HP-71B: 1^Inf Error

[Albert Chan]

(06-10-2023 01:47 PM)EdS2 Wrote:  I notice IEEE floating point specifies the result of 0 × ∞ as NaN.
If 1^x is computed as exp(log(1)*x) then that would be the same.

IEEE goal is not always mathematically correct.
It tried to avoid NaN if possible, with a default numerical value to (hopefully) simplify code.

https://www.open-std.org/jtc1/sc22/wg14/...eV5.10.pdf

page 182 Wrote:pow(x,0) is specified to return 1 for any x, because there are significant applications where 1 is
more useful than NaN. pow(f(t),g(t)) approaches 1 in all cases where f and g are analytic
functions and g(t) approaches zero. The result 1 better supports applications where the second
argument is integral. pow(NAN,0) returns 1.0 on the general principle that if a result is independent
of the numerical value of an argument, then that result is appropriate if that argument is a NaN.

(06-09-2023 11:38 PM)rprosperi Wrote:  I think this proves that Infinity is even, right?

Yes!

page 180 Wrote:The result of pow(–2,∞) is +∞, because all large positive floating-point values are even integers.

--> pow(-1, ∞) = pow(1, ∞) = 1

Again, this is not mathematically sound, but only for "convenience"