4
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I have

y[x_] := 1/Sqrt[2] (5 - x - Sqrt[8 - (x + 1)^2])
{x0, x1} = {-1 - Sqrt[8], -1 + Sqrt[8]};

and I want to calculate

{y'[x0], y'[1], y'[x1]}

but Mathematica cannot evaluate the left and right values, nor can Limit.

What is the proper syntax/method to get the answer {-Infty, 0, Infty}?

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  • $\begingroup$ Limit[y'[z], z -> #] & /@ {x0, 1, x1} gives {-\[Infinity],0,(-I) \[Infinity]} $\endgroup$ – kglr Aug 1 at 22:52
  • $\begingroup$ I get {Indeterminate, 0, Indeterminate} for the above command. I use version 11.2.0.0. $\endgroup$ – mf67 Aug 1 at 23:06
  • $\begingroup$ mf67, it works in v9. In v12 I also get {Indeterminate, 0, Indeterminate} $\endgroup$ – kglr Aug 1 at 23:09
  • $\begingroup$ So I might have to wait for next major update to get a v9-result? Using {y'[x0 + 10.^-5], y'[1], y'[x0 - 10.^-5]} to try to "go around" the problem I get {-266.621, 0, -0.707107 + 265.915 I}. Is there some way to have only real values returned? $\endgroup$ – mf67 Aug 1 at 23:16
  • $\begingroup$ mf67, i posted an answer that works in v12. $\endgroup$ – kglr Aug 1 at 23:21
10
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You can use Limit with the option Direction:

MapThread[ Limit[y'[z], z -> #, Direction -> #2] &, 
 {{x0, 1, x1}, {"FromAbove", "TwoSided", "FromBelow"}}]

{-∞, 0, ∞}

Much simpler form (from mf67's comment below):

Limit[y'[z], z -> {x0, 1, x1}, Direction -> {"FromAbove", "TwoSided", "FromBelow"}]

same result

$VersionNumber

12.

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  • 1
    $\begingroup$ The older syntax (for people using older versions) goes like MapThread[Limit[y'[z], z -> #, Direction -> #2] &, {{x0, 1, x1}, {-1, 0, 1}}] $\endgroup$ – J. M. will be back soon Aug 2 at 7:58
  • 1
    $\begingroup$ I'm trying to understand the command. I got the same result with Limit[y'[z], z -> {x0, 1, x1}, Direction -> {"FromAbove", "TwoSided", "FromBelow"}]. What are the differences compared to the original MapThreadcommand? $\endgroup$ – mf67 Aug 2 at 9:46
  • $\begingroup$ @mf67, the difference is my ignorance of the fact that Limit is Listable:) $\endgroup$ – kglr Aug 2 at 9:49

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