How do I get the nth derivative in my scheme program?

Question

I can't seem to make the correct outcome, but I don't know how to set up my nth value any way else... It does the 0th and 1st derivative correctly then it gives me a crazy negative number.. do you know what could be the problem?

Code:

    (define (der f h) 
    (lambda (x) (/ (- (f (+ x h)) (f x)) 
                           h)
                      )
                )
   (define (cube x) (* x x x))

   (define (many-der f h n)
       (if (= n 0)
            f
            (many-der (der f h) h (- n 1))))

  (define der-of-cube-n (many-der cube .00000000000001 2))
(der-of-cube-n 5)

-142108547152020.03

I've attempted to rearange it so then the else statement starts with der but I get the same output when n=2...

Any help would be greatly appreciated!!

Answer 1

Your h of .00000000000001 is too small; so small that you are running into rounding errors. Here is a result with another h

(define der-of-cube-n (der-n cube 0.0001 2))
> (der-of-cube-n 5)
30.000597917023697

Note: second derivative of x^3 is 6x.

Of course, one of the important attributes of Scheme is that it supports exact numbers of arbitrary precision. So if you really want h to be that small you can formulate your inputs to be 'exact'. Like this:

> (define der-of-cube-n (der-n cube (/ 10000000000000) 2))
> (der-of-cube-n 5)
150000000000003/5000000000000
> (rationalize (der-of-cube-n 5) 0.01)
3e1