Question

can someone show the process for solving this partial fraction? http://math.feld.cvut.cz/mt/exd/1/gif1/ecd1aj.gif

Answer 1

is the denominator in prime factors already?

Answer 2

regradless, it look like you have degree 2 factors; which would require us to account for at best, linear numerators

Answer 3

\[\frac{1}{f^2g^2}=\frac{a_1x+b_1}{f^2}+\frac{a_2x+b_2}{g^2}\]
then its just a matter of redoing the addition and comparing parts

Answer 4

Hello, Castiel,

The rational function you present is

1
--------------------- and this can be re-written as
(x^2 + 1)(x^2 + 2x + 1)


1
---------------------- .
(x^2 + 1) (x + 1)^2

In words, you have a "repeated root" in that (x + 1)^2.

I have little idea where you're coming from and am prepared to explain in more detail if need be But for now, you can break up your rational expression/ function as follows:

1 Ax + B C D
----------------- = ----------- + ------------- + -------------
(x^2 + 1) (x + 1)^2 (x^2 + 1) x + 1 (x + 1)^2


The next task is to determine the values of A, B, C and D that make this equation true.

A good place to start is to multiply each of the three fractions on the right by the quantity necessary to make every denominator = to (x^2 + 1) (x + 1)^2 .

Then you can cross out all of the (equal) denominators and concentrate on the equation in x, A, B, C and D that remains.

Need help in determining A, B, C and D? If so, just ask.

Best to you.

Answer 5

Sorry I haven't been more accurate about the problem I'm having. First of all, thank you for your explanation, it has been really helpful. I can do some simpler fractions of this sort but I'm having trouble finding a,b,c,d when I have repeated root and Ax+B. In what order should I start determining them? Thank you.

Answer 6

Castiel,

If you do as I've suggested, you'll get the following equation:

1=(Ax+B)(x+1)^2 + C(x^2+1)(x+1) + D(x^2+1).

I'd suggest you verify this for yourself.

There are several ways to solve for the coefficients A, B, C and D.
I believe you'll end up in every case with a system of four equations in four unknowns, which could be solved with matrices, determinants, addition/subtraction and/or substitution.

I note that substituting the x value -1 will result in the following equation:
1 = (Ax+B)(0) + C(2)(0) + D(3). Therefore, D = 1/3.

This would leave you with a system of 3 equations in 3 unknowns.

Since there is no other convenient x value at which the terms of my equation (above) would be mostly zero, we're left with the possibility of choosing other, distinct, simple values of x, such as x=0, x=1, x=-2, x=2, and so on.

Each time you substitute such a value, you'll get a new set of linear equations in A, B, C and D (or fewer, if you've already found values such as D = 1/3).

Want to give this a try?

Just ask if you have more questions. Good luck!

Answer 7

PS: The order in which you determine your A, B, C, D is immaterial. It just happened that I was able to determine that D = 1/3 simply by letting x = 1. As I said earlier, there is no other convenient x value ... (see above).