How do you use Newton's method to find the approximate solution to the equation #x^5+x^3+x=1#?

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Answer 1 · 2016-12-17T00:21:39

#x=0.636883# to 6dp

Let #f(x) = x^5+x^3+x-1# Then our aim is to solve #f(x)=0#

First let us look at the graphs:
graph{x^5+x^3+x-1 [-10, 10, -5, 4.995]}

We can see there is one solution in the interval # 0 < x < 1 #.

We can find the solution numerically, using Newton-Rhapson method

# f(x) = x^5+x^3+x-1 => f'(x) = 5x^4+3x^2+1 #, and using the Newton-Rhapson method we use the following iterative sequence

# { (x_0=1), ( x_(n+1)=x_n - f(x_n)/(f'(x_n)) ) :} #

Then using excel working to 6dp we can tabulate the iterations as follows:

enter image source here

And we conclude that the remaining solution is #x=0.636883# to 6dp