3rd square root

113

Mar 14, 2003 at 12:52 AM

Whats the best way to calulate the 3rd squareroot of a number a. -CProgrammer

511

Mar 14, 2003 at 12:59 AM

It''s just the third root, not the third square root

Use pow to raise the number to a third (1.0f / 3.0f). The same goes for any root, which is just the reciprocal of the associated power, i.e squared is x² and square root is x^{1 / 2}.

Agentidd

Mar 14, 2003 at 1:10 AM

  #include <iostream>
#include "math.h"

using std::cout;



void main()
{
    int a = 5;

    cout << sqrt(sqrt(sqrt(a)));
    return;
}

That would be the 3rd squarroot of a.

But... maybe you mean the "CUBE ROOT" of a number.
Remember you can raise any number by a fraction is giving the same number the nth root.

So a^(1/3) will give you the Cube root of a.

-DD

uutee

Mar 14, 2003 at 1:20 AM

pow(x,1.0/3.0) isn''t sufficient - you can''t do pow if x is negative. It should be:

#define cbrt(x) ((x)>=0 ? pow((x),1.0/3.0)

ow(-(x),1.0/3.0))

Or if you''re using some other programming language than C, adapt from this:

void cbrt( float x )
{
if( x >= 0 )
return pow(x,1.0/3.0);
else
return pow(-x,1.0/3.0);
}

- Mikko Kauppila

511

Mar 14, 2003 at 6:50 PM

quote:
Original post by Anonymous Poster
you can''t return if the function is void...can you?

You can return, you just can''t return a value. It''s an error if you try

Author

113

Mar 16, 2003 at 12:52 AM

Pow is a good solution but say i dont have pow how would i make my own effective! 3rd root function. I made one that just finds a smaller number close to the result and a bigger one and then it just tries all the ones between according to a detail level.

Any ideas for a good algorithm
-CProgrammer

122

Mar 16, 2003 at 2:16 AM

Okay. Here''s a cbrt without pow:

float cbrt(float f) { return exp(log(f)/3); }

Author

113

Mar 16, 2003 at 2:36 AM

thanks Beer hunter thats exactly what i need.

-CProgrammer

116

Mar 16, 2003 at 7:13 AM

This should also work:

x/3 - (2*x^2)/(9*2) + (10*x^3)/(27*6) - (80*x^4)/(81*24) + ... + ((-1)^(n+1)*(2*5*8* .. * (2 + 3(n-2)))*x^n)/(n!*3^n) ..
and so on.. this should converge towards the third root of x.

Just added this since it is a way to do it using only basic functions like +, -, * and /.

You are probably better off using Beer Hunters version though since the exp() and log() functions are way better optimized than you can do with this function (unless you know machine language really well)

YoshiN

Mar 16, 2003 at 12:50 PM

quote:
Original post by Jesper T
This should also work:

x/3 - (2*x^2)/(9*2) + (10*x^3)/(27*6) - (80*x^4)/(81*24) + ... + ((-1)^(n+1)*(2*5*8* .. * (2 + 3(n-2)))*x^n)/(n!*3^n) ..
and so on.. this should converge towards the third root of x.

Just added this since it is a way to do it using only basic functions like +, -, * and /.

You are probably better off using Beer Hunters version though since the exp() and log() functions are way better optimized than you can do with this function (unless you know machine language really well)

My question is what is the radius of convergence for that series? I''m too lazy to actually figure it out, but my guess is that it is not infinite...

116

Mar 16, 2003 at 1:56 PM

Well, it should be infinite, unless I did a calculation error.

122