Show that x_n = x_{n-1}/2 + 1/x_{n-1} converges for x_0 > √2

[╔(σ_σ)╝]

Homework Statement

Show that

$$x_n = \frac{x_{n-1}}{2} + \frac{1}{x_{n-1}}$$ converges.

$$x_0 > \sqrt2$$

Homework Equations

The Attempt at a Solution

I am semi-stumped.

I guess I can use the AM-GM on $x_n$ to show that it is greater than root 2.

Then show that $x_n - x_{n-1} < 0$


I think that solutions uses too much stuff.


I am looking hints at other solutions.

 

[Dick]

Your recursive sequence is closely related to the function f(x)=x/2+1/x, in that x_n=f(x_(n-1)). It might be interesting to know where f is increasing and decreasing. Hint?

 

[╔(σ_σ)╝]

Your recursive sequence is closely related to the function f(x)=x/2+1/x, in that x_n=f(x_(n-1)). It might be interesting to know where f is increasing and decreasing. Hint?

Cool, I guess calculus is not that useless (joke). :P

So f(x) is increasing for $|x| > \sqrt2$.
f(x) is decreasing when |x| < sqrt(2)

That is no good.

 

[╔(σ_σ)╝]

Any further hints ?

 

[jbunniii]

Any further hints ?

If $x_n$ does converge, what must its limit be? You can easily determine this by taking limits of both sides of the defining recurrence relation.

 

[Dick]

Any further hints ?

I meant to say look at f(x)-x. That's what tells you whether x_n is greater than or less than f(x_n-1). Sorry.

 

[╔(σ_σ)╝]

If $x_n$ does converge, what must its limit be? You can easily determine this by taking limits of both sides of the defining recurrence relation.

I know what the limit should be. In fact, I already know the solution to the problem but I want a different solution.
I don't like this solution

$$x_n - x_{n-1} = \frac{1}{x_{n-1}} - \frac{x_{n-1}}{2}$$

$$\frac{1}{x_{n-1}} - \frac{x_{n-1}}{2} < 0$$

Implies

$${x_{n-1}}^2 > 2$$

Now I still need to use the fact that $x_n > \sqrt2$:/.

I meant to say look at f(x)-x. That's what tells you whether x_n is greater than or less than f(x_n-1). Sorry.

It's okay, it's late at night. :)

So

f(x) -x = -x/2 + 1/x

f(x) -x < 0 when |x| > sqrt(2)This helps me out.

Anyway, at some point I would need to show that x_n > sqrt(2) :$. Any suggestions that do not use AM-GM.

 

[Dick]

I know what the limit should be. In fact, I already know the solution to the problem but I want a different solution.
I don't like this solution

$$x_n - x_{n-1} = \frac{1}{x_{n-1}} - \frac{x_{n-1}}{2}$$

$$\frac{1}{x_{n-1}} - \frac{x_{n-1}}{2} < 0$$

Implies

$${x_{n-1}}^2 > 2$$

Now I still need to use the fact that $x_n > \sqrt2$:/.




It's okay, it's late at night. :)

So

f(x) -x = -x/2 + 1/x

f(x) -x < 0 when |x| > sqrt(2)


This helps me out.

Anyway, at some point I would need to show that x_n > sqrt(2) :$. Any suggestions that do not use AM-GM.

Ok then. You ARE interested in where f(x) is increasing and decreasing. f(x)-sqrt(2) is 0 for x=sqrt(2) and increasing for x>sqrt(2). So if x0>sqrt(2), then f(x0)-sqrt(2)>0.

 

[╔(σ_σ)╝]

Ok then. You ARE interested in where f(x) is increasing and decreasing. f(x)-sqrt(2) is 0 for x=sqrt(2) and increasing for x>sqrt(2). So if x0>sqrt(2), then f(x0)-sqrt(2)>0.

Okay thanks a lot for the help.