# Homework Help: 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.

21. Dec 19, 2013

### Ray Vickson

So what? It can be used for all r < a, so you get a formula that is 100% correct for r < a. It is a separate issue altogether whether or not you can extend the answer to the case r = a. If you cannot, your beginning formulation must either be (i) applicable only to r < a; or (ii) in error somehow. There are separate issues that you are confounding, and that is leading to confusion on your part, or so it seems.

Anyway, if you do what I suggested you will get exactly what the book says. If you don't agree with the method and the answer, you are saying that you disagree with the book as well.

Last edited: Dec 19, 2013
22. Dec 20, 2013

### Intrastellar

Ok, if you insist.
There is a common factor in the numerator of both fractions, can you locate it ?

23. Dec 20, 2013

### yungman

I know I can pull out $\frac {r^{n+1}}{a^n}$. But that does not help to make the whole thing to zero.

24. Dec 20, 2013

### Ray Vickson

It is NOT zero, and there is no reason why it should be. All you need is
$$\lim_{n \to \infty} \frac{r^{n+1}}{a^n} = 0$$
and that is true if $r < a.$ Have you forgotten what is meant by the sum of an infinite series?

25. Dec 20, 2013

### yungman

No I did not forget, I need to have the formula work for $r=a$. This is the solution of Laplace with boundary condition at $r=a$. You cannot say the OP was wrong from the book and only look at $r<a$. Solution has to be continuous at the boundary.

Also, the method in the OP was suggested by the book. I just cannot cancel it out.

Last edited: Dec 20, 2013