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Partial fractions- repeated linear factors

by nirvana1990
Tags: factors, fractions, linear, partial, repeated
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nirvana1990
#1
Oct1-07, 02:26 PM
P: 46
1. The problem statement, all variables and given/known data


I don't understand something I have read about partial fractions so I wonder if anyone can help!

To each repeated linear factor in the denominator of the form (x-a)^2, there correspond partial fractions of the form : A/(x-a) + B/(x-a)^2

Is this true if we have (a+x)^2? Also why is this true? The book I'm using doesn't have an explanation and I can't find much on the internet (well nothing that I understand!)
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malawi_glenn
#2
Oct1-07, 02:32 PM
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you mean if we have like:

[tex] \dfrac{3z + 1}{(3+z)^2} [/tex]

you do the same thing as if you would have:

[tex] \dfrac{5z+ 2}{(z-8)^2} [/tex]

The form you are talking about, is that it is costumary to write so, scince you can factorize a polynomial as this:
P(x) = (x-a)(x-b)*..*(x-t)

where: a,b,..,t are zeroths to the polynomial.

In an example maybe a is 3 and b = -4, then you'll have :
P(x) = (x-3)(x-(-4)) = (x-3)(x+4)




And if you have things like this:

[tex] \dfrac{something}{(z-8)^3} [/tex]

Then you make this ansatz:

[tex] \dfrac{A}{z-8} + \dfrac{B}{(z-8)^2} + \dfrac{C}{(z-8)^3} [/tex]

And if you have like:
[tex] \dfrac{something}{z^2 + 3z-8} [/tex]

You make this ansatz:
[tex] \dfrac{Az+B}{z^2 + 3z-8} [/tex]

And so on.
nirvana1990
#3
Oct1-07, 02:43 PM
P: 46
Thanks for the reply.
When we're splitting up the fraction into partial fractions why do we then split up the denominator like this: A/(z-8)+B/(z-8)^2+C/(z-8)^3?
Why do we not do : A/(z-8)+B/(z-8)+c/(z-8)?[/quote]
Add them! That's just (A+B+C)/(z-8) and since A, B, C are just some constants, their sum is just "some constant"- that would be exactly the same as A/(z+8) with different A.

I have seen an example showing that the method I just wrote is inconsistent (although in the example they used a squared term) but I don't see why we have squared terms and cubed terms as the denominators since (z-8)^3 is (z-8)(z-8)(z-8) why don't we just form the partial fractions using these denominators?
They are not "independent". You need the different powers so that you can get z2 and z in the numerator:
A/(z-8)+ B/(z-8)2+ C/(z-8)3= A(z-8)2/(z-8)^3+ B(z-8)/(z-8)^2+ C/(z-8)3= (Az2-16Az+ 64A+ Bz- 8B+ C)/(z-8)3= (Az2-(16A+ B)z+ (64A-8B+ C)/(z-8)3. That way, you can choose A, B, and C to make the coeffients anything you want. If you used only A/(z-8), you can't!

malawi_glenn
#4
Oct1-07, 02:49 PM
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Partial fractions- repeated linear factors

Quote Quote by nirvana1990 View Post
Thanks for the reply.
When we're splitting up the fraction into partial fractions why do we then split up the denominator like this: A/(z-8)+B/(z-8)^2+C/(z-8)^3?
Why do we not do : A/(z-8)+B/(z-8)+c/(z-8)?

I have seen an example showing that the method I just wrote is inconsistent (although in the example they used a squared term) but I don't see why we have squared terms and cubed terms as the denominators since (z-8)^3 is (z-8)(z-8)(z-8) why don't we just form the partial fractions using these denominators?

Just because what you said, that leads to an inconsistency. I may not have time to give you a full mathematical proof now, I will just say that sometimes we must accept things as they are and move on. maybe later you can go back and prove it yourself :)

As with the fundamental theorem of algebra, that theoreme is used in on your first algebra courses, but you have no ability to prove it until you reach Complex analysis.

maybe some other guy have the time to give you the full proof. Partial fractions is used quite often, so it is very good to know the technuiqes.

But I can give you a clue; you want to have:
(z^2 + 5z + 4)/(z-8)^3 = A/(z-8)+B/(z-8)^2+C/(z-8)^3
Then you can not have only (z-8) terms in the denominator; that will give you:
(z^2 + 5z + 4)/(z-8)^3 = (A + B + C)/(z-8)

And that does not (in general) work; thats why we do the ansatz : A/(z-8)+B/(z-8)^2+C/(z-8)^3
nirvana1990
#5
Oct1-07, 03:45 PM
P: 46
Aahh it says I wrote that reply up there! Anyway thanks for the help- so the only reason we do that is so that we are able to compare the coefficients? I don't think I'll look for the proof online! Maybe I'll just move on to the next chapter....
Thanks again!
YuUZoe
#6
Oct10-07, 08:34 PM
P: 2
how about this :

something/(x^2-1)^2

how do you solve the denominator?

Thanks
malawi_glenn
#7
Oct11-07, 12:04 AM
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(x^2 - 1) = (x +1)(x - 1)

Now what do you suggest?

Hint: read all the posts done by be in this thread.
BicolXpress
#8
Oct11-07, 04:41 AM
P: 1
Quote Quote by YuUZoe View Post
how about this :

something/(x^2-1)^2

how do you solve the denominator?

Thanks

u mean the numerator?

y/(x^2-1)^2 = A/(x+1) + B/(x-1)
malawi_glenn
#9
Oct11-07, 04:43 AM
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Quote Quote by BicolXpress View Post
u mean the numerator?

y/(x^2-1)^2 = A/(x+1) + B/(x-1)

No that is completly wrong.
Gib Z
#10
Oct11-07, 05:29 AM
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http://www.physicsforums.com/showthread.php?t=149559


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