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

Answer 1

#x=0.2698741376# to 10dp.

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.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

See below.

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.))#

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

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.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

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.

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.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7