/**
* Safe, platform consistent implementations of some Math functions
*
* These functions are implemented in JS to avoid observed differences
* between results of different floating point libraries, see
* https://bugzilla.mozilla.org/show_bug.cgi?id=531915
*
* They mostly meet the ECMAScript Edition 5 spec, see
* http://www.ecma-international.org/publications/files/ECMA-ST/Ecma-262.pdf
*
* See simulation/components/tests/test_Math.js for tests.
*/
/**
* Approximation of cosine of a (radians)
*/
Math.cos = function(a)
{
// Bring a into the 0 to +pi range without expensive branching.
// Uses the symmetry that cos is even.
a = (a + Math.PI) % (2*Math.PI);
a = Math.abs((2*Math.PI + a) % (2*Math.PI) - Math.PI);
// make b = 0 if a < pi/2 and b=1 if a > pi/2
var b = (a-Math.PI/2) + Math.abs(a-Math.PI/2);
b = b/(b+1e-30); // normalize b to one while avoiding divide by zero errors.
// if a > pi/2 send a to pi-a, otherwise just send a to -a which has no effect
// Using the symmetry cos(x) = -cos(pi-x) to bring a to the 0 to pi/2 range.
a = b*Math.PI - a;
var c = 1 - 2*b; // sign of the output
// Taylor expansion about 0 with a correction term in the quadratic to make cos(pi/2)=0
return c * (1 - a*a*(0.5000000025619951 - a*a*(1/24 - a*a*(1/720 - a*a*(1/40320 - a*a*(1/3628800 - a*a/479001600))))));
};
/**
* Approximation of sine of a (radians)
*/
Math.sin = function(a)
{
return Math.cos(a - Math.PI/2);
};
/**
* Approximation of arctangent of a, returns angle from -pi/2 to pi/2
*/
Math.atan = function(a)
{
var tanPiBy6 = 0.5773502691896257;
var tanPiBy12 = 0.2679491924311227;
var sign = 1;
var inverted = false;
var tanPiBy6Shift = 0;
if (a < 0 || 1/a === -Infinity)
{
// tan(x) = -tan(-x) so remove sign now and put it back at the end
sign = -1;
a *= -1;
}
if (a > 1)
{
// tan(pi/2 - x) = 1/tan(x)
inverted = true;
a = 1/a;
}
if (a > tanPiBy12)
{
// tan(x-pi/6) = (tan(x) - tan(pi/6)) / (1 + tan(pi/6)tan(x))
tanPiBy6Shift = Math.PI/6;
a = (a - tanPiBy6) / (1 + tanPiBy6*a);
}
// Now a will be in the range [-tan(pi/12), tan(pi/12)]
// Use the taylor expansion around 0 with a correction to the linear term to match the pi/12 boundary
// atan(x) = x - x^3/3 + x^5/5 - ...
