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 r_{o}
is a first approximation of the root r. Now point P(r_{o},f(r_{o}))
is on the curve of the function. The slope of the tangent line
in P is f'(r_{o}). This tangent line intersects the x-axis in a
point with a x-value r_{1}. The value r_{1} is a better approximation
of the root than ro. From r_{1} you can find a better approximation
r_{2}. ...
Taking t_{1} as a new approximation we find t_{2}= 0.472834787153
Taking t_{2} as a new approximation we find t_{3}= 0.472833908996
Taking t_{3} as a new approximation we find t_{4}= 0.472833908996
You see that this method is a very powerful method to approximate
the root.
Again it is superfluous to say that it is easy to computerize this method.
Then we have the root in less than a second.
This method requires that the first approximation is sufficiently close
t_{o} the root r.
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,
using iteration.
Start with an approximation x_{0} of the root.
Calculate x_{0}, x_{1}, ..., x_{n}, ... such that
Consider the equation x = f(x).
If the sequence x_{0}, x_{1}, ... , x_{n}, ... 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 x^{3} + 4 x^{2} - 1 = 0 with the
iteration method.
We can rewrite this equation as x = x + (x^{3} + 4 x^{2} - 1) .
x_{o} = 0.5 is a first approximation of the root.
The f(x) from the previous theorem is here x + x^{3} + 4 x^{2} - 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.(x^{3} + 4 x^{2} - 1)
Now, we can choose the value of r without changing the root.
The f(x) from the theorem is now x + r.(x^{3} + 4 x^{2} - 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.(3x^{2} + 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(x^{3} + 4 x^{2} - 1) starting with x = 0.5
0.47375
0.472892306269531
0.472837688600127
0.472834153854809
0.472833924859327
0.472833910023068
0.472833909061846
0.47283390899957
0.472833908995535
We want the roots of x ln(x) - ln(0.7) = 0.
Using a plot we see that there are two roots and we can choose the
approximations 0.28 and 0.46.
We will first apply the method to 0.28
We write x = x + r(x ln(x) - ln(0.7))
We choose an r-value such that 1+r(1+ln(0.28))=0. r_{o}=3.66 is a good approximation.
We apply iteration to x = x + 3.66 (x ln(x) - ln(0.7))
In a few steps we find very good results.
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 t_{o} = 1 radian as a first approximation of t.
t = t + r( t - (1/2)sin(2t) - pi/4)
We take r = r_{o} and r_{o} is approximately the root of
1 + r (1 - cos(2t_{o}) = 0
<=> r_{o} = 1/(cos(2)-1)
Choose r_{o} = -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).
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