Is there Anybody can help me on this? Bless you. Express (2x^4-2x^3+x)/((2x-1)^2(x-2)) in partial fraction.

Answer 1

The simplest way to do this is as follows. We have

f(x) = (2x^4-2x^3+x)/((2x-1)^2(x-2))

and want to write f(x) in partial fractions. Now, each partial fraction is a term like 1/(a + b x) and then it is clear where this function has a singularity.

If you then expand the function around each singularity and subtract all the singular terms from
f(x), you get a function without any sungularities. One can then show that you are left with a polynomial. If the degree of the numerator is smaller than the degree of the denominator, that polynomial is zero. So, the singular terms are just the partial fractions.

The polynomial is nothing more than the "singularity at infinity". I.e. if you let x tend to infinity, the function can blow up too. This part can be treated in exactly the same way as the other partial fraction contributions by expanding the function f(x) around infinity (i.e. in powers of 1/x) and keeping only the terms with positive powers of x.

In this case f(x) has a singularity at x= 1/2, x = 2 and at x = infinity.

Expansion around x = 1/2:

f(1/2 + u) = u^(-2) (1/2 + u)/(u-3/2)
[2 (1/2 + u)^3 -2 (1/2 + u)^2 + 1] =

-1/4 u^(-2)- 1/2 u^(-1) + nonsingular terms =

-1/4 (x-1/2)^(-2)- 1/2 (x-1/2)^(-1) + nonsingular terms

The expansion around x = 2 is trivial to find, because we can write:

f(x) = 1/(x-2) g(x)

with g(x) = (2x^4-2x^3+x)/(2x-1)^2

So, around x = 2, the expansion is:

g(2)/(x-2) + nonsingular terms =

2/(x-2)

Finally, you expand f(x) around infinity. We have:

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

1/(2x) [1 + 2/(2x) + 3/(2x)^2 +
4/(2x)^3 + ...] =

1/2 x^(-1) + 1/2 x^(-2) + 3/8 x^(-3) + 1/4 x^(-4) + ...

1/(x-2) = 1/x 1/(1 - 2/x) =

x^(-1) [1 + 2/x + 4/x^2 + 8/x^3 + ...] =

x^(-1) + 2 x^(-2) + 4x^(-3) + 8 x^(-4) + ...

So, around infinity, the expansion is:

x(2x^3-2x^2+1)x^(-2) [1/2 + 1/2 x^(-1) + 3/8 x^(-2) + 1/4 x^(-3) + ...]
[1 + 2 x^(-1) + 4x^(-2) + 8 x^(-3) + ...] =

(2x^2-2x)(1/2 + 3/2 x^(-1)) =

x^2 + 2x - 3 + O(1/x)

Note that the O(1/x) terms are nonsigular terms when expanding around infinity.

So, the partial fraction expansion is:

f(x) =

x^2 + 2x - 3 + 2/(x-2) -1/4 1/(x-1/2)^2
- 1/2 1/(x-1/2)