

3. Dec 18, 2013

### yungman

It is given $r\leq a$ and also $|e^{j(\theta-\phi)}|\leq 1$. So that fits $|x|\leq 1$

$$\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}\;\Rightarrow\;x=\left(\frac{r}{a}\right) e^{j(\theta-\phi)}\;\Rightarrow\;\frac{1}{1-x}=\frac{a}{a-re^{j(\theta-\phi)}}$$

The same steps are use:
$$\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}\;\Rightarrow\;x=\left(\frac{r}{a}\right) e^{-j(\theta-\phi)} \;\Rightarrow\;\frac{1}{1-x}=\frac{a}{a-re^{-j(\theta-\phi)}}$$

But that will give:
$$1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{a}{a-re^{j(\theta-\phi)}}+\frac{a}{a-re^{-j(\theta-\phi)}}$$

4. Dec 18, 2013

### R136a1

Notice that the summation in clamtrox' post starts at $0$ and yours starts at $1$.

5. Dec 18, 2013

### yungman

Thanks, I missed that. I have to see whether it's making any difference.

6. Dec 18, 2013

### yungman

So with the extra term of $n=0$, both will give a 1 for $n=0$.

$$1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{a}{a-re^{j(\theta-\phi)}}+\frac{a}{a-re^{-j(\theta-\phi)}}+2$$

I then factor 1 into one of the fraction:

$$1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{a}{a-re^{j(\theta-\phi)}}+\frac{a}{a-re^{-j(\theta-\phi)}}+2=1+\frac{2a-re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{2a-re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}$$

I still cannot get the right answer.

7. Dec 18, 2013

### Saitama

Don't you think that should be -2 instead of +2?

You can then give -1 to each of the fraction to get the right answer.

8. Dec 18, 2013

### yungman

Can you explain how you can make it to -1. If you look at post #3, it's all positive result. n=0 then the result is +1. I don't understand why you can assign -1 to it. Please help.

Thanks

9. Dec 19, 2013

### Saitama

You agree that
$$\sum_{i=0}^{\infty} x^i=\frac{1}{1-x}$$

for $|x| < 1$.

The above series is:
$$1+x+x^2+x^3+.......=\frac{1}{1-x}$$
Subtract 1 from both the sides to get
$$\sum_{i=1}^{\infty} x^i =\frac{1}{1-x}-1$$

Do you agree and does this clear your doubt?

Last edited: Dec 19, 2013
10. Dec 19, 2013

### clamtrox

$$\sum_{n=1}^\infty x^n = x \sum_{n=0}^\infty x^n$$

11. Dec 19, 2013

### Intrastellar

$$\sum_{k=m}^\infty ar^k=\frac{ar^m}{1-r}$$

Here is the more general formula, it should be easier to go from here.

12. Dec 19, 2013

### yungman

The book suggested the method in my original post and I just don't see how that work!!

13. Dec 19, 2013

### yungman

Do you have a link that explain a little more on this? What is the limit of $r$ in this?

14. Dec 19, 2013

### Saitama

15. Dec 19, 2013

### yungman

Thanks for your help. Lastly I just want to verify this:

$$\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}\;\Rightarrow\;x=\left(\frac{r}{a}\right) e^{j(\theta-\phi)}$$

It is given $r\leq a$ so $\left(\frac{r}{a}\right)^n\leq 1$

$$|e^{jn(\theta-\phi)}|=e^{jn(\theta-\phi)}e^{-jn(\theta-\phi)}=1$$

$$\Rightarrow\;|x|=\left|\left(\frac{r}{a}\right) e^{j(\theta-\phi)}\right|\leq 1$$
Thanks

16. Dec 19, 2013

### yungman

I just read through the article. I don't think this can work for my problem. As posted, the $r\leq 1$ in my case. This does not work for $r=1$ as the denominator become zero and the solution is unbounded.

So I have to go back to my original post #1. Maybe that's the reason the book used a complicated formula.

Thanks

17. Dec 19, 2013

### Saitama

You are right, and that is also the reason I edited my previous post.

Can you perhaps post the complete problem statement? I don't see the bounds stated anywhere in the OP.

18. Dec 19, 2013

### yungman

This is part of the derivation of Laplace equation on a disk. The books show these three steps. I am trying to verify this by deriving from beginning to the end. The important thing is the boundary of the disk is $r=a$. So the solution has to be finite when $r=a$ which rules out the easy solution of the few suggestions above.

It is given by the book:
$$1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}=\frac{a^2-r^2}{a^2-2at\cos(\theta-\phi)+r^2}$$

This is my attempt:
For $\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}$, multiply by $\frac{a-re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}$

$$\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}(a-re^{j(\theta-\phi)})$$

$$=\frac{r^n e^{jn(\theta-\phi)}}{a^{n-1}}+\frac{r^{n-1} e^{j(n-1)(\theta-\phi)}}{a^{n-2}}+\frac{r^{n-2} e^{j(n-2)(\theta-\phi)}}{a^{n-3}}\cdot \cdot \cdot \cdot\cdot \cdot+\frac{r^2e^{j2(\theta-\phi)}}{a}+re^{j(\theta-\phi)}$$

$$-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}-\frac{r^{n}e^{j(n(\theta-\phi)}}{a^{n-1}}-\frac{r^{n-1}e^{j(n-1)(\theta-\phi)}}{a^{n-2}}\cdot \cdot \cdot \cdot\cdot \cdot-\frac{r^3e^{-j3(\theta-\phi)}}{a^2}-\frac{r^2e^{j2(\theta-\phi)}}{a}$$
As you can see, all except two terms get cancelled out giving:
$$=re^{j(\theta-\phi)}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}$$
$$\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}=\frac{re^{j(\theta-\phi)}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}}{a-re^{j(\theta-\phi)}}$$

The same steps are use:
$$\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=\frac{re^{-j(\theta-\phi)}-\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}}{a-re^{-j(\theta-\phi)}}$$

I expand the series out. You can see most cancel each other and leaving only two terms which factored out into 4 simpler terms. You can see two of the terms match the answer:
$$1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}-\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}} +\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}-\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}$$

Compare with $1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}$

$$\Rightarrow\;\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}}+\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}=0$$

Last edited: Dec 19, 2013
19. Dec 19, 2013

### Ray Vickson

I cannot figure out what you are trying to do, but it does seem you are making the problem 100 times more difficult than it is. You want to evaluate
$$1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)} = 1 + 2\, \text{Re}\left[ \sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)} \right]$$
The summation can be easily computed using
$$\sum_{n=1}^{\infty} x^n = \frac{x}{1-x}, \;\; x = (r/a)\, e^{j(\theta-\phi)}.$$

20. Dec 19, 2013

### yungman

Yes, you are correct, but this formula cannot accommodate when $r=a$. This is a Laplace equation on a disk where boundary condition is $r=a$. So this formula cannot be used.