Find the solution of the equation 2x^3+e^x=0

Find the solution of the equation 2x^3+e^x=0 using the 3 iterations of Newton's method.

Newton's method says that successively better approximations to
the solution of f(x) = 0 may be had using the following algorithm:

x(n+1) = x(n) - f [x(n)] / f '[x(n)]

It's always good to start as close as you can to where you think the solution is.

2x^3 + e^x = 0 OR

e^x = -2x^3

I look at that last equation as a means to make a reasonable estimate of where to start.

The left side is an exponential that is always positive and the right side is a decreasing cubic that crosses the x-axis at x = 0. The solution to the equation (if there be one) must be an x < 0.

Also notice in the original equation both the cubic and the exponential are increasing functions for all x and 2x^3 is zero at x = 0 so if you add another increasing function to it, the y value will reach zero BEFORE x = 0.

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

f '(x) = 6x^2 + e^x

and x - f(x) / f ' (x) = [4x^3 + e^x(x - 1)] / [6x^2 + e^x]

Storing g(x) = [4x^3 + e^x(x - 1)] / [6x^2 + e^x] into a calculator capable of that function and using an educated guess of x = -0.5:

First approximation is about -0.669250134

second approximation is about -0.64192804

third approximation is about -0.64100635

fourth approximation is about -0.64100533182

Fifth really hones-in at -0.641005331819

And the sixth is so close to the fifth that the digits the calculator gives are all the same to 12 decimals.

So first of all, it should be clear that this doesn't have a simple analytic solution. You can use Newton's method to get successively better approximations of x:


2x^3 = -e^x
x^3 = -1/2 e^x
x = cuberoot (-1/2 e^x)

Plug in x = 0 on the right, and you get:

x_1 = cuberoot (-1/2 e^0) = cuberoot (-1/2) ~= -.7937
x_2 = cuberoot (-1/2 e^(-.7937)) ~= -.6092
x_3 = cuberoot (-1/2 e^(-.6092)) ~= -.6478

You can keep going to get better and better approximations.

Note:

Some equations like this have a closed form in terms of something called the "Lambert W() Function." Lambert W is the function such that x = W(x) e^W(x). In this case:

2x^3 = e^x
cuberoot(2) x = e^(x/3)
cuberoot(2) x e^(-x/3) = 1
-x/3 e^(-x/3) = -1/(3cuberoot(2)) = -cuberoot(4) / 6
-x/3 = W(-cuberoot(4)/6)
x = -3 W(-cuberoot(4)/6)

You would not be expected to find this answer though, and this wasn't what the question asked.

1+1=2