Definite integration yields a strange answer. (Solved)
01-31-2019, 12:51 AM
Post: #19
 Joe Horn Senior Member Posts: 2,008 Joined: Dec 2013
RE: Definite integration yields a strange answer. (Solved)
(01-30-2019 08:21 PM)Aries Wrote:
(11-14-2015 08:13 PM)parisse Wrote:  ... For fractional powers, you must understand that x^(1/3) *is* a complex number if x<0, whatever the complex mode setting is (on the hp49, the system would ask you to switch to complex mode or error, on the Prime it will return a complex number). If you want the real 3rd root, just use NTHROOT or surd.

Why x^(1/3) is a complex number if x<0 ?
q is odd …

Simple answer: Graph all three cube roots, and you'll see why.

More complete answer: Because (1) every non-zero number has three distinct cube roots, and (2) calculators which allow complex results return the "principal root" which is the one with the least ARG (that is, the smallest angle from the real axis). The ARG of the real cube root of a negative number is 180°, but the ARG of one of the complex cube roots is 60°, which makes it the principal root.

<0|ɸ|0>
-Joe-
 « Next Oldest | Next Newest »