Calculus- Early Transcendentals, 2021a

5.4. Linear and Higher Order Approximations 189
We can now clearly see how (x 2 ,0) is a better estimate of the root of f (x), rather than any of the
previous points. Moving forward, we will get:
x 3 = x 2 − f (x 2)
f ′ (x 2 )
as:
Rest assured, (x 3 ,0) will be an even better estimate of the root! We express the general iterative step
x n+1 = x n − f (x n)
f ′ (x n )
The idea is to iterate these steps to obtain the desired accuracy. Here is an example.
Example 5.32: Newton method to approximate roots
Use Newton’s method to approximate the roots of f (x)=x 3 − x + 1.
Solution. You can try to find solve the equation algebraically to see that this is a difficult task, and thus it
make sense to try a numerical method such as Newton’s.
To find an initial value x 0 , note that f (−1)=−5and f (0)=1, and by the Intermediate Value Theorem
this f has a root between these two values, and we decide to start with x 0 = −1 (you can try other values
to see what happens).
Note that f ′ (x)=3x 2 − 1, and thus we get
Thus we can produce the following values (try it):
x n+1 = x n − f (x n)
f ′ (x n ) = x + n − x3 n − x n + 1
3x 2 n − 1
x 0 = −1
x 1 = −1.5000
x 2 = −1.347826..
x 3 = −1.325200..
x 4 = −1.324718..
x 5 = −1.324717..
x 6 = −1.324717..
···
and we can now approximate the root as −1.324717.
♣
As with any numerical method, we need to be aware of the weaknesses of any technique we are using.