Programming with HP50g

[Allen]

(08-30-2015 07:24 AM)Gerson W. Barbosa Wrote:  These make for nice HP-42S solutions:
Program LBL A
Program LBL B

Gerson, yes, nicely written! Your program LBL A is exactly the approach used by the "less complex" approach on the 48g. I was looking for a way to calculate \(A^3B-AB^3 = AB(A^2-B^2) \) and thought.. wouldn't it be nice if we could make \(B\) imaginary before squaring to save a subtraction. Happily, both \(A^2-B^2 \) and \(AB\) appear in the real and imaginary part of \( z^2=(A+iB)^2 = A^2-B^2 +i2AB \), however it is necessary to halve the product just as you did in Program LBL A:
\begin{equation} \frac{\operatorname{Im}(z^2) \operatorname{Re}(z^2)}{2} =A^3B-AB^3 \end{equation}
I had explored a way to use the matrix determinant to calculate
\begin{equation} \operatorname{det} \begin{bmatrix}
A^2 & B^2 \\
AB & AB
\end{bmatrix} = A^3B-AB^3
\end{equation}

But, alas, the "stack stuff" needed to get A^2, B^2 and AB onto the stack was going to exceed 1 command.

Code:


<<"stack stuff"  DUP 2 IDN SIZE ->ARR DET>>

I also found if one expands \(z^n\) in the manner of this solution, it creates a multi-dimensional, multi-variate complex pascal's triangle. The solutions here only deal with a few elements:



Image from: "Pascal's triangle 5" by User:Conrad.Irwin originally User:Drini - Extracted from Image:PascalSimetria.svg with minor alterations. Licensed under CC BY-SA 3.0 via Commons - https://commons.wikimedia.org/wiki/File:...ngle_5.svg