Use Newton's Method to approximate a solution of the equation e^(-x)=5x+9, starting with the initial guess indicated. x1=-4. x2=? x3=? The solution found by Newton's Method is x=?

Answer 1

Let f(x) =e^-x - 5x - 9

Then, f'(x) = -e^-x - 5

Thus, the Newton step is x_n+1 = x_n - f(x_n)/f'(x_n)

Substituting, as x₁ = -4, x₂ = -4 - f(-4)/f'(-4) = -4 - (e^-x - 5x - 9)/(-e^-x - 5) =

-4 - (e^--4 - 5*-4 - 9)/(-e^--4 - 5) = -4 - (e⁴ - 5*-4 - 9)/(-e⁴ - 5) = -4 + 1.101 =

-2.899

You will see the quadratic convergence of Newton's method is such that x₆ is 1.2 * 10-5 from the solution and x₇ is 3.3 * 10-12 from the solution.

Answer 2

f(x) = e^(-x) -5x-9 is the function in whose solutions we are interested...

the derivative is f'(x) = - e^(-x) - 5

iteration x accuracy f(x)

1 -2.899325734160150473396 -3 65.59815003314423907811

2 -1.877884109263400163284 -2 23.65852396522040914896

3 -1.277426404429911724331 -1 6.929073567936435487293

4 -1.163942942338895457811 -1 0.974527352423740196418

5 -1.161230300725386416651 -1.16 0.0222505400242261293398

6 -1.161228864020128561427 -1.16123 1.17721621299355417E-5

7 -1.161228864019726277426 -1.16122886402 3.2962570411E-12

8 -1.161228864019726277426 -1.161228864019726277426 3.E-22

-1.161228864019726277426