Using Newton's Method to approximate the root of a function
Understanding Newton’s Method
Newton’s method lets us approximate the solution of a function, which is the point where the function crosses the ???x???-axis. Keep the following in mind when you use Newton’s method:
The function must be in the form ???f(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.
Hi! I'm krista.
I create online courses to help you rock your math class. Read more.
The Newton’s method formula is
???x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}???
where ???x_n??? is our starting approximation and ???x_{n+1}??? is our next approximation. Once ???x_n??? and ???x_{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
Take the course
Want to learn more about Calculus 1? I have a step-by-step course for that. :)
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 ???x_0=-1???.
???x^2=x???
First we verify that our equation is in the form ???f(x)=0???.
???x^2-x=0???
Next we take the derivative of our function.
???2x-1=0???
Since we were given ???x_0???, we can set ???x_n=x_0??? and ???x_{n+1}=x_1???, and we get
???x_1=x_0-\frac{(x_0)^2-x_0}{2x_0-1}???
We were given ???x_0=-1???, so we’ll plug that in and get
???x_1=-1-\frac{(-1)^2-(-1)}{2(-1)-1}???
???x_1=-1-\frac{1+1}{-2-1}???
???x_1=-1-\frac{2}{-3}???
???x_1=-1+\frac{2}{3}???
???x_1=-\frac13\approx-0.3333???
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, ???x_1=-1/3??? in our next approximation, where we’ll solve for ???x_2???.
???x_2=x_1-\frac{(x_1)^2-x_1}{2x_1-1}???
???x_2=-\frac13-\frac{\left(-\frac13\right)^2-\left(-\frac13\right)}{2\left(-\frac13\right)-1}???
???x_2=-\frac13-\frac{\frac19+\frac13}{-\frac23-1}???
???x_2=-\frac13-\frac{\frac19+\frac39}{-\frac23-\frac33}???
???x_2=-\frac13-\frac{\frac49}{-\frac53}???
???x_2=-\frac13+\frac{\frac49}{\frac53}???
???x_2=-\frac13+\frac49\left(\frac35\right)???
???x_2=-\frac13+\frac43\left(\frac15\right)???
???x_2=-\frac13+\frac{4}{15}???
???x_2=-\frac{5}{15}+\frac{4}{15}???
???x_2=-\frac{1}{15}\approx -0.0667???
We’ll use ???x_2=-1/15??? to solve for ???x_3???.
???x_3=x_2-\frac{(x_2)^2-x_2}{2x_2-1}???
???x_3=-\frac{1}{15}-\frac{\left(-\frac{1}{15}\right)^2-\left(-\frac{1}{15}\right)}{2\left(-\frac{1}{15}\right)-1}???
???x_3=-\frac{1}{15}-\frac{\frac{1}{225}+\frac{1}{15}}{-\frac{2}{15}-1}???
???x_3=-\frac{1}{15}-\frac{\frac{1}{225}+\frac{15}{225}}{-\frac{2}{15}-\frac{15}{15}}???
???x_3=-\frac{1}{15}-\frac{\frac{16}{225}}{-\frac{17}{15}}???
???x_3=-\frac{1}{15}+\frac{\frac{16}{225}}{\frac{17}{15}}???
???x_3=-\frac{1}{15}+\frac{16}{225}\left(\frac{15}{17}\right)???
???x_3=-\frac{1}{15}+\frac{16}{15}\left(\frac{1}{17}\right)???
???x_3=-\frac{1}{15}+\frac{16}{255}???
???x_3=-\frac{17}{255}+\frac{16}{255}???
???x_3=-\frac{1}{255}\approx -0.0039???
We’ll use ???x_3=-1/255??? to solve for ???x_4???.
???x_4=x_3-\frac{(x_3)^2-x_3}{2x_3-1}???
???x_4=-\frac{1}{255}-\frac{\left(-\frac{1}{255}\right)^2-\left(-\frac{1}{255}\right)}{2\left(-\frac{1}{255}\right)-1}???
???x_4=-\frac{1}{255}-\frac{\frac{1}{255^2}+\frac{1}{255}}{-\frac{2}{255}-1}???
???x_4=-\frac{1}{255}-\frac{\frac{1}{255^2}+\frac{255}{255^2}}{-\frac{2}{255}-\frac{255}{255}}???
???x_4=-\frac{1}{255}-\frac{\frac{256}{255^2}}{-\frac{257}{255}}???
???x_4=-\frac{1}{255}+\frac{\frac{256}{255^2}}{\frac{257}{255}}???
???x_4=-\frac{1}{255}+\frac{256}{255^2}\left(\frac{255}{257}\right)???
???x_4=-\frac{1}{255}+\frac{256}{255}\left(\frac{1}{257}\right)???
???x_4=-\frac{1}{255}+\frac{256}{255\cdot257}???
???x_4=-\frac{257}{255\cdot257}+\frac{256}{255\cdot257}???
???x_4=-\frac{1}{255\cdot257}???
???x_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 ???0???.
