newtons method to locate t

**dat1611** · August 9th 2010, 02:39 PM

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

**pickslides** · August 9th 2010, 02:52 PM

Hi there, this is quite hard to follow, maybe its the order on how is it written.

I get given $x_0 = 10$ with $f(x) = x^2-216 \implies f'(x) = 2x$

$x_1 = x_0-\frac{f(x_0)}{f'(x_0)} = 10-\frac{f(10)}{f'(10)} = 10-\frac{10^2-216}{2\times 10} =10-\frac{-116}{20} =4.2$

$x_2 = x_1-\frac{f(x_1)}{f'(x_1)} = 4.2-\frac{f(4.2)}{f'(4.2)} = \dots$

not sure where $a$ or $t$ comes from? Are you confusing this with the bi-section method?

**Plato** · August 9th 2010, 02:52 PM

Originally Posted by dat1611

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

As written, that question is meaningless.
The Newton’s method is not even correct.

**dat1611** · August 9th 2010, 02:54 PM

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

**pickslides** · August 9th 2010, 02:57 PM

Originally Posted by dat1611

thats excactly how its written

The second time you have supplied this you have divided the function and derivative as required. Additional information is required to answer the question as is so i'm picking option 7.

**Plato** · August 9th 2010, 03:21 PM

I agree with option 7.
Consider: $f(x)=x^3-3x+6,~\7~b=1$

**chisigma** · August 9th 2010, 11:52 PM

The Newton's method find the solutions of an equation of the type $f(x)=0$ starting fron an $x_{1}$ and generating the sequence defined by...

$\displaystyle x_{n+1} = x_{n} - \frac{f(x_{n})}{f^{'} (x_{n})}$ (1)

A simple question: how to select $x_{1}$ in order to have convergence?...

The simple answer : it depends from the specific $f(x)$ ...

As easy example lets consider the equation $x^{2} - 4 =0$ . In that case the (1) becomes...

$\displaystyle \Delta_{n} = x_{n+1} - x_{n} = \frac{4-x^{2}_{n}}{2\ x_{n}} = \varphi (x_{n})$ (2)

The function $\varphi(*)$ is illustrated in the figure...

The solution is $x_{0} = 2$ and the 'red line' is $l(x)$ , i.e. the line with slope -1 crossing the x axis in $x=x_{0}$ . Because for $x>0$ is $\varphi (x) \ge l(x)$ any initial value $x_{1} > 0$ will produce convergence and that is easy to be verified. Not for all $f(x)$ however the matter is so easy

...

Kind regards

$\chi$ $\sigma$

**HallsofIvy** · August 10th 2010, 03:03 AM

Originally Posted by dat1611

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

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.

3.u>a
4.u<a

These don't make sense. There was no "a" in your problem.

5.u>t
6.u<t

These don't make sense. "t" is a variable and can have any value.

7.impossible to determine

thats excactly how its written

I guess 7 is the only one left.

**chisigma** · August 10th 2010, 06:53 AM

Originally Posted by Plato

I agree with option 7.
Consider: $f(x)=x^3-3x+6,~\7~b=1$

It is a very interesting case

... the equation is...

$\displaystyle x^{3} - 3\ x + 6 = 0$ (1)

...and its only real solution is $x_{0} = -2.35530139761...$ . The corresponding $\varphi(*)$ [see my previous post...] is...

$\displaystyle \varphi(x)= - \frac{1}{3}\ \frac{x^{3} - 3\ x + 6}{x^{2}-1}$ (2)

... and it is represented here...

The function has two singularities in $x=1$ and $x=-1$ and it is evident [see my previous post...] that any initial value $x_{1} < -1$ will produce a sequence converging to $x_{0}$ . For $x_{1}>-1$ and $x_{1} \ne 1$ the analysis is quite complex and the convergence is not guaranted...

Kind regards

$\chi$ $\sigma$

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