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Newton’s Method

April 10th, 2009 |

Author: Merrin

Newton’s method can be used to find roots of equations. It is based on the following observation from the following graph of a function that intersects the x axis at the point x*. If you extrapolate the linearization of the function at a point x_n then it can approximate finding the root via x_(n+1). By the next extrapolation from the location of the new root x_(n+1), an improved estimation results which may eventually converge to the actual root x*.

newton

Newton’s method uses linearization to repeatedly approximate the root of an equation

$f'(x_n)= \cfrac{f(x_{n})-0}{x_{n}-x_{n+1}}$

$x_{n+1} =x_n - \cfrac{f(x_n)}{f'(x_n)}$

Newton’s method doesn’t always work, sometimes the iterations can oscillate or grow towards infinity. Studying how different initial conditions in the complex plane diverge or converge can lead to interesting fractal pictures. A coloring scheme is usually used depending on how quickly the initial condition converges or diverges. Newton’s method is an important tool for numerical calculation of solutions of equations or even nth roots and so forth.

Example 1. Use Newton’s method to find the square root of two.

Solution 1. We need to find the root of

$x^2= 2$

so we use the function

$f(x) = x^2 - 2$

Now take the derivative

$f'(x) = 2x$

and apply Newton’s method

$x_{n+1} = x_n - \cfrac{x_{n}^2-2}{2x_n} = \cfrac{x_n}{2} + \cfrac{1}{x_n}$

Now iterate from an initial guess of 1.3

$\begin{array}{l} x_1 = 1.3 \\ x_2 = 1.419230769 \\ x_3 = 1.414222431 \\x_4 = 1.414213562 \\ x_5 = 1.414213562\end{array}$

Example 2. Find the root of

$\cfrac{xe^x}{e^x-1}=5$

Solution2. Form the function

$f(x) = \cfrac{xe^x}{e^x-1}-5$

then takes its derivative

$f'(x) = \cfrac{u'v - v'u}{v^2} = \cfrac{(e^x+xe^x)(e^x-1) -xe^x(e^x-1)}{(e^x-1)^2}$

$f'(x)=\cfrac{e^x}{e^x-1}$

Now apply Newtons method

$x_{n+1} = x_{n} + 5(1-e^{-x_n})-x_n$

With an initial guess of x=3.5 the value converges to x = 4.965114232. There is only one root of f(x).

Example 3. Find some roots of the equation

$x = \tan(2x)$

Solution 3. Form the function

$f(x) = x-\tan(2x)$

Then take its derivative

$f'(x) = 1-2\sec^2(2x)$

Now apply Newton’s method

$x_{n+1} = x_n - \cfrac{x_n-\tan(2x_n)}{1-2\sec^2(2x_n)}$

The initial guess has to be pretty close to where x intersects tan(2x) for Newton’s method to converge. There are multiple roots. We can choose one initial guess to be x=2.0 the value converges to x=2.137391136 . If we choose x = 3.7 the value converges to x = 3.798273010, and if we choose x= 5.35 the value converges to x= 5.40633666. This is not surprising since the root is near 13pi/8, a location of a vertical asymptote.

Example 4. Find the root of

$x=\cosh (x/3)$

Solution 4. First for the function

$f(x) = x-\cosh(x/3)$

Then take its derivative

$f'(x) = 1 - \cfrac{1}{3}\sinh(x/3)$

Now apply Newton’s method

$x_{n+1} = x_n - \cfrac{x_n-\cosh(x_n/3)}{1-\cfrac{1}{3}\sinh(x_n/3)}$

Now try an initial guess of x = 1 and we find that the root converges to x = 1.063495371. With an initial guess of x=7 we find that the root converges to x = 8.483243866. There are only two roots as can be seen by solving for the roots graphically.

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