Advanced Graphing error?

[Jan_D]

The results I mentioned in post #7 I obtained in the Advanced Graphing app, firmware version 11226, on my Android tablet.

As already pointed out defining (-1)^1/3 as (-1)^1/3= -1 has some serious disadvantages.
I will outline it once more.
When we follow the kind of reasoning which leads to this definition, (-1)^0.3333 would not exist, even though it differs only slightly from (-1)^1/3 because 0.3333=3333/10000, which is a fraction with an even denominator and an odd numerator, in which case the exponential would not exist.

On the other hand (-1)^(3334/10001) would exist, but is not –1 but +1.

So with this definition (-1)^a is not defined for general real numbers, nor for general rational numbers, but only on a subset of the rational numbers, Q, and the function is extremely discontinuous at every point.

This is not something which we like.

A mathematically more satisfying way is the complex definition, which makes use of the imaginary unit i, which satisfies the equation i*i= -1.

In that case the definition is:
(-1)^a=cos(π *a)+i*sin(π *a)

Now we have no discontinuities, and (-1)^a is defined for every a in R (even for every a in C)

This definition makes sense because when a in Z we get:
(-1)^n= cos(π *n)+i*sin(π *n)= cos(π *n)=(-1)^n, which is what we want.

Moreover it satisfies the exponential rule
(-1)^a * (-1)^b=(-1)^(a+b)

because
(-1)^a * (-1)^b=[cos(π *a)+i*sin(π *a)]* [cos(π *b)+i*sin(π *b)]=
[cos(π *a)* cos(π *b)- sin(π *a)* sin(π b)]+i*[ cos(π a)* sin(π b)+ sin(π a)* cos(π *b)]=
cos(π*(a+b))+i*sin(π*(a+b))=(-1)^(a+b)

We can now define the exponential for a general negative base as:
(-b)^a=(-1)^a * b^a.