I can't understand this in partial fractions

  • Thread starter dilan
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  • #51
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Oh I see. So this [itex]\lim_{N\to\infty}S_{N}=S[/itex] 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?
 
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  • #52
arildno
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Okay, you said that resolving into partial fractions is no problem, right?
 
  • #53
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Ya no problem at all.
 
  • #54
arildno
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All right, then!
Now, consider a fraction of the form [itex]\frac{1}{1+2x}[/itex]
If we write this as:
[tex]\frac{1}{1+2x}=\frac{1}{1-(-2x)}[/itex]
then from the above, we can write:
[tex]\frac{1}{1-(-2x)}=\sum_{n=0}^{\infty}(-2x)^{n}[/tex]
as long as |-2x|<1, that is |x|<1/2 (Remember that a number x and its negative, -x, has the same ABSOLUTE value!)

Agreed?
 
  • #55
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Yep agreed
 
  • #56
arildno
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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 [itex](-2x)^{4}[/itex], i.e, the coefficient is [itex](-2)^{4}=16[/itex]
Agreed?
 
  • #57
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oh ya I see. Yep agreed.
 
  • #58
arildno
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Okay, suppose we are to find the coefficient for the fourth power of the series expansion of the fraction [itex]\frac{1}{(1+2x)^{2}}[/itex]
As long as |x|<1/2, we may write:
[tex]\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})[/tex]

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 [itex]x^{4}[/itex]

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:
[tex](-2x)^{0}*(-2x)^{4}+(-2x)^{1}*(-2x)^{3}+(-2x)^{2}*(-2x)^{2}+(-2x)^{3}*(-2x)+(-2x)^{4}*(-2x)^{0}=5*16x^{4}=80x^{4}[/tex]

That is, the coefficent is 80.
Agreed?
 
  • #59
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Is this like the pascal trangle. I mean some what related to the binomial theorem?
 
  • #60
arildno
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Somewhat similar to that, yes! :smile:

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.
 
  • #61
arildno
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Hmm..I just reviewed your original question:
It seems to want the general termof the expansion, rather than the specific one.

1.Have you learnt about the Cauchy product of series yet?

2. Have you learnt about Taylorseries, and how to compute them?
 
  • #62
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you mean like |x|<1 right?
 
  • #63
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I am sort of

arildno said:
Hmm..I just reviewed your original question:
It seems to want the general termof the expansion, rather than the specific one.

1.Have you learnt about the Cauchy product of series yet?

2. Have you learnt about Taylorseries, and how to compute them?


Sorry arildno :frown: I havn't learnt any of them. Will it be very difficult for me to learn it?:smile:
 
  • #64
arildno
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What sort of class are you taking?
I'm suddenly unsure as to what method your teacher had in mind when giving that exercise.
 
  • #65
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I am just in the A/L class. Just finished my Local O/Ls. I just swiched to the english medium now. I don't know, realy that teacher is too advance sort of. Even the other students also sort of get fedup of maths like. But I am trying to somehow keep up. He told us that there is a big gap between O/Ls and A/Ls. He told that it is very difficult to cover this gap. He thinks that we have done everything. I don't know whether in the London sylluba it's there. But in our O/L syllubas non of this is there that you thought above except for the equation

Sn = n/2 (a+l)
and
Sn = n/2 (2a + (n-1)d)

Well that's the way. In our country only from arround 20% - #8% like pass Maths.
I am just trying to come into the standard. Well I believe that PF was the best place for me.
 
  • #66
arildno
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Seems that you've got a tough time ahead, but the right attitude to face it then! :smile:

Now, have you learnt that a function f(x) may be written in series form:
[itex]f(x)=f(0)+f'(0)x+f''(0)\frac{x^{2}}{1*2}+f'''(0)\frac{x^{3}}{1*2*3}+++[/tex]
where, say, f''(0) means the 2nd derivative of f, evaluated at x=0?
 
  • #67
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Well

arildno said:
Seems that you've got a tough time ahead, but the right attitude to face it then! :smile:

Now, have you learnt that a function f(x) may be written in series form:
[itex]f(x)=f(0)+f'(0)x+f''(0)\frac{x^{2}}{1*2}+f'''(0)\frac{x^{3}}{1*2*3}+++[/tex]
where, say, f''(0) means the 2nd derivative of f, evaluated at x=0?

I am sort of like stuck. But I have to face these. TO be honest arildno I don't know that also. :cry:
What am I to do? have any website that I can learn all these and come back to you?
 
  • #68
arildno
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Well, it isn't as bad as you think, because what I've tried to find out was if there was an ALTERNATIVE method of solving your problem than the one I initially thought of, and that you knew about.
That is; It isn't loads and loads of new stuff you have to learn in order to tackle your original question, from what we've covered in this thread, we may well find a way to answer your question without using the concept of derivatives, Taylor series and so on.

