# Calculus

## Applications of the Derivative # Newton’s Method

## Solved Problems

Click or tap a problem to see the solution.

### 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 $$\left[ {0,\frac{\pi }{2}} \right].$$

### 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 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 $$\left[ {0,\frac{\pi }{2}} \right].$$

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 $${x_4} = 1.2207$$