Post Reply 
Problem with fractional exponents and negative numbers
12-06-2016, 12:54 AM
Post: #1
Problem with fractional exponents and negative numbers
Hi, I tried to elevate a negative number to a fractional number and I get an error in home and a complex number on CAS. Do you know how to fix it? Is it a configuration problem? Thank you.


Attached File(s) Thumbnail(s)
   
Find all posts by this user
Quote this message in a reply
12-06-2016, 01:27 AM (This post was last modified: 12-06-2016 01:30 AM by Han.)
Post: #2
RE: Problem with fractional exponents and negative numbers
Consider using NTHROOT.

Some discussion of the issue can be found here: http://www.hpmuseum.org/forum/thread-357...ht=nthroot

Graph 3D | QPI | SolveSys
Find all posts by this user
Quote this message in a reply
12-06-2016, 01:44 AM
Post: #3
RE: Problem with fractional exponents and negative numbers
Tried to use it but I cant, It just returns a common root.
Find all posts by this user
Quote this message in a reply
12-06-2016, 05:33 AM
Post: #4
RE: Problem with fractional exponents and negative numbers
Can you be a bit more clear on what the issue is, then? You have listed what you typed into the calculator, but it is not clear what you were expecting as far as results go.

Graph 3D | QPI | SolveSys
Find all posts by this user
Quote this message in a reply
12-06-2016, 07:36 AM
Post: #5
RE: Problem with fractional exponents and negative numbers
Hi German90,
Home settings; switch on "allow complex output from real input"

— Dirk Hartland
Find all posts by this user
Quote this message in a reply
12-08-2016, 04:53 PM
Post: #6
RE: Problem with fractional exponents and negative numbers
First we have to know how g^b is defined when b is not an integer.

So what would 0.47^3.1 be?

It is defined by means of the well-known functions e^x and ln(x), as follows:

g^b=e^[b*ln(g)]

This seems reasonable because e^[b*ln(g)]=e^[ln(g^b)]=g^b.

But when we use this definition in the case of g being negative we have to calculate the logarithm of a negative number, which does not exist within the realm of real numbers.

Within the realm of complex numbers such a logarithm exists though.

That is the reason why the result is a complex number, written as r+s*i, where r and s are real numbers, and i is the imaginary unit.
Find all posts by this user
Quote this message in a reply
Post Reply 




User(s) browsing this thread: 1 Guest(s)