# 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 $\cssId{element2}{{x_{n + 1}} = {x_n} – \frac{{f\left( {{x_n}} \right)}}{{f^\prime\left( {{x_n}} \right)}}}$ 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 5

Let $$f\left( x \right) = {x^3} – 7.$$ Using Newton’s method, compute 3 iterations for this function with the initial guess $${x_0} = 2.$$

### Example 6

Approximate the solution of the equation $$x\ln x = 1$$ with an accuracy of 4 decimal places. Use the initial guess $${x_0} = 2.$$

### Example 7

Using Newton’s method, find the solution of the equation $$x + {e^x} = 0$$ with an accuracy of 3 decimal places.

### Example 8

Find an approximate solution, accurate to 5 decimal places, to the equation $$\cos x = {x^2}$$ that lies in the interval $$\left[ {0,\large{\frac{\pi }{2}}\normalsize} \right].$$

### Example 9

Find an approximate solution, accurate to 4 decimal places, to the equation $$\sin \left( {{e^x}} \right) = 1$$ on the interval $$\left[ {0,\pi } \right].$$

### Example 10

Given the equation $${x^4} – x – 1 = 0.$$ It is known that this equation has a root in the interval $$\left( {1,2} \right).$$ Find an approximate value of the root with an accuracy of 4 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

$\ln 2 \approx 0.69315$

### Example 5.

Let $$f\left( x \right) = {x^3} – 7.$$ Using Newton’s method, compute 3 iterations for this function with the initial guess $${x_0} = 2.$$

Solution.

The iterative formula for Newton’s method is given as

${x_{n + 1}} = {x_n} – \frac{{f\left( {{x_n}} \right)}}{{f^\prime\left( {{x_n}} \right)}}.$

Find out the derivative:

${f^\prime\left( x \right) = \left( {{x^3} – 7} \right)^\prime }={ 3{x^2}.}$

The first iteration is equal to

${{x_1} = {x_0} – \frac{{f\left( {{x_0}} \right)}}{{f^\prime\left( {{x_0}} \right)}}} ={ 2 – \frac{{{2^3} – 7}}{{3 \cdot {2^2}}} }={ 1.916667}$

Next, we perform two more iterations:

${{x_2} = {x_1} – \frac{{f\left( {{x_1}} \right)}}{{f^\prime\left( {{x_1}} \right)}}} ={ 1.916667 – \frac{{{{1.916667}^3} – 7}}{{3 \cdot {{1.916667}^2}}} }={ 1.912938}$

${{x_3} = {x_2} – \frac{{f\left( {{x_2}} \right)}}{{f^\prime\left( {{x_2}} \right)}}} ={ 1.912938 – \frac{{{{1.912938}^3} – 7}}{{3 \cdot {{1.912938}^2}}} }={ 1.912933}$

After $$3$$ iterations we’ve got the approximate solution with an accuracy of $$5$$ decimal places.

Answer: $${x_3} = 1.91293$$

### Example 6.

Approximate the solution of the equation $$x\ln x = 1$$ with an accuracy of 4 decimal places. Use the initial guess $${x_0} = 2.$$

Solution.

Consider the function

$f\left( x \right) = x\ln x – 1$

and apply Newton’s method to find zero of the function.

Find the derivative by the product rule:

${f^\prime\left( x \right) = \left( {x\ln x – 1} \right)^\prime }={ 1 \cdot \ln x + x \cdot \frac{1}{x} }={ \ln x + 1.}$

Calculate the first approximation:

${{x_1} = {x_0} – \frac{{f\left( {{x_0}} \right)}}{{f^\prime\left( {{x_0}} \right)}}} ={ 2 – \frac{{2\ln 2 – 1}}{{\ln 2 + 1}} }={ 1.77184}$

Continue the iterative process until we reach an accuracy of $$4$$ decimal places.

