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Assume we are working with complex numbers . To reduce a polynomial fraction to partial fractions is to rearrange it so it is the sum of terms of the form a/(x-b)n Suppose we have two polynomials P(x) and Q(x), and we consider the fraction F(x)=P(x)/Q(x). The Fundamental theorem of algebra tells us we can factorise Q(x) completly, so
Q(x)=(x-p0)n0(x-p1)n1...(x-pk)nk
p0,... pk are the poles of the fraction : they are where Q(x) is zero and thus where the fraction blows up. The aim of the game is to find numbers ai,j and a polynomial R(x) of degree less than that of Q such that :
F(x)=a0,1/(x-p0) + ... a0,n0/(x-p0)n0+...+ak,1/(x-pk) + ... ak,nk/(x-pk)nk + R(x)
R(x) is just the remainder from the division of P(x) by Q(x). Obtaining the ai,ni is fairly easy. Just multiply the whole fraction by (x-pi)ni. All the terms in the expression will now have a power of (x-pi) in front of them, except the term the term with ai,ni. Evaluating (x-pi)niF(x) at pi allows you to get ai,ni.

The trouble starts when you want to get the other coefficients. You can't just multiply by the appropriate polynomial and evaluate at the appropriate pole because some (x-p) terms will still be left in the denominator. One solution is to subtract ai,ni/(x-pi)ni from F(x). You can then multiply the resulting fraction G(x) by (x-pi)ni-1 and evaluate it at pi to obtain ai,ni-1. Repeat for all the multiplicities of all the poles and you will have F expressed in terms of partial fractions.

A simple example is F(x)=1/(x2-1). Its poles are 1 and -1.

F(x)(x-1)=1/(x+1)
F(x)(x+1)=1/(x-1)
Evaluating these 2 at 1 and -1 respectively gives 1/2 and -1/2.
Thus F(x) = 1/(2(x-1)) - 1/(2(x+1))

Although finding the coefficients is not always easy for large fractions, it makes many operations easier. In particular, integration is much easier, since we know the antiderivative of a/(x-b)^n.

The above writeup by Brontosaurus is most excellent. It is a sterling explanation of partial fractions by a mathematician, for mathematicians. However, the rest of us might find it a little difficult to follow - I likes my poles North, South, or in Poland, so I do. So I present to you, without further ado:

Partial Fractions For The Masses
A subsidiary of Maths for the Masses

A little background knowledge is required. I'm guessing, if you're reading this node, you already know a little bit about maths in general - specifically, quadratic equations. If you don't know about them, don't panic, just take a breath, go swot up on factoring quadratic equations. This node will still be here when you come back probably.

Finished? Good, then I'll go on. If you read that node above (good noder, here is a bun) then you're all set - if you didn't, then you've got some nerve, mister. Either way, let's begin.

Partial fractions have a few uses throughout maths - most notably integration and summing infinite series. For more information on integrating with partial fractions, read to the end of this node. But anyway, let's begin. For real this time.

1. Simple Partial Fractions

Now. The most basic algebraic fraction involving a quadratic you're likely to come across is:

       Y     
  ------------
  ax2 + bx + c
Now, using your newfound (or longstanding) knowledge of quadratic equations, we can factorise that bottom line into the form
       Y
  ------------      Y = any constant
 (x + a)(x + b)

This can be split into its partial fractions as shown:

       Y                A         B
  ------------   =    -----  +  -----
 (x + a)(x + b)      (x + a)   (x + b)

    ORIGINAL         PARTIAL FRACTIONS

    where A and B are constants.

Now here's the clever part. You wind the film in the reel backwards- no, sorry, wrong node. What you really do is compare coefficients. To do this, you need to multiply each of the partial fractions by a value that will make the respective denominators the same as the original denominator.

Confused? Don't worry, here's a lovely example:
       Y        <--- numerator
  ------------                      ORIGINAL
 (x + a)(x + b) <--- denominator
So the denominator is (x + a)(x + b). If we look at our two partial fractions, we see that the 'A' denominator needs to be multiplied by (x + b), and the 'B' denominator needs to be multiplied by (x + a). Thus:
    A       (x + b)        A(x + b)
  -----  *   -----   =   ------------
 (x + a)    (x + b)     (x + a)(x + b)

    B       (x + a)        B(x + a)
  -----  *   -----   =   ------------
 (x + b)    (x + a)     (x + a)(x + b)
Now we can really get to work. Since the denominators are the same across the board, for the moment we can ignore them and concentrate on the numerators, or coefficients. Thus, we have Y = A(x + b) + B(x + a). All we need to do now is work out what A and B are equal to, and we do this by substituting in values for x that result in one of the equations equalling zero. Like so:
   Y = A(x + b) + B(x + a)

 Sub in x = -b:
   Y = A(-b + b) + B(-b + a)
   Y = A(0) + B(a - b)
   Y = B(a - b)

   B = Y / (a - b)

 Similarly, to find A, sub in x = -a:
   Y = A(-a + b) + B(0)
   Y = A(b - a)

   A = Y / (b - a)
And that's nearly it. You now have values for A and B. Sub them into the original equation and you're laughing.
The following is a worked example, using real proper numbers instead of silly incomprehensible letters.

