Question

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

Answer 1

`x=0.636883` to 6dp

Explanation:

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:

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