CODEWITHSUNDEEP
c#.netmathfloating-point
JK.
JK.

Reputation: 1154

Nth root of small number return an unexpected result in C#

When I try to take the N th root of a small number using C# I get a wrong number.

For example, when I try to take the third root of 1.07, I get 1, which is clearly not true.

Here is the exact code I am using to get the third root.

MessageBox.Show(Math.Pow(1.07,(1/3)).toString());

How do I solve this problem?

I would guess that this is a floating point arithmetic issue, but I don't know how to handle it.

Upvotes: 5

Views: 5401

Answers (4)

I. J. Kennedy
I. J. Kennedy

Reputation: 25819

I'm pretty sure the "exact code" you give doesn't compile. 

MessageBox.Show(Math.Pow(1.07,(1/3).toString()));

The call to toString is at the wrong nesting level, needs to be ToString, and (1/3) is integer division, which is probably the real problem you're having. (1/3) is 0 and anything to the zeroth power is 1. You need to use (1.0/3.0) or (1d/3d) or ...

Upvotes: 9

Doug
Doug

Reputation: 331

This may help in case you have a real nth root precision problem, but my experiance is that the builtin Math.Pow(double, int) is more precise:

    private static decimal NthRoot(decimal baseValue, int N)
    {
        if (N == 1)
            return baseValue;
        decimal deltaX;
        decimal x = 1M;
        do
        {
            deltaX = (baseValue / Pow(x, N - 1) - x) / N;
            x = x + deltaX;
        } while (Math.Abs(deltaX) > 0);
        return x;
    }

    private static decimal Pow(decimal a, int b)
    {
        if (b == 0) return 1;
        if (a == 0) return 0;
        if (b == 1) return a;
        if (b % 2 == 0)
            return Pow(a * a, b / 2);
        else if (b % 2 == 1)
            return a * Pow(a * a, b / 2);
        return 0;
    }

Upvotes: 1

Yuriy Faktorovich
Yuriy Faktorovich

Reputation: 68687

C# is treating the 1 and the 3 as integers, you need to do the following:

Math.Pow(1.07,(1d/3d))

or 

Math.Pow(1.07,(1.0/3.0))

It is actually interesting because the implicit widening conversion makes you make a mistake.

Upvotes: 13

paxdiablo
paxdiablo

Reputation: 881443

First things first: if that's the exact code you're using, there's likely something wrong with your compiler :-)

MessageBox.Show(Math.Pow(1.07,(1/3).toString()));

will evaluate (1/3).toString() first then try and raise 1.07 to the power of that string.

I think you mean:

MessageBox.Show(Math.Pow(1.07,(1/3)).ToString());

As to the problem, (1/3) is being treated as an integer division returning 0 and n0 is 1 for all values of n.

You need to force it to a floating point division with something like 1.0/3.0.

Upvotes: 3

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