Newton’s Method
Solved Problems
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.
Figure 3. 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 \({x_4} = 1.2207\)
