Derivatives on HP 42S

[Thomas Klemm]

(08-24-2018 02:51 AM)lrdheat Wrote:  When trying a more complicated equation: x^(4/3) + 4(x^(1/3)), I get a correct f(x) for all values of x (x=-6, f(x)=3.63, x=-4, f(x)=0 x=0, f(x)=0 x=2, f(x)=7.56)

The function can be written as

\(y=\sqrt[3]{x}(x+4)\)

Even though you might think otherwise the domain of this function is

\(\{x\in\mathbb{R}:x\geqslant0\}\)

The problem with the cubic root of negative numbers is which value to choose as the principal value. The advantage of choosing the negative, real value is that you don't have to deal with complex numbers. That's probably the reason for this schoolbook definition.

The problem however is that this choice is not continuous: as soon as the exponent is slightly off from \(\frac{1}{3}\) you get a complex value anyway and then the principal value jumps to the branch of the value with the smallest argument.

The HP-42S doesn't provide a cubic root function and thus we have to use \(y^x\) with an approximation to \(\frac{1}{3}\) which is the reason you get a complex result for a negative value.

This is the program I used to calculate the values of the function:

Function

Code:

LBL "Fx"
RCL ST X
3
1/X
Y↑X
X<>Y
4
+
*
END

It returns complex values for negative real values.
Nothing can hinder us to still try to find a critical point for which I got:

x=-4.00000057735

But for \(x=-4\) the function will return

0 i0

Which is the reason we get that result.

I haven't stressed that but there are two conditions that the function has to meet:

1. it has to be real valued
2. it must be analytical
   

The 2nd condition means you can't use operations like ABS or SIGN in your program to define the function. These functions aren't analytical.

For the given function the minimum is at (0, 0).

HTH
Thomas