You are using Newton's method to solve
e^x - 2. If you first guess is x1=1, what value will you calculate for the next approximation x2?
...I don't know what Newton's method is, so I looked it up...and I'm still confused. Help plz?
Thx
first of all, to "solve" we need an equation.
You have the expression e^x - 2
Are you solving e^x - 2 = 0 ?
I will assume you are.
let f(x) = e^x - 2
f ' (x) = e^x
Newton's method says ...
xn+1 = xn - f(xn)/f ' (xn)
so x2 = 1 - (e^1 - 2)/e^1
= .73576
x3 = .73576 - (e^.73576 - 2)/e^.73576 = .69404
x4 = .69404 - (e^.69404 - 2)/e^.69404 = .69314758
notice we are rapidly approaching some kind of fixed value.
If the value of x of our input returns as output, we have the solution.
A most remarkable procedure.
Using logs I get x = .69314718 , so our method was accurate to 6 digits after only 3 iterations.
Sure! I'd be happy to explain Newton's method and help you solve this problem.
Newton's method is an iterative numerical approximation method used to find roots of equations. It involves repeatedly refining an initial guess to get closer and closer to the actual root.
Here's how Newton's method works:
1. Start with an initial guess, x1.
2. Calculate the value of the function at x1 (in this case, e^x1 - 2) and its derivative.
3. Use the formula x2 = x1 - (f(x1) / f'(x1)) to calculate the next approximation, x2.
4. Repeat steps 2 and 3 until the desired level of accuracy is achieved.
Now, let's apply Newton's method to solve the equation e^x - 2.
Given that x1 = 1, let's calculate the next approximation, x2:
1. Start with x1 = 1.
2. Calculate f(x1) and f'(x1):
f(x) = e^x - 2
f'(x) = e^x
f(x1) = e^1 - 2 = e - 2
f'(x1) = e^1 = e
3. Use the formula x2 = x1 - (f(x1) / f'(x1)):
x2 = 1 - ((e - 2) / e)
Now, you can calculate the value of x2 using the above formula, and that will be your next approximation.