Help with solving for the
Root of an Equation
Prof. Richard B. Goldstein

More detailed
help with

JavaScript Syntax

(x2 + 2y - 1)5/3 is Math.pow((x*x+2*y-1),5/3)
eqsample is Math.sqrt(Math.sin(x)/(Math.pow(2,x)+3*Math.log(x)))
BISECTION METHOD
The root is found by starting with an interval [a, b] that brackets the root. The midpoint, p, is then tested to see if f(p) is of the same sign as f(a) or f(b). Either a or b is then replaced and the interval is bisected - split in half. root1bi
FALSE POSITION METHOD
The root is found by starting with an interval [a, b] that brackets the root. The secant connecting (a,f(a)) and (b,f(b) is found and its x-axis intercept, p, is then tested to see if f(p) is of the same sign as f(a) or f(b). Either a or b is then replaced and the interval is then shrunk. root1fp
SECANT METHOD
Two initial estimates p0 and p1 of the root are given. The points on the curve are then connected by a secant. The x-axis intercept is called p2. Next, p1 and p2 are used to produce p3. This method converges more rapidly then the ones above - in fact, it is almost as fast as the Newton-Raphson method but requires fewer function evaluations. There are problems if the first two points are not well chosen. The first two methods are more robust - that is, if [a, b] brackets a single root, then they will always be located by the above two methods for a continuous function. root1se

Return to 1 Variable Root