CODEWITHSUNDEEP
javafloating-pointnanfloating-point-precision
SuSaNN
SuSaNN

Reputation: 48

Floating point - NaN

I am working on a research project on java in which some tough calculations have to be done. However I am done with most part but stuck at a point.. I have to calculate the following :

(2.1-2.3) raised to power 0.3.

But I get the answer NaN.. I have tried store result with both float and double variables, however it shows same result. Strangely when I did the same with a calculator , it showed result -0.430512

I can not figure out how to make it work with java
Code for same is: 

Math.pow((provider1[k][a][m]-provider1[k][j][m]),prior[k][m]);.

When the values of above are 2.1,2.3 and 0.3 - NaN is the output.

However when the values are 2.1, 2.3 and 3, it gets the correct value (-0.08). Can someone please help how to do this.

Upvotes: 0

Views: 737

Answers (4)

Susan
Susan

Reputation: 40

It must be an error. Check the calculator you are using.

Upvotes: 0

tricknology
tricknology

Reputation: 1138

I hope this helps you or someone else in the future:

The link below uses the example below to explain the mathematics behind the complex number solutions you get when you raise a negative base to a fractional number and how abs() can help you. When you try to do with with standard Math.* functions it doesn't know how to handle the complex portion of the solution. Using Java as an example, to make a calculation with a negative base and a positive fractional exponent, normally one could use:

double d = Math.pow(base,exp);

However, with a negative base and fractional exponent this results in NaN because complex solutions exist.

If the complex portion of a solution doesn't matter to you (like it didn't in my case), you can use this approach.

The equation used in this example is a parameter "Phi" in the Yamada and Gunn modification to the Rackett equation:

phi = (1-T/Tc)^(2/7) - (1-Tr/Tc)^(2/7)

double calculatePhi(double T, double Tr, double Tc){
  double l = T/Tc;
  double m = Tr/Tc;
  double n = 2/7.0;
  double q = (1-l);  //possibly negative
  double r = (1-m);  //possibly negative
  q = abs(q);
    double numa = Math.exp(n * Math.log(q));
  r = abs(r)
    double numb = Math.exp(n * Math.log(r));
  double phi = numa - numb;
  return phi;
}

See this link for mathematical proof:

A Programmatic Approach to Finding a Real Solution to a Function Having a Potentially Negative Base and Fractional Exponent

Upvotes: 0

rgettman
rgettman

Reputation: 178263

According to the Math.pow Javadocs:

  • If the first argument is finite and less than zero
    • if the second argument is a finite even integer, the result is equal to the result of raising the absolute value of the first argument to the power of the second argument
    • if the second argument is a finite odd integer, the result is equal to the negative of the result of raising the absolute value of the first argument to the power of the second argument
    • if the second argument is finite and not an integer, then the result is NaN.

(emphasis mine)

That is because raising a negative number to a non-integral power yields a complex number. One case is not being able to take the square root of a number: (-1)^0.5. The result doesn't exist in real numbers, and it inspired the imaginary number i which in math is defined to be the square root of -1. In general, raising a negative number to a non-integral power yields a complex number, which is the sum of a real number and an imaginary number.

Upvotes: 4

duffymo
duffymo

Reputation: 308763

Looks like a complex number to me.

My HP 15C scientific calculator tells me that (-0.2)^0.3 is an error.

Time for you to buy a new calculator.

Upvotes: 0

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