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f' = 2x + 1 f(0) = -3 x1 = x0 - f(0) / f'(0) shoot - i'm not sure thats right - i'll have to check
I am also not sure, if you find anything about it pls post it
newtons method is finding the equation of the tangent line; getting its root for a new x value to repeat the iteration right?
Xn = f'(x)(x-x0)+y0 = 0
I think so
(1,2) Xn = (2(1)+1)(x-1)+2
Xn = 3x - 1; Xn = 1/3 then right?
what do you label as Xn?
new X value; Xn just seemed appropriate
I coulda went with Larry, but.... i know a cucumber named Larry lol
sorry still not clear, I am just working it out, but I dont see how you get y0=2
x^2 +x = 3 is that a typo?
x^2 +x +3 maybe?
that is it
x^2 +x = 3 ; when x=1 this is simply a false statement then 1+1 = 3???
yep but the method will tend to the right answer
x^2 +x -3 = 0 then
that is the same
then remove the 0 to find some solutions
when x = 1; y = ? (1,-1) then right?
use this point for your tangent equation that follows: y-y0 = m(x-x0) y = m(x-x0)+y0
m = f'(x)
y = (2(1)+1)(x-1)-1
y = 3x-4; the root is x = 4/3 then right?
that is why i did not see how that 2 appeared
:) right concept, wrong thought lol
right and I should do this again to get the result, thanks for the help!
youre welcome :)
andras:what/where are you studying?
pure maths, just finished 1st year
are you on facebook? pardon me if posting this here is a breach of ettiquet
:) I dont mind. Yes I am on facebook
you can add me if you wish but I dont use facebook that often. Andras Honyek
will do...I just sent you a request...neat...most people i know are glued to their fb accounts.