Performing Math operations on decimal datatype in C#?

Question

I was wondering if the above was at all possible. For example:

Math.Sqrt(myVariableHere);

When looking at the overload, it requires a doub

Answer 1

I don't understand why all the answers to that question are the same.

There are several ways to calculate the square root from a number. One of them was proposed by Isaac Newton. I'll only write one of the simplest implementations of this method. I use it to improve the accuracy of double's square root.

// x - a number, from which we need to calculate the square root
// epsilon - an accuracy of calculation of the root from our number.
// The result of the calculations will differ from an actual value
// of the root on less than epslion.
public static decimal Sqrt(decimal x, decimal epsilon = 0.0M)
{
    if (x < 0) throw new OverflowException("Cannot calculate square root from a negative number");

    decimal current = (decimal)Math.Sqrt((double)x), previous;
    do
    {
        previous = current;
        if (previous == 0.0M) return 0;
        current = (previous + x / previous) / 2;
    }
    while (Math.Abs(previous - current) > epsilon);
    return current;
}

About speed: in the worst case (epsilon = 0 and number is decimal.MaxValue) the loop repeats less than a three times.

If you want to know more, read this (Hacker's Delight by Henry S. Warren, Jr.)

Answer 2

Simple: Cast your decimal to a double and call the function, get the result and cast that back to a decimal. That will probably be faster than any sqrt function you could make on your own, and save a lot of effort.

Answer 3

Math.Sqrt((double)myVariableHere);

Will give you back a double that's the square root of your decimal myVariableHere.

Answer 4

I just came across this question, and I'd suggest a different algorithm than the one SLenik proposed. This is based on the Babylonian Method.

public static decimal Sqrt(decimal x, decimal? guess = null)
{
    var ourGuess = guess.GetValueOrDefault(x / 2m);
    var result = x / ourGuess;
    var average = (ourGuess + result) / 2m;

    if (average == ourGuess) // This checks for the maximum precision possible with a decimal.
        return average;
    else
        return Sqrt(x, average);
}

It doesn't require using the existing Sqrt function, and thus avoids converting to double and back, with the accompanying loss of precision.

Answer 5

In most cases involving a decimal (currency etc), it isn't useful to take a root; and the root won't have anything like the expected precision that you might expect a decimal to have. You can of course force it by casting (assuming we aren't dealing with extreme ends of the decimal range):

decimal root = (decimal)Math.Sqrt((double)myVariableHere);

which forces you to at least acknowledge the inherent rounding issues.