# Binomial expansion, general coefficient

1. Jul 19, 2014

### Appleton

1. The problem statement, all variables and given/known data
Find the coefficient of x^n in the expansion of each of the following functions as a series of ascending powers of x.

$\frac{1}{(1+2x)(3-x)}$

2. Relevant equations

3. The attempt at a solution

$(1+2x)^{-1} = 1 + (-1)2x + \frac{(-1)(-2)}{2!}(2x)^2 + \frac{(-1)(-2)(-3)}{3!}(2x)^3... -1<2x<1$

$= 1 - 2x + 4x^2 - 8x^3... -\frac{1}{2}<x<\frac{1}{2}$

$\frac{1}{3}(1-\frac{x}{3})^{-1} = \frac{1}{3}(1 + (-1)(-\frac{x}{3}) + \frac{(-1)(-2)}{2!}(-\frac{x}{3})^2 + \frac{(-1)(-2)(-3)}{3!}(-\frac{x}{3})^3...) -1<-\frac{x}{3}<1$

$= \frac{1}{3} + \frac{x}{9} + \frac{x^2}{27} + \frac{x^3}{81}... -3<x<3$

$\frac{1}{(1+2x)(3-x)}=\frac{1}{3}-\frac{5}{9}x+\frac{31}{27}x^2-\frac{185}{81}x^3... -\frac{1}{2}<x<\frac{1}{2}$

I can't see an obvious pattern in the numerators of the coefficients, perhaps I should be able to. So I look at the coefficient of each binomial where I see that the coefficient of $x^n$ in the expansion of $(1+2x)^{-1}$ is $(-2)^n$ and the coefficient of $x^n$ in the expansion of $(3-x)^{-1}$ is $(\frac{1}{3})^{n+1}$. So $\frac{1}{(1+2x)(3-x)}$ could also be expressed as $(\sum^{∞}_{r=0}(-2x)^r)(\sum^{∞}_{r=0}(\frac{1}{3})^{r+1}(x)^r) -\frac{1}{2}<x<\frac{1}{2}$. However as I feel I am getting tantalisingly close to a solution I realise that their multiplication complicates matters.

So I start to look at patterns in the process of multiplying the 2 expansions together and find that
$\frac{1}{(1+2x)(3-x)}=\frac{1}{3}(\sum^{∞}_{r=0} (-1)^r(2x)^r(\sum^{∞}_{k=0} (\frac{x}{3})^k)) -\frac{1}{2}<x<\frac{1}{2}$

By now I've strayed into territory outside of my text book, ie the recursive summation, this might well be meaningless as it looks like it never bottoms out of the k loop, but I hope it illustrates my thought process, however Ill conceived it might be.
I thought there might be a way to eliminate the k in the last expression to yield my general coefficient.
Any guidance on the approach to tackling this kind of problem would be much appreciated.

2. Jul 19, 2014

### pasmith

You can swap multiplication by $(-2x)^r$ and summation over $k$ to obtain
$$\frac{1}{(1 + 2x)(3 - x)} = \frac13 \sum_{r=0}^\infty \sum_{k=0}^\infty \frac{(-1)^r2^r}{3^k}x^{r+k}$$
and now the change of variables to $n = r + k$ and $m = k$ where $0 \leq n \leq \infty$ and $0 \leq m \leq n$ suggests itself.

3. Jul 20, 2014

### Appleton

Thanks for your reply pasmith. Trying to follow your suggestion I got the coefficient of x^n as $\sum ^{∞}_{m} \frac{(-2)^{n-m}}{3^{m+1}}$ which I struggled to make sense of.
But by using your distribution of (-2x)^r I was able to visualise the whole summation as a 2 dimensional array of terms and by adding diagonal terms come up with an expression for the coefficient of every nth power of x:
$\sum ^{n}_{m=0} \frac{(-2)^m3^{m-1} }{3^n}$
However I am still unable to express this coefficient in terms of just n and without the summation. Something tells me I'm making more of a meal of this than is necessary.

4. Jul 20, 2014

### pasmith

This is indeed the idea.

Your sum here is a geometric series: $$\sum ^{n}_{m=0} \frac{(-2)^m3^{m-1}}{3^n} = \frac1{3^{n+1}} \sum_{m=0}^n (-6)^m.$$ There is a formula for summing such series.

The stricter constraint is that $|x| < \frac12$, so it makes sense to express the series in powers of $2x$ rather than $x/3$. Hence $$\frac13 \sum_{r=0}^\infty \sum_{k=0}^\infty (-2x)^r \frac{x^k}{3^k} = \frac13 \sum_{r=0}^\infty \sum_{k=0}^\infty \frac{1}{3^k(-2)^k} (-2x)^{r+k} = \frac13 \sum_{r=0}^\infty \sum_{k=0}^\infty \left(-\frac{1}{6}\right)^k (-2x)^{r+k} = \frac13 \sum_{n=0}^\infty (-2x)^n \sum_{k=0}^n \left(-\frac{1}{6}\right)^k$$ where the sum over $k$ is again a geometric series.

One can show that both expressions for the coefficient of $x^n$ are equal, as they must be.

5. Jul 20, 2014

### Appleton

So the coefficient is $\frac{1}{21}(3^{-1}+(-1)^n(2)^{n+1}(3))$.
I still don't follow the substitution of n for r+k. How is that justified?

6. Jul 21, 2014

### pasmith

The series is absolutely convergent, so its terms can be re-ordered. Instead of taking them in the order (0,0), (0,1), (0,2), ..., (1,0), (1,1), .... one takes them in the order (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

7. Jul 21, 2014

### Ray Vickson

Originally you have a sum over all pairs (r,k) with 0 ≤ r < ∞ and 0 ≤ k < ∞. This is just all the points in the lattice {(0,0), (1,0), (0,1), (1,1), (0,2) ,..... } in 2-dimensional (r,k)-space. Because of absolute convergence (at least, in a certain x-range) you can re-express the sum: sum over each of the $45^o$ lines $r+k = 0, 1, 2, \ldots$, and within each such line, sum over the lattice points contained in that portion of the line between the k and r axes. That is, for fixed $r+k = n$ you need to do the sum for (say) $k = 0,1,2, \ldots, n$, with $r = n - k$ for each such $k$. Why would you do that? The answer is: because you have terms $x^{r+k}$ in the sum, and you want to isolate the coefficient of $x^{r+k}$ for each fixed value $r+k = n$.

8. Jul 21, 2014

### Appleton

Thank you both very much for your help, very interesting and illuminating.