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
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&amp;gt; \sqrt{2} then \{x_n\} is a decreasing sequence with lower bound.<br /> <br /> 2) If x_0&amp;lt; \sqrt{2} then \{x_n\} is an increasing sequence with upper bound.

 

[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&lt; \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&lt;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}&gt;x_2, x_4=\frac{11}{8}&lt;x_3 and x_5=\frac{27}{19}&gt;x_4, so the sequence is neither monotonic increasing from x_2&lt;\sqrt{2} nor monotonic decreasing from x_3&gt;\sqrt{2}.

 

[Mark44]

Mark, the iteration is x_{n+1}= y(x_n) or
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&amp;gt; \sqrt{2} then \{x_n\} is a decreasing sequence with lower bound.<br /> <br /> 2) If x_0&amp;lt; \sqrt{2} then \{x_n\} is an increasing sequence with upper bound.

<br /> Thanks!