Question

Find the solution of the equation `2x3+ex=0` using three iterations of Newton's method. Also solve the equation using the graphics calculator and compare your results.?

Answer 1

Three iterations gives:

`x_1= -0.743700`
`x_2= -0.652154`
`x_3= -0.641152`

We get convergence to `6 \ dp` with `5` iterations.

And we conclude that the solution is (to 6dp):

`x = -0.641005`

Explanation:

Assuming we want to solve:

`2x^3+e^x = 0`

Let:

`f(x) = 2x^3+e^x`

It is important to understand that we cannot use Newton's Method to determine the existence of a root, or establish the number of roots of `f(x)=0` in any particular interval. A graphical approach is recommended initially to establish this:

graph{2x^3+e^x [-5, 5, -10, 10]}

From which we can be confident that there is a single root, `alpha` where `-1 lt alpha lt 0`. Our aim is to solve `f(x)=0`. To find the solution numerically, using Newton-Rhapson method we use the following iterative sequence

# { (x_1,=x_0,), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)),n gt 1 ) :} #

Therefore we need the derivative:

`\ \ \ \ \ \ f(x) = 2x^3+e^x`
`:. f'(x) = 6x^2+e^x`

So our iterative formula is:

# { (x_1,=x_0,), ( x_(n+1), = x_n - (2x^3+e^x)/(6x^2+e^x), x gt 1 ) :} #

Then using excel working to 6dp with `x_0=-1`, we can tabulate the iterations as follows:

We get convergence to `6 \ dp` with `6` iterations.

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 (to 5dp):

`x = -0.641005`

The use of a graphical calculator cannot solve the equation, only estimate it