Calculus

Applications of the Derivative

Applications of Derivative Logo

Newton’s Method

Newton's method (or Newton-Raphson method) is an iterative procedure used to find the roots of a function.

Newton's method
Figure 1.

Suppose we need to solve the equation \(f\left( x \right) = 0\) and \(x=c\) is the actual root of \(f\left( x \right).\) We assume that the function \(f\left( x \right)\) is differentiable in an open interval that contains \(c.\)

To find an approximate value for \(c:\)

  1. Start with an initial approximation \({x_0}\) close to \(c.\)
  2. Determine the next approximation by the formula
    \[{x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f^\prime\left( {{x_0}} \right)}}.\]
  3. Continue the iterative process using the formula
    \[{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f^\prime\left( {{x_n}} \right)}}\]
    until the root is found to the desired accuracy.

Let's apply Newton's method to approximate \(\sqrt 3.\) Suppose that we need to solve the equation

\[f\left( x \right) = {x^2} - 3 = 0,\]

where the root \(c \gt 0.\)

Take the derivative of the function:

\[f^\prime\left( x \right) = \left( {{x^2} - 3} \right)^\prime = 2x.\]

Let \({x_0} = 2.\) Calculate the next approximation \({x_1}:\)

\[{{x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f^\prime\left( {{x_0}} \right)}} = 2 - \frac{{{2^2} - 3}}{{2 \cdot 2}} = 2 - \frac{1}{4} = 1.75}\]

In the next step, we get

\[{x_2} = {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f^\prime\left( {{x_1}} \right)}} = 1.75 - \frac{{{{1.75}^2} - 3}}{{2 \cdot 1.75}} = 1.732143\]

Similarly, we find the approximate value \({x_3}:\)

\[{x_3} = {x_2} - \frac{{f\left( {{x_2}} \right)}}{{f^\prime\left( {{x_2}} \right)}} = 1.732143 - \frac{{{{1.732143}^2} - 3}}{{2 \cdot 1.732143}} = 1.73205081\]

In this iteration, the approximation accuracy is \(8\) decimal places like in a smartphone calculator.

The square root of 3.
Figure 2.

So, we were able to compute the square root of \(3\) with the accuracy to \(8\) decimal places just for \(3\) steps!

Solved Problems

Example 1.

Approximate \(\sqrt[3]{2}\) to 6 decimal places.

Solution.

We apply Newton's method to the function \(f\left( x \right) = {x^3} - 2\) assuming \(x \ge 0\) and perform several successive iterations using the formula

\[{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f^\prime\left( {{x_n}} \right)}}.\]

Let \({x_0} = 1.\) This yields the following results:

\[{x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f^\prime\left( {{x_0}} \right)}} = 1 - \frac{{{1^3} - 2}}{{3 \cdot {1^2}}} = 1.3333333\]
\[{x_2} = {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f^\prime\left( {{x_1}} \right)}} = 1.3333333 - \frac{{{{1.3333333}^3} - 2}}{{3 \cdot {{1.3333333}^2}}} = 1.2638889\]

Similarly, we get

\[{x_3} = 1.2599335\]
\[{x_4} = 1.2599211\]
\[{x_5} = 1.2599211\]

We see that the \(4\)th iteration gives the approximation to \(6\) decimal places, so the answer is \({x_4} = 1.259921\)

Example 2.

Determine how many iterations does it take to compute \(\sqrt 5\) to 8 decimal places using Newton's method with the initial value \({x_0} = 2?\)

Solution.

We apply Newton's method to the function

\[f\left( x \right) = {x^2} - 5.\]

The iterations are given by the formula

\[{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f^\prime\left( {{x_n}} \right)}}.\]

The first approximation is equal to

\[{x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f^\prime\left( {{x_0}} \right)}} = 2 - \frac{{{2^2} - 5}}{{2 \cdot 2}} = 2.25\]

Continue the process to get the following approximations:

\[{x_2} = 2.236111111\]
\[{x_3} = 2.236067978\]
\[{x_4} = 2.236067977\]

Hence, it takes \(3\) iterations to get the approximate value of \(\sqrt 5\) to \(8\) decimal places (not too bad!)

Example 3.

Approximate the solution of the equation \[{x^2} + x - 3 = 0\] to 7 decimal places with the initial guess \({x_0} = 2.\)

Solution.

We apply the recurrent formula given by Newton's method:

\[x_{n + 1} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f^\prime\left( {{x_n}} \right)}}.\]

In the first step we get

\[{x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f^\prime\left( {{x_0}} \right)}} = 2 - \frac{{{2^2} + 2 - 3}}{{2 \cdot 2 + 1}} = 1.4\]

The next approximations are given by

\[{x_2} = 1.30526316\]
\[{x_3} = 1.30277735\]
\[{x_4} = 1.30277564\]
\[{x_5} = 1.30277564\]

Thus, we were able to get the approximate solution with an accuracy of \(7\) decimal places after \(4\) iterations. It is equal to

\[{x_4} = 1.30277564\]

Example 4.

Approximate \(\ln 2\) to 5 decimal places.

Solution.

To find an approximate value of \(\ln 2,\) we use the recurrent formula

\[x_{n + 1} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f^\prime\left( {{x_n}} \right)}}.\]

Starting from \({x_0} = 1,\) we obtain the following successive approximate values for \(\ln 2:\)

\[{x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f^\prime\left( {{x_0}} \right)}} = 1 - \frac{{{e^1} - 2}}{{{e^1}}} = 0.735759\]
\[{x_2} = {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f^\prime\left( {{x_1}} \right)}} = 0.735759 - \frac{{{e^{0.735759}} - 2}}{{{e^{0.735759}}}} = 0.694042\]

The next calculations produce

\[{x_3} = 0.693148\]
\[{x_4} = 0.693147\]

One can see that we've got the approximation to \(5\) decimal places on the \(3\)rd step. So the answer is

\[\ln 2 \approx 0.69315\]

See more problems on Page 2.

Page 1 Page 2