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When I do a binomial expansion on $\frac{1}{(1-2x)(1+3x)}$ about $0$, I can do it in 2 ways.

Method 1 $$\frac{1}{(1-2x)(1+3x)}=\frac{1}{1-2x}\frac{1}{1+3x}$$ Thus, getting $(1+y_1)^{-1}(1+y_2)^{-1}$. This method would give me a radius of convergence $|2x|<1$ AND $|3x|<1$. So, the radius of convergence is $-\frac{1}{3}< x < \frac{1}{3}$.

Method 2 $$\frac{1}{(1-2x)(1+3x)}=\frac{1}{1+x-6x^2}$$ I can let $y=x-6x^2$, thus getting $(1+y)^{-1}$.

Now, the convergence will be for $|y|<1$, or $|x-6x^2|<1$. This actually gives me $-\frac{1}{3} < x < \frac{1}{2}$. So, I have $\frac{1}{3}<x<\frac{1}{2}$ in my convergence as well.

Question: So which is the correct method and why is there such an inconsistency?

I have my own theories, but I'll see what everyone has to say before weighing in.

Poseidonium
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1 Answers1

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Assuming you are doing the expansion around $0$, you get convergence out to the nearest root (in the complex plane). In both cases this is $\frac 13$. The error is assuming that the radius of convergence of $\frac 1{1+x-6x^2}$ is that with three terms in the denominator, they pollute each other in the power series. They do it in just the way to make Method 1 correct.

Ross Millikan
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  • If I expand using $$\frac{1}{(1-2x)(1+3x)}=\frac{1}{1+x-6x^2}$$ I can let $y=x-6x^2$, thus getting $(1+y)^{-1}$.

    Now, the convergence will be for $|y|<1$, or $|x-6x^2|<1$. This actually gives me $-\frac{1}{3} < x < \frac{1}{2}$. So, I have $\frac{1}{3}<x<\frac{1}{2}$ in my convergence as well.

     – Poseidonium Jan 28 '13 at 05:21