
#1
Oct107, 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 (xa)^2, there correspond partial fractions of the form : A/(xa) + B/(xa)^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!) 



#2
Oct107, 02:32 PM

Sci Advisor
HW Helper
P: 4,739

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}{(z8)^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) = (xa)(xb)*..*(xt) 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) = (x3)(x(4)) = (x3)(x+4) And if you have things like this: [tex] \dfrac{something}{(z8)^3} [/tex] Then you make this ansatz: [tex] \dfrac{A}{z8} + \dfrac{B}{(z8)^2} + \dfrac{C}{(z8)^3} [/tex] And if you have like: [tex] \dfrac{something}{z^2 + 3z8} [/tex] You make this ansatz: [tex] \dfrac{Az+B}{z^2 + 3z8} [/tex] And so on. 



#3
Oct107, 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/(z8)+B/(z8)^2+C/(z8)^3? Why do we not do : A/(z8)+B/(z8)+c/(z8)?[/quote] Add them! That's just (A+B+C)/(z8) 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. A/(z8)+ B/(z8)^{2}+ C/(z8)^{3}= A(z8)^{2}/(z8)^3+ B(z8)/(z8)^2+ C/(z8)^{3}= (Az^{2}16Az+ 64A+ Bz 8B+ C)/(z8)^{3}= (Az^{2}(16A+ B)z+ (64A8B+ C)/(z8)^{3}. That way, you can choose A, B, and C to make the coeffients anything you want. If you used only A/(z8), you can't! 



#4
Oct107, 02:49 PM

Sci Advisor
HW Helper
P: 4,739

Partial fractions repeated linear factorsJust 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)/(z8)^3 = A/(z8)+B/(z8)^2+C/(z8)^3 Then you can not have only (z8) terms in the denominator; that will give you: (z^2 + 5z + 4)/(z8)^3 = (A + B + C)/(z8) And that does not (in general) work; thats why we do the ansatz : A/(z8)+B/(z8)^2+C/(z8)^3 



#5
Oct107, 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! 



#6
Oct1007, 08:34 PM

P: 2

how about this :
something/(x^21)^2 how do you solve the denominator? Thanks 



#7
Oct1107, 12:04 AM

Sci Advisor
HW Helper
P: 4,739

(x^2  1) = (x +1)(x  1)
Now what do you suggest? Hint: read all the posts done by be in this thread. 



#8
Oct1107, 04:41 AM

P: 1

u mean the numerator? y/(x^21)^2 = A/(x+1) + B/(x1) 



#10
Oct1107, 05:29 AM

HW Helper
P: 3,353




Register to reply 
Related Discussions  
Partial Fractions  Calculus & Beyond Homework  11  
Partial Fractions, Irreducible quadratic factors  Calculus & Beyond Homework  2  
Partial fractions  Precalculus Mathematics Homework  1  
Please i need some help in Partial Fractions  Calculus & Beyond Homework  1  
Partial Fractions..  Linear & Abstract Algebra  9 