I can't understand this in partial fractions

[dilan]

I really 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 really 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 into my mind.

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


Thanks a lot. I really appreciate if anyone can help me.

Thanks again.
Dilan

 

[arildno]

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

 

[dilan]

hi arildno,

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

Thanks

 

[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:
$$S_{N}=\frac{1-x^{N}}{1-x}$$

Now, ask yourself:
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, [itex]S=\frac{1}{1-x}[/tex].<br /> <br /> Now, let us apply this to a factor like [itex]\frac{1}{1+4x^{2}}=\frac{1}{1-(-(2x)^{2})}[/tex]<br /> Clearly, the radius of convergence of the associated power series is given by $|-(2x)^{2}|<1\to{4x^{2}}<1\to|x|<\frac{1}{2}$<br /> That is, the radius of convergence is 1/2.[/itex][/itex]

 

[dilan]

hi arildno,

Thanks a lot for your interest in trying to teach me this. You really are very intereted to teach me this and I am really thankful for that. But the problem is still I find it a little difficult because I haven't done these javascript:;
LaTeX graphic is being generated. Reload this page in a moment.

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

Thanks

 

[arildno]

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

 

[dilan]

hi arildno,

Ok I am trying my best again to understan this. I had a problem in using the Latex (how do you really 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

 

[arildno]

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

 

[VietDao29]

Ok I am trying my best again to understan this. I had a problem in using the Latex (how do you really use that).

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. :)

 

[dilan]

Thanks

Hi,

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

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

 

[arildno]

Are you unsure at why the partial sum [itex]S_{N}=1+x+x^{2}++++x^{N-1}[/tex] can be rewritten as: $$S_{N}=\frac{1-x^{N}}{1-x}$$ ?[/itex]

 

[dilan]

ya you that's right

Hi,

Absolutely right. That's the place where I am stuck a littl also.

 

[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?

 

[dilan]

Ok understand that. THe once left is 1 - uhh some X should come here.

Am I right?

 

[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?

 

[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?

 

[arildno]

Note that in the left-hand expression, $S_{N}-xS_{N}$, 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?

 

[dilan]

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

 

[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?

 

[dilan]

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

 

[arildno]

Right!
But the number will be positive, because 1-x will be negative as well (negative divided by negative is positive).

So, if you let N grow without bounds, then $S_{N}$ will grow without bounds as well, when x>1

Now, consider the situation when x<-1:
What happens if N is a really big EVEN number?
What happens if N is a really big ODD number?

 

[dilan]

Now, consider the situation when x<-1:
What happens if N is a really big EVEN number?

The value for X will be positive right?
and the answer will be negative. I mean Sn

What happens if N is a really big ODD number?
The value for X will be negative right?
and the answer will be positive. I mean Sn

 

[arildno]

Right!
And as N grows bigger, the Sn will switch between a whopping big positive number and to an even "bigger" negative number. Agreed?

 

[dilan]

Ya agreed. Because the n is changing right? from odd to even and even to odd like that it will go on right?

 

[arildno]

Right!
So, neither in the case x>1 or x<-1 is it meaningful to say that a limit exists for Sn as N goes to infinity?

 

[dilan]

ya agreed.

 

[arildno]

But, consider that -1<x<1, that is, |x|<1
What happens to the value of Sn as N grows bigger and bigger?

 

[dilan]

What does |x|<1 this mean? i mean by using |x|

 

[dilan]

The value gets smaller and smaller right?

 

[dilan]

Either positive or negative values right?