Using Newton's Method to approximate the root of a function

 
 
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Understanding Newton’s Method

Newton’s method lets us approximate the solution of a function, which is the point where the function crosses the xx-axis. Keep the following in mind when you use Newton’s method:

  • The function must be in the form f(x)=0f(x)=0.

  • The more approximations we take, the closer we’ll get to the actual solution.

  • For each approximation, we have to use our answer from the previous approximation.

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The Newton’s method formula is

xn+1=xn−f(xn)f′(xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}

where xnx_n is our starting approximation and xn+1x_{n+1} is our next approximation. Once xnx_n and xn+1x_{n+1} are exactly equal or equal to a specified number of decimal places, we can stop taking approximations.

 
 

Using Newton’s Method to approximate the root in a particular interval

 
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Applying Newton’s Method with four steps to find the value of a root in the interval

Example

Use Newton’s method to find an approximation of the root of the function to four decimal places, when x0=−1x_0=-1.

x2=xx^2=x

First we verify that our equation is in the form f(x)=0f(x)=0.

x2−x=0x^2-x=0

Next we take the derivative of our function.

2x−1=02x-1=0

Since we were given x0x_0, we can set xn=x0x_n=x_0 and xn+1=x1x_{n+1}=x_1, and we get

x1=x0−(x0)2−x02x0−1x_1=x_0-\frac{(x_0)^2-x_0}{2x_0-1}

We were given x0=−1x_0=-1, so we’ll plug that in and get

x1=−1−(−1)2−(−1)2(−1)−1x_1=-1-\frac{(-1)^2-(-1)}{2(-1)-1}

x1=−1−1+1−2−1x_1=-1-\frac{1+1}{-2-1}

x1=−1−2−3x_1=-1-\frac{2}{-3}

x1=−1+23x_1=-1+\frac{2}{3}

x1=−13≈−0.3333x_1=-\frac13\approx-0.3333

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Newton’s method lets us approximate the solution of a function, which is the point where the function crosses the x-axis.

We’ll use the value we just found, x1=−1/3x_1=-1/3 in our next approximation, where we’ll solve for x2x_2.

x2=x1−(x1)2−x12x1−1x_2=x_1-\frac{(x_1)^2-x_1}{2x_1-1}

x2=−13−(−13)2−(−13)2(−13)−1x_2=-\frac13-\frac{\left(-\frac13\right)^2-\left(-\frac13\right)}{2\left(-\frac13\right)-1}

x2=−13−19+13−23−1x_2=-\frac13-\frac{\frac19+\frac13}{-\frac23-1}

x2=−13−19+39−23−33x_2=-\frac13-\frac{\frac19+\frac39}{-\frac23-\frac33}

x2=−13−49−53x_2=-\frac13-\frac{\frac49}{-\frac53}

x2=−13+4953x_2=-\frac13+\frac{\frac49}{\frac53}

x2=−13+49(35)x_2=-\frac13+\frac49\left(\frac35\right)

x2=−13+43(15)x_2=-\frac13+\frac43\left(\frac15\right)

x2=−13+415x_2=-\frac13+\frac{4}{15}

x2=−515+415x_2=-\frac{5}{15}+\frac{4}{15}

x2=−115≈−0.0667x_2=-\frac{1}{15}\approx -0.0667

We’ll use x2=−1/15x_2=-1/15 to solve for x3x_3.

x3=x2−(x2)2−x22x2−1x_3=x_2-\frac{(x_2)^2-x_2}{2x_2-1}

x3=−115−(−115)2−(−115)2(−115)−1x_3=-\frac{1}{15}-\frac{\left(-\frac{1}{15}\right)^2-\left(-\frac{1}{15}\right)}{2\left(-\frac{1}{15}\right)-1}

x3=−115−1225+115−215−1x_3=-\frac{1}{15}-\frac{\frac{1}{225}+\frac{1}{15}}{-\frac{2}{15}-1}

x3=−115−1225+15225−215−1515x_3=-\frac{1}{15}-\frac{\frac{1}{225}+\frac{15}{225}}{-\frac{2}{15}-\frac{15}{15}}

x3=−115−16225−1715x_3=-\frac{1}{15}-\frac{\frac{16}{225}}{-\frac{17}{15}}

x3=−115+162251715x_3=-\frac{1}{15}+\frac{\frac{16}{225}}{\frac{17}{15}}

x3=−115+16225(1517)x_3=-\frac{1}{15}+\frac{16}{225}\left(\frac{15}{17}\right)

x3=−115+1615(117)x_3=-\frac{1}{15}+\frac{16}{15}\left(\frac{1}{17}\right)

x3=−115+16255x_3=-\frac{1}{15}+\frac{16}{255}

x3=−17255+16255x_3=-\frac{17}{255}+\frac{16}{255}

x3=−1255≈−0.0039x_3=-\frac{1}{255}\approx -0.0039

We’ll use x3=−1/255x_3=-1/255 to solve for x4x_4.

x4=x3−(x3)2−x32x3−1x_4=x_3-\frac{(x_3)^2-x_3}{2x_3-1}

x4=−1255−(−1255)2−(−1255)2(−1255)−1x_4=-\frac{1}{255}-\frac{\left(-\frac{1}{255}\right)^2-\left(-\frac{1}{255}\right)}{2\left(-\frac{1}{255}\right)-1}

x4=−1255−12552+1255−2255−1x_4=-\frac{1}{255}-\frac{\frac{1}{255^2}+\frac{1}{255}}{-\frac{2}{255}-1}

x4=−1255−12552+2552552−2255−255255x_4=-\frac{1}{255}-\frac{\frac{1}{255^2}+\frac{255}{255^2}}{-\frac{2}{255}-\frac{255}{255}}

x4=−1255−2562552−257255x_4=-\frac{1}{255}-\frac{\frac{256}{255^2}}{-\frac{257}{255}}

x4=−1255+2562552257255x_4=-\frac{1}{255}+\frac{\frac{256}{255^2}}{\frac{257}{255}}

x4=−1255+2562552(255257)x_4=-\frac{1}{255}+\frac{256}{255^2}\left(\frac{255}{257}\right)

x4=−1255+256255(1257)x_4=-\frac{1}{255}+\frac{256}{255}\left(\frac{1}{257}\right)

x4=−1255+256255⋅257x_4=-\frac{1}{255}+\frac{256}{255\cdot257}

x4=−257255⋅257+256255⋅257x_4=-\frac{257}{255\cdot257}+\frac{256}{255\cdot257}

x4=−1255⋅257x_4=-\frac{1}{255\cdot257}

x4=−165,535≈−0.0000x_4=-\frac{1}{65,535}\approx-0.0000

The solution is only getting smaller, which means that the next approximation will also be all zeros to the first four decimal places. Therefore, the approximation of the root of the function to four decimal places is 00.

 
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