Binomial expansion, general coefficient

[Appleton]

Homework Statement

Find the coefficient of x^n in the expansion of each of the following functions as a series of ascending powers of x.

[itex]\frac{1}{(1+2x)(3-x)}[/itex]

Homework Equations

The Attempt at a Solution

[itex](1+2x)^{-1} = 1 + (-1)2x + \frac{(-1)(-2)}{2!}(2x)^2 + \frac{(-1)(-2)(-3)}{3!}(2x)^3... -1<2x<1[/itex]

[itex]= 1 - 2x + 4x^2 - 8x^3... -\frac{1}{2}<x<\frac{1}{2}[/itex]

[itex]\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[/itex]

[itex]= \frac{1}{3} + \frac{x}{9} + \frac{x^2}{27} + \frac{x^3}{81}... -3<x<3[/itex]

[itex]\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}[/itex]

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 [itex]x^n[/itex] in the expansion of [itex](1+2x)^{-1}[/itex] is [itex](-2)^n[/itex] and the coefficient of [itex]x^n[/itex] in the expansion of [itex](3-x)^{-1}[/itex] is [itex](\frac{1}{3})^{n+1}[/itex]. So [itex]\frac{1}{(1+2x)(3-x)}[/itex] could also be expressed as [itex](\sum^{∞}_{r=0}(-2x)^r)(\sum^{∞}_{r=0}(\frac{1}{3})^{r+1}(x)^r) -\frac{1}{2}<x<\frac{1}{2}[/itex]. However as I feel I am getting tantalisingly close to a solution I realize that their multiplication complicates matters.

So I start to look at patterns in the process of multiplying the 2 expansions together and find that
[itex]\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}[/itex]

By now I've strayed into territory outside of my textbook, 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.

 

[pasmith]

Homework Statement

Find the coefficient of x^n in the expansion of each of the following functions as a series of ascending powers of x.

[itex]\frac{1}{(1+2x)(3-x)}[/itex]

Homework Equations

The Attempt at a Solution

[itex](1+2x)^{-1} = 1 + (-1)2x + \frac{(-1)(-2)}{2!}(2x)^2 + \frac{(-1)(-2)(-3)}{3!}(2x)^3... -1<2x<1[/itex]

[itex]= 1 - 2x + 4x^2 - 8x^3... -\frac{1}{2}<x<\frac{1}{2}[/itex]

[itex]\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[/itex]

[itex]= \frac{1}{3} + \frac{x}{9} + \frac{x^2}{27} + \frac{x^3}{81}... -3<x<3[/itex]

[itex]\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}[/itex]

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 [itex]x^n[/itex] in the expansion of [itex](1+2x)^{-1}[/itex] is [itex](-2)^n[/itex] and the coefficient of [itex]x^n[/itex] in the expansion of [itex](3-x)^{-1}[/itex] is [itex](\frac{1}{3})^{n+1}[/itex]. So [itex]\frac{1}{(1+2x)(3-x)}[/itex] could also be expressed as [itex](\sum^{∞}_{r=0}(-2x)^r)(\sum^{∞}_{r=0}(\frac{1}{3})^{r+1}(x)^r) -\frac{1}{2}<x<\frac{1}{2}[/itex]. However as I feel I am getting tantalisingly close to a solution I realize that their multiplication complicates matters.

So I start to look at patterns in the process of multiplying the 2 expansions together and find that
[itex]\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}[/itex]

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

 

[Appleton]

Thanks for your reply pasmith. Trying to follow your suggestion I got the coefficient of x^n as [itex]\sum ^{∞}_{m} \frac{(-2)^{n-m}}{3^{m+1}}[/itex] 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:
[itex]\sum ^{n}_{m=0} \frac{(-2)^m3^{m-1}
}{3^n}[/itex]
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.

 

[pasmith]

Thanks for your reply pasmith. Trying to follow your suggestion I got the coefficient of x^n as [itex]\sum ^{∞}_{m} \frac{(-2)^{n-m}}{3^{m+1}}[/itex] 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

This is indeed the idea.

come up with an expression for the coefficient of every nth power of x:
[itex]\sum ^{n}_{m=0} \frac{(-2)^m3^{m-1}
}{3^n}[/itex]
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.

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

The stricter constraint is that [itex]|x| < \frac12[/itex], so it makes sense to express the series in powers of [itex]2x[/itex] rather than [itex]x/3[/itex]. Hence [tex]
\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
[/tex] where the sum over [itex]k[/itex] is again a geometric series.

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

 

[Appleton]

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

 

[pasmith]

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

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), ...

 

[Ray Vickson]

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

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##.

 

[Appleton]

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