Newton's method

Exercises · Derivatives · with worked solutions · PDF

Exercise 1

Consider

\[ f(x) = e^{-x} - x \]

and verify whether Newton's method can be applied on the interval \([0,1]\). If so, find an estimate of \(x \in[0,1]\) such that:

\[ e^{-x} - x = 0 \]

Solution

We consider the function:

\[ f(x) = e^{-x} - x, ~~~f'(x) = -e^{-x}-1, ~~~f''(x) = e^{-x} \]

We have:

\[ f(0) =1>0 {\rm ~~~and~~~} f(1) <0 \]

We consider the interval \([0,1]\); for \(x \in [0,1]\) we have:

\[ f'(x) < 0 {\rm ~~~and~~~} f''(x) > 0 \]

Solution

Hence on the interval \([0,1]\) the function \(f(x) = e^{-x} - x\) satisfies the hypotheses of the theorem.

We are in the case of a function that is strictly decreasing on \([0,1]\), since \(f'(x)<0, x \in [0,1]\), and convex on \([0,1]\), since \(f''(x)>0, x \in [0,1]\). The endpoints of the chosen interval are: \(a=0\) and \(b=1\).

Hypothesis 3 holds, that is \(f(0) \cdot f''(0) >0\), and the sequence becomes:

\[ x_0 =0, \qquad x_{n+1} = x_n - \frac{e^{-x_n} - x_n}{-e^{-x_n}-1} = \frac{x_n+1}{1+e^{x_n}} {\rm ~~with~~} n \in \N \]

Since

\[\begin{align*} x_n - \frac{e^{-x_n} - x_n}{-e^{-x_n}-1} &= x_n + \frac{e^{-x_n} - x_n}{e^{-x_n}+1}=x_n + \frac{e^{-x_n}}{e^{-x_n}} \frac{1 - e^{x_n}\;x_n}{1+e^{x_n}} \\[2ex] &= \frac{x_n(1+e^{x_n})+1-e^{x_n}x_n}{1+e^{x_n}} = \frac{x_n+1}{1+e^{x_n}} \end{align*}\]

we therefore have:

\[ x_0=0,~~~x_1=0.5,~~~x_2=0.5663\dots,~~~x_3=0.5671\dots,~~~x_4=0.5671\dots,~~~x_5=0.5671\dots \]

Solution

At the first iteration the tangent line is:

\[ y = 1 - 2 \; x {\rm ~~and~~} x_1 = 0 - \frac{1}{-2}=\frac{1}{2} \]