# Homework Help: I can't understand this in partial fractions

1. Feb 23, 2006

### dilan

I realy find in difficult to solve the second part of these type of questions,

Here are two questions of them

Question number 1

Resolve into partial fractions

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

For what range of values of "x" can this function be expanded as a series in ascending powers of "x"? Write down the coefficients of "x^n" in this expansion.

Question number 2

Resolve into partial fractions

2/(1-2x)^2(1+4x^2)

And hence obtain the coefficients of "x^4n" and "x^4n+1" in the expansion of this function in ascending powers of "x". State the range of values of "x" for which the expansion is valid

Ok these are the two types that I want to show. Resolving to partial fractions is not a difficult task for me. But the problem here is that I can't understand the second part where they ask for coefficients of "x^n" and the range of "x" for which the expansion is valid and so on. I realy can't understrand the second part of these type of questions.

I just want to ask is there anyone who can provide me some links to learn how to solve these type of questions. I just need to get this very clear in to my mind.

So please is there anyone who can help me to learn about the second part of this question.

Thanks alot. I realy appreciate if anyone can help me.

Thanks again.
Dilan

2. Feb 23, 2006

### arildno

1. The range of validity concerns the radius of convergence of the power series expansions.

3. Feb 23, 2006

### dilan

hi arildno,

Thanks for the reply. But I don't think by mentioning like that I will realy get it. Do you know any web site that can teach me these?

Thanks

4. Feb 23, 2006

### arildno

Okay, let's look at a simple example:
Consider the finite partial sum:
$$S_{N}=\sum_{n=0}^{N-1}x^{n}=1+x+x^{2}+++x^{N-1}$$
This can be rewritten as, for ANY value of x:
$$S_{N}=\frac{1-x^{N}}{1-x}$$

For what values of x does the limit $\lim_{N\to\infty}S_{N}$ exist?

Clearly, we must require |x|<1

That is, the radius of convergence of the INFINITE series $S\equiv\lim_{N\to\infty}S_{N}$ is 1, and in our case, $S=\frac{1}{1-x}[/tex]. Now, let us apply this to a factor like [itex]\frac{1}{1+4x^{2}}=\frac{1}{1-(-(2x)^{2})}[/tex] Clearly, the radius of convergence of the associated power series is given by [itex]|-(2x)^{2}|<1\to{4x^{2}}<1\to|x|<\frac{1}{2}$
That is, the radius of convergence is 1/2.

5. Feb 23, 2006

### dilan

hi arildno,

Thanks alot for your interest in trying to teach me this. You realy are very intereted to teach me this and I am realy thankful for that. But the problem is still I find it a little difficult because I haven't done these javascript:;

Well if you can get a website of this that would teach step by step it would be realy easy for me.

Thanks

6. Feb 23, 2006

### arildno

Pinpoint what is difficult for you.
This is the best website to learn from.

7. Feb 23, 2006

### dilan

hi arildno,

Ok I am trying my best again to understan this. I had a problem in using the Latex (how do you realy use that).
Again I will try. There is a piece that I want to post but can't because the keys are not there in my keyboard. The thing with lim. I don't know anything about that.
I am trying to get it in the best way again. I will post here when I am stuck. Just going step by step again

8. Feb 23, 2006

### arildno

You can see the LATEX code behind a particular expression by clicking on it.

9. Feb 24, 2006

### VietDao29

You may want to have a glance at:
Introducing LaTeX Math Typesetting. It's one of the stickys in the board Math & Science Tutorials (the first board from the top in the Forums Home).
Just read the 3 .PDF files in the first post. They are all short guides to LaTeX.
Another thing to remember is that one can always click on any LaTeX image to see its code. :)

10. Feb 24, 2006

### dilan

Thanks

Hi,

Thanks for introducing me to Latex. It's realy great. Realy easy.

Ok back to the topic. Does my sum deal with induction?:uhh:

11. Feb 24, 2006

Are you unsure at why the partial sum $S_{N}=1+x+x^{2}++++x^{N-1}[/tex] can be rewritten as: $$S_{N}=\frac{1-x^{N}}{1-x}$$ ? 12. Feb 24, 2006 ### dilan ya ya that's right Hi, Absolutely right. That's the place where I am stuck a littl also. 13. Feb 24, 2006 ### arildno Okay! To derive the non-obvious result, let's multiply the sum with x: $$xS_{N}=x+x^{2}+x^{3}++++x^$$ Agreed? Okay, now comes the cool move: Regard the difference: $$S_{N}-xS_{N}$$ A LOT of terms cancel here; can you see which are retained? 14. Feb 24, 2006 ### dilan Ok understand that. THe once left is 1 - uhh some X should come here. Am I right? 15. Feb 24, 2006 ### arildno Yes, you are left with: $$S_{N}-xS_{N}=1-x^{N}\to(1-x)S_{N}=1-x^{N}\to{S}_{N}=\frac{1-x^{N}}{1-x}$$ Do you agree with this? 16. Feb 24, 2006 ### dilan hi arildno, We put the (1-x) Sn = 1-x^n the 1 in front of -x (the one in front of Sn) come because all get canceled right? Last edited: Feb 24, 2006 17. Feb 24, 2006 ### arildno Note that in the left-hand expression, [itex]S_{N}-xS_{N}$, we may regard $S_{N}$ as the COMMON factor in the identical expression $1*S_{N}-x*S_{N}$.
Clearly, by the distributive law, we have the identity: $1*S_{N}-x*{S}_{N}\equiv(1-x)*S_{N}$
Agreed?

18. Feb 24, 2006

### dilan

Oh ya ya. i see. That's very clear. Thanks for expressing it like that. ok agreed.

19. Feb 24, 2006

### arildno

Okay, now you are ready to tackle the concept of radius of convergence for the INFINITE series!

If you let the value of x be greater than 1, what will happen to the value of the series $$S_{N}=\frac{1-x^{N}}{1-x}$$ if you let N be a really big number?

20. Feb 24, 2006

### dilan

Oh my you will get a very big answer. Right?