Consider finding the root of
= ex - 4x
Try as you may, there is no algebraic technique that finds this root.
We will approximate the solution as follows:
The graph shows that a solution lies between 0
Our initial guess is
Now draw a tangent line through (1,f(1)). Next
see where the tangent line crosses the x-axis. The tangent line is a close
approximation to the curve for nearby values, hence the x-intercept of the
tangent line is close to the x-intercept of the curve. The tangent line
y - f(1) = f '(1)(x - 1)
The x intercept occurs when y = 0, hence
= f '(1)(x - 1)
solving for x,
= 1 -
This x will not be the true root, but will
be a better guess than x = 1. We will use
this (call it x2) as our
second guess. Next play the same game:
= x2 -
The graph below shows this construction. The blue line is
the first tangent line and the purple line is the second tangent line.
Continue this process to get
xn+1 = xn -
For our example this expression is
Use a calculator or computer to find the values.
x1 = 1, x2 = 0,
x3 =.3333, x4 = .3572,
x5 =.3574, x6 = .3574
We see that .3574 is the root accurate to 4
using Newton's method.
Hint: Find the root of x2 - 5.
When Newton's Method Fails
If our first guess (or any guesses thereafter) is a
point at which there is a horizontal tangent line, then this line will never
hit the x-axis, and Newton's Method will fail to locate a root. If
there is a horizontal tangent line then the derivative is zero, and we
cannot divide by f '(x) as the formula
If our guesses oscillate back and forth then Newton's method
will not work.
If there are two roots, we must have a first guess
near the root that we are interested in, otherwise Newton's method will find
the wrong root.
If there are no roots, then Newton's method will fail to
find it. (This can be frustrating when you are using your calculator
to find a root.
Explain why Newton's method fails to find the root of
f(x) = x1/3
with an initial guess of x = 1.
f '(x) = 1/3 x -2/3
xn+1 = xn -
= xn - 3xn = -2xn
1/3 x -2/3
This gives us
x1 = 1, x2
= -2(1) = -2
x3 = -2(-2) = 4,
x4 = -2(4) = -8
These numbers are growing (in absolute value) instead of
converging. In fact, we have
xn = (-1)n 2n-1
Hence Newton's method fails. However, it is clear that
there is a root at x = 0. Notice that at
0, the derivative is undefined.
An app to explore
A proof that Newton's Method will
always converge for a cubic with a positive derivative.
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