related maths special series

Partial fraction decomposition or partial fraction expansion is used to reduce the degree of either the numerator or the denominator of a rational function.

The outcome of partial fraction expansion expresses that function as a sum of fractions, where:

* the denominator of each term is a power of an irreducible (not factorable) polynomial and
* the numerator is a polynomial of smaller degree than that irreducible polynomial.

(This is taken from Wikipedia!)

I will put more explanation for this thru examples below

Explanation by an example
To express (x-1)/((3x-5)(x-3) as a partial fraction,
we write..

x - 1 A B
--------------- = -------- + -------
(3x - 5)(x - 3) (3x - 5) (x - 3)

Now multiply back and compare coefficients. RHS =

A(x-3) + B(3x-5)
-----------------------
(x-3)(3x-5)

Ax + B3x-3A-5B
=-----------------------
(x-3)(3x-5)

so A+3B=1
3A+5B=1

now solve these to get A=-1/2 and B=1/2

So that is the partial fraction for the given question..

We will take a little bit mroe complex example to see this one

multiply this whoe thing by (x+1) and then put x=-1

So (3.1-12+11)/(-1+2)(-1+3) = A = 1

Similarly on multiplyin the first equation by (x+2) and substuting x=-2 we get B=1 and C=1

so the partial fractin reduces to

Another type:

To solve for A: Set the denominator of the first fraction to zero, 1 − 2x = 0. Solving for x gives the Parāvartya value for A, when x = ½. When we substitute this value, x = ½, into the relation of numerators we have 3(1/2) + 5 = A + B(0). Solving for A gives us A = 3/2 + 5 = 13/2. Hence, numerator A equals six and one-half.

To solve for B: Since the equation of the numerators, here, 3x + 5 = A + B(1 − 2x), is true for all values of x, pick a value for x and use it to solve for B. As we have solved for the value of A above, A = 13/2, we may use that value to solve for B.

Using this method,

Example 1

Here, the denominator splits into two distinct linear factors:

q(x) = x2 + 2x − 3 = (x + 3)(x − 1)

so we have the partial fraction decomposition

Multiplying through by x2 + 2x - 3, we have the polynomial identity

1 = A(x − 1) + B(x + 3)

Substituting x = -3 and x = 1 into this equation gives A = -1/4 and B = 1/4, so that

Example 2

After long-division, we have

Since (-4)2-4(8) = -16 < 0, x2 - 4x + 8 is irreducible, and so

Multiplying through by x3 - 4x2 + 8x, we have the polynomial identity

4x2 − 8x + 16 = A(x2 − 4x + 8) + (Bx + C)x

Taking x = 0, we see that 16 = 8A, so A = 2. Comparing the x2 coefficients, we see that 4 = A + B = 2 + B, so B = 2. Comparing linear coefficients, we see that -8 = -4A + C = -8 + C, so C = 0. Altogether,

The following example illustrates almost all the "tricks" one would need to use short of consulting a computer algebra system.

Example 3

After long-division and factoring, we have

The partial fraction decomposition takes the form

Multiplying through by (x - 1)3(x2 + 1)2 we have the polynomial identity

x5 − 2x4 + 5x3 − 5x2 + 6x − 1
= A(x − 1)2(x2 + 1)2 + B(x − 1)(x2 + 1)2 + C(x2 + 1)2 + (Dx + E)(x − 1)3(x2 + 1) + (Fx + G)(x − 1)3

Taking x = 1 gives 4 = 4C, so C = 1. Similarly, taking x = i gives 2 + 2i = (Fi + G)(2 + 2i), so Fi + G = 1, so F = 0 and G = 1 by equating real and imaginary parts. We now have the identity

x5 − 2x4 + 5x3 − 5x2 + 6x − 1
= A(x − 1)2(x2 + 1)2 + B(x − 1)(x2 + 1)2 + (x2 + 1)2 + (Dx + E)(x − 1)3(x2 + 1) + (x − 1)3

Taking constant terms gives E = A - B + 1, taking leading coefficients gives A = -D, and taking x-coefficients gives B = 3 - D - 3E. Putting all of this together, E = A - B + 1 = -D - (3 - D - 3E) + 1 = 3E - 2, so E = 1 and A = B = -D. Now,

x5 − 2x4 + 5x3 − 5x2 + 6x − 1
= A(x − 1)2(x2 + 1)2 + A(x − 1)(x2 + 1)2 + (x2 + 1)2 + ( − Ax + 1)(x − 1)3(x2 + 1) + (x − 1)3

Taking x = -1 gives -20 = -8A - 20, so A = B = D = 0. The partial fraction decomposition of f(x) is thus