${{x_2} = {x_1} – \frac{{f\left( {{x_1}} \right)}}{{f^\prime\left( {{x_1}} \right)}}} ={ 1.77184 – \frac{{1.77184 \cdot \ln \left( {1.77184} \right) – 1}}{{\ln \left( {1.77184} \right) + 1}} }={ 1.76323}$

${{x_3} = {x_2} – \frac{{f\left( {{x_2}} \right)}}{{f^\prime\left( {{x_2}} \right)}}} ={ 1.76323 – \frac{{1.76323 \cdot \ln \left( {1.76323} \right) – 1}}{{\ln \left( {1.76323} \right) + 1}} }={ 1.76322}$

As you can see, we have obtained the required accuracy after only $$2$$ steps.

Answer: $${x_2} = 1.7632$$

### Example 7.

Using Newton’s method, find the solution of the equation $$x + {e^x} = 0$$ with an accuracy of 3 decimal places.

Solution.

We choose $${x_0} = – 1$$ and compute the first approximation:

${{x_1} = {x_0} – \frac{{f\left( {{x_0}} \right)}}{{f^\prime\left( {{x_0}} \right)}}} ={ – 1 – \frac{{ – 1 + {e^{ – 1}}}}{{1 + {e^{ – 1}}}} }={ – 0.5379}$

Continue the process until we get the result with the required accuracy.

${{x_2} = {x_1} – \frac{{f\left( {{x_1}} \right)}}{{f^\prime\left( {{x_1}} \right)}} }={ – 0.5379 – \frac{{ – 0.5379 + {e^{ – 0.5379}}}}{{1 + {e^{ – 0.5379}}}} }={ – 0.5670}$

${{x_3} = {x_2} – \frac{{f\left( {{x_2}} \right)}}{{f^\prime\left( {{x_2}} \right)}} }={ – 0.5670 – \frac{{ – 0.5670 + {e^{ – 0.5670}}}}{{1 + {e^{ – 0.5670}}}} }={ – 0.5671}$

We see that we’ve already got a stable result to $$3$$ decimal places, so the approximate solution is $$x \approx – 0.567$$

### Example 8.

Find an approximate solution, accurate to 5 decimal places, to the equation $$\cos x = {x^2}$$ that lies in the interval $$\left[ {0,\large{\frac{\pi }{2}}\normalsize} \right].$$

Solution.

First we rewrite this equation in the form

$\cos x – {x^2} = 0.$

Suppose that the initial value of the root is $${x_0} = 1.$$ Let’s get the first approximation using Newton’s method:

${{x_1} = {x_0} – \frac{{f\left( {{x_0}} \right)}}{{f^\prime\left( {{x_0}} \right)}}} ={ {x_0} – \frac{{\cos {x_0} – x_0^2}}{{ – \sin {x_0} – 2{x_0}}} }={ 1 – \frac{{\cos 1 – {1^2}}}{{ – \sin 1 – 2 \cdot 1}} }={ 0.838218}$

Here and further we write approximate values with 6 decimal places to track the convergence of the result.

In the next step, we have

${{x_2} = {x_1} – \frac{{\cos {x_1} – x_1^2}}{{ – \sin {x_1} – 2{x_1}}} }={ 0.838218 }-{ \frac{{\cos \left( {0.838218} \right) – {{0.838218}^2}}}{{ – \sin \left( {0.838218} \right) – 2 \cdot 0.838218}} }={ 0.824242}$

The third approximation gives the following value of the root:

${{x_3} = {x_2} – \frac{{\cos {x_2} – x_2^2}}{{ – \sin {x_2} – 2{x_2}}} }={ 0.824242 }-{ \frac{{\cos \left( {0.824242} \right) – {{0.824242}^2}}}{{ – \sin \left( {0.824242} \right) – 2 \cdot 0.824242}} }={ 0.824132}$

Continue computations:

${{x_4} = {x_3} – \frac{{\cos {x_3} – x_3^2}}{{ – \sin {x_3} – 2{x_3}}} }={ 0.824132 }-{ \frac{{\cos \left( {0.824132} \right) – {{0.824132}^2}}}{{ – \sin \left( {0.824132} \right) – 2 \cdot 0.824132}} }={ 0.824132}$

The $$4$$th iteration preserves the first $$6$$ decimal places. This means that the required accuracy of $$5$$ decimal places was reached at the $$3$$rd step, so the answer is

${x_3} = 0.82413$

### Example 9.

Find an approximate solution, accurate to 4 decimal places, to the equation $$\sin \left( {{e^x}} \right) = 1$$ on the interval $$\left[ {0,\pi } \right].$$

Solution.

