Partial fractions- repeated linear factors

In summary, the conversation is about partial fractions and how to split up a fraction into its partial fractions. It is mentioned that for repeated linear factors in the denominator, there correspond partial fractions of the form A/(x-a) + B/(x-a)^2. The question is asked if this is also true for factors like (a+x)^2 and it is explained that it is because it follows the same pattern. Another example is given to illustrate the method. The conversation then moves on to discussing why the denominators are written in the form A/(z-8) + B/(z-8)^2 + C/(z-8)^3 instead of just A/(z-8) + B/(z-8) +
  • #1
nirvana1990
46
0

Homework Statement




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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  • #2
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, science 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.
 
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  • #3
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!
 
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  • #4
nirvana1990 said:
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; that's why we do the ansatz : A/(z-8)+B/(z-8)^2+C/(z-8)^3
 
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  • #5
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
how about this :

something/(x^2-1)^2

how do you solve the denominator?

Thanks
 
  • #7
(x^2 - 1) = (x +1)(x - 1)

Now what do you suggest?

Hint: read all the posts done by be in this thread.
 
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  • #8
YuUZoe said:
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)
 
  • #9
BicolXpress said:
u mean the numerator?

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


No that is completely wrong.
 

1. What are partial fractions and how are they used?

Partial fractions are a method used to simplify and solve rational expressions. They involve breaking down a complex fraction into smaller, simpler fractions to make it easier to solve. This method is often used in integration and in solving systems of linear equations.

2. What are repeated linear factors in partial fractions?

Repeated linear factors in partial fractions refer to the situation where the denominator of a rational expression has the same linear factor repeated multiple times. For example, (x+2)^2 would be a repeated linear factor in the expression 3/(x+2)^2. This can occur in both proper and improper fractions.

3. How do you solve for coefficients in partial fractions with repeated linear factors?

To solve for coefficients in partial fractions with repeated linear factors, you first need to write out the partial fraction decomposition with unknown coefficients. Then, you can equate the numerators of the original expression and the decomposed expression, and solve for the unknown coefficients using algebraic methods.

4. Can partial fractions with repeated linear factors have imaginary solutions?

Yes, partial fractions with repeated linear factors can have imaginary solutions. This can occur when the original rational expression has complex coefficients and/or when there are repeated linear factors with complex roots. In these cases, the unknown coefficients in the partial fraction decomposition may also be complex numbers.

5. Are there any special cases or shortcuts when dealing with partial fractions with repeated linear factors?

Yes, there are a few special cases and shortcuts that can be used when dealing with partial fractions with repeated linear factors. For example, if the degree of the numerator is equal to or greater than the degree of the denominator, you can use polynomial long division to simplify the expression before breaking it down into partial fractions. Additionally, if there are only two repeated linear factors, you can use the method of undetermined coefficients to solve for the unknown coefficients.

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