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

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Answer 1 · 2016-12-17T00:21:39

$x = 0.636883$ $x=0.636883$ to 6dp

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

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 $0 < x < 1$ $0 < x < 1$ .

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

How do you use Newton's method to find the approximate solution to the equation x^5+x^3+x=1x5+x3+x=1x^5+x^3+x=1?