Mathematics
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OpenStudy (anonymous):
Use Newton's method to find an approximate solution for the equation
x^2+x=3
Start with x0=1 and iterate Newton's method twice to find x2.
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OpenStudy (anonymous):
f' = 2x + 1
f(0) = -3
x1 = x0 - f(0) / f'(0)
shoot - i'm not sure thats right - i'll have to check
OpenStudy (anonymous):
I am also not sure, if you find anything about it pls post it
OpenStudy (amistre64):
newtons method is finding the equation of the tangent line; getting its root for a new x value to repeat the iteration right?
OpenStudy (amistre64):
Xn = f'(x)(x-x0)+y0 = 0
OpenStudy (anonymous):
I think so
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OpenStudy (amistre64):
(1,2)
Xn = (2(1)+1)(x-1)+2
OpenStudy (amistre64):
Xn = 3x - 1; Xn = 1/3 then right?
OpenStudy (anonymous):
what do you label as Xn?
OpenStudy (amistre64):
new X value; Xn just seemed appropriate
OpenStudy (anonymous):
ok
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OpenStudy (amistre64):
I coulda went with Larry, but.... i know a cucumber named Larry lol
OpenStudy (anonymous):
:)
OpenStudy (anonymous):
sorry still not clear, I am just working it out, but I dont see how you get y0=2
OpenStudy (amistre64):
x^2 +x = 3 is that a typo?
OpenStudy (amistre64):
x^2 +x +3 maybe?
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OpenStudy (anonymous):
that is it
OpenStudy (anonymous):
first one
OpenStudy (amistre64):
x^2 +x = 3 ; when x=1 this is simply a false statement then
1+1 = 3???
OpenStudy (anonymous):
yep but the method will tend to the right answer
OpenStudy (amistre64):
x^2 +x -3 = 0 then
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OpenStudy (anonymous):
that is the same
OpenStudy (amistre64):
then remove the 0 to find some solutions
OpenStudy (amistre64):
when x = 1; y = ?
(1,-1) then right?
OpenStudy (anonymous):
-1 right
OpenStudy (amistre64):
use this point for your tangent equation that follows:
y-y0 = m(x-x0)
y = m(x-x0)+y0
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OpenStudy (amistre64):
m = f'(x)
OpenStudy (amistre64):
y = (2(1)+1)(x-1)-1
OpenStudy (anonymous):
clear
OpenStudy (amistre64):
y = 3x-4; the root is x = 4/3 then right?
OpenStudy (anonymous):
that is why i did not see how that 2 appeared
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OpenStudy (amistre64):
:) right concept, wrong thought lol
OpenStudy (anonymous):
right and I should do this again to get the result, thanks for the help!
OpenStudy (amistre64):
youre welcome :)
OpenStudy (anonymous):
A mathematica solution with comments. Refer to the pdf attachment.
OpenStudy (anonymous):
andras:what/where are you studying?
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OpenStudy (anonymous):
pure maths, just finished 1st year
OpenStudy (anonymous):
are you on facebook?
pardon me if posting this here is a breach of ettiquet
OpenStudy (anonymous):
:) I dont mind. Yes I am on facebook
OpenStudy (anonymous):
you can add me if you wish but I dont use facebook that often. Andras Honyek
OpenStudy (anonymous):
will do...I just sent you a request...neat...most people i know are glued to their fb accounts.