Question

Use Newton's Method to solve the equation? `lnx+e^x=0`

Answer 1

See below.

Explanation:

According to Taylor expansion for `f(x)` near `x_0`

`f(x)=f(x_0)+((df)/(dx))_(x_0)(x-x_0)+O(abs(x-x_0)^2)`

Now supposing that `x_0 approx 0` then we can assume

`f(x_1)=f(x_0)+((df)/(dx))_(x_0)(x_1-x_0) =0`

This gives rise to an iterative approximation procedure

`f(x_k) = ((df)/(dx))_(x_k)(x_(k+1)-x_k) =0` or

`x_(k+1) = x_k - ((df)/(dx))_(x_k)^(-1)f(x_k)`

In this case we have

`((df)/(dx))_(x_k) = 1/x_k+e^(x_k)` so

`x_(k+1)=x_k-x_k/(1+x_ke^(x_k))(log(x_k)+e^(x_k))`

or

`x_(k+1)=(1-(log(x_k)+e^(x_k))/(1+x_ke^(x_k)))x_k`

With `x_0 = 0.5` the iteration history is

`((x_k, f(x_k)),(0.5, 0.955574),(0.238107, -0.166189),(0.268497, -0.00692001),(0.269872, -0.0000118326),(0.269874, -3.4569*10^-11),(0.269874, 0.))`

Answer 2

`x=0.2698741376` to 10dp.

Explanation:

We have:

`f(x) = lnx+e^x`

Our aim is to solve `f(x)=0`. First let us look at the graph:
graph{ln(x)+e^x [-5, 5, -10, 10]}

We can see that there is one solution in the interval `0 lt x lt 1`. Let us start with an initial approximation `x=1`.

To find the solution numerically, using Newton-Rhapson method we use the following iterative sequence

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

Therefore we need the derivative:

`\ \ \ \ \ \ \f(x) = lnx+e^x`
`:. f'(x) = 1/x+e^x`

Then using excel working to 10dp 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 we have very rapid convergence, and the solution is `x=0.2698741376` to 10dp.