Solved

# cube root of x

Posted on 1997-11-09
971 Views
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!
0

LVL 2

Expert Comment

ID: 1216617
function cube(x:real):integer;
begin
x=x*x*x;
cube=round(x);
end

Hope this helps.
0

LVL 2

Expert Comment

ID: 1216618
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

Wilfred

0

LVL 2

Expert Comment

ID: 1216619
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.
0

Author Comment

ID: 1216620
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.
0

Accepted Solution

mwdejager earned 200 total points
ID: 1216621
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;

Begin
If x = 0 Then
CubeRoot := 0
Else
CubeRoot := Exp(Ln(x) / 3);
End;
0

LVL 2

Expert Comment

ID: 1216622

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.

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;
Begin
If x = 0 Then
CubeRoot := 0
Else
Loop
{
Yn1 = ((Yn + Yn) + x / Yn / Yn) / 3;
if (absolute(Yn1 - Yn) < 0.00001)
exit loop;

Yn = Yn1;
}
CubeRoot := Yn1;
End;

Hope this helps

Wilfred
0

## Featured Post

Are you unable to connect or configure Hotmail email account in Microsoft Outlook 2010, 2007? Or Outlook.com emails are not downloading to Outlook? Lets’ see the problem and resolve Outlook Connector error syncing folder hierarchy (0x8004102A).
What is Backup? Backup software creates one or more copies of the data on your digital devices in case your original data is lost or damaged. Different backup solutions protect different kinds of data and different combinations of devices. For e…
This demo shows you how to set up the containerized NetScaler CPX with NetScaler Management and Analytics System in a non-routable Mesos/Marathon environment for use with Micro-Services applications.
You have products, that come in variants and want to set different prices for them? Watch this micro tutorial that describes how to configure prices for Magento super attributes. Assigning simple products to configurable: We assigned simple products…