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2.7.67 Programming Tutorial Exercise 10

Taking the derivative of a term of the form ‘x^n’ will produce a term like ‘n x^(n-1)’. Taking the derivative of a constant produces zero. From this it is easy to see that the ‘n’th derivative of a polynomial, evaluated at ‘x = 0’, will equal the coefficient on the ‘x^n’ term times ‘n!’.

(Because this definition is long, it will be repeated in concise form below. You can use C-x * m to load it from there. While you are entering a Z ` Z ' body in a macro, Calc simply collects keystrokes without executing them. In the following diagrams we’ll pretend Calc actually executed the keystrokes as you typed them, just for purposes of illustration.)

2:  5 x^4 + (x + 1)^2          3:  5 x^4 + (x + 1)^2
1:  6                          2:  0
    .                          1:  6
                                   .

  ' 5 x^4 + (x+1)^2 RET 6        C-x ( Z `  [ ] t 1  0 TAB

Variable 1 will accumulate the vector of coefficients.

2:  0              3:  0                  2:  5 x^4 + ...
1:  5 x^4 + ...    2:  5 x^4 + ...        1:  1
    .              1:  1                      .
                       .

   Z ( TAB         RET 0 s l x RET            M-TAB ! /  s | 1

Note that s | 1 appends the top-of-stack value to the vector in a variable; it is completely analogous to s + 1. We could have written instead, r 1 TAB | t 1.

1:  20 x^3 + 2 x + 2      1:  0         1:  [1, 2, 1, 0, 5, 0, 0]
    .                         .             .

    a d x RET                 1 Z )         DEL r 1  Z ' C-x )

To convert back, a simple method is just to map the coefficients against a table of powers of ‘x’.

2:  [1, 2, 1, 0, 5, 0, 0]    2:  [1, 2, 1, 0, 5, 0, 0]
1:  6                        1:  [0, 1, 2, 3, 4, 5, 6]
    .                            .

    6 RET                        1 + 0 RET 1 C-u v x

2:  [1, 2, 1, 0, 5, 0, 0]    2:  1 + 2 x + x^2 + 5 x^4
1:  [1, x, x^2, x^3, ... ]       .
    .

    ' x RET TAB V M ^            *

Once again, here are the whole polynomial to/from vector programs:

C-x ( Z `  [ ] t 1  0 TAB
           Z (  TAB RET 0 s l x RET M-TAB ! /  s | 1
                a d x RET
         1 Z ) r 1
      Z '
C-x )

C-x (  1 + 0 RET 1 C-u v x ' x RET TAB V M ^ *  C-x )

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