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!

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Solved Problems

Click or tap a problem to see the solution.

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 \({x_4} = 1.259921\)

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!)

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

\[{x_4} = 1.30277564\]

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

\[\ln 2 \approx 0.69315\]

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: \({x_3} = 1.91293\)

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: \({x_2} = 1.7632\)

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 \(x \approx – 0.567\)

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

\[{x_3} = 0.82413\]

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 \(x \approx 0.4516\)

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 \({x_4} = 1.2207\)