# Textbook question on power series

1. Dec 2, 2005

### Swatch

My textbook has an example on multiplication of power series.
" Multiply the geometric series $$x^n$$ by itself to get a power series for $$1/(1-x)^2$$ for |x|<1 "
from this we get the $$c_{n}$$=n+1
O.K. I get that the coefficients are 1 for all n but why +1.
Could someone please explain this to me if possible.

2. Dec 2, 2005

### Physics Monkey

You want to multiply $$(1 + x + x^2 + x^3 + ...)(1 + x + x^2 + x^3 + ...)$$, so just try few examples first to see what's going on. The $$x^0$$ term is just $$1*1 = 1$$, the $$x^1$$ term is $$x*1+1*x = 2x$$, the $$x^2$$ term is $$x^2*1 + x*x + 1*x^2 = 3x^2$$, and the $$x^3$$ is $$x^3*1 + x^2*x+x*x^2+1*x^3 =4x^3$$. Now, can you begin to see a pattern forming?

Alternatively, you can just apply the formula for the coeffecients of the product series in terms of the coeffecients of the two original series. It looks something like $$c_k = \sum^k_{r=0} a_r b_{k-r}$$ where the a's and b's are the coeffecients of the original series, and the c's are coeffecients of the product series.

Last edited: Dec 2, 2005
3. Dec 2, 2005

### benorin

Right call Physics Monkey:
The formula for the Cauchy product of series as it is presented in this http://mwt.e-technik.uni-ulm.de/world/lehre/basic_mathematics/di/node14.php3 [Broken].
A quick version is:
Suppose $\sum_{n=0}^{\infty} a_n$ and $\sum_{n=0}^{\infty} b_n$ converge absolutely. Then
$$\left( \sum_{n=0}^{\infty} a_n\right) \left( \sum_{n=0}^{\infty} b_n\right) = \sum_{n=0}^{\infty} \sum_{k=0}^{n} a_{k}b_{n-k}$$ also converges absolutely.
Alternately, look here, under the heading A Variant.

Last edited by a moderator: May 2, 2017
4. Dec 3, 2005

### Swatch

O.K. I see what's going on now, thanks. But I still don't get how to get n+1 from $$c_k = \sum^k_{r=0} a_r b_{k-r}$$ without doing some multiplication. Please explain to me if you can.

5. Dec 3, 2005

### benorin

the answer to you above question

Using

$$\left( \sum_{n=0}^{\infty} a_n\right) \left( \sum_{n=0}^{\infty} b_n\right) = \sum_{n=0}^{\infty} \sum_{k=0}^{n} a_{k}b_{n-k}$$

we have

$$\left( \sum_{n=0}^{\infty} x^n\right) \left( \sum_{n=0}^{\infty} x^n\right) = \sum_{n=0}^{\infty} \sum_{k=0}^{n} x^{k}x^{n-k} = \sum_{n=0}^{\infty} \sum_{k=0}^{n} x^{k+(n-k)} = \sum_{n=0}^{\infty} x^{n} \sum_{k=0}^{n} 1 = \sum_{n=0}^{\infty}(n+1) x^{n}$$

where $\sum_{k=0}^{n} 1 =n+1$ is the answer to you above question.

6. Dec 3, 2005

### Swatch

O.K. I getting close to understanding this. What puzzles me is the last sum. Do you put n+1 so the first term isn't 0? I think I'm missing some information to understand this.

7. Dec 3, 2005

### matt grime

We put n+1 because that is what it is... You're adding up the number 1, n+1 times, so the answer is n+1.

But what's wrong with just multiplying out the power series?

You can multiply (1+x+x^2+.....)(1+x+x^2+...)

and count and think and, well, it's just true.... there is nothing clever going on. to end up with x^n in the product you can only get it from multiplying x^r in the first and x^{n-r} and each of those multiplications contributes 1 to the coefficiant of x^n and there is one contributrion from each r as r goes from 0 to n so you get 1 added up n+1 times.

Last edited: Dec 3, 2005
8. Dec 3, 2005

### Swatch

I think I understand this no, just hvae to get some practice solving these kind of problems. Thanks.