Bisection faster than Newton?
bisection, Newton, Horner
JavaScript
// Horner method for degree-n polynomial
function eval (a, t) {
// f(x) = a0+ a1x + ... + anxn
var n = a.length - 1;// degree (n)
var b = [];
var c = [];
var i, k;
for (i = 0; i <= n; i++)
b.push(0), c.push(0);
b[n] = a[n];
c[n] = b[n];
for (k = n-1; k >= 1; k--) {
b[k] = a[k] + t*b[k+1];
c[k] = b[k] + t*c[k+1];
}
b[0] = a[0] + t*b[1];
return [b[0],c[1]];
}
// simple bisection
function bisection (func, interval, eps) {
var xLo = interval[0];
var xHi = interval[1];
fHi = func(coeff,xHi)[0]; // fb
fLo = func(coeff,xLo)[0]; // fa
if (fLo * fHi > 0)
return undefined;
var xMid, fHi, fLo, fMid;
var iter = 0;
while (xHi - xLo > eps) {
++iter;
xMid = (xLo+xHi)/2;
fMid = func(coeff,xMid)[0]; // fc
if (Math.abs(fMid) < eps)
return [xMid, iter];
else if (fMid*fLo < 0) { // fa*fc < 0 --> [a,c]
xHi = xMid;
fHi = fMid;
} else { // fc*fb < 0 --> [c,b]
xLo = xMid;
fLo = fMid;
}
}
return [(xLo+xHi)/2, iter];
}
// f(x) = 5x^3 - 27x^2 + 60x - 20
// = 5*(x-0.4)*(x^2 - 5x + 10)
var coeff = [-20,60,-27,5];
var t0 = performance.now();
var sol1 = Newton (eval, 0.5, 1e-4);
var t1 = performance.now();
var sol0 = bisection (eval, [0,1], 1e-4);
var t2 = performance.now();
console.log ('Newton time: '+ (t1-t0).toFixed(3) + ': ' + sol1);
console.log ('bisection time: '+ (t2-t1).toFixed(3) + ': ' + sol0);
/*
INPUT: Function f, endpoint values a, b, tolerance TOL, maximum iterations NMAX
CONDITIONS: a < b, either f(a) < 0 and f(b) > 0 or f(a) > 0 and f(b) < 0
OUTPUT: value which differs from a root of f(x)=0 by less than TOL
N ← 1
While N ≤ NMAX # limit iterations to prevent infinite loop
c ← (a + b)/2 # new midpoint
If f(c) = 0 or (b – a)/2 < TOL then # solution found
Output(c)
Stop
EndIf
N ← N + 1 # increment step counter
If sign(f(c)) = sign(f(a)) then a ← c else b ← c # new...