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: <math>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>
Després de fer la divisió i factoritzar, s'obté:
After long-division and factoring, we have
: <math>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>
La descomposició en fraccions parcials pren la forma següent:
The partial fraction decomposition takes the form
: <math>\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>
MultiplyingMultiplicant througha byles dues bandes per (''x'' − 1)<sup>3</sup>(''x''<sup>2</sup> + 1)<sup>2</sup> we haves'obté thela polynomialidentitat identitypolinomial:
: <math>
\begin{align}
& {} \quad 2x^6-4x^5+5x^4-3x^3+x^2+3x \\
</math>
TakingPrenent ''x'' = 1 giveses té 4 = 4''C'', soi per tant ''C'' = 1. D'una manera Similarlysimilar, takingprenent ''x'' = [[complexnombre numbercomplex|''i'']] givess'obté 2 + 2''i'' = (''Fi'' + ''G'')(2 + 2''i''), sode tal manera que ''Fi'' + ''G'' = 1, soper tant ''F'' = 0 andi ''G'' = 1 by(igualant equatingles realparts andreals [[complexi number|imaginary]] partscomplexes). WithAmb ''C'' = ''G'' = 1 andi ''F'' = 0, takingprenent ''x'' = 0 we gets'obté ''A'' - ''B'' + 1 - ''E'' - 1 = 0, thusi per tant ''E'' = ''A'' - ''B''.
Ara es té la identitat:
We now have the identity
: <math>
</math>
Si s'expandeix i s'ordena segons l'exponent de ''x'' s'obté:
Expanding and sorting by exponents of x we get
: <math>
</math>
Ara es poden comparar els coeficients i veure que:
We can now compare the coefficients and see that
: <math>
</math>
withAmb ''A'' = 2 - ''D'' andi -''A'' -3 ''D'' =-4 we gets'obté ''A'' = ''D'' = 1 andi, soper tant, ''B'' = 0; continuant, furthermore is ''C'' = 1, ''E'' = ''A'' - ''B'' = 1, ''F'' = 0 andi ''G'' = 1.
TheLlavors, partialla fractiondescomposició decompositionen offraccions parcials de ''ƒ''(''x'') is thusés:
: <math>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 ''x=1'' and at ''x=i'' 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 it is just ''p(a)'' if ''m=1''.)
Thus, for instance the first derivative at ''x=1'' gives
: <math> 2\cdot6-4\cdot5+5\cdot4-3\cdot3+2+3 = A\cdot(0+0) + B\cdot( 2+ 0) + 8 + D\cdot0 </math>
that is ''8 = 2B + 8'' so ''B=0''.
===Exemple 4 (mètode dels residus)===
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