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I would like Mathematica to print $\arctan(4/0)$ as $\frac{\pi}{2}$ but it does not because in mathematica 1/0 is complexinfinity. How do I make it print n/0 as Infinity so that Arctan will give me what I want?

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    $\begingroup$ Why positive infinity? Why not DirectedInfinity[-1]? For ArcTan, why not -Pi/2? $\endgroup$
    – Michael E2
    Jan 28, 2021 at 23:23
  • $\begingroup$ Actually yeah I would like it to be positive or negative infinity depending on sign of n. $\endgroup$
    – user824530
    Jan 29, 2021 at 5:13
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    $\begingroup$ $\frac{4}{0} = {\infty}_{+}$ when $`` 0 "$ is infinitesimally positive ("positive zero"), but not when it's infinitesimally negative ("negative zero") nor perfectly zero (integer-zero). So it'd seem like, for this to be a correct result, you'd need to clarify what $`` 0 "$ means in that expression. $\endgroup$
    – Nat
    Jan 29, 2021 at 19:35
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    $\begingroup$ @Nat, this is where it'd have been nice to have a "signed zero" like in IEEE's model, but alas... $\endgroup$
    – J. M.'s eventual burnout ♦
    Jan 30, 2021 at 2:04
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    $\begingroup$ @user, then you should indeed be using two-argument arctangent in the first place. $\endgroup$
    – J. M.'s eventual burnout ♦
    Jan 30, 2021 at 2:05

3 Answers 3

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On the off chance that what is desired in a computational environment in which n/0 evaluates to (positive) Infinity and t = ArcTan[..] falls in the range -Pi/2 <= t <= Pi/2. One should use caution when overwriting built-in functions. I doubt this will cause problems in computations in which n/0 should always mean positive Infinity and not ComplexInfinity.

ClearAll[evaluateWithRealPower];
SetAttributes[evaluateWithRealPower, HoldFirst];
evaluateWithRealPower[code_] :=
  Internal`InheritedBlock[{Power},
   Unprotect[Power];
   Power[0, p_?Negative] := Infinity;
   Power[0., p_?Negative] := Infinity; (* optional *)
   Protect[Power];
   code];

evaluateWithRealPower[ArcTan[4/0]]
(*  \[Pi]/2  *)
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  • $\begingroup$ Perfect. Thank you! $\endgroup$
    – user824530
    Jan 29, 2021 at 5:11
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Use the two-arg version of ArcTan instead:

ArcTan[0, 4]

π/2

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    $\begingroup$ ...and if you also need to evaluate at 0, 0, you could use Arg[x + I y] instead. $\endgroup$
    – J. M.'s eventual burnout ♦
    Jan 29, 2021 at 1:15
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You can convert expressions of ComplexInfinity to Infinity by using the ReplaceAll operator /..

ArcTan[4/0 /. ComplexInfinity -> Infinity]

which gives Pi/2 as output.

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