# Partial Fraction Decomposition with numerator degree more than denominators

• Feb 19th 2013, 02:51 PM
Partial Fraction Decomposition with numerator degree more than denominators
Typically, if the numerator's degree is more than the denominator, I think you do polynomial division, first (right?). But what if that is not convenient (or if we're just too lazy to distribute). Say, for example you want to partial fraction decompose:

$\frac{x^4}{(x-1)^3}$

I could FOIL the bottom, but we want to find if there is another way to do this. We tried

$\frac{x^4}{(x-1)^3} = \frac Ax + \frac B{x^2} + \frac C{x^3} = \frac {Ax^2 +(-2A+B)x +(A-B+C)}{(x-1)^3}$.

But, there are no $x^4$ terms on the right and so it ends up with A, B, and C all equal to 0 (not surprisingly).

So is the only way to do this to expand the bottom? What if we had to do partial fractions on $\frac{x^9}{(x-1)^8}$ (or bigger)? How do you partially fraction decompose that (sans a CAS)?
• Feb 19th 2013, 03:13 PM
chiro
Re: Partial Fraction Decomposition with numerator degree more than denominators

Are you trying to integrate this function? If so try the substitution u = x-1 and then simplify this integral to get a set of x^n type expressions which are easy to integrate.
• Feb 19th 2013, 03:15 PM
peruvian
Re: Partial Fraction Decomposition with numerator degree more than denominators
I haven't done this in ages but shouldn't it be:
x^4/(x-1)^3= A/(x-1) + B/(x-1)^2 + C/(x-1)^3

Then you have
x^4= A(x-1)^2 + B(x-1) + C
• Feb 20th 2013, 12:15 AM
BobP
Re: Partial Fraction Decomposition with numerator degree more than denominators
If you were to carry out the division, you would get an x term, a constant term, and a remainder which would be (in general) a quadratic divided by (x-1)^3.
The usual rules for partial fractions are only needed for the remainder (fraction) part.
So, assume that

$\frac{x^{4}}{(x-1)^{3}}=Ax+B+\frac{C}{(x-1)}+\frac{D}{(x-1)^{2}}+\frac{E}{(x-1)^{3}}.$

(In this particular case, we can see by inspection that A=1.)

With the x^9 case, the assumed expansion would be

$\frac{x^{9}}{(x-1)^{8}}=Ax+B+\frac{C}{(x-1)}+\frac{D}{(x-1)^{2}}+\dots + \frac{J}{(x-1)^{8}}.$
• Feb 20th 2013, 12:30 AM
Prove It
Re: Partial Fraction Decomposition with numerator degree more than denominators
Quote:

Typically, if the numerator's degree is more than the denominator, I think you do polynomial division, first (right?). But what if that is not convenient (or if we're just too lazy to distribute). Say, for example you want to partial fraction decompose:

$\frac{x^4}{(x-1)^3}$

I could FOIL the bottom, but we want to find if there is another way to do this. We tried

$\frac{x^4}{(x-1)^3} = \frac Ax + \frac B{x^2} + \frac C{x^3} = \frac {Ax^2 +(-2A+B)x +(A-B+C)}{(x-1)^3}$.

But, there are no $x^4$ terms on the right and so it ends up with A, B, and C all equal to 0 (not surprisingly).

So is the only way to do this to expand the bottom? What if we had to do partial fractions on $\frac{x^9}{(x-1)^8}$ (or bigger)? How do you partially fraction decompose that (sans a CAS)?

\displaystyle \begin{align*} \frac{x^4}{(x - 1)^3} &= \frac{x^4 - 4x^3 + 6x^2 - 4x + 1 + 4x^3 - 6x^2 + 4x - 1}{(x - 1)^3} \\ &= \frac{(x - 1)^4 + 4x^3 - 6x^2 + 4x - 1}{(x - 1)^3} \\ &= x - 1 + \frac{4x^3 - 6x^2 + 4x - 1}{(x - 1)^3} \\ &= x - 1 + \frac{ 4x^3 - 12x^2 + 12x - 4 + 6x^2 - 8x + 3 }{(x - 1)^3} \\ &= x - 1 + \frac{4(x - 1)^3 + 6x^2 - 8x + 3}{(x - 1)^3} \\ &= x - 1 + 4 + \frac{6x^2 - 8x + 3}{(x - 1)^3} \\ &= x + 3 + \frac{6x^2 - 8x + 3}{(x - 1)^3} \end{align*}

Now apply the partial fraction decomposition \displaystyle \begin{align*} \frac{A}{x - 1} + \frac{B}{(x - 1)^2} + \frac{C}{(x - 1)^3} \end{align*} to the remainder.