Solve Reduction Problem for (1+x)^1/3 in Mathcad

[PoFon]

I have a function : (1+x)^1/3 and I need to do reduction in mathcad.

I wrote this :

But the problem is that : it does not work with negative numbers.
How to fix it?

 

[Mark44]

does not work with negative numbers.

How so? What does your code do or not do?

 

[PoFon]

How so? What does your code do or not do?

This code does reduction with positive numbers ( function (x+1)^1/3), but doesn't work with negative. And i don't know how to fix it.

 

[Mark44]

This code does reduction with positive numbers ( function (x+1)^1/3), but doesn't work with negative. And i don't know how to fix it.

Does the problem occur when your code attempts to raise a negative number to the 1/3 power? If so, algebraically there shouldn't be a problem, but MathCad is probably using logs to compute powers, in which case the expression x + 1 needs to be positive. One workaround would be to have logic that determines the sign of x + 1. If x + 1 turns out to be negative, compute (-x - 1) to the 1/3 power. That should give you a positive number, which you'll need to multiply by -1.

 

[DavidHume]

Except that raising a negative number to a fractional power generally results in a complex number. For instance, here are the results we get raising 2 and -2 to the 1/3 power (in J since I don't have MathCAD):

   2 _2 ^ %3
1.25992 0.629961j1.09112

As you can see, the result of the negative number raised to the reciprocal of 3 power is a complex number with real part 0.629961 and imaginary part 1.09112. In MathCAD, this would be represented as 0.629961 + 1.09112i .

 

[Mark44]

Except that raising a negative number to a fractional power generally results in a complex number. For instance, here are the results we get raising 2 and -2 to the 1/3 power (in J since I don't have MathCAD):

   2 _2 ^ %3
1.25992 0.629961j1.09112

As you can see, the result of the negative number raised to the reciprocal of 3 power is a complex number with real part 0.629961 and imaginary part 1.09112. In MathCAD, this would be represented as 0.629961 + 1.09112i .

I understand what you're saying, but the problem is that the result should be a real number, not a complex number. The function $f(x) = \sqrt[3]{x} = x^{1/3}$ is defined for all real numbers, and the range is all real numbers, so taking the cube root of -2 (or raising -2 to the 1/3 power) should result in a negative real number. In my previous post I explained why many computer systems produce results that are mathematically incorrect.

 

[DavidHume]

It's not true that the range of cube root is real numbers - it's not even true for the cube root of one. Take a look at this MathCAD paper - http://gekor-it.de/media/81174be56cf4ccc5ffff870fac144233.pdf - where the author shows the cube roots of one to be (1, (-1/2)+(√3 i)/2, (-1/2)-(√3 i)/2). You can see the same result in the Wikipedia article on the root of unity: https://en.wikipedia.org/wiki/Root_of_unity .

The point for the original poster is that MathCAD can handle complex numbers but I don't know how well it does this or what it does in the context of a reduction.

 

[Mark44]

It's not true that the range of cube root is real numbers - it's not even true for the cube root of one.

$\sqrt[3]{-1} = -1$ and $\sqrt[3]{-8} = -2$, which you can verify by raising -1 and -2 to the third power, respectively.
All of the odd roots (cube root, fifth root, seventh root, etc.) have real values for any real numbers.

Take a look at this MathCAD paper - http://gekor-it.de/media/81174be56cf4ccc5ffff870fac144233.pdf - where the author shows the cube roots of one to be (1, (-1/2)+(√3 i)/2, (-1/2)-(√3 i)/2).

Sure, there are n values for the nth root of any real number, but the principal cube root (or fifth root or seventh root, etc.) of any real number is a real number. The principal cube root of 1 is 1, just as the principal square root of 4 (denoted $\sqrt{4}$) is 2, even though -2 is also a square root of 4. All of the odd roots have one real number principal root, so if you take the cube root of -2, you should get a real number back.

You can see the same result in the Wikipedia article on the root of unity: https://en.wikipedia.org/wiki/Root_of_unity .

The point for the original poster is that MathCAD can handle complex numbers but I don't know how well it does this or what it does in the context of a reduction.

My point is that MathCAD (or whatever) is able to find the square root of a positive number, but should also be able to take an odd root of a negative number, producing a negative real number. Due to the algorithm used, it isn't able to do so, and I gave a workaround for this inability.

 

[DavidHume]

I see - your workaround should be fine as long as you only care about the principal root.