I couldn't find any function for the cubic root in the System.Math class, only the Math.Sqrt() function, I decided to write my own function. After some searching I found out about the formula for Newton approximation method and implemented it in my own function. As you might understand, you can only approximate it.
If you are in need of computing the cubic root, please feel free to copy and paste the code, or change it if you want. I see no reason to protect this code that isn't even optimized yet. If you anyone has any suggestions how to improve it, please post it here.
/// <summary>
/// Computes an approximate cube root of a number,
/// by using the Newton approximation for next guess.
/// </summary>
/// <param name="x">The number to compute the cube root from.</param>
/// <returns></returns>
public static double Cbrt(double x)
{
double y; // Guess
double d; // Last difference of y3 and x
double l; // The limit for optimal guess
// Check for simple cases:
#region
if (x == 0)
return 0;
else if (x == 1)
return 1;
else if (x == -1)
return -1;
else
#endregion
{
l = Math.Abs(x * 1E-14); // Set the limit appropriately
// the multiplication with x (its magnitude) should
// ensure no infinite loops, at the cost
// of some precision on high numbers.
// Make initial guess:
#region
double g = Math.Abs(x); // Do this guess on a positive number
if (g < 1)
y = x;
else if (g < 10)
y = x / 3;
else if (g < 20)
y = x / 6;
else if (g < 50)
y = x / 10;
else if (g < 100)
y = x / 20;
else if (g < 1000)
y = x / 50;
else if (g < 5000)
y = x / 100;
else if (g < 10000)
y = x / 500;
else if (g < 50000)
y = x / 1000;
else if (g < 100000)
y = x / 50000;
else
y = x / 100000;
#endregion
// Improve guess immediately:
y = ((x / (y * y)) + 2 * y) / 3; // Newton's approx. for new guess
d = Math.Abs(y * y * y - x); // Calculate difference
#region
while (l < d)
{
y = ((x / (y * y)) + 2 * y) / 3; // Newton's approx. for new guess
d = Math.Abs(y * y * y - x); // Calculate difference
}
#endregion
return y;
}
}
The biggest flaw of code would probably be the initial guess. Another
way to do it would just be to find x^(1/3), but you never know what
initial guesses happens inside either. The initial guess branching can't be too big, as bigger numbers will make the function take longer time because of more testing, and it can't be too small as bigger numbers will require more improvements.
I wrote a profiled version of the function, and found the following averages:
Number of improvements: Ranging from 5~8 for smaller numbers, to over a hundred
Time spent: 5000 ticks over 3.5million ticks (using System.Diagnostics.Stopwatch) on first run (because of JIT), which dropped down to ~150 on all following trials.
Finding the cubic root
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Finding the cubic root
Sekhmet
ashish taralekar
you have it. . . x^(1/3). . . how would you calculate x^5
hmmmmm
Rturlapati
private double CubicRoot(double x)
{
return Math.Pow(x, (1.0 / 3.0));
}