Asked by Anonymous
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
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
Answers
Answered by
Reiny
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 ...
x<sub>n+1</sub> = x<sub>n</sub> - f(x<sub>n</sub>)/f ' (x<sub>n</sub>)
so x<sub>2</sub> = 1 - (e^1 - 2)/e^1
= .73576
x<sub>3</sub> = .73576 - (e^.73576 - 2)/e^.73576 = .69404
x<sub>4</sub> = .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.
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 ...
x<sub>n+1</sub> = x<sub>n</sub> - f(x<sub>n</sub>)/f ' (x<sub>n</sub>)
so x<sub>2</sub> = 1 - (e^1 - 2)/e^1
= .73576
x<sub>3</sub> = .73576 - (e^.73576 - 2)/e^.73576 = .69404
x<sub>4</sub> = .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.
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