Actual timed Mathomatic output from the demo script

Mathomatic version 15.7.2
Copyright (C) 1987-2012 George Gesslein II.
200 equation spaces available in memory,
1920 kilobytes per equation space.
HTML color mode enabled; manage by typing "help color".
1−> clear all
1−> ; Some symbolic differentiation examples follow.
1−> 
1−> ; Take the derivative of the absolute value function:
1−> |x|

          1
#1: (x^2)^–
          2

1−> derivative ; The result is the sign function sgn(x), which gives the sign of x.
Differentiating with respect to (x) and simplifying...

        x
#2: –––––––––
           1
    ((x^2)^–)
           2

2−> repeat echo *
*******************************************************************************
2−> ; Mathomatic can differentiate anything that doesn't require symbolic logarithms.
2−> y=e^(1+1/x)

               1
#3: y = ê^(1 + –)
               x

3−> derivative ; The first order derivative is:
Differentiating the RHS with respect to (x) and simplifying...

                  1
         -(ê^(1 + –))
                  x
#4: y' = ––––––––––––
             x^2

4−> derivative ; The second order derivative is:
Differentiating the RHS with respect to (x) and simplifying...

              1        1
          (ê^(– + 1))·(– + 2)
              x        x
#5: y'' = –––––––––––––––––––
                  x^3

5−> expand fraction ; Perhaps easier to read:

              1               1
          (ê^(– + 1))   2·(ê^(– + 1))
              x               x
#5: y'' = ––––––––––– + –––––––––––––
              x^4            x^3

5−> repeat echo *
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5−> ; A Taylor series demonstration:
5−> y=x_new^n ; x_new is what we want, without using the root operator.

#6: y = x_new^n

6−> x_new ; It is easily solved for in Mathomatic.

              1
#6: x_new = y^–
              n

6−> y ; But we want an algorithm to compute it without using non-integer exponentiation.

#6: y = x_new^n

6−> taylor x_new 1 x_old ; build the (nth root of y) iterative approximation formula
Computing the Taylor series of the RHS and simplifying...
1 derivative applied.

#7: y = x_old^n + (n·(x_old^(n − 1))·x_new) − (n·x_old^n)

7−> solve verify x_new ; solve for the output variable, verifying the result

             (y − x_old^n)
#7: x_new = (––––––––––––– + 1)·x_old
              (x_old^n·n)

Solution verified.
7−> simplify ; convergent nth root approximation formula:

                       y
                   (––––––– − 1)
                    x_old^n
#7: x_new = x_old·(––––––––––––– + 1)
                         n

7−> copy ; "calculate x_old 10000" tests this formula, if you would like to see for yourself.

                       y
                   (––––––– − 1)
                    x_old^n
#8: x_new = x_old·(––––––––––––– + 1)
                         n

7−> replace x_old x_new with x ; make x_old (input) and x_new (output) the same

             y
           (––– − 1)
            x^n
#7: x = x·(––––––––– + 1)
               n

7−> x ; make sure the formula was correct by solving for x
Removing possible solution: "x = 0".

          1
#7: x = y^–
          n

7−> repeat echo *
*******************************************************************************
7−> ; Another Taylor series demo:
7−> e^x ; enter the exponential function

#9: ê^x

9−> taylor x 10 0 ; generate a 10th order taylor series of the exponential function
Computing the Taylor series and simplifying...
10 derivatives applied.

             x^2   x^3   x^4   x^5   x^6   x^7     x^8     x^9      x^10
#10: 1 + x + ––– + ––– + ––– + ––– + ––– + –––– + ––––– + –––––– + –––––––
              2     6    24    120   720   5040   40320   362880   3628800

10−> laplace x ; do a Laplace transform on it

     1    1     1     1     1     1     1     1     1     1      1
#11: – + ––– + ––– + ––– + ––– + ––– + ––– + ––– + ––– + –––– + ––––
     x   x^2   x^3   x^4   x^5   x^6   x^7   x^8   x^9   x^10   x^11

11−> simplify ; show the structure of the result

                                                       1
                                                  (1 + –)
                                                       x
                                             (1 + –––––––)
                                                     x
                                        (1 + –––––––––––––)
                                                   x
                                   (1 + –––––––––––––––––––)
                                                 x
                              (1 + –––––––––––––––––––––––––)
                                               x
                         (1 + –––––––––––––––––––––––––––––––)
                                             x
                    (1 + –––––––––––––––––––––––––––––––––––––)
                                           x
               (1 + –––––––––––––––––––––––––––––––––––––––––––)
                                         x
          (1 + –––––––––––––––––––––––––––––––––––––––––––––––––)
                                       x
     (1 + –––––––––––––––––––––––––––––––––––––––––––––––––––––––)
                                     x
#11: –––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––
                                   x

11−> laplace inverse x ; undo the Laplace transform

             x^2   x^3   x^4   x^5   x^6   x^7     x^8     x^9      x^10
#12: 1 + x + ––– + ––– + ––– + ––– + ––– + –––– + ––––– + –––––– + –––––––
              2     6    24    120   720   5040   40320   362880   3628800

12−> compare with 10 ; check the result
Comparing #10 with #12...
Expressions are identical.
Finished reading file "demo.in".
12−> 
End of input.