I can't understand this in partial fractions

[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 a lot.

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).

 

[dilan]

Oh I see. So this \lim_{N\to\infty}S_{N}=S is something about a limit. Realy great to learn about this. I didn't know about this. Actualy I haven't done this yet. Ok I get it now.

ok arildno I get it upto here without any problem. Now my next step will be?

 

[arildno]

Okay, you said that resolving into partial fractions is no problem, right?

 

[dilan]

Ya no problem at all.

 

[arildno]

All right, then!
Now, consider a fraction of the form \frac{1}{1+2x}
If we write this as:
\frac{1}{1+2x}=\frac{1}{1-(-2x)}[/itex]<br /> then from the above, we can write:<br /> \frac{1}{1-(-2x)}=\sum_{n=0}^{\infty}(-2x)^{n}<br /> as long as |-2x|&lt;1, that is |x|&lt;1/2 (Remember that a number x and its negative, -x, has the same ABSOLUTE value!)<br /> <br /> Agreed?

 

[dilan]

Yep agreed

 

[arildno]

Furthermore, if you are to find the coefficient for the fourth power of x in this series expansion of 1/(1+2x), then that would be found from the term (-2x)^{4}, i.e, the coefficient is (-2)^{4}=16
Agreed?

 

[dilan]

oh you I see. Yep agreed.

 

[arildno]

Okay, suppose we are to find the coefficient for the fourth power of the series expansion of the fraction \frac{1}{(1+2x)^{2}}
As long as |x|<1/2, we may write:
\frac{1}{(1+2x)^{2}}=\frac{1}{1+2x}*\frac{1}{1+2x}=(\sum_{n=0}^{\infty}(-2x)^{n})*(\sum_{n=0}^{\infty}(-2x)^{n})

Now, as we are only interested in determining the coefficients to the fourth power of x, then we need only care about those products of terms from the first series with terms from the second series that yield something with x^{4}

In particular, those products will be:
1. Multiplying the 0th power term in the first series with the 4th power term in the second series
2.Multiplying the 1st power term in the first series with the 3rd power term in the second series
3.Multiplying the 2nd power term in the first series with the 2nd power term in the second series
4.Multiplying the 3rd power term in the first series with the 1st power term in the second series
5.Multiplying the 4th power term in the first series with the 0th power term in the second

That is, we will find the coefficient of the 4th power of x in the series product by calculating:
(-2x)^{0}*(-2x)^{4}+(-2x)^{1}*(-2x)^{3}+(-2x)^{2}*(-2x)^{2}+(-2x)^{3}*(-2x)+(-2x)^{4}*(-2x)^{0}=5*16x^{4}=80x^{4}

That is, the coefficient is 80.
Agreed?

 

[dilan]

Is this like the pascal trangle. I mean some what related to the binomial theorem?

 

[arildno]

Somewhat similar to that, yes!

But, the main thing is, now you know how to find the power series representation of any fraction of the form 1/(1-ax), and also, how to find the coefficient of a particular power for products like 1/(1-ax)^2
That's basically all you need in order to answer your original question.
Finally, remember that EVERY power series you make must be within its own radius of convergence in order to be meaningful; thus, the interval for which ALL your power series are valid on is that which is less than the LEAST radius of convergence.