Heres a nice algorithm for positive integer powers, it starts by dealing with some simple cases and then uses a loop testing the binary bits of the exponent. For example to find 3^11 11 in binary is 1011 so the stages in the loop are
- bitMask = 1011, evenPower = 3, result = 3
- bitMask = 101, evenPower = 3*3 = 9, result = 3*9 = 27
- bitMask = 10, evenPower = 9*9 = 81, result = 27
- bitMask = 1, evenPower = 81*81 = 6561, result = 27*6561 = 177147
That is the evenPower squares at each loop, and the result gets multiplied by the evenPower if the bottom bit is 1. The code has been lifted from Patricia Shanahan’s method http://mindprod.com/jgloss/power.html which in turn has its roots in Kunth and can be traced back to 200 BC in india.
/**
* A fast routine for computing integer powers.
* Code adapted from {@link <a href="http://mindprod.com/jgloss/power.html">efficient power</a>} by Patricia Shanahan pats@acm.org
* Almost identical to the method Knuth gives on page 462 of The Art of Computer Programming Volume 2 Seminumerical Algorithms.
* @param l number to be taken to a power.
* @param n power to take x to. 0 <= n <= Integer.MAX_VALUE
* Negative numbers will be treated as unsigned positives.
* @return x to the power n
*
*/
public static final double power(double l,int n)
{
assert n>=0;
double x=l;
switch(n){
case 0: x = 1.0; break;
case 1: break;
case 2: x *= x; break;
case 3: x *= x*x; break;
case 4: x *= x; x *= x; break;
case 5: { double y = x*x; x *= y*y; } break;
case 6: { double y = x*x; x = y*y*y; } break;
case 7: { double y = x*x; x *= y*y*y; } break;
case 8: x *= x; x *= x; x *= x; break;
default:
{
int bitMask = n;
double evenPower = x;
double result;
if ( (bitMask & 1) != 0 )
result = x;
else
result = 1;
bitMask >>>= 1;
while ( bitMask != 0 ) {
evenPower *= evenPower;
if ( (bitMask & 1) != 0 )
result *= evenPower;
bitMask >>>= 1;
} // end while
x = result;
}
}
return x;
}
For a real exponent you will basically need ways of finding exp and log. You can use Taylor series which are the simplest to get but there are much better method. We have
exp(x) = 1 + x + x^2/2 + x^3/6 + x^4/24 + x^5/120 + x^6/6! + ...
ln(1+x) = x - x^2 /2 + x^3 /3 - x^4 / 4 + x^5 / 5 - x^6/6 + ... |x|<1
To find x^y note ln(x^y) = y*ln(x). Now we need to get the argument in the right range so we can use our power series. Let x = m * 2^ex, the mantissa and exponent chosen so 1/sqrt(2)<= m < sqrt(2) and ln(m*2^ex) = ln(m) + ex*ln(2). Let h = m-1 and find ln(1+h).
Using java and floats as there is an easy way to find the internals of the IEEE representation (I've used float as there are fewer bits to cope with)
int b = Float.floatToIntBits(x);
int sign = (b & 0x80000000) == 0 ? 1 : -1;
int mattissa = b & 0x007fffff;
int ex = ((b & 0x7f800000) >> 23 ) - 127;
in javascript it might be easiest to us Number.toExponential and parse the results.
Next construct a number z in the desired range 1/sqrt(2) < z < sqrt(2)
int bits = mattissa | 0x3f800000;
float z = Float.intBitsToFloat(bits);
if(z>root2) {
z = z/2;
++ex;
}
Use this function to find the log of 1+x using a taylor series
static float ln1px(float x) {
float x_2 = x*x; // powers of x
float x_3 = x_2 * x;
float x_4 = x_3 * x;
float x_5 = x_4 * x;
float x_6 = x_5 * x;
float res = x - x_2 /2 + x_3 /3 - x_4 / 4 + x_5 / 5 - x_6/6;
return res;
}
this seems to be good to three significant figures, often much better when x is close to 0.
The log of our number x can then be found
float w = z - 1;
float ln_z = ln1px(w);
float ln_x = ln_z + ln2 * ex;
System.out.println("ln "+ln_x+"\t"+Math.log(x));
Now to the actual power if we write y = n + a where n is an integer and a is fractional. So
x^y=x^(n+a) = x^n * x^a. use the first algorithm in this answer to find the x^n. Writing x=m*2^ex then ln((m*2^ex)^a) = yln(m) + yex*ln(2) and
x^a=exp(ln((m*2^ex)^a)) = exp(a * ln(m)) * exp(a * ln(2))^ex
the two exponential terms have fairly small values so the taylor series should be good.
We need one function for the taylor series of the exponential function
static float exp(float x) {
float x_2 = x*x; // powers of x
float x_3 = x_2 * x;
float x_4 = x_3 * x;
float x_5 = x_4 * x;
float x_6 = x_5 * x;
float res = 1+ x + x_2 /2 + x_3 /6 + x_4 / 24 + x_5 / 120 + x_6/ 720;
return res;
}
finally we can put the pieces together
// Get integer and fractional parts of y
int n = (int) Math.floor(y);
float a = y-n;
float x_n = power(x,n); // x^n
float a_ln_m = a * ln_z; // ln(m^a) = a ln(m)
float a_ln_2 = a * ln2; // ln(2^a) = a ln(2)
float m_a = exp(a_ln_m); // m^a = exp(a ln(m))
float _2_a = exp(a_ln_2); // 2^a = exp(a ln(2))
float _2_a_ex = power(_2_a,ex); // (2^ex)^a = 2^(a*ex) = (2^a)^ex
float x_a = m_a * _2_a_ex; // x^a = m^a * 2^(a*ex)
float x_y = x_n * x_a; // x^y = x^n * x^a
System.out.println("x^y "+x_y+"\t"+Math.pow(x,y));
That should be the complete program, you need some smarts to cope with negative arguments etc.
Note this is not particularly accurate as I've only used a few terms of the taylor series. Other SO questions have more detailed answers How can I write a power function myself?
