Graph with exponent not correct
10-26-2017, 05:45 PM
Post: #1
 MullenJohn Junior Member Posts: 43 Joined: Sep 2017
Graph with exponent not correct
When I graph the function (X^2 - 4)^(2/3) on the HP Prime there is nothing showing between -2 and +2. (This also happens on my HP 50g calculator)

However when I rearrange the function to the form 3√(x^2-4)^2/3 the graph display properly between -2 and +2.

It appears the calculator cannot graph a fractional exponent to an expression within a set of parentheses.

Does anyone know why and if so is there a setting that need to be done?

Thanks - Cheers!
10-26-2017, 06:54 PM (This post was last modified: 10-26-2017 06:57 PM by Dieter.)
Post: #2
 Dieter Senior Member Posts: 2,397 Joined: Dec 2013
RE: Graph with exponent not correct
(10-26-2017 05:45 PM)MullenJohn Wrote:  When I graph the function (X^2 - 4)^(2/3) on the HP Prime there is nothing showing between -2 and +2. (This also happens on my HP 50g calculator)

However when I rearrange the function to the form 3√(x^2-4)^2/3 the graph display properly between -2 and +2.

First of all, I assume you mean the cube root of (x²–4)² here.

Between –2 and +2 the value for x²–4 becomes negative, e.g. for x=0 it is –4. Now, what is –4^(2/3)? There is no real result, the answer is –1,2599+2,1822 i, a complex number. This cannot be shown in a graph with a real y-axis only.

On the other hand the second way of writing the formula causes less problems: (x²–4)² is always positive (or at least zero). So the cube root can be calculated easily. But this is not exactly the same as the first formula.

Finally there is another pitfall: Fractional powers of a negative base usually cause problems. This is because – maybe except in exact mode – the calculator takes the numeric 12-digit value 0,66666 66666 67. Which is not 2/3 but only an approximation, so the root is not defined.

Dieter
10-27-2017, 06:51 AM
Post: #3
 cyrille de brébisson Senior Member Posts: 1,047 Joined: Dec 2013
RE: Graph with exponent not correct
Hello,

negative^non integer is a function (when it can be calculated) that has more than one solution.
By convention, the first (as in the one that forms the smallest angle with the horizontal axis in the positiive direction) is the one chosen.

The nthroot = ^1/N correspondance is not a true matematical equality (despite what most of us have learned at school, but if this was the only falsehood that is taught at school, I would be happy to let it go, my 10 year old kid's science teacher is teaching him right now that planets orbits are circular and when he said otherwise, he got reprimended!!!!)...

Anyway, a NTHROOT b's definition is: if b>=0, b^1/a. if b<0 and a odd -(-b)^1/a else it is non defined (or the complex number b^1/a if complex numbers are allowed, but normally NTHROOT's application domain does not include complexes, it only does by extention).

But when first taught of nthroot, it is just as an oposit to powers, and at that time, power is limited to simple cases with a being a natural number. And teachers never tell the limit of the application of the function.

Anyhow, now you do know how to draw the function that you are interested in, which is the most important.

Cyrille

Although I work for the HP calculator group, the views and opinions I post here are my own. I do not speak for HP.
10-27-2017, 08:23 AM (This post was last modified: 10-27-2017 08:25 AM by StephenG1CMZ.)
Post: #4
 StephenG1CMZ Senior Member Posts: 945 Joined: May 2015
RE: Graph with exponent not correct
Cyrille,

I am sorry to hear that and I do hope that teacher gets reprimanded too.
I still recall an examination - it might have been the "11-plus" that determined which school you went to - with a multiple choice question regarding the distance to the nearest star.

The choices were 4 years and 3 random numbers, so I did not answer... Until an invigilator saw that and encouraged me to select the expected wrong answer.

Stephen Lewkowicz (G1CMZ)
10-27-2017, 02:54 PM
Post: #5
 MullenJohn Junior Member Posts: 43 Joined: Sep 2017
RE: Graph with exponent not correct
Thank you Cyrille and Dieter for explaining the reason. Cheers!
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