(35s) y^x for y < 0 and x < 1 (complex roots)

04292016, 06:28 AM
(This post was last modified: 04292016 11:39 AM by brianddk.)
Post: #1




(35s) y^x for y < 0 and x < 1 (complex roots)
Howdy,
I stumbled across a thread a while back discussing the following well documented behavior on the 35s Code: 2 Obviously if 'x' can be expressed as a rational number and if both the numerator and the denominator are both odd, then the sign of y is irrelevant and it will carry through the exponentiation. To this end, I wrote a program that would take 'x' and 'y' and, if y < 0 and FP(x) > 0, then it would try to find an 'acceptable' rational representation of 'x' that had an odd numerator and denominator to yield a negative real result. Otherwise, it gives up and simply produces the answer in complex form. 'Acceptable' answers are given in the form of a user provided delta that that ratio must be within for the approximation to be accepted. My question, is... is there a better way than brute force 'rationalization' to determine if an exponent will yield a real (nonimaginary) answer. I seem to recall a method using natural logs, but I don't know if it would get me closer to the question of realvsimaginary results. PS... what I ended up with is effectively Code: Given: MyCalcs: Physical: {hp48gx, hp50g, hp35s} Emu: {hp42s(Free42), hp41c(v41)} Blog: https://brianddk.github.io/ 

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(35s) y^x for y < 0 and x < 1 (complex roots)  brianddk  04292016 06:28 AM
RE: (35s) y^x for y < 0 and x < 1 (complex roots)  Dieter  04292016, 09:09 AM
RE: (35s) y^x for y < 0 and x < 1 (complex roots)  brianddk  04292016, 11:29 AM

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