Editing Partial fraction decomposition (section)

== Examples ==

=== Example 1 ===

<math display="block">f(x)=\frac{1}{x^2+2x-3}</math>

Here, the denominator splits into two distinct linear factors:

<math display="block">q(x)=x^2+2x-3=(x+3)(x-1)</math>

so we have the partial fraction decomposition

<math display="block">f(x)=\frac{1}{x^2+2x-3} =\frac{A}{x+3}+\frac{B}{x-1}</math>

Multiplying through by the denominator on the left-hand side gives us the polynomial identity

<math display="block">1=A(x-1)+B(x+3)</math>

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

<math display="block">f(x) =\frac{1}{x^2+2x-3} =\frac{1}{4}\left(\frac{-1}{x+3}+\frac{1}{x-1}\right)</math>

=== Example 2 ===

<math display="block">f(x)=\frac{x^3+16}{x^3-4x^2+8x}</math>

After [[Polynomial long division|long division]], we have

<math display="block">f(x)=1+\frac{4x^2-8x+16}{x^3-4x^2+8x}=1+\frac{4x^2-8x+16}{x(x^2-4x+8)}</math>

The factor ''x''<sup>2</sup> − 4''x'' + 8 is irreducible over the reals, as its [[discriminant]] {{math|1=(−4)<sup>2</sup> − 4×8 = −16}} is negative. Thus the partial fraction decomposition over the reals has the shape

<math display="block">\frac{4x^2-8x+16}{x(x^2-4x+8)}=\frac{A}{x}+\frac{Bx+C}{x^2-4x+8}</math>

Multiplying through by ''x''<sup>3</sup> − 4''x''<sup>2</sup> + 8''x'', we have the polynomial identity

<math display="block">4x^2-8x+16 = A \left(x^2-4x+8\right) + \left(Bx+C\right)x</math>

Taking ''x'' = 0, we see that 16 = 8''A'', so ''A'' = 2.  Comparing the ''x''<sup>2</sup> coefficients, we see that 4 = ''A'' + ''B'' = 2 + ''B'', so ''B'' = 2.  Comparing linear coefficients, we see that −8 = −4''A'' + ''C'' = −8 + ''C'', so ''C'' = 0.  Altogether,

<math display="block">f(x)=1+2\left(\frac{1}{x}+\frac{x}{x^2-4x+8}\right)</math>

The fraction can be completely decomposed using [[complex numbers]]. According to the [[fundamental theorem of algebra]] every complex polynomial of degree ''n'' has ''n'' (complex) roots (some of which can be repeated). The second fraction can be decomposed to:

<math display="block">\frac{x}{x^2-4x+8}=\frac{D}{x-(2+2i)}+\frac{E}{x-(2-2i)}</math>

Multiplying through by the denominator gives:

<math display="block">x=D(x-(2-2i))+E(x-(2+2i)) </math>

Equating the coefficients of {{math|''x''}} and the constant (with respect to {{math|''x''}}) coefficients of both sides of this equation, one gets a system of two linear equations in {{math|''D''}} and {{math|''E''}}, whose solution is

<math display="block">D=\frac{1+i}{2i}=\frac{1-i}{2}, \qquad E=\frac{1-i}{-2i}=\frac{1+i}{2}.</math>

Thus we have a complete decomposition:

<math display="block">f(x)=\frac{x^3+16}{x^3-4x^2+8x}=1+\frac{2}{x}+\frac{1-i}{x-(2+2i)}+\frac{1+i}{x-(2-2i)}</math>

One may also compute directly {{math|''A'', ''D''}} and {{math|''E''}} with the residue method (see also example 4 below).

=== Example 3 ===

This example illustrates almost all the "tricks" we might need to use, short of consulting a [[computer algebra system]].

