1. ## sequence related question...

hello all,

I have a tough past exam question that I can not seem to fully understand. If we use newtons method to approx $\sqrt(a)$ by looking for the 0 of the function $f$ defined by $f(x) = x^2 - a$, the sequence $\{x_n\}$ as $n=1$ to $\infty$ is generated, where $x_1$ = 1 and $x_{n+1}$ = $\frac{1}{2}(x_n+\frac{a}{x_n})$.

The question states that show that if:
$0, then $x_{m+1}>\sqrt(a)$.

• Hint: add $x_{m+1}-\sqrt(a)$, and complete the square.

Not sure how to approach this one, but using the hint would I do something like..
$\frac{1}{2}(x_m+\frac{a}{m_n})-\sqrt(a)$ = $\frac{x_m}{2}+\frac{a}{m_n} - \sqrt(a)$ ..

Thanks..

2. ## Re: sequence related question...

Originally Posted by Oiler
hello all,

I have a tough past exam question that I can not seem to fully understand. If we use newtons method to approx $\sqrt(a)$ by looking for the 0 of the function $f$ defined by $f(x) = x^2 - a$, the sequence $\{x_n\}$ as $n=1$ to $\infty$ is generated, where $x_1$ = 1 and $x_{n+1}$ = $\frac{1}{2}(x_n+\frac{a}{x_n})$.

The question states that show that if:

$0, then $x_{m+1}>\sqrt(a)$.

• Hint: add $x_{m+1}-\sqrt(a)$, and complete the square.
Not sure how to approach this one, but using the hint would I do something like..
$\frac{1}{2}(x_m+\frac{a}{m_n})-\sqrt(a)$ = $\frac{x_m}{2}+\frac{a}{m_n} - \sqrt(a)$ ..

Thanks..
The procedure to be adopted in this type of problem is illustrated in...

http://www.mathhelpforum.com/math-he...-i-188482.html

The difference equation can be written in the form...

$\Delta_{n}= x_{n+1}-x_{n}= -\frac{1}{2}\ (x_{n}-\frac{a}{x_{n}})= f(x_{n})$ (1)

If we limit our attention to x>0, is easy to see that f(x) has only one 'attractive fixed point' [a point $x_{0}$ where is $f(x_{0})=0$ and $f^{'}(x_{0})<0$...] in $x_{0}=\sqrt{a}$ and, considering that is...

$f(x)= -\frac{1}{2}\ (x-\frac{a}{x}) \implies f^{'}(x)= -\frac{1}{2}\ (1+\frac{a}{x^{2}})$ (2)

... we can conclude that...

a) for all the $x_{1} >0$ the sequence defined by (1) will converge to $\sqrt{a}$ no matter which is a...

b) if $x_{1}>\sqrt{a}$ then the condition $|f(x)|<|x-x_{0}|$ is verified and the convergence is 'monotonic'...

c)if $x_{1}<\sqrt{a}$ then the condition $|f(x)|<|x-x_{0}|$ isn't verified and the convergence is 'oscillating'...

Kind regards

$\chi$ $\sigma$