If anyone here is a math expert on the Newton-Raphson method please post some information on it.?

Question

I need information specifically with regards to when the Newton-Raphson method (in calculus) will converge (mathematical explanations will be helpful) and when it will not and why. Thanks.

Don't post random crap please. There is another place for that.

Answer 1

Consider y = f(x).

The equation of the tangent at (x0,y0) is given by

y - y0 =f'(x0) (x - x0).

This line will intersect the x axis at y=0. This leads to

-y0 =f'(x0) (x - x0) and solving for x leads to

x = x0 - y0/f'(x0).

Geometrically, the intercept of the tangent at x0 will be a better approximation to a root of f(x)=0 than x0. The process is used iteratively.

Clearly it will fail when

1. f'(x) = 0
2. It may sometimes not converge dpending on the initial choice of x0 and the function f eg

f(x) = sqrt(x-1) for x>=1
= -sqrt(x-1) for x<1

Try this with x0 = 0.5. It cycle behind two values and never converge.

Answer 2

Only in the Bronx. I tried it several times around the Upper-W Side, seems like those cats have been hip to it for a minute, Yao Ming?

Anyway, my advice would be stick it to 'em greasy, but not WD-40. Squeaking is OK.. when it's 3:30 AM and everybody is passed out drunk.

Yodel at ya boyscout.

Answer 3

you can google for this.

this is the formula :

Fn+1 = D*Fn

F is the result. D is a Derivative it is usually a MAtrix.

It will converge when | D | <1 .

because Fn = D^nF0 ; iif |D|>1 the Fn is unbounded,

Answer 4

If u have -one million/(x^2) then u have no soln and if u have one million/(x^2) then u have a soln 0.7245 aprx. to tutor it take function f=x^2+one million/x-(one million/x^2) at x=0.5 u have f=-one million.seventy 5 and at x=one million f=one million so for sure graph of f cuts x axis bw 0.5 and 0.7 (part1) Now use formulation x1=x0-f(x)/f '(x) and save on iterating till u get the ans :) (part2)