Question

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

Answer 1

The solution is `x=0.3820` to 4dp.

Explanation:

We want to solve:

`x+sqrt(x) = 1 => x+sqrt(x) -1 =0`

Let `f(x) = x+sqrt(x) -1` Then our aim is to solve `f(x)=0`.

First let us look at the graphs:
graph{x+sqrt(x) -1 [-2, 3, -2, 2]}

We can see there is one solution in the interval `0 le x le 1`, so will use `x=1` as our initial approximation

To find the solution numerically, using Newton-Rhapson method we will need the derivative `f'(x)`.

`\ \ \ \ \ \ \f(x) = x+sqrt(x) -1`
`:. f'(x) = 1+1/2x^(-1/2)`
`\ \ \ \ \ \ \ \ \ \ \ \ \ \= 1+1/(2sqrt(x))`

The Newton-Rhapson method uses the following iterative sequence

`{ (x_1,=1), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :}`

Then using excel working to 8dp we can tabulate the iterations as follows:

We could equally use a modern scientific graphing calculator as most new calculators have an " Ans " button that allows the last calculated result to be used as the input of an iterated expression.

And we conclude that the solution is `x=0.3820` to 4dp.