How to show a sequence converges

[henry22]

Hey guys,

I have a function y=(x+2)/(x+1) and I have performed iterations to show that for any initial value other than -sqrt{2} the sequence converges to sqrt{2}.

So I have found that sqrt{2} is a stable fixed point and -sqrt{2} is an unstable fixed point.

Now I have to prove my findings using a standard test for convergence?

Can anyone help.

 

[defunc]

hi there.

a sufficient condition is that
the derivative in the fixed point has to be smaller than one in absolute value

 

[Mark44]

Hey guys,

I have a function y=(x+2)/(x+1) and I have performed iterations to show that for any initial value other than -sqrt{2} the sequence converges to sqrt{2}.

So I have found that sqrt{2} is a stable fixed point and -sqrt{2} is an unstable fixed point.

Now I have to prove my findings using a standard test for convergence?

Can anyone help.

How are you getting sqrt(2)? As x --> infinity, (x + 2)/(x + 1) --> 1.

 

[defunc]

How are you getting sqrt(2)? As x --> infinity, (x + 2)/(x + 1) --> 1.

its a fixed point iteration scheme. this specific case you can derive from Newtons method to find the zeros of x^2-2.

 

[Mark44]

But what do the zeroes of x^2 - 2 have to do with (x + 2)/(x - 1)?

 

[M.Alastair]

Hi all,

I need help to show that the following sequence converges:

n^n/ ((n + 3)^ (n + 1))

I can't seem to find any standard limits which can help prove that it is convergent.

now consider the limit as

 

[HallsofIvy]

Mark, the iteration is $x_{n+1}= y(x_n)$ or
[tex]x_{n+1}= \frac{x_n+ 2}{x_n+1}[/itex]<br /> <br /> Assuming that has a limit, L, taking the limit on both sides gives<br /> $$L= \frac{L+2}{L+ 1}$$<br /> and from that, $L(L+1)= L^2+ L= L+ 2$ so that $L^2= 2$.<br /> <br /> henry22, you should be able to show that<br /> 1) If $x_0> \sqrt{2}$ then $\{x_n\}$ is a decreasing sequence with lower bound.<br /> <br /> 2) If $x_0< \sqrt{2}$ then $\{x_n\}$ is an increasing sequence with upper bound.[/tex]

 

[Martin Rattigan]

Hey guys,

I have a function y=(x+2)/(x+1) and I have performed iterations to show that for any initial value other than -sqrt{2} the sequence converges to sqrt{2}.

Wouldn't an initial value of -1 also be a problem? Also the sequence of starting values -3/2, -7/5, -17/12 ... which reach the value -1 at some stage of iteration.

 

[Martin Rattigan]

... If $x_0< \sqrt{2}$ then $\{x_n\}$ is an increasing sequence with upper bound.

No, if $x_0=-\sqrt{2}$ then $x_n=-\sqrt{2}$. Also if $x_0=-5/4$ then $x_1=-3<x_0$.

Neither do positive terms behave as suggested. E.g. from the second example $x_2=\frac{1}{2}$, $x_3=\frac{5}{3}>x_2$, $x_4=\frac{11}{8}<x_3$ and $x_5=\frac{27}{19}>x_4$, so the sequence is neither monotonic increasing from $x_2<\sqrt{2}$ nor monotonic decreasing from $x_3>\sqrt{2}$.

 

[Mark44]

Mark, the iteration is $x_{n+1}= y(x_n)$ or
[tex]x_{n+1}= \frac{x_n+ 2}{x_n+1}[/itex]<br /> <br /> Assuming that has a limit, L, taking the limit on both sides gives<br /> $$L= \frac{L+2}{L+ 1}$$<br /> and from that, $L(L+1)= L^2+ L= L+ 2$ so that $L^2= 2$.<br /> <br /> henry22, you should be able to show that<br /> 1) If $x_0> \sqrt{2}$ then $\{x_n\}$ is a decreasing sequence with lower bound.<br /> <br /> 2) If $x_0< \sqrt{2}$ then $\{x_n\}$ is an increasing sequence with upper bound.[/tex]

$$Thanks!$$