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

 

[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 S_{N}=1+x+x^{2}++++x^{N-1}[/tex] can be rewritten as: S_{N}=\frac{1-x^{N}}{1-x} ?

 

[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?

 

[arildno]

|x| means the ABSOLUTE value of the number x, and if x is non-zero, the absolute value is always positive.

So the absolute value of "2" is 2, and the absolute value of "-2" is also 2.
The absolute value is simply the distance of a number on the number line from the origin.

 

[dilan]

So because of the absolute value are we getting positive values for X?

 

[arildno]

Hmm..not sure what you mean:
The term x^{N} will approach zero as N towards to infinity if |x|<1.
Do you agree to that?

 

[dilan]

I think I am a little confused. Does this happen because |x| < 1 will at a certain stage reach 0 I mean like x = 0

 

[arildno]

No, think of x as a FIXED number, lying between -1 and 1.
If you multiply a positive number that is less than one with itself, will the product be less than or greater than the number itself?

 

[dilan]

Well it say the number = y

and |y|<1

then the product will be less than the number right?

 

[dilan]

I mean when you multiply it by itself.

 

[arildno]

Right!
So when you multiply itself with itself N times, where N is some big number, then x^{N} will be very close to zero,right?

 

[dilan]

Right right it will be very close to zero. I mean it will go on like 0.0000000000001323 like that right?

 

[arildno]

You've got it.

So, if we have S_{N}=\frac{1-x^{N}}{1-x}, |x|<1 and N is really big, what will S_{N} be approximately equal to?

 

[dilan]

Will Sn approximately be equal to = 1/1-x

Am I right?

 

[arildno]

Perfectly! (that is 1/(1-x), remember parentheses..)

Thus, it gives perfect meaning to say that as N goes to infinity, Sn converges to a number S=1/(1-x), or that the INFINITE series S is a meaningful concept. Agreed?

 

[dilan]

uhh arildno does converge means like bringing it to one place like? Sorry to ask this because I am from a non-english country?

Ya I agree that now the S has a meaningful value.

 

[dilan]

Oh please continue please? I can really understand what you teach me than my school teacher.

 

[arildno]

So, what we have found out, is that the INFINITE series,
S=\sum_{n=0}^{\infty}x^{n}
is a meaningful concept, as long as |x|<1.
We call 1 here to be the radius of convergence for the infinite series, that is the bound we must put on x, in order for the infinite series to have any meaning (I.e, being some number).
Okay?

 

[dilan]

Ya right. 1 is the bound because if it gose beyond 1 then we won't find a bound because we can get large numbers right?

Anyway this is the first time I used this symbol next to the "Sn ="
Can you teach me about this also?
S=\sum_{n=0}^{\infty}x^{n}

 

[VietDao29]

Ya right. 1 is the bound because if it gose beyond 1 then we won't find a bound because we can get large numbers right?

Anyway this is the first time I used this symbol next to the "Sn ="
Can you teach me about this also?
S=\sum_{n=0}^{\infty}x^{n}

\sum_{i = m} ^ n, this is the summation symbol. It's a capital Sigma.
i represents the index of summation; m is the lower bound of summation, and n is the upper bound of summation.
----------
\sum_{i = m} ^ n (a_i)
means that all you need is just to sum from a_m to a_n
\sum_{i = m} ^ n (a_i) = a_m + a_{m + 1} + a_{m + 2} + ... + a_{n - 1} + a_n
----------------
Example:
\sum_{k = 1} ^ 3 \left( \frac{k}{k + 1} \right) = \frac{1}{1 + 1} + \frac{2}{2 + 1} + \frac{3}{3 + 1} = \frac{23}{12}
----------------
For Convergent series, you may want to have a look here: Convergent series.
If you say some series S_N = \sum_{k = 1} ^ N (a_k) converges to some number L, then it means that: \lim_{N \rightarrow \infty} S_N = L, \ N \in \mathbb{N}
Can you get this? :)

 

[dilan]

Ok I got it now

So, what we have found out, is that the INFINITE series,
S=\sum_{n=0}^{\infty}x^{n}
is a meaningful concept, as long as |x|<1.
We call 1 here to be the radius of convergence for the infinite series, that is the bound we must put on x, in order for the infinite series to have any meaning (I.e, being some number).
Okay?

Hi VietDao29 thanks for your post about the meaning of the Capital letter of zigma with the other letters. Realy useful. Thanks alot.

Ya now I undertand. Okay we have now got a meaningful concept for Sn. Earlier I had a little problem in undertanding the symbol, but now I get it. Okay now we have a meaningful concept as long as |x|<1

But I don't know anything about this
\lim_{N \rightarrow \infty} S_N = L, \ N \in \mathbb{N}

Anyway arildno I hope that this will not be needed for now. If yes I would like to know about that also.
Okay upto here now I can get it. What's the next step?

 

[VietDao29]

But I don't know anything about this
\lim_{N \rightarrow \infty} S_N = L, \ N \in \mathbb{N}

Have you learned limit by the way? Something looks like:
\lim_{x \rightarrow 3} x ^ 2 = 9 (this is an example of limit of a function)
or:
\lim_{n \rightarrow \infty} \frac{1}{n} = 0 (an example of limit of a sequence)?

 

[arildno]

Okay, just substitute S for L, and the statement \lim_{N\to\infty}S_{N}=S means that the value of S_{N} approaches S as N trundles off into infinity. (Something you already know).
N\in\mathbb{N} just means that N is a natural number (1, 2, 350000)and so on (not a fraction or decimal number).