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

We have:

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

Our aim is to solve

graph{ln(x)+e^x [-5, 5, -10, 10]}

We can see that there is one solution in the interval

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

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See below.

This gives rise to an iterative approximation procedure

In this case we have

or

#((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.))#

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To solve the equation ( \ln(x) + e^x = 0 ) using Newton's Method, you would need to start with an initial guess ( x_0 ), then use the iterative formula:

[ x_{n+1} = x_n - \frac{\ln(x_n) + e^{x_n}}{\frac{1}{x_n} + e^{x_n}} ]

Continue iterating until the solution converges to the desired level of accuracy.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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