power function to find the power where exponent is in decimal and less than 1

Question

I was trying to create a program that finds the power of a real number . The problem is that exponent is in decimal and less than 1 but not negative.

suppose we have to find the power of

5^0.76

what i really tried was i wrote 0.76 as 76/100 and it would be 5^76/100 and after that i wrote

here is the code if you want to see what i did

public class Struct23 {


public static void main(String[] args) {
    double x = 45;
    int c=0;
    StringBuffer y =new StringBuffer("0.23");

    //checking whether the number is valid or not
    for(int i =0;i<y.length();i++){
        String subs = y.substring(i,i+1);


        if(subs.equals(".")){
            c=c+1;
        }     
    }
     if(c>1){ 
         System.out.println("the input is wrong");
             }
     else{
       String nep= y.delete(0, 2).toString();
        double store = Double.parseDouble(nep);
        int length = nep.length();
        double rootnum = Math.pow(10, length);
        double skit = power(x,store,rootnum);
        System.out.println(skit);

}

}
 static double power(double x,double store,double rootnum){
    //to find the nth root of number
    double number =  Math.pow(x, 1/rootnum);

     double power = Math.pow(number, store);
return power;
}

}

the answer would come but the main problem is that i cannot use pow function to do that

i can't also use exp() and log() functions.

 i can only use 

   +
   -
   *
   /

help me suggest your ideas .

thanks in advance

Please add your code to the question (minimal example) rather than linking — FujiApple, Aug 24 '16 at 04:24
If the exponent part is integer, writing your own power function is pretty easy. A lot of example are available online. — Naved Alam, Aug 24 '16 at 04:52
@NavedAlam im pretty sure he explicitly states they are not ints ... but you are definately correct it becomes quite trivial then — Joran Beasley, Aug 24 '16 at 04:55
@JoranBeasley In the example he provided, he is basically trying to find the numerator and denominator and calling the power function on each of them. These numerator and denominator values would be basically integer. — Naved Alam, Aug 24 '16 at 05:03
the fraction is not an integer... and finding the nth root is not nearly as trivial as finding the power of x^n when n is an int (although not that hard either) — Joran Beasley, Aug 24 '16 at 05:18

score 4 · Accepted Answer · edited May 23 '17 at 12:33

def newtons_sqrt(initial_guess, x, threshold=0.0001):
    guess = initial_guess
    new_guess = (guess+float(x)/guess)/2
    while abs(guess-new_guess) > threshold :
        guess=new_guess
        new_guess = (guess+float(x)/guess)/2
    return new_guess




def power(base, exp,threshold=0.00001):
    if(exp >= 1): # first go fast!
        temp = power(base, exp / 2);
        return temp * temp
    else: # now deal with the fractional part
        low = 0
        high = 1.0
        sqr = newtons_sqrt(base/2,base)
        acc = sqr
        mid = high / 2

        while(abs(mid - exp) > threshold):
            sqr = newtons_sqrt(sqr/2.0,sqr)

            if (mid <= exp):
                low = mid
                acc *= sqr
            else:
                high = mid
                acc *= (1/sqr)
            mid = (low + high) / 2;
        return acc

print newtons_sqrt(1,8)
print 8**0.5

print power(5,0.76)
print 5**0.76

I reapropriated most of this answer from https://stackoverflow.com/a/7710097/541038

you could also expound on newtons_sqrt to give newtons_nth_root ... but then you have to figure out that 0.76 == 76/100 (which im sure isnt too hard really)

score 2 · Answer 2 · answered Aug 24 '16 at 04:59

2

you can convert your number to complex form of it and then use de Moivre' formula to compute the nth root of your number using your legal oprations.

answered Aug 24 '16 at 04:59

Mohammad Mahmoudi

57
3

power function to find the power where exponent is in decimal and less than 1

2 Answers2