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+--- Thread: (35s) y^x for y < 0 and x < 1 (complex roots) (/thread-6172.html)
(35s) y^x for y < 0 and x < 1 (complex roots) - brianddk - 04-29-2016 06:28 AM
Howdy,
I stumbled across a thread a while back discussing the following well documented behavior on the 35s
Code:
2
+/-
3
y^x
LASTx
1/x
y^x
**INVALID y^xObviously 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 (non-imaginary) 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 real-vs-imaginary results.
PS... what I ended up with is effectively
Code:
Given:
x = n/d where n and d are odd
y where y < 0
f(x,y) = abs((y+0i)^(n/d)) * -1 ; where n is odd
f(x,y) = abs((y+0i)^(n/d)) ; where n is evenRE: (35s) y^x for y < 0 and x < 1 (complex roots) - Dieter - 04-29-2016 09:09 AM
(04-29-2016 06:28 AM)brianddk Wrote: Howdy,
I stumbled across a thread a while back discussing the following well documented behavior on the 35sCode:
2
+/-
3
y^x
LASTx
1/x
y^x
**INVALID y^x
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.
According to this rule –8^(3/5) should work, but it doesn't => INVALID yx.
I think it's much simpler: the error in your example occurs because you want the calculator to evaluate (–8)^0,333333333333.
But this does NOT equal (–8)^(1/3), so no (real) result exists.
That's why there is a XROOT function (x√y, on the K key). –8 [ENTER] 3 [x√y] returns –2, as expected. Here no errors occur for y<0 if x is an odd integer.
So instead of –8^(3/5) you can use –8^3 = –512 and then the 5th root of this yields –3,4822.
Dieter
RE: (35s) y^x for y < 0 and x < 1 (complex roots) - brianddk - 04-29-2016 11:29 AM
(04-29-2016 09:09 AM)Dieter Wrote: According to this rule –8^(3/5) should work, but it doesn't => INVALID yx.
I think it's much simpler: the error in your example occurs because you want the calculator to evaluate (–8)^0,333333333333.
But this does NOT equal (–8)^(1/3), so no (real) result exists.
That's why there is a XROOT function (x√y, on the K key). –8 [ENTER] 3 [x√y] returns –2, as expected. Here no errors occur for y<0 if x is an odd integer.
So instead of –8^(3/5) you can use –8^3 = –512 and then the 5th root of this yields –3,4822.
Your correct, I wasn't clear. What I meant was
Code:
Given:
x = n/d where d is odd
y where y < 0
f(x,y) = abs((y+0i)^(n/d)) * -1 ; where n is odd
f(x,y) = abs((y+0i)^(n/d)) ; where n is even