Editing Partial fraction decomposition (section)

=== 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.