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 \(c:\)
- Start with an initial approximation \({x_0}\) close to \(c.\)
- Determine the next approximation by the formula \[{x_1} = {x_0} – \frac{{f\left( {{x_0}} \right)}}{{f^\prime\left( {{x_0}} \right)}}.\]
- 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 \(\sqrt[3]{2}\) 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 \(\sqrt[3]{2}\) 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
\[\ln 2 \approx 0.69315\]
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 \({x_4} = 1.2207\)