1
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I've defined the two following functions:

logit[x_] := Module[{},
   Log[x/(1 - x)]
   ];
invLogit[x_] := Module[{},
   E^x/(1 + E^x)
   ];

One is the inverse of the other. However,

In[37]:= logit[invLogit[34.55555]]

Out[37]= 34.6574

Is there a way to increase the precision of the calculations?

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  • 1
    $\begingroup$ logit[invLogit[34.55555`20]]. $\endgroup$ – AccidentalFourierTransform Dec 2 '18 at 14:46
  • $\begingroup$ @AccidentalFourierTransform thanks for the comment. However, if these functions have to receive a variable, how would I do then? e.g. logit[invLogit[s]], where s is some value computed in another function. $\endgroup$ – An old man in the sea. Dec 2 '18 at 14:48
  • $\begingroup$ The inverse relationship is only valid for real values of x. Evaluate logit[invLogit[2 + 8 I]] // Simplify $\endgroup$ – Bob Hanlon Dec 2 '18 at 16:10
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Something like this works

logit[x_] := Module[{}, Log[x/(1 - x)]];
invLogit[u_] := Module[{x}, x = SetPrecision[u, 30]; E^x/(1 + E^x)];

logit[invLogit[34.55555]]
(* 34.5555500000000 *)

The issue is that the gradient of logit approaches zero for large arguments. Therefore there is a loss of precision in evaluating it. Enforcing a larger precision calculation helps get back to where you started.

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  • $\begingroup$ Mikado, thanks for your answer. In the set precision command, is the $MaxExtraPrecision the best numerical approximation we can have in general terms in mathematica? $\endgroup$ – An old man in the sea. Dec 2 '18 at 15:22
  • 1
    $\begingroup$ From the help for $MaxExtraPrecision I don't think that it is relevant here. I think you can specify as many extra digits of precision as you need. $\endgroup$ – mikado Dec 2 '18 at 15:36

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