1. ## correct? (sequences)

exercise:

A sequence is recursively defined by:
a(0) = 5
a(n+1) = 3 - 2/a(n)

Show that the sequence converges and calculate the limit.

my soution:
Imageshack - img0001swm.jpg

would you say it is correct?

2. ## Re: correct? (sequences)

Originally Posted by nappysnake
exercise:
A sequence is recursively defined by:
a(0) = 5
a(n+1) = 3 - 2/a(n)
Show that the sequence converges and calculate the limit.
my soution:
Imageshack - img0001swm.jpg
would you say it is correct?

3. ## Re: correct? (sequences)

Great!
Now, second exercise:
A sequence is recursively defined by a(0) = 0 and a(n+1) = a(n)^2 + 1/4
Show that the sequence converges and calculate the limit.

I tried applying a similar solution to the above, but although this sequence is monotone increasing it has only a lower bound = 0.
Is there any theorem i can apply or any help from you guys on how to show that the sequence converges, hence prove that the limit can be calculated?

4. ## Re: correct? (sequences)

Originally Posted by nappysnake
Great!
Now, second exercise:
A sequence is recursively defined by a(0) = 0 and a(n+1) = a(n)^2 + 1/4
Show that the sequence converges and calculate the limit.

I tried applying a similar solution to the above, but although this sequence is monotone increasing it has only a lower bound = 0.
Is there any theorem i can apply or any help from you guys on how to show that the sequence converges, hence prove that the limit can be calculated?
The difference equation can be written as...

$\Delta_{n}= a_{n+1}-a_{n}= a^{2}_{n} - a_{n} + \frac{1}{4}= f(a_{n}),\ a_{0}=0$ (1)

In...

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

... it has been defined the condition for which e sequence produced by (1) converges or not and the procedure for find the limit if exist. In particular the sequence $a_{n}$ converges to an $x_{0}$ only if is $f(x_{0})=0$ and in that case we call $x_{0}$ 'fixed point'. If in $x=x_{0}$ f(x) has positive slope, then $x_{0}$ is a 'repulsive fixed point' and only the initial value $a_{0}=x_{0}$ [and a very restricted set of points] produces a sequence converging to it. If in $x=x_{0}$ f(x) has negative slope, then $x_{0}$ is an 'attractive fixed point' and in most cases there is convergence if $a_{0}$ belongs to a set of points around $x_{0}$. In Your case $f(x)=x^{2}-x+\frac{1}{4}$ has only one zero [even if with multelicity two...] in $x_{0}= \frac{1}{2}$. In this case f(x) has in $x_{0}$ 'zero slope' so that we can define $x_{0}= \frac{1}{2}$ 'half- attracting fixed point' and that means that only initial values $a_{0} \le x_{0}$ can produce a sequence converging at $x_{0}$. According to the Theorem 4.1 of the post I indicated the values of $a_{0}$ that satisfy the condition $|f(x)| \le |x_{0}-x|$ will produce a sequence converging at $x_{0}$ and it is easy to verify that any value $-\frac{1}{2} \le a_{0} \le \frac{1}{2}$ will produce a sequence converging at $\frac{1}{2}$ and any other value of $a_{0}$ will produce a diverging sequence...

Kind regards

$\chi$ $\sigma$