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Suppose, f[x] has a pole at a.

How do I quickly find

Limit[(f[a+h]+f[a-h])/2,h->0]

? I mean, the expression for the function can be complicated, and changing all a's to a+h and a-h may be boring.

Improve this question
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    $\begingroup$ Have a look at Residue. $\endgroup$ – Hugh Feb 14 at 17:45
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    $\begingroup$ @Hugh: Looking in Residue, one understands this deals nothing with the question. I think Limit[(f[a+h]+f[a-h])/2,h->0] is meaningless. $\endgroup$ – user64494 Feb 14 at 18:03
  • $\begingroup$ @Anixx: Think of f[x_]:=1/x^2;Limit[(1/h^2+1/(-h)^2)/2,h->0] $\endgroup$ – user64494 Feb 14 at 18:06
  • $\begingroup$ @Anixx: That limit equals $0$ if $a$ is a pole of odd order and equals ComplexInfinity if $a$ ia a pole of even order. $\endgroup$ – user64494 Feb 14 at 18:21
  • $\begingroup$ @user64494 no, that limit not always equal to 0. Take Gamma function at 0 or Zeta function at 1 for instance. $\endgroup$ – Anixx Feb 14 at 18:23

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