Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

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.

share|improve this answer
    
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 5:21

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.