1. ## Binomial Expansion help

Three ways to expand 1/(1+3x+2x^2) = 1/[(1+x)(1+2x)] = 2/(1+2x) - 1/(1+x)

(a) (1+y)^-1 where y=3x+2x^2

(b) 2(1+2x)^-1 - (1+x)^-1

(c) (1+x)^-1 * (1+2x)^ -1

These methods all give the correct polynomial.

Now with (b) and (c) the range of validity, -0.5<x<0.5 drops out clearly.
However, for (a) the range of validity is abs(y)<1 which gives a different set of values for x.

What is the resolution? Abandon method (a)? Why doesn't it work?

Any thoughts?!

2. ## Re: Binomial Expansion help

Both are correct.

3. ## Re: Binomial Expansion help

Originally Posted by SlipEternal
Both are correct.
Sorry what do you mean? The range of validity in (a) doesn't agree with (b) & (c)

4. ## Re: Binomial Expansion help

Originally Posted by rodders
Sorry what do you mean? The range of validity in (a) doesn't agree with (b) & (c)
They also produce different power series. The fraction $\dfrac{1}{1+3x+2x^2}$ exists for any $x \neq -1, x\neq -\dfrac{1}{2}$. You are discussing the radius of convergence of the geometric series expansions of the fraction. A different series (or the product of two series) will have a different radius of convergence for the same fraction.

5. ## Re: Binomial Expansion help

The thing i don't get, is this is the same series but just expanded using three different methods yet in case (a) i get a different radius of convergence. Apologies if i am being dim!

6. ## Re: Binomial Expansion help

Originally Posted by rodders
The thing i don't get, is this is the same series but just expanded using three different methods yet in case (a) i get a different radius of convergence. Apologies if i am being dim!
Yes, a different expansion will have a different radius of convergence. That is correct. I am not sure I understand your question. Are you just surprised that is true? Or are you questioning whether it is true? And if you question whether it is true, why would you think it is not true?

7. ## Re: Binomial Expansion help

No, So my original question is about trying the expand the same thing but using three different methods.
The three methods work and the expansions are all the same as the original question is the same.
However, in method (a) above, how do you determine what the radius of convergence is?

8. ## Re: Binomial Expansion help

Originally Posted by rodders
No, So my original question is about trying the expand the same thing but using three different methods.
The three methods work and the expansions are all the same as the original question is the same.
However, in method (a) above, how do you determine what the radius of convergence is?
The expansions are NOT the same. They may be equivalent in their respective radii of convergence, but they are not the same.

Method (a):
$\displaystyle \dfrac{1}{1+3x+2x^2} = \sum_{n\ge 0}(-1)^n(3x+2x^2)^n = 1-(3x+2x^2)+(3x+2x^2)^2-(3x+2x^2)^3\pm \cdots = 1-3x+7x^2-15x^3+31x^4-63x^5\pm \cdots$

Method (c):
$\displaystyle \dfrac{1}{1+x}\cdot \dfrac{1}{1+2x} = \left( \sum_{n\ge 0}(-x)^n \right)\left(\sum_{n \ge 0}(-2x)^n \right)$
This product only converges to the result of Method (a) when $|x|<\dfrac{1}{2}$

Method (b):
$\displaystyle \dfrac{2}{1+2x}-\dfrac{1}{1+x} = \left(2\sum_{n\ge 0}(-2x)^n\right) - \left(\sum_{n\ge 0}(-x)^n \right)$
This difference only converges to the result of Method (a) when $|x| < \dfrac{1}{2}$.

They are not all equivalent.

9. ## Re: Binomial Expansion help

Ok.. this is messing with my brain.
What is the radius of convergence for (a) and how do you establish it?

10. ## Re: Binomial Expansion help

Originally Posted by rodders
Ok.. this is messing with my brain.
What is the radius of convergence for (a) and how do you establish it?
You use the fact that a geometric series $\displaystyle \sum_{n\ge 0} r^n$ converges when $|r|<1$. So, the radius of convergence for (a) is $|-3x-2x^2| < 1$.

11. ## Re: Binomial Expansion help

Ok thanks. I will do some more thinking!

12. ## Re: Binomial Expansion help

Ok. here is my problem. You are saying the expansions are not all the same.. I say they are.
Looking at (c) above

If you continue to expand (1+x)^-1 and multiply it with (1+2x)^-1 you will get the expanded terms for what you did above for method (a).

The same happens for method (b)

13. ## Re: Binomial Expansion help

Originally Posted by rodders
Ok. here is my problem. You are saying the expansions are not all the same.. I say they are.
Looking at (c) above

If you continue to expand (1+x)^-1 and multiply it with (1+2x)^-1 you will get the expanded terms for what you did above for method (a).

The same happens for method (b)
They are only the same in their radius of convergence. You cannot multiply them out until you "reach infinity". You can only "reach infinity" if they converge (reaching infinity is not a mathematical term. I do not recall the correct term. Basically, you have to look at the entire series as one complete unit before you can consider multiplying it out). The series you get in (b) and (c) only converge in their respective radii. None of this contradicts anything that you are looking at, so the fact that this is confusing you means that you are not understanding what it all means. Radius of convergence is the range of values over which a specific series converges. Proving that it is even possible to multiply series is a fairly complicated process (definitely not pre-university algebra). The specific circumstances under which you can multiply infinite series is a very difficult concept. For me to fully grasp it, it took an understanding of point-set topology and ring theory. Once you are exposed to more sophisticated mathematical tools, you will be introduced to when and how the concept of multiplying and adding infinite series even makes sense mathematically.