Hi there, this is quite hard to follow, maybe its the order on how is it written.
I get given with
not sure where or comes from? Are you confusing this with the bi-section method?
use newtons method to locate t starting with b as an initial guess.
let u=b - f(b)f'(b) denote the first approximation supplied by newtons method
which of the followin are true
1.u>b
2.u<b
3.u>a
4.u<a
5.u>t
6.u<t
7.impossible to determine
so i had
x^2=612
f(x)=x^2-612
f'(x)=2x
b=10
t=sqrt(612)
10-(10^2-612)/(2*10)=35.6
35.6-(35.6^2-612/(2*35.6)=26.395
1.u>b true
2.u<b false
im not ure what a is supposed to be
suppose you use newtons method to locate t starting with b as an initial guess.
let u=b - f(b)/f'(b) denote the first approximation supplied by newtons method
ii) which of the followin are true?
1.u>b
2.u<b
3.u>a
4.u<a
5.u>t
6.u<t
7.impossible to determine
thats excactly how its written
The Newton's method find the solutions of an equation of the type starting fron an and generating the sequence defined by...
(1)
A simple question: how to select in order to have convergence?...
The simple answer : it depends from the specific ...
As easy example lets consider the equation . In that case the (1) becomes...
(2)
The function is illustrated in the figure...
The solution is and the 'red line' is , i.e. the line with slope -1 crossing the x axis in . Because for is any initial value will produce convergence and that is easy to be verified. Not for all however the matter is so easy ...
Kind regards
If f(b) and f'(b) have the same sign, the u< b. If f(b) and f'(b) have different sign, then u> b. Since it is possible to give examples of both, neither of these can be true.
These don't make sense. There was no "a" in your problem.3.u>a
4.u<a
These don't make sense. "t" is a variable and can have any value.5.u>t
6.u<t
I guess 7 is the only one left.7.impossible to determine
thats excactly how its written
It is a very interesting case ... the equation is...
(1)
...and its only real solution is . The corresponding [see my previous post...] is...
(2)
... and it is represented here...
The function has two singularities in and and it is evident [see my previous post...] that any initial value will produce a sequence converging to . For and the analysis is quite complex and the convergence is not guaranted...
Kind regards