Just give me a bit of time, OK?
 
  • #69
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Ok no problem. Realy appreciate your intrest in helping me. Can't believe, but realy you are great person.
Thanks alot
Hope I will find a solution.
 
  • #70
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hi arildno,

Uhh were you able to get anything? Well it's ok take your time, I am not in a hurry.
 
  • #71
arildno
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Okay, I'll post something on the Cauchy product of infinite series first:
Suppose you've got two series [itex]S_{1}=\sum_{n=0}^{\infty}a_{n}x^{n}, S_{2}=\sum_{m=0}^{\infty}b_{m}x^{m}[/itex]
where [itex]a_{n},b_{m}[/itex] are coefficients for the n'th and m'th terms respectively (it is smart, and allowable, to use a different letter for the indices for the two series)

Now, suppose we wish to multiply the two series together to a single series S; how are the coefficients of S related to the coefficients of [itex]S_{1},S_{2}[/itex] ?

Now, let's just use a finite amount usual numbers, and see how the result OUGHT to be:
If we are to multiply the number (2+3+1) (that is, 6) with (5+7+8) (that is 20), we may do it as follows:
[itex](2+3+1)(5+7+8)=2*5+2*7+2*8+2*8+3*5+3*7+3*8+1*5+1*7+1*8[/itex]
The crucial point to notice that we multiply each term in the first parenthesis with each term in the other parenthesis, and then sum all the 3*3=9 terms together (3*3 terms since each parenthesis has 3 numbers).

Thus, generalizing to an infinite series, we could write our product as:
[tex]S=S_{1}*S_{2}=\sum_{n=0}^{\infty}a_{n}x^{n}*\sum_{m=0}^{\infty}b_{m}x^{m}=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_{n}b_{m}x^{n+m}[/tex]
where the last DOUBLE sum should be thought of as saying:
1. Fix n=0, and calculate the sum [itex]\sum_{m=0}^{\infty}a_{0}b_{m}x^{m+0}[/itex]
that is, multiply the term [itex]a_{0}x^{0}[/itex] with EACH term in [itex]S_{2}[/itex], and sum them together

2. ADD to your result from 1. the series you get by fixing n=1, i.e, by first calculating the sum [itex]\sum_{m=0}^{\infty}a_{1}b_{m}x^{m+1}[/itex], and then adding all of this to the result you got from 1.
(That is, multiplying the term [itex]a_{1}x^{1}[/itex] with EACH term in [itex]S_{2}[/itex], add up the result, and add this result with that you got from 1.)
3. And so on..

Got that?
 
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  • #72
arildno
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I'll assume you do, and continue:
Now, as with any sum of numbers, it shouldn't matter in which succession you add them together, right? (That is: 1+2+3=2+1+3=3+1+2 and so on)

We wish now to add up all terms in our double sum in ASCENDING powers of x.
We introduce a new index, j=n+m
Rewriting this, we have n=j-m
Now, for FIXED j, m can have any value between 0 and j.
Thus, we get the upper and lower limits on j and m:
[itex]0\leq{j}\leq\infty,0\leq{m}\leq{j}[/itex]
and we get, by substituting j=n+m or n=j-m:
[tex]\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}a_{n}b_{m}x^{n+m}=\sum_{j=0}^{\infty}\sum_{m=0}^{j}a_{j-m}b_{m}x^{j}=\sum_{j=0}^{\infty}c_{j}x^{j}, c_{j}=\sum_{m=0}^{j}a_{j-m}b_{m}[/tex]
That is, cj is the coefficient of the j'th power of x.

Example:
You were to multiply together:
[tex](1+(-2x)+(-2x)^{2}+++)(1+(-2x)+(-2x)^{2}+++)[/tex]
We can write this as the product of the series:
[tex]S=\sum_{n=0}^{\infty}(-2)^{n}x^{n}\sum_{m=0}^{\infty}(-2)^{m}x^{m}=\sum_{j=0}^{\infty}c_{j}x^{j}, c_{j}=\sum_{m=0}^{j}(-2)^{j-m}(-2)^{m}=\sum_{m=0}^{j}(-2)^{j}=(-2)^{j}\sum_{m=0}^{j}1=(j+1)(-2)^{j}[/tex]

Thus, we have the series representation:
[tex]\frac{1}{(1+2x)^{2}}=\sum_{j=0}^{\infty}(j+1)(-2)^{j}x^{j}[/tex]
That is, the coefficient of the j'th power of x is [itex](j+1)(-2)^{j}[/tex]
Note that this is in tune for j=4 which we already have calculated:
[tex]c_{4}=(4+1)(-2)^{4}=5*16=80[/tex]
 
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