We apply Newton’s method with the initial guess $${x_0} = 0.5$$

Consider the function

$f\left( x \right) = \sin \left( {{e^x}} \right) – 1.$

The derivative is written as

$f^\prime\left( x \right) = {e^x}\cos \left( {{e^x}} \right).$

Then the first approximation is given by

${{x_1} = {x_0} – \frac{{f\left( {{x_0}} \right)}}{{f^\prime\left( {{x_0}} \right)}}} ={ 0.5 – \frac{{\sin \left( {{e^{0.5}}} \right) – 1}}{{{e^{0.5}}\cos \left( {{e^{0.5}}} \right)}} }={ 0.4764}$

Continue the iteration process until we get an accuracy of $$3$$ decimal places.

${{x_2} = {x_1} – \frac{{f\left( {{x_1}} \right)}}{{f^\prime\left( {{x_1}} \right)}}} ={ 0.4764 – \frac{{\sin \left( {{e^{0.4764}}} \right) – 1}}{{{e^{0.4764}}\cos \left( {{e^{0.4764}}} \right)}} }={ 0.4641}$

${x_3} = 0.4579$

${x_4} = 0.4548$

${x_5} = 0.4532$

${x_6} = 0.4524$

${x_7} = 0.4520$

${x_8} = 0.4518$

${x_9} = 0.4517$

${x_{10}} = 0.4516$

${x_{11}} = 0.4516$

As you can see, the process converges slowly enough, so it took $$10$$ steps to obtain a stable result to $$4$$ decimal places.

The answer is $$x \approx 0.4516$$

### Example 10.

Given the equation $${x^4} – x – 1 = 0.$$ It is known that this equation has a root in the interval $$\left( {1,2} \right).$$ Find an approximate value of the root with an accuracy of 4 decimal places.

Solution.

Take the derivative:

${f^\prime\left( x \right) = \left( {{x^4} – x – 1} \right)^\prime }={ 4{x^3} – 1.}$

Notice that $$f\left( 1 \right) = – 1,$$ $$f\left( 2 \right) = 13,$$ so we choose $${x_0} = 1$$ as the initial guess.

Using the recurrent relation

${x_{n + 1}} = {x_n} – \frac{{f\left( {{x_n}} \right)}}{{f^\prime\left( {{x_n}} \right)}},$

we compute several successive approximations:

${{x_1} = {x_0} – \frac{{f\left( {{x_0}} \right)}}{{f^\prime\left( {{x_0}} \right)}} }={ 1 – \frac{{{1^4} – 1 – 1}}{{4 \cdot {1^3} – 1}} }={ 1.3333}$

${{x_2} = {x_1} – \frac{{f\left( {{x_1}} \right)}}{{f^\prime\left( {{x_1}} \right)}} }={ 1.3333 – \frac{{{{1.3333}^4} – 1.3333 – 1}}{{4 \cdot {{1.3333}^3} – 1}} }={ 1.2358}$

${{x_3} = {x_2} – \frac{{f\left( {{x_2}} \right)}}{{f^\prime\left( {{x_2}} \right)}} }={ 1.2358 – \frac{{{{1.2358}^4} – 1.2358 – 1}}{{4 \cdot {{1.2358}^3} – 1}} }={ 1.2211}$

${{x_4} = {x_3} – \frac{{f\left( {{x_3}} \right)}}{{f^\prime\left( {{x_3}} \right)}} }={ 1.2211 – \frac{{{{1.2211}^4} – 1.2211 – 1}}{{4 \cdot {{1.2211}^3} – 1}} }={ 1.2207}$

${{x_5} = {x_4} – \frac{{f\left( {{x_4}} \right)}}{{f^\prime\left( {{x_4}} \right)}} }={ 1.2207 – \frac{{{{1.2207}^4} – 1.2207 – 1}}{{4 \cdot {{1.2207}^3} – 1}} }={ 1.2207}$

You can see that we’ve got the solution accurate up to $$4$$ decimal places in the $$4$$th step. Therefore, we can write the answer as $${x_4} = 1.2207$$