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== Procediment ==
{{traducció|en|Partial fraction}}
GivenDonats twodos polynomialspolinomis <math>P(x)</math> andi <math>Q(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_n)</math>, whereon theles ''α''<sub>''i''</sub> aresón distinctdiferents constants andi el deggrau ''P'' < ''n'', partialles fractionsfraccions areparcials generallygeneralment obtained bys'obtenen supposingsuposant thatque
: <math>\frac{P(x)}{Q(x)} = \frac{c_1}{x-\alpha_1} + \frac{c_2}{x-\alpha_2} + \cdots + \frac{c_n}{x-\alpha_n}</math>
andi solvingresolent forper theles constants ''c''<sub>''i''</sub> constants,(per by substitution, bysubstitució) [[equatingigualació de coeficients|igualant theels coefficientscoeficients]] ofde termstermes involvingque thevan powersacomponyats ofde les potències de ''x'', or otherwise. Es (Thistracta, isde afet, variantd'una ofvariant thedel [[methodmètode ofdels undeterminedcoeficients coefficientsindeterminats]].)
Un càlcul més directe molt relacionat amb la [[interpolació de Langrange]] consisteix en escriure
A more direct computation, which is strongly related with [[Lagrange interpolation]] consists in writing
: <math>\frac{P(x)}{Q(x)} = \sum_{i=1}^n \frac{P(\alpha_i)}{Q'(\alpha_i)}\frac{1}{(x-\alpha_i)} </math>
where <math>Q'</math> is the derivative of the polynomial <math>Q</math>. ▼
▲whereon <math>Q'</math> isés thela derivativederivada ofdel the polynomialpolinomi <math>Q</math>.
This approach does not account for several other cases, but can be modified accordingly:
Aquesta aproximació no justifica molts altres casos, però es pot modificar:
* IfSi deggrau ''P'' <math> \ge </math> deggrau ''Q'', thenllavors ités isnecessari necessaryfer to perform thela [[ Polynomialpolinomi# DivisibilityDivisibilitat| Euclideandivisió divisioneuclidiana]] ofde ''P'' byper ''Q'' , using(vegeu ''[[ polynomialDivisió longde divisionpolinomis]] ''), givingla qual cosa dóna ''P''(''x'') = ''E''(''x'') ''Q''(''x'') + ''R''(''x'') withamb deggrau ''R''</sub> < ''n''. DividingDividint byper ''Q''(''x'') this givess'obté▼
▲* If deg ''P'' <math> \ge </math> deg ''Q'', then it is necessary to perform the [[Polynomial#Divisibility|Euclidean division]] of ''P'' by ''Q'', using [[polynomial long division]], giving ''P''(''x'') = ''E''(''x'') ''Q''(''x'') + ''R''(''x'') with deg ''R''</sub> < ''n''. Dividing by ''Q''(''x'') this gives
:: <math>\frac{P(x)}{Q(x)} = E(x) + \frac{R(x)}{Q(x)},</math>
:and then seek partial fractions for the remainder fraction (which by definition satisfies deg ''R''</sub> < deg ''Q'').
:i llavors s'intenten buscar les fraccions parcials per la fracció residu (que, per definició, satisfà grau ''R''</sub> < deg ''Q'').
* If ''Q''(''x'') contains factors which are irreducible over the given field, then the numerator ''N''(''x'') of each partial fraction with such a factor ''F''(''x'') in the denominator must be sought as a polynomial with deg ''N'' < deg ''F'', rather than as a constant. For example, take the following decomposition over '''R''':
* Si ''Q''(''x'') conté factors que són irreductibles en el cos donat, llavors el numerador ''N''(''x'') de cada fracció parcial amb tal factor ''F''(''x'') al denominador s'ha de buscar com un polinomi amb grau ''N'' < grau ''F'', i no pas com una constant. Per exemple, si es pren la següent descomposició sobre '''R''':
:: <math>\frac{x^2 + 1}{(x+2)(x-1)\color{Blue}(x^2+x+1)} = \frac{a}{x+2} + \frac{b}{x-1} + \frac{\color{OliveGreen}cx + d}{\color{Blue}x^2 + x + 1}.</math>
* SupposeEs suposa ''Q''(''x'') = (''x'' − ''α'')<sup>''r''</sup>''S''(''x'') andi ''S''(''α'') ≠ 0. ThenLlavors ''Q''(''x'') hasté aun zero ''α'' ofde [[Multiplicitymultiplicitat (mathematicsmatemàtiques)#Multiplicity of a root of a polynomial|multiplicitymultiplicitat]] ''r'', andi inen thela partialdescomposició fractionen decompositionfraccions parcials, ''r'' offraccions theparcials partialinvolucraran fractionsles willpotències involve the powers ofde (''x'' − ''α''). ForPer illustrationexemple, takesi es pren ''S''(''x'') = 1 to gets'obté thela followingsegüent decompositiondescomposició:
:: <math>\frac{P(x)}{Q(x)} = \frac{P(x)}{(x-\alpha)^r} = \frac{c_1}{x-\alpha} + \frac{c_2}{(x-\alpha)^2} + \cdots + \frac{c_r}{(x-\alpha)^r}.</math>
=== IllustrationIl·lustració ===
In an example application of this procedure, {{nowrap|(3''x'' + 5)/(1 − 2''x'')<sup>2</sup>}} can be decomposed in the form
: <math>\frac{3x + 5}{(1-2x)^2} = \frac{13/2}{(1-2x)^2} + \frac{-3/2}{(1-2x)}.</math>
=== ResidueMètode methoddels residus ===
{{see also|Heaviside cover-up method}}
Over the complex numbers, suppose ''ƒ''(''x'') is a rational proper fraction, and can be decomposed into
Note that ''P''(''x'') and ''Q''(''x'') may or may not be polynomials.
== Vegeu també ==
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