Residue method

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In Partial fractions I introduced the trick of carrying out the method of partial fractions by evaluating 1/P(x) numerically at $x=r_i+\epsilon$ , near where 1/P blows up. Sometimes we would like to have an exact result rather than a numerical approximation. We can accomplish this by using an infinitesimal number dx rather than a small but finite $\epsilon$ . For simplicity, let's assume that all of the roots r_i are distinct, and that 's highest-order term is x^n . We can then write as the product P(x)=(x-r_1)(x-r_2)...(x-r_n) . For products like this, there is a notation $\prod$ (capital Greek letter "pi") that works like $\sum$ does for sums:

$P(x)=\prod_{i=1}^{n}(x-r_i)$

It’s not necessary that the roots be real, but for now we assume that they are. We want to find the coefficients A_i such that $\frac{1}{P(x)}=\sum \frac{A_i}{x-r_i}$

We then have $\frac{1}{P(r_i+dx)}\\ \begin{align*} &=\frac{1}{dx\prod _{j\neq i}(r_i-r_j+dx)} \\ &= \frac{1}{dx\prod _{j\neq i}(r_i-r_j)}+\cdots \\&=\frac{A_i}{dx}+\cdots \end{align*}$

where … represents finite terms that are negligible compared to the infinite ones. Multiplying on both sides by , we have

$\frac{1}{P'(r_i)}+\cdots =A_i+\cdots$

where the … now stand for infinitesimals which must in fact cancel out, since both A_i i and 1/P' are real numbers.

Example

The partial-fraction decomposition of the function

$\frac{1}{x^4-5x^3-25x^2+65x+84}$ was found numerically on Partial fractions. The coefficient of the 1/(x-3) term was found numerically to be $A_1\approx -8.930\times 10^{-3}$ . Determine it exactly using the residue method.

Differentiation gives P'(x)=4x^3-15x^2-50x+65 . We then have A_1=1/P'(3)=-1/112 .

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Residue method

Example