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Partial fraction decomposition
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== The role of the Taylor polynomial == The partial fraction decomposition of a rational function can be related to [[Taylor's theorem]] as follows. Let <math display="block">P(x), Q(x), A_1(x),\ldots, A_r(x)</math> be real or complex polynomials assume that <math display="block">Q=\prod_{j=1}^{r}(x-\lambda_j)^{\nu_j},</math> satisfies <math display="block">\deg A_1<\nu_1, \ldots, \deg A_r<\nu_r, \quad \text{and} \quad \deg(P)<\deg(Q)=\sum_{j=1}^{r}\nu_j.</math> Also define <math display="block">Q_i=\prod_{j\neq i}(x-\lambda_j)^{\nu_j}=\frac{Q}{(x-\lambda_i)^{\nu_i}}, \qquad 1 \leqslant i \leqslant r.</math> Then we have <math display="block">\frac{P}{Q}=\sum_{j=1}^{r}\frac{A_j}{(x-\lambda_j)^{\nu_j}}</math> if, and only if, each polynomial <math>A_i(x)</math> is the Taylor polynomial of <math>\tfrac{P}{Q_i}</math> of order <math>\nu_i-1</math> at the point <math>\lambda_i</math>: <math display="block">A_i(x):=\sum_{k=0}^{\nu_i-1} \frac{1}{k!}\left(\frac{P}{Q_i}\right)^{(k)}(\lambda_i)\ (x-\lambda_i)^k. </math> Taylor's theorem (in the real or complex case) then provides a proof of the existence and uniqueness of the partial fraction decomposition, and a characterization of the coefficients. ===Sketch of the proof=== The above partial fraction decomposition implies, for each 1 β€ ''i'' β€ ''r'', a polynomial expansion <math display="block">\frac{P}{Q_i}=A_i + O((x-\lambda_i)^{\nu_i}), \qquad \text{for } x\to\lambda_i,</math> so <math>A_i</math> is the Taylor polynomial of <math>\tfrac{P}{Q_i}</math>, because of the unicity of the polynomial expansion of order <math>\nu_i-1</math>, and by assumption <math>\deg A_i<\nu_i</math>. Conversely, if the <math>A_i</math> are the Taylor polynomials, the above expansions at each <math>\lambda_i</math> hold, therefore we also have <math display="block">P-Q_i A_i = O((x-\lambda_i)^{\nu_i}), \qquad \text{for } x\to\lambda_i,</math> which implies that the polynomial <math> P-Q_iA_i</math> is divisible by <math> (x-\lambda_i)^{\nu_i}.</math> For <math> j\neq i, Q_jA_j</math> is also divisible by <math>(x-\lambda_i)^{\nu_i}</math>, so <math display="block"> P- \sum_{j=1}^{r}Q_jA_j</math> is divisible by <math>Q</math>. Since <math display="block"> \deg\left( P- \sum_{j=1}^{r}Q_jA_j \right) < \deg(Q)</math> we then have <math display="block"> P- \sum_{j=1}^{r}Q_jA_j=0,</math> and we find the partial fraction decomposition dividing by <math> Q</math>.
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