       10
 -------------
 (x2 + 3x - 4)

Convert the quadratic into its factors:
 x2 + 3x - 4 = (x + 4)(x - 1)


Therefore

       10             A         B
 -------------  =   -----  +  -----
 (x2 + 3x - 4)     (x + 4)   (x - 1)


Cross-multiply, and take coefficents:
 10 = A(x - 1) + B(x + 4)


Sub in x = -4 to negate B:
 10 = A(-4 - 1) + B(0)
 10 = A(-5)

 A = 10 / -5
 A = -2


Now sub in x = 1 to negate A:
 10 = A(1 - 1) + B(1 + 4)
 10 = A(0) + B(5)

 B = 10 / 5
 B = 2

Put these back into the original, and we get:

       10            -2         2
 -------------  =   -----  +  -----
 (x2 + 3x - 4)     (x + 4)   (x - 1)

That wasn't so bad, was it? "But," I hear you say, "what if I have 10x instead of 10 as my original coefficient? Or 5x? Or 10998521x?" No problem, bucky! Simply substitute your value for x into that x as well. Say, for the example above, it was 10x instead of just 10:

 10x = A(x - 1) + B(x + 4)

Sub in x = -4 to negate B:
 10(-4) = A(-4 - 1) + B(0)
 -40 = A(-5)

 A = -40 / -5
 A = 8

Now sub in x = 1 to negate A:
 10(1) = A(0) + B(1 + 4)
 10 = B(5)

 B = 10 / 5
 B = 2

So our equation becomes

      10x            40         2
 -------------  =   -----  +  -----
 (x2 + 3x - 4)     (x + 4)   (x - 1)
Simple as that.
There is a slight complication.. if you are making partial fractions so you can integrate them, then the numerator must be of a lower degree than the denominator. To achieve this, use long division.
Now, let's make things slightly trickier.

2. Trickier Partial Fractions

So, you think you're a hotshot now. You is da partial fractions mastah. Here's a new trick, then; how do you solve a fraction like this?

       20
  ------------
 (x + 5)(x + 7)2
It is not - and that is NOT in big-assed capital letters - this equation:
       20            A         B         C
  ------------  =  -----  +  -----  +  -----
 (x + 5)(x + 7)2   (x + 5)   (x + 7)   (x + 7)
The correct equation, and yes, I know it doesn't make any fucking sense, is this:
       20            A         B         C
  ------------  =  -----  +  -----  +  -----
 (x + 5)(x + 7)2   (x + 5)   (x + 7)   (x + 7)2
I know that seems a bit illogical; just accept it. To solve this equation, let's consider each P.F. individually, like in the first example way up there.
    A      (x + 7)2      A(x + 7)2
  -----  *  -----   =  ------------
 (x + 5)   (x + 7)2   (x + 5)(x + 7)2

    C      (x + 5)      C(x + 5)
  -----  *  -----   =  ------------
 (x + 7)2  (x + 5)    (x + 5)(x + 7)2
Now, the B part is oddish. Since it's over (x + 7), and the denominator to be achieved is (x + 5)(x + 7)2, we therefore need to multiply by (x + 5)(x + 7).
    B      (x + 5)(x + 7)   B(x + 5)(x + 7)
  -----  *  ------------  =  -------------
 (x + 7)   (x + 5)(x + 7)   (x + 5)(x + 7)2
Now all that remains is to solve that mother. It's much the same as that first method above (noticing a pattern?) but with the added complication of simultaneous equations. God, I love those.
Taking coefficients:
 20 = A(x + 7)2 + B(x + 5)(x + 7) + C(x + 5)

Let x = -7:
 20 = A(0)2 + B(-7 + 5)(0) + C(-7 + 5)
 20 = -2C

 C = 20 / -2
 C = -10
Note that both A and B were negated.