var r = a*(1.0000000000390272 - a*a*(1/3 - a*a*(1/5 - a*a*(1/7 - a*a*(1/9 - a*a*(1/11 - a*a*(1/13 - a*a/15)))))));
// shift the result back where necessary
r += tanPiBy6Shift;
if (inverted)
r = Math.PI/2 - r;
return sign * r;
};
/**
* Approximation of arctangent of y/x, returns angle from -pi to pi
*/
Math.atan2 = function(y,x)
{
// get unsigned x,y for ease of calculation, this means all angles are in the range [0, pi/2]
var ux = Math.abs(x);
var uy = Math.abs(y);
// holds the result in the upper right quadrant
var r;
// Handle all edges cases to match the spec
if (uy === 0)
r = 0;
else
{
if (ux === 0)
r = Math.PI / 2;
if (uy === Infinity)
{
if (ux === Infinity)
r = Math.PI / 4;
else
r = Math.PI / 2;
}
else
{
if (ux === Infinity)
r = 0;
else
r = Math.atan(uy/ux);
}
}
// puts the result into the correct quadrant
// 1/(-0) is the only way to determine the sign for a 0 value
if (x < 0 || 1/x === -Infinity)
{
if (y < 0 || 1/y === -Infinity)
return -Math.PI + r;
else
return Math.PI - r;
}
else
{
if (y < 0 || 1/y === -Infinity)
return -r;
else
return r;
}
};
Math.acos = function()
{
error("Math.acos() does not yet have a synchronization safe implementation");
};
Math.asin = function()
{
error("Math.asin() does not yet have a synchronization safe implementation");
};
Math.tan = function()
{
error("Math.tan() does not yet have a synchronization safe implementation");
};
/**
* Approximation of raising x to the power y
*/
Math.pow = function(x, y)
{
if (Math.round(y) === y)
{
if (y >= 0)
return Math.intPow(x, y);
return 1 / Math.intPow(x, -y);
}
// log has the biggest error when x ~=~ 1
// exp has the biggest error when y*log(x) ~<~ 0
// so the biggest error happens around numbers like pow(0.9999,0.0001),
// that has an error of 10^-17. So I think we're safe
return Math.exp(y*Math.log(x));
};
/**
* Approximation of the exponential function, e raised to the power x
*/
Math.exp = function(x)
{
if (x < 0)
var iPart = 1/Math.intPow(Math.E, -Math.floor(x));
else
var iPart = Math.intPow(Math.E, Math.floor(x));
if (x === Math.floor(x))
// no need to loop if we know the answer
return iPart;
// the integer part is known, work further with the decimal part of x
x = x - Math.floor(x); // x \in [0,1)
// taylor series around 0
// max error ~=~ 10^(-16)
var dPart = 1;
for (var i = 22; i > 0; i--)
dPart = 1+x*dPart/i;
// total precision ~=~ 17 decimal digits
return iPart*dPart;
};
/**
* Approximation of the natural logarithm of x
*
* For values very close to 1, the error of 10^-16 could become bigger than the actual value
* But this also happens with the native log function
*/
Math.log = function(x)
{
if (!(x >= 0))
return NaN;
if (x === 0)
return -Infinity;
if (x === Infinity)
return x;
// start with calculating the binary logarithm
// based on http://en.wikipedia.org/wiki/Binary_logarithm#Real_number
// calculate to 50 fractional bits -> error ~=~ 10^-16
var precisionBits = 50;
// calculate integer log, rounded down
// when implemented in C, just count the number of bits before the fraction
// without leading zeros. This may be negative.
var log = 0;
if (x >= 1)
{
for (var i = 1; i <= x; i *= 2)
log++;
log--;
i /= 2;
}
else
{
for (var i = 1; i > x; i /= 2)
log--;
}
// now lb(x) = log + lb(y) with y = x/i. So y \in [1,2)
var y = x/i;
// if we're done, or there's a minimal rounding error and we should be done
// convert to natural logarithm
if (y <= 1)
return log / Math.LOG2E;
var m = 0;
var add = 1;
while (true)
{
while (m <= precisionBits && y < 4)
{
m++;
y *= y;
add /= 2;
}
if (m > precisionBits)
break;
log += add;
y /= 2;
}
// convert binary logarithm to natural logarithm;
return log / Math.LOG2E;
};
/**
* Calculate the power for positive integer exponents
*/
Math.intPow = function(x, y)
{
if (Math.abs(y) === Infinity)
{
if (Math.abs(x) === 1)
return NaN;
if (Math.abs(x) < 1 && y > 0 || Math.abs(x) > 1 && y < 0)
return 0;
return Infinity;
}
var powers = [x];
var binary = [1];
var i = 0;
for (var e = 2; e <= y; e *= 2)
{
// calculate x^i, using x^(i/2)
powers.push(powers[i]*powers[i]);
binary.push(e);
i++;
}
var result = 1;
var i = binary.length;
while (y > 0)
{
if (binary[--i] <= y)
{
result *= powers[i];
y -= binary[i];
}
}
// error margin = 0 (default JS error)
return result;
};