nth_root of num?

nth_root of num?

[N Hur]

Hi,

I am wondering if anyone has a function "nth_root" for big_int in
camllight. As far as I know only "sqrt_big_int" is provided in the num
library. Could anyone send me the code if it is possible?

Regards,

	Namhyun Hur.

Re: nth_root of num?

[Claude Marche]

>>>>> "Namhyun Hur" == N Hur <mapnh@maths.bath.ac.uk> writes:

    Namhyun Hur> Hi,

    Namhyun Hur> I am wondering if anyone has a function "nth_root"
    Namhyun Hur> for big_int in camllight. As far as I know only
    Namhyun Hur> "sqrt_big_int" is provided in the num library. Could
    Namhyun Hur> anyone send me the code if it is possible?

    Namhyun Hur> Regards,

    Namhyun Hur> 	Namhyun Hur.

I have written one, here is the code. It works with a personal version
of power, where the second argument is an int, not a num.

There is one function that rounds the result to the floor, one to the
ceiling. Of course, it is working only on big_ints, not ratios. 

Hope this helps, (Any comments welcome)

-- 
| Claude Marché           | mailto:Claude.Marche@lri.fr |
| LRI - Bât. 490          | http://www.lri.fr/~marche/  |
| Université de Paris-Sud | phoneto: +33 1 69 15 64 85  |
| F-91405 ORSAY Cedex     | faxto: +33 1 69 15 65 86    |

open Num;;

let num_one = Int 1;;

let rec power_num x = function
    0 -> num_one
  | 1 -> x
  | n ->
      let y = square_num (power_num x (n/2))
      in
      	if (n mod 2)=0
      	then y
	else mult_num y x
;;

let root_floor n x =

  if le_num x num_one
  then x
  else
    let predn = pred n
    in
    let numpredn = Int predn
    and numn = Int n
    in
    let rec approx y =
      let ypnm1 = (power_num y predn)
      in      
      let new_y = 
      	quo_num 
      	  (add_num (mult_num numpredn (mult_num y ypnm1)) x) 
	  (mult_num numn ypnm1)
      in
      	if ge_num new_y y
      	then y
      	else approx new_y
    in 
      approx x
;;

let root_ceil n x =
  succ_num (root_floor n (pred_num x))
;;

Re: nth_root of num?

[Claude Marche]

>>>>> "N Hur" == N Hur <mapnh@maths.bath.ac.uk> writes:

    N Hur> Thank you.

    N Hur> But running your code for example,

    N Hur> 	root_floor 3 (Int 2) (* menaing third_root of 2 *);;
	
    N Hur> gives 1, while what we want is 1.2599210498 ...

Sorry, I misunderstood your question. I though you where interested in
computing the nth root of a big_num. If you are interested in
computing the nth root of a float, I really don't understand why you
are using big nums. 

An algorithm for computing the nth root of a float is easy :

let nth_root n x = exp (log(x)/.(float_of_int n));;

Hope this helps. If it doesn't answer to your question, please explain
your problem more precisely.

- Claude



-- 
| Claude Marché           | mailto:Claude.Marche@lri.fr |
| LRI - Bât. 490          | http://www.lri.fr/~marche/  |
| Université de Paris-Sud | phoneto: +33 1 69 15 64 85  |
| F-91405 ORSAY Cedex     | faxto: +33 1 69 15 65 86    |

Re: nth_root of num?

[John Whitley]

Claude Marche writes:
 > An algorithm for computing the nth root of a float is easy :
 > 
 > let nth_root n x = exp (log(x)/.(float_of_int n));;

Alternatively, one could use the (**) float exponentiation operator:

let nth_root n x = x ** (1.0 /. n);;

-- John