<math display="block">f(x)=\frac{x^9-2x^6+2x^5-7x^4+13x^3-11x^2+12x-4}{x^7-3x^6+5x^5-7x^4+7x^3-5x^2+3x-1}</math>

After [[Polynomial long division|long division]] and [[polynomial factorization|factoring]] the denominator, we have

<math display="block">f(x)=x^2+3x+4+\frac{2x^6-4x^5+5x^4-3x^3+x^2+3x}{(x-1)^3(x^2+1)^2}</math>

The partial fraction decomposition takes the form

<math display="block">\frac{2x^6-4x^5+5x^4-3x^3+x^2+3x}{(x-1)^3(x^2+1)^2} = \frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{(x-1)^3}+\frac{Dx+E}{x^2+1}+\frac{Fx+G}{(x^2+1)^2}.</math>

Multiplying through by the denominator on the left-hand side we have the polynomial identity

<math display="block">\begin{align}
&2x^6 - 4x^5 + 5x^4 - 3x^3 + x^2 + 3x \\[4pt] 
={}&A\left(x-1\right)^2 \left(x^2+1\right)^2+B\left(x-1\right)\left(x^2+1\right)^2 +C\left(x^2+1\right)^2 + \left(Dx+E\right)\left(x-1\right)^3\left(x^2+1\right)+\left(Fx+G\right)\left(x-1\right)^3
\end{align}</math>

Now we use different values of ''x'' to compute the coefficients:

<math display="block">\begin{cases} 4 = 4C & x =1 \\ 2 + 2i = (Fi + G) (2+ 2i) & x = i \\ 0 = A- B +C - E - G & x = 0 \end{cases}</math>

Solving this we have:

<math display="block">\begin{cases} C = 1 \\ F =0, G =1 \\ E = A-B\end{cases}</math>

Using these values we can write:

<math display="block">\begin{align}
&2x^6-4x^5+5x^4-3x^3+x^2+3x \\[4pt]
={}& A\left(x-1\right)^2 \left(x^2+1\right)^2 + B\left(x-1\right)\left(x^2+1\right)^2 + \left(x^2 + 1\right)^2 + \left(Dx + \left(A-B\right)\right)\left(x-1\right)^3 \left(x^2+1\right) + \left(x-1\right)^3 \\[4pt]
={}& \left(A + D\right) x^6 + \left(-A - 3D\right) x^5 + \left(2B + 4D + 1\right) x^4 + \left(-2B - 4D + 1\right) x^3 + \left(-A + 2B + 3D - 1\right) x^2 + \left(A - 2B - D + 3\right) x
\end{align}</math>

We compare the coefficients of ''x''<sup>6</sup> and ''x''<sup>5</sup> on both side and we have:

<math display="block">\begin{cases} A+D=2 \\ -A-3D = -4 \end{cases} \quad \Rightarrow \quad A= D = 1.</math>

Therefore:

<math display="block">2x^6-4x^5+5x^4-3x^3+x^2+3x = 2x^6 -4x^5 + (2B + 5) x^4 + (-2B - 3) x^3 + (2B +1) x^2 + (- 2B + 3) x</math>

which gives us ''B'' = 0. Thus the partial fraction decomposition is given by:

<math display="block">f(x)=x^2+3x+4+\frac{1}{(x-1)} + \frac{1}{(x - 1)^3} + \frac{x + 1}{x^2+1}+\frac{1}{(x^2+1)^2}.</math>

Alternatively, instead of expanding, one can obtain other linear dependences on the coefficients computing some derivatives at <math>x = 1, \imath</math> in the above polynomial identity. (To this end, recall that the derivative at ''x'' = ''a'' of (''x'' − ''a'')<sup>''m''</sup>''p''(''x'') vanishes if ''m'' > 1 and is just ''p''(''a'') for ''m'' = 1.) For instance the first derivative at ''x'' = 1 gives

<math display="block"> 2\cdot6-4\cdot5+5\cdot4-3\cdot3+2+3   = A\cdot(0+0) + B\cdot( 4+ 0) + 8 + D\cdot0 </math>

that is 8 = 4''B'' + 8 so ''B'' = 0.

===Example 4 (residue method)===

<math display="block"> f(z)=\frac{z^{2}-5}{(z^2-1)(z^2+1)}=\frac{z^{2}-5}{(z+1)(z-1)(z+i)(z-i)}</math>

Thus, ''f''(''z'') can be decomposed into rational functions whose denominators are ''z''+1, ''z''−1, ''z''+i, ''z''−i. Since each term is of power one, −1, 1, −''i'' and ''i'' are simple poles.