Now let x = -5:
 20 = A(-5 + 7)2 + B(0)(-5 + 7) + C(0)
 20 = 4A

 A = 20 / 4
 A = 5
This time, B and C were negated.
Now, we have values for A and C, but not B. To find B, we need to sub in the A and C values, and any value for x that does not negate B. For the sake of argument, let's pick 0 as our value.
Sub in A = 5, C = -10 and x = 0:
 20 = 5(0 + 7)2 + B(0 + 5)(0 + 7) + (-10)(0 + 5)
 20 = 245 + 35B - 50

 20 - 245 + 50 = 35B
 -175 = 35B

 B = -175 / 35
 B = -5

Now bring it all together:
       20             5        -5        -10
  ------------  =   -----  +  -----  +  -----
 (x + 5)(x + 7)2   (x + 5)   (x + 7)   (x + 7)2

       OR

       20             5         5        10 
  ------------  =   -----  -  -----  -  -----
 (x + 5)(x + 7)2   (x + 5)   (x + 7)   (x + 7)2
And that's it. Now you can solve the vast majority of partial fractions.

Just before you go, there is one final example I'd like to show you.

3. Quadratic Partial Fractions

This is a quick quick quick explanation. By now, you should be able to create partial fractions very easily. However, sometimes you'll end up with a P.F with a denominator of (ax2 + bx + c) which for one reason or another you cannot factorise further. Have no fear: instead of crying, dry those eyes and use the numerator Ax + B. Example:

             Y
  ----------------------
 (ax2 + bx + c)(ex + f)

becomes

     Ax + B          C
  ------------  +  ------
 (ax2 + bx + c)   (ex + f)
And this can be solved simply by using the methods detailed above. Magick, eh?

Class dismissed. There will be a test next Thursday, in the normal tutorial slot.




Partial fractions and integration
As promised, here's a brief example of how partial fractions are used in integration. Only read if you know the basics of integration.

Say we have the integral
  /\
  |    (x + 1)
  |  ----------- dx
  | (x2 + 3x + 2)
 \/

One of the rules when integrating fractions is that the numerator has to equal the differential of the denominator, or numerator = d/dx(denominator). Since this is not true in this case, we can split up the fraction into.. wait for it.. partial fractions.
 (x2 - 3x + 2) = (x - 1)(x - 2)

    (x + 1)         A         B
  -----------  =  -----  +  -----
 (x2 + 3x + 2)   (x - 1)   (x - 2)

Now, if we were to integrate both sides, we would get:

  /\                    /\              /\
  |    (x + 1)          |    A          |    B
  |  ----------- dx  =  |  ----- dx  +  |  ----- dx
  | (x2 + 3x + 2)       | (x - 1)       | (x - 2)
 \/                    \/              \/

Solve for A and B, using the above methods, and we get
  A = -2  and  B = 3.

Sub these values in:

  /\                    /\              /\
  |    (x + 1)          |   -2          |    3
  |  ----------- dx  =  |  ----- dx  +  |  ----- dx
  | (x2 + 3x + 2)       | (x - 1)       | (x - 2)
 \/                    \/              \/

Rearrange:

  /\                      /\               /\
  |    (x + 1)            |    1           |    1
  |  ----------- dx  =  -2|  ----- dx  +  3|  ----- dx
  | (x2 + 3x + 2)         | (x - 1)        | (x - 2)
  \/                     \/               \/

And finally, integrating using the fraction rule:

 (Integral of algebraic fraction) = -2loge(x - 1) + 3loge(x - 2) + c

  = 3loge(x - 2) - 2loge(x - 1) + c


Rules for Integrating Partial Fractions

  • The numerator must be of a lower degree than the denominator. (If not, use long division).
  • Factorise the denominator into its prime factors.
  • Linear factor (ax + b) gives a PF of the form (A / (ax + b)).
  • Factors (ax + b)2 give P.Fs (A / (ax + b)) + (B / (ax + b)2).
  • Factors (ax + b)3 give P.Fs (A / (ax + b)) + (B / (ax + b)2) + (C / (ax + b)3).
  • Quadratic factor (ax2 + bx + c) gives a P.F ((Ax + B) / (ax2 + bx + c)).


Information from class notes / study. Integration example from Lecture 14, Maths 1M(ii), QUB Electrical Engineering.
This node sponsored by Maths for the Masses, the letter j and the number 3.1415926535.

Lurid thanky dance at:
  montecarlo for suggesting I give some uses for PFs;
  10998521 says "With regard to applications, you can also use them to help sum infinite series, e.g. 1/n(n+1)";
  jrn for various stylistic nitpicks

I'm coming for you, downvoter! You cannot escape me for long!

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