Fractions are giving me more problems

[K.I.L.E.R]

A B(x + c)
_____ + ______ = HOW DID 'C' COME INTO THE PICTURE?
(x + b) (x + (b^2))

This is impossible.
How is this possible?

Orginal form:

A + B
---------------------
(a + b)(a + (b^2))

They call this crud "Partial fractions" but give no real explanations to the phantom variables.

 

[KimB]

Okay, something is very wrong with what you wrote.

First of all, I'd recommend using a

 block so that it comes out right.
Edit: Looks like it'd be best to also use a constant-width text editor, too, as the editor on the forums is not constant-width.  Oh, well.

Secondly, the "original form" you wrote down just cannot result in the partial fraction you wrote.

The idea with partial fractions is that if you have a multiple between two things on the denominator, you can consider that this original fraction is actually the sum of two fractions, and this denominator is the common denominator between these two fractions.

For example, if we have:
[code]
   1
--------
x(x+1)

...then we can write this as:

 A      B
--- + ----
 x     x+1

Here, A and B are unkown numbers that can be determined by enforcing that when you sum the second set of fractions, you get the first fraction for all possible values of x.

If you require that your parameters A and B be pure numbers, you may find this impossible to do if the denominators are higher-order polynomials. So you may instead make the numerator of one of the partial fractions into something like: Ax + B.

I'll perform a small example. Take the fraction:

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

Now, I want to write something of the form:

 A         B
--- + ---------
 x      x^2+1

...But I expect from experience that this will likely not work, as I'll have too few equations, so I instead write:

 A     Bx + C
--- + --------
 x      x^2+1

Now, I set this equal to the original fraction to find the values of A, B, and C. To do this, I multiply both sides of the equation by x(x^2 + 1), which results in:

A(x^2 + 1) + (Bx + C)x = 1

I'll now rewrite the above in a form that isolates each power of x:

x^2(A + B) + x(C) + A = 1

Now, the only way for the above equation to be valid for any possible value of x is if I have:

A + B = 0
C = 0
A = 1

...which sets:

A = 1
B = -1
C = 0

So finally I have:

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

...which can be checked easily.

Note that if you don't have high enough polynomials on the numerator of your partial fraction, or put them in the wrong place, you'll get contradictory answers for your values, such as A = 1 and A = 0.

 

[K.I.L.E.R]

Thanks.
This is the identity partial fraction?

I'm having a little trouble gathering where did the numbers come from.
I assume you solved them by sumultaneous equations?

 

[Mintmaster]

I'm having a little trouble gathering where did the numbers come from.
I assume you solved them by sumultaneous equations?

That's one way, but there's a nice trick you can use. Say you're doing

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

You want to rewrite this in the following form:

  2x - 3       A         B
---------- = -----  +  -----
(x+1)(x+2)   (x+1)     (x+2)

Multiply through by (x+1)(x+2), and you get:
2x-3 = A(x+2) + B(x+1)

Method 1:
The task at hand is to find A and B so that this holds for all x. So it must work for x = -1.
Plug that in, and you get 2(-1)-3 = A(-1+2) ==> A = -5
It must work for x = -2, also so plug that in and you get 2(-2)-3 = B(-2+1) ==> B = 7

So your final result is:

  2x - 3      -5         7
---------- = -----  +  -----
(x+1)(x+2)   (x+1)     (x+2)

Method 2:
The simultaneous equation method would expand and group 2x-3 = A(x+2) + B(x+1), so you wind up with:
(2 - A - B)x - 3 - 2A - B = 0.

The only way this holds for all x is if (2 - A - B) equals zero and (-3 - 2A - B) also equals zero. So you get the system of equations:
A + B = 2
2A + B = -3

Solving this gives you the same result, but requires a bit more work IMO.

The first method is pretty neat, but unfortunately it doesn't work for more complicated problems like the one Chalnoth showed you.