Hence, the residues associated with each pole, given by
<math display="block">\frac{P(z_i)}{Q'(z_i)} = \frac{z_i^2 - 5}{4z_i^3},</math>
are 
<math display="block"> 1, -1, \tfrac{3i}{2}, -\tfrac{3i}{2},</math>
respectively, and

<math display="block"> f(z)=\frac{1}{z+1}-\frac{1}{z-1}+\frac{3i}{2}\frac{1}{z+i}-\frac{3i}{2}\frac{1}{z-i}. </math>

===Example 5 (limit method)===

[[Limit (mathematics)|Limits]] can be used to find a partial fraction decomposition.<ref>{{cite book|last=Bluman|first=George W.| title=Problem Book for First Year Calculus|year=1984|publisher=Springer-Verlag|location=New York|pages=250–251}}</ref> Consider the following example:

<math display="block"> \frac{1}{x^3 - 1}</math>

First, factor the denominator which determines the decomposition:

<math display="block"> \frac{1}{x^3 - 1} = \frac{1}{(x - 1)(x^2 + x + 1)} = \frac{A}{x - 1} + \frac{Bx + C}{x^2 + x + 1}.</math>

Multiplying everything by <math>x-1</math>, and taking the limit when <math>x \to 1</math>, we get

<math display="block">\lim_{x \to 1} \left((x-1)\left ( \frac{A}{x-1} + \frac{Bx + C}{x^2 + x + 1} \right )\right) = \lim_{x \to 1} A + \lim_{x \to 1}\frac{(x-1)(Bx + C)}{x^2 + x + 1} =A.</math>

On the other hand,

<math display="block">\lim_{x \to 1} \frac{(x-1)}{(x - 1)(x^2 + x + 1)} =  \lim_{x \to 1}\frac{1}{x^2 + x + 1} = \frac13,</math>

and thus:

<math display="block">A = \frac{1}{3}.</math>

Multiplying by {{math|''x''}} and taking the limit when <math>x \to \infty</math>, we have

<math display="block">\lim_{x \to \infty} x\left( \frac{A}{x-1} + \frac{Bx + C}{x^2 + x + 1} \right )= \lim_{x \to \infty} \frac{Ax}{x-1} + \lim_{x \to \infty} \frac{Bx^2+Cx}{x^2+x+1}= A+B,</math>

and

<math display="block">\lim_{x \to \infty} \frac{x}{(x - 1)(x^2 + x + 1)} =0.</math>

This implies {{math|1=''A'' + ''B'' = 0}} and so <math>B = -\frac{1}{3}</math>.

For {{math|1=''x'' = 0}}, we get  <math>-1 = -A + C,</math> and thus  <math>C = -\tfrac{2}{3}</math>.

Putting everything together, we get the decomposition

<math display="block">\frac{1}{x^3 -1} = \frac{1}{3} \left( \frac{1}{x - 1} + \frac{-x -2}{x^2 + x + 1} \right ).</math>

===Example 6 (integral)===
Suppose we have the indefinite [[integral]]:

<math display="block">\int \frac{x^4+x^3+x^2+1}{x^2+x-2} \,dx</math>

Before performing decomposition, it is obvious we must perform polynomial long division and [[Factorization|factor]] the denominator. Doing this would result in:

<math display="block">\int \left(x^2 + 3 + \frac{-3x+7}{(x+2)(x-1)}\right) dx</math>

Upon this, we may now perform partial fraction decomposition.

<math display="block">\int \left(x^2+3+ \frac{-3x+7}{(x+2)(x-1)}\right) dx = \int \left(x^2+3+ \frac{A}{(x+2)}+\frac{B}{(x-1)}\right) dx</math>
so:
<math display="block">A(x-1)+B(x+2)=-3x+7</math>.
Upon substituting our values, in this case, where x=1 to solve for B and x=-2 to solve for A, we will result in:

<math display="block">A=\frac{-13}{3} \ , B=\frac{4}{3} </math>

Plugging all of this back into our integral allows us to find the answer:

<math display="block">\int \left(x^2+3+ \frac{-13/3}{(x+2)}+\frac{4/3}{(x-1)}\right) \,dx = \frac{x^3}{3} \ + 3x-\frac{13}{3} \ln(|x+2|)+\frac{4}{3} \ln(|x-1|)+C </math>