Should 1^x == 1?

Dearests,

using Symbolics
@variables p, α, N
D = Differential(α)
D((p^α + 1 - p)^N) / log(p) |> expand_derivatives |> x -> substitute(x, α => 1) |> simplify

correctly returns N*p*(1^(N - 1)).

I was wondering if a substitution rule should be implemented such that 1^x -> 1 no matter what x is. BTW, Mathematica does it automatically.

f[ α_, p_, N_] := (p^α_ + 1 - p)^N
D[f[a, p, N], a]/(Log[p]) /. a -> 1  #  returns N * p 
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It might be a good idea. Other simplifications area already done by default (you can check them here).

I’ll add a PR for this.

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What if x=∞ or -∞?

1 is still a reasonable result. (math people think of it as an indeterminate form, but that tends to be less useful than defining it to equal 1.)

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Open an issue. Note that Symbolics is a typed symbolic system, so @variables p is @variables p::Real which is the “standard” type and can make these tricks. I forget the name, but it’s like @variables p::SafeReal has different rule sets, for example omitting a/a == 1, which is safe for the case of a=0. The default is what users generally want, but the safe mode gives people a way to opt in to a slightly slower system that will work for all edge cases, giving the best of both worlds.

Done!

Issue #437 on SymbolicsUtils.

Thanks
A