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I have to show that for $f,g$ analytic on some domain and $a$ a double zero of $g$, we have:

$$\operatorname{Res} \left(\frac{f(z)}{g(z)}, z=a\right) = \frac{6f'(a)g''(a)-2f(a)g'''(a)}{3[g''(a)]^2}.$$

The problem is that direct calculation using the formula (for pole of order $2$):

$$\operatorname{Res}(h(z),z=a)=\lim_{z \to a} \frac{d}{dz}\left( (z-a)^2h(z) \right)$$

is extremely ugly, given that we're dealing with a quotient. Is there some sort of trick to make the calculation more manageable?

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  • $\begingroup$ Well...either what you see or else to write down the relevant terms of the Laurent Series around $\;z=a\;$ of $\;f/g\;$ ... $\endgroup$
    – DonAntonio
    Nov 1, 2013 at 14:34
  • $\begingroup$ IMO there ought to be at least one or two up-votes besides mine here by now. $\endgroup$ Nov 1, 2013 at 16:41
  • $\begingroup$ Aside: "Res" is also used for "resultant", which is a useful function for dealing with polynomials and their roots. While what you written is surely standard (or at least very reasonable) when in context, it took me a moment to figure out what the context was! $\endgroup$
    – user14972
    Nov 1, 2013 at 17:02

2 Answers 2

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A little Taylor expansion takes you a long way. Say

$$g(z) = (z-a)^2\left(a_2 + a_3(z-a) + (z-a)^2\cdot \tilde{g}(z)\right)$$

with $a_2 \neq 0$, and

$$f(z) = b_0 + b_1(z-a) + (z-a)^2\cdot \tilde{f}(z).$$

Then

$$\begin{align} \frac{f(z)}{g(z)} &= \frac{b_0 + b_1(z-a) +(z-a)^2\tilde{f}(z)}{(z-a)^2\left(a_2 + a_3(z-a) + (z-a)^2\tilde{g}(z)\right)}\\ &= \frac{1}{a_2(z-a)^2}\frac{b_0 + b_1(z-a) + (z-a)^2\tilde{f}(z)}{1 + \frac{a_3}{a_2}(z-a) + (z-a)^2h(z)}\\ &= \frac{1}{a_2(z-a)^2}\left(b_0 + b_1(z-a)\right)\left(1-\frac{a_3}{a_2}(z-a)\right) + \tilde{h}(z)\\ &= \frac{c}{(z-a)^2} + \frac{b_1 - (b_0a_3)/a_2}{a_2(z-a)} + k(z), \end{align}$$

so the residue is

$$\frac{b_1a_2 - b_0a_3}{a_2^2} = \frac{f'(a)\frac12g''(a) - f(a)\frac16g'''(a)}{\left(\frac12g''(a)\right)^2} = \frac{6f'(a)g''(a) - 2f(a)g'''(a)}{3g''(a)^2}.$$

That didn't hurt, doctor.

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  • $\begingroup$ I don't understand most of the manipulations you wrote (particularily the 3rd equality sign after $f/g$), but the first sentence of your answer got me on the right track, so I divided the series and compared coefficients and got the right answer. $\endgroup$ Nov 1, 2013 at 15:38
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    $\begingroup$ The denominator has the form $1 + q$, so I expanded it in a geometric series $\dfrac{1}{1+q} = 1 - q + q^2 - \dotsb$. Since $a$ is a zero of order two, we can ignore everything that has a $(z-a)^2$ in it in the expansion (and in the Taylor expansion of $f$), and that leaves only $1-q$. $\endgroup$ Nov 1, 2013 at 15:41
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Because $a$ is a double zero of $g(z)$, write

$$g(z) = (z-a)^2 p(z)$$

where $p(a) \ne 0$ and is analytic, etc. etc.

Then

$$\operatorname*{Res}_{z=a} \frac{f(z)}{g(z)} = \left [\frac{d}{dz} \frac{f(z)}{p(z)} \right ]_{z=a}$$

Now,

$$\frac{d}{dz} \frac{f(z)}{p(z)} = \frac{f'(z) p(z)-f(z) p'(z)}{p(z)^2}$$

Also, note that

$$g(z) = \frac12 g''(a) (z-a)^2 + \frac16 g'''(a) (z-a)^3+\cdots = p(a) (z-a)^2 + p'(a) (z-a)^3+\cdots$$

Therefore

$$p(a) = \frac12 g''(a)$$

and

$$p'(a) = \frac16 g'''(a)$$

Thus

$$\operatorname*{Res}_{z=a} \frac{f(z)}{g(z)} = \frac{f'(a) \frac12 g''(a) - f(a) \frac16 g'''(a)}{\frac14 [g''(a)]^2}$$

which is equivalent to the formula you seek.

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  • $\begingroup$ I don't get this bit: $\operatorname*{Res}_{z=a} \frac{f(z)}{g(z)} = \left [\frac{d}{dz} \frac{f(z)}{p(z)} \right ]_{z=a}$. Why can we just forget the $(z-a)^2$ term in the limit? $\endgroup$ Nov 1, 2013 at 15:38
  • $\begingroup$ @DepeHb: The presence of the $p(z)$ solo in the denominator implies that I have already canceled that term out. $\endgroup$
    – Ron Gordon
    Nov 1, 2013 at 16:59
  • $\begingroup$ Of course! Thanks for clearing it up. $\endgroup$ Nov 1, 2013 at 17:15

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