# Calculus

## Applications of the Derivative # Newton’s Method

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

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.

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

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Approximate $$\sqrt{2}$$ to 6 decimal places.

### 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 $${x_0} = 2?$$

### Example 3

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

### Example 4

Approximate $$\ln 2$$ to 5 decimal places.

### Example 1.

Approximate $$\sqrt{2}$$ 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.