Newton's Method
Newton's Method
Consider finding the root of
y
= e^{x}  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
and 2.
Our initial guess is
x
= 1
Now draw a tangent line through (1,f(1)). Next
see where the tangent line crosses the xaxis. The tangent line is a close
approximation to the curve for nearby values, hence the xintercept of the
tangent line is close to the xintercept of the curve. The tangent line
has equation
y  f(1) = f '(1)(x  1)
The x intercept occurs when y = 0, hence
f(1)
= f '(1)(x  1)
solving for x,
f(1)
x
= 1 
f '(1)
This x will not be the true root, but will
be a better guess than x = 1. We will use
this (call it x_{2}) as our
second guess. Next play the same game:
f(x_{2})
x_{3}
= x_{2} 
f '(x_{2})
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
f(x_{n})
x_{n+1} = x_{n} 
f
'(x_{n}) 
For our example this expression is
Use a calculator or computer to find the values.
x_{1} = 1, x_{2} = 0,
x_{3} =.3333, x_{4} = .3572,
x_{5} =.3574, x_{6} = .3574
We see that .3574 is the root accurate to 4
decimal places.
Exercise
Estimate
using Newton's method.
Hint: Find the root of x^{2}  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 xaxis, 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
requires.

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.
Example
Explain why Newton's method fails to find the root of
f(x) = x^{1/3}
with an initial guess of x = 1.
Solution
We have
f '(x) = 1/3 x ^{2/3}
so that
x_{n}^{1/3}
x_{n+1} = x_{n} 
= x_{n}  3x_{n} = 2x_{n}
1/3 x ^{2/3}
This gives us
x_{1} = 1, x_{2}
= 2(1) = 2
x_{3} = 2(2) = 4,
x_{4} = 2(4) = 8
These numbers are growing (in absolute value) instead of
converging. In fact, we have
x_{n} = (1)^{n} 2^{n1}
Hence Newton's method fails. However, it is clear that
there is a root at x = 0. Notice that at
x =
0, the derivative is undefined.
An app to explore
Newton's Method
A proof that Newton's Method will
always converge for a cubic with a positive derivative.
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