When f(x) is continuous in the environment of a single root r,
then f(x) changes sign through r. There is a small interval
[a,b] including r such that f(a).f(b) < 0.
Taking the midpoint m of [a,b], there are three possibilities.
f(m) = 0 ; then m is the root r.
f(m).f(a) < 0 ; then the root r is in [a,m]
f(m).f(b) < 0 ; then the root r is in [m,b]
Now we can restart the procedure with the smaller interval
[a,m] or [m,b]. The interval becomes smaller and smaller,
so we can find an approximation of the root r.
has a positive root r in [0,1]. f(0)<0 and f(1)>0.
Since f(0.5) = 0.125 the root is in [0, 0.5].
Since f(0.25) = -0.73 the root is in [0.25, 0.5].
Since f(0.375) = -0.38 the root is in [0.375, 0.5].
Since f(0.4375) = -0.15 the root is in [0.4375, 0.5].
Since f(0.46875) = -0.018 the root is in [0.46875, 0.5].
Since f(0.484375) = 0.05 the root is in [0.46875, 0.484375].
and so we approach the root 0.472834
It is superfluous to say that it is easy to computerize this method.
Then we have the root in less than a second.
Say f(x) is differentiable in the environment of a root r and ro
is a first approximation of the root r. Now point P(ro,f(ro))
is on the curve of the function. The slope of the tangent line
in P is f'(ro). This tangent line intersects the x-axis in a
point with a x-value r1. The value r1 is a better approximation
of the root than ro. From r1 you can find a better approximation
Suppose that you can bring an equation g(x)=0 in the form x = f(x).
We'll show that you can solve this equation , on certain conditions,
Start with an approximation x0 of the root.
Calculate x0, x1, ..., xn, ... such that
Consider the equation x = f(x).
If the sequence x0, x1, ... , xn, ... belongs to an interval I, where |f'(x)| < k < 1 ,
then the sequence has a limit L and L is the only root
of x=f(x) in the interval I.
We try to solve the equation x3 + 4 x2 - 1 = 0 with the
We can rewrite this equation as x = x + (x3 + 4 x2 - 1) .
xo = 0.5 is a first approximation of the root.
The f(x) from the previous theorem is here x + x3 + 4 x2 - 1 .
But the condition |f'(x)| < k < 1 is not satisfied in the
environment of the root.
Therefore, we transform the equation to
x = x + r.(x3 + 4 x2 - 1)
Now, we can choose the value of r without changing the root.
The f(x) from the theorem is now x + r.(x3 + 4 x2 - 1).
From the proof of the theorem above, we know that the convergence is very fast
if f '(x) has a value near 0 in the environment of the root.
f'(x) = 1 + r.(3x2 + 8t)
Since the root is near 0.5, we choose r such that
f'(0.5) = 1 + r.(3 (0.5)2 + 8.(0.5)) = 0
r = -0.21052631579 but we choose r = -0.21.
Now, f'(x) is near 0 in the environment of the root.
We apply iteration to x = x - 0.21(x3 + 4 x2 - 1) starting with x = 0.5
Two intersecting circles have radius one. There arise three areas.
Find the distance between the center points such that the three areas have the same size.
The specified angle is t radians.
The area of the triangle OAB is sin(t)cos(t).
The area of the circular sector OAB is t.
The area of the circular segment to the right of the chord AB is
t - sin(t)cos(t) = t - (1/2)sin(2t)
De area of this segment should be equal to one quarter of the circle area.
t - (1/2)sin(2t) = pi/4
<=> t - (1/2)sin(2t) - pi/4 = 0
We calculate t with the efficient iteration method.
We choose to = 1 radian as a first approximation of t.
t = t + r( t - (1/2)sin(2t) - pi/4)
We take r = ro and ro is approximately the root of
1 + r (1 - cos(2to) = 0
<=> ro = 1/(cos(2)-1)
Choose ro = -0.7
We apply the iteration to the equation
t = t - 0.7( t - (1/2)sin(2t) - pi/4)
We find t = 1.1549407
The distance between the center points is 2.cos(t) = 0.807945
Remark : If the two circles have a radius R, t does not change and
the distance between the center points is 2.R.cos(t).