Thread: Finding the Constants in partial fractions

Originally Posted by Mr_Green

Find constants A, B, and C such that:


1/(x^3 - x) = A/(x-1) + B/(x) + C/(x+1)




Thanks

Note: 1/(x^3 -x)=1/x(x^2 - 1) = 1/x(x+1)(x-1) = A/(x-1) + B/x + C/(x+1)

The Lcm of A/(x-1) + B/x + C/(x+1) is x(x+1)(x-1), so we can rewrite A/(x-1) + B/x + C/(x+1) as [A(x^2 + x) + B(x^2 -1) + C(x^2 -x)]/x(x+1)(x-1)

Expanding the numerator we get:
(Ax^2 + Ax + Bx^2 - B + Cx^2 - Cx)/x(x+1)(x-1)
Grouping like powers of x we get:
[(A+B+C)x^2 + (A-C)x - B]/x(x+1)(x-1)

So now we have 1/x(x+1)(x-1) = [(A+B+C)x^2 + (A-C)x - B]/x(x+1)(x-1)
Since the denominators are equal, the numerators must be equal to maintain the equation.

=> 1 = (A+B+C)x^2 + (A-C)x - B
Now we equate like powers of x (Note we can think of 1 as 0x^2 + 0x +1)

Then A+B+C=0-------------(1)
A-C=0----------------(2)
B = -1 ----------------(3)

=> A+C=1 -------------------Plugged in the value of B
A-C=0 --------------------rewrite equation 2

Adding these equations eliminates C, so we end up with:

2A=1 => A=1/2.
But A+C = 1, => C=1/2.

Thus the constants are A=1/2, B=-1, C=1/2

So 1/(x^3 - x) = (1/2)/(x-1) - 1/(x) + (1/2)/(x+1), and you can check this.


There is another major method used to come up with the 3 equations, but this is the mainstream one I think.

Hello, Mr_Green!

This is a problem in Partial Fractions.

Find constants A, B, and C such that:
. . 1/(x³ - x) = A/(x - 1) + B/(x) + C/(x + 1)

. . . . . . . . . . . . . .1 . . . . . . . .A . . . .B . . . . C
We have: . ----------------- . = . --- + ------ + -------
. . . . . . . . x(x - 1)(x + 1) . . . . x . . .x - 1 . . x + 1


Multiply through by x(x - 1)(x + 1):

. . 1 .= .A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1)


Let x = 0: . 1 .= .A(-1)(1) + B(0)(1) + C(0)(-1) . → . A = -1

Let x = 1: . 1 .= .A(0)(2) + B(1)(2) + C(1)(0) . → . B = ½

Let x = -1: .1 .= .A(-2)(0) + B(-1)(0) + C(-1)(-2) . → . C = ½

Yup, that's a much nicer way to do it, gets to the point quickly, good for exams. (I think you switched the values for A and B though).