 Fractional exponents vs. radical form ... - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: HP Prime (/forum-5.html) +--- Thread: Fractional exponents vs. radical form ... (/thread-14625.html) Fractional exponents vs. radical form ... - DrD - 03-10-2020 09:40 AM Pardon my English, but math is not my native language: Frustration: Expressions, (or equations), similar to x^([even numbered numerator]/denominator), when x<0. An Example: (-1)^(2/3); ==> (-1/2)+i*sqrt(3)/2 (3) NTHROOT ((-1)^2); ==> 1, the desired result f(x):=(x^(2/3)); g(x):=((3) NTHROOT (x^2)); subst({f(x), g(x)}, x=(-1)); ==> {((1/2)+(i*sqrt(3)/2))^2,1} A setting would be nice, that would force fractional exponents, to be the same as their radical form ... not sure how to best define it, but like obscenity, "I know it when I see it!" -Dale- RE: Fractional exponents vs. radical form ... - Albert Chan - 03-10-2020 02:46 PM XCas> radpow(a,b) := surd(a^numer(b), denom(b)) numer(b), denom(b) had b simplified first, which is exactly what we wanted Note: this assumed b is not symbolic variable XCas> radpow(-64, 2/6) ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿// = radpow(-64, 1/3) = surd(-64,3) = -4 XCas> radpow(-64, 3/6) ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿// = radpow(-64, 1/2) = surd(-64,2) = 8*i XCas> radpow(-64, 4/6) ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿// = radpow(-64, 2/3) = surd(4096,3) = 16 see https://www.math.ubc.ca/~israel/advisor/advisor5/a37r1.htm RE: Fractional exponents vs. radical form ... - CyberAngel - 03-10-2020 02:51 PM (03-10-2020 09:40 AM)DrD Wrote:  Pardon my English, but math is not my native language: Frustration: Expressions, (or equations), similar to x^([even numbered numerator]/denominator), when x<0. An Example: (-1)^(2/3); ==> (-1/2)+i*sqrt(3)/2 (3) NTHROOT ((-1)^2); ==> 1, the desired result f(x):=(x^(2/3)); g(x):=((3) NTHROOT (x^2)); subst({f(x), g(x)}, x=(-1)); ==> {((1/2)+(i*sqrt(3)/2))^2,1} A setting would be nice, that would force fractional exponents, to be the same as their radical form ... not sure how to best define it, but like obscenity, "I know it when I see it!" -Dale- The root function has a different defining area/range from rational or reciprocal exponent function. That according to high school math. RE: Fractional exponents vs. radical form ... - lrdheat - 03-10-2020 04:53 PM Aside from the pure mathematical rigor, a setting allowing the user this option looks useful. RE: Fractional exponents vs. radical form ... - CyberAngel - 03-10-2020 07:34 PM (03-10-2020 04:53 PM)lrdheat Wrote:  Aside from the pure mathematical rigor, a setting allowing the user this option looks useful. nthroot allows what is supposed to be allowed. For wrong math turn to Casio or TI. RE: Fractional exponents vs. radical form ... - lrdheat - 03-11-2020 12:56 AM I agree that the decimal approximation would produce an answer of the sort that was objected to (or at least asked for a choice in how the fractional power was interpreted). A choice would be nice for a fractional exponent that could be interpreted as symbolically equivalent to a surf/nroot as an option. RE: Fractional exponents vs. radical form ... - CyberAngel - 03-11-2020 01:01 AM (03-11-2020 12:56 AM)lrdheat Wrote:  I agree that the decimal approximation would produce an answer of the sort that was objected to (or at least asked for a choice in how the fractional power was interpreted). A choice would be nice for a fractional exponent that could be interpreted as symbolically equivalent to a surf/nroot as an option. no RE: Fractional exponents vs. radical form ... - DrD - 03-11-2020 08:50 AM (03-10-2020 02:51 PM)CyberAngel Wrote:  The root function has a different defining area/range from rational or reciprocal exponent function. That according to high school math. The use of parentheses, (-1) in my original example, was an important distinction, (to clarify the order of operations). My thinking was that: x^[even numerator]/denominator; would result in a real-value, regardless of the fractional vs. radical form, (for real-value inputs). Example: x^(2/1) where x=(-n) Processed as: [(-n) * (-n)] = (+)n^2, not: -1 * (n * n) = (-)n^2.