Use Newton's method to find all roots of the equation correct to six decimal places?

Question

#1/x=1+x^3# there should be 2 roots

Ezra Adams · Accepted Answer

See below.

Defining #f(x) = 1/x-(1+x^3)# if #x_0# is such that

#f(x_0) approx 0# then

#f(x_1) = f(x_0) + f'(x_0)(x_1-x_0)+O_2(abs(x_1-x_0))#

where #O_2(abs(x_1-x_0))# is something that goes to #0# with the order of #abs(x_1-x_0)^2#. In those circumstances we can assume that #f(x_1) approx 0# so

#0=f(x_0)+f'(x_0)(x_1-x_0)# which is an iterative procedure used for root computation. So

#x_(k+1)=x_k-f(x_k)/(f'(x_k))#

In our case we have

#f(x_k) = 1/x_k-(1+x_k^3)# and

#f'(x_k) = -(1/x^2 + 3 x^2)#

With this procedure we will obtain all real roots.

Now beginning with #x_0 = -1# we have

#((k,x_k),(1,-1.0),(2,-1.25),(3,-1.22126),(4,-1.22074),(5,-1.22074))#

Trying now #x_0=1#

#((k,x_k),(1,1.0),(2,0.75),(3,0.724449),(4,0.724492),(5,0.724492))#

Attached a plot of #y=1/x# in blue and #y = 1+x^3# in red.

Isaiah Allen · Answer

undefined To use Newton's method to find all roots of an equation correct to six decimal places, follow these steps:

1. Start with an initial guess $ x_0 $ for the root.
2. Use the formula $ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $ to iterate and update the guess until the desired accuracy is achieved.
3. Repeat the iteration process until the difference between successive approximations is within the desired tolerance level.

Here's a basic algorithm:

1. Choose an initial guess $ x_0 $.
2. Calculate $ f(x_0) $ and $ f'(x_0) $.
3. Update the guess using the formula $ x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $.
4. Repeat steps 2 and 3 until the desired accuracy is achieved.

Repeat this process for each initial guess to find all roots of the equation.

Please provide the specific equation for which you want to find the roots using Newton's method.