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#1 2005-10-24 03:05:27

Andorin
Member
Registered: 2005-10-24
Posts: 4

Partial Fractions with Repeated Factors

Hi,

   I'm trying to solve a particular partial fraction namely:

         2X^2 - X + 1
      -------------------
      (X + 1)(X - 1)^2

   I am aware of the fractions that it is meant to split into... the numerators are all meant to be 1.

   Its how on earth they get there that confuses me.

   I can work C out (in terms of there being three constants/numerators named A, B and C) as being 1 but then I get A as being 1/2... what am I doing wrong??? Or is the book I'm working from barkers?

   Could someone run me through this one?

Cheers in advance

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#2 2005-10-24 03:59:15

ryos
Member
Registered: 2005-08-04
Posts: 394

Re: Partial Fractions with Repeated Factors

This one doesn't need A's and C's. You can just break it up, like so:

      2x²                   x                       1
-------------  -  --------------  +  ---------------
(x+1)(x-1)²     (x+1)(x-1)²       (x+1)(x-1)²

Expand the denominator:

          2x²                        x                         1
------------------  -  -------------------  +  ---------------
x³ - 3x² - x + 1      x³ - 3x² - x + 1       (x+1)(x-1)²

                         2x²
------------------------------------------
2x²[ (x/2) - (3/2) - (1/2x) + (1/2x²) ]

               x
--------------------------
x[ x² - 3x + (1/x) - 1 ]


                      1                                       1                            1
------------------------------------  -  ----------------------  +  --------------- 
(x/2) - (3/2) - (1/2x) + (1/2x²)      x² - 3x + (1/x) - 1        (x+1)(x-1)²

Ugly but true.

Last edited by ryos (2005-10-24 04:05:17)


El que pega primero pega dos veces.

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#3 2005-10-24 07:44:32

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Partial Fractions with Repeated Factors

I think that's far more complicated than it needs to be.

  2X² - X + 1
-------------------
(X+1)(X - 1)²

Split up into partial fractions:      A           B           C
                                             -----  +  -----  +  -------
                                             X+1       X-1       (X-1)²

Multiply out:  A(X-1)² + B(X+1)(X-1) + C(X+1) = 2X² - X + 1

Substitute in X = 1 : 2C = 2   ∴  C = 1
Substitute in X = -1: 4A = 4   ∴  A = 1
Find B: X² - 2X + 1 + B(X² - 1) + X + 1 = 2X² - X + 1
Simplify: B(X² - 1) = X² - 1  ∴  B = 1

Therefore,   2X² - X + 1             1           1           1
                -----------------        -----  +  -----  +  -------
                 ( X+1)(X-1)²          X+1       X-1       (X-1)²


Not as ugly, still true. smile


Why did the vector cross the road?
It wanted to be normal.

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