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cube root of x

Assignment: Write a program to read in a real number x and output the integer n closest to the cube root of x. Assume that x is always non-negative. The program should allow the user to repeat this calculation as often as desired.

We just covered functions, loops (not yet at arrays). Please provide the code. Thanks!
1 Solution
function cube(x:real):integer;

Hope this helps.  
kellyjj, I believe chadd is looking for a function that will return the cube root of x, not the cube of x

Chadd, I could give you the code but it goes against my principle of completing another persons assignment. This way you won't learn anything. If you wish, I could give you the pseudo-code based on the Newton-Raphson method. You could then try to implement it & if you face any problem you can ask me more specific questions.

If you are still interested only in the code, I hope someone else can help you


wpinto :  

Opps!! I guess I don't know how to read. hehe  =]   Darn it.  Oh well, it will teach chaddd to get others to do his work.
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chadd082197Author Commented:
To kellyjj: This will help me very little. And I am not getting others to do my work, I simply wanted a little help (I thought that's what this site was all about). Thanks anyway tho.

To wpinto: I respect that. Any algorithm that you could provide would be helpful I'm sure. The 200 points is still at large.
This fuction assumes that x >= 0.
If a^n = e^(n log a)  where the base is e,
then the following function should work.

Function CubeRoot(x : Real) : Real;

      If x = 0 Then      
            CubeRoot := 0
            CubeRoot := Exp(Ln(x) / 3);
Hi chadd,

The above code will give you the cube root if you have access to functions such as log & exp. Also note that cube root of x is equivalent to x^(1/3), so if you have access to a power function its even easier.

The below algoritm needs only access to the addition & division operators

Here's the algorithm I promised

To compute y = x^(1/3)    
y^3 = x    
y^3-x = 0    
Let f(y) = y^3-x
f'(y) = 3y^2    
By Newton iteration:
y[n+1] = y[n] - f(y[n])/f'(y[n]) -------- f'(y[n]) <> 0
y[n+1] = y[n] - (y[n]^3-x)/(3y[n]^2) ---- 3y[n]^2 <> 0
y[n+1] = y[n] - (y[n]-x/y[n]^2)/3 ------- y[n] <> 0
y[n+1] = (2y[n] + x/y[n]^2) / 3 --------- x <> 0    

Now the answer is in y[n] if y[n+1] = y[n]. Since we are dealing with floating point nos, its highly unlikely that you will reach the above condition. So you will have to check for a small difference between them eg. y[n+1] - y[n] < 0.000001. Note, the smaller the constant, the more precise the answer but the number of iterations will be more!

Here's the pseudo code

Function CubeRoot(x : Real) : Real;
VAR Yn : Real;
VAR Yn1 : Real;
   If x = 0 Then
   CubeRoot := 0
     Yn1 = ((Yn + Yn) + x / Yn / Yn) / 3;
     if (absolute(Yn1 - Yn) < 0.00001)
     exit loop;

     Yn = Yn1;
   CubeRoot := Yn1;

Hope this helps


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