Thread: 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?

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?

Your work is sound.

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?

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...

$\displaystyle \Delta_{n}= a_{n+1}-a_{n}= a^{2}_{n} - a_{n} + \frac{1}{4}= f(a_{n}),\ a_{0}=0$ (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 $\displaystyle a_{n}$ converges to an $\displaystyle x_{0}$ only if is $\displaystyle f(x_{0})=0$ and in that case we call $\displaystyle x_{0}$ 'fixed point'. If in $\displaystyle x=x_{0}$ f(x) has positive slope, then $\displaystyle x_{0}$ is a 'repulsive fixed point' and only the initial value $\displaystyle a_{0}=x_{0}$ [and a very restricted set of points] produces a sequence converging to it. If in $\displaystyle x=x_{0}$ f(x) has negative slope, then $\displaystyle x_{0}$ is an 'attractive fixed point' and in most cases there is convergence if $\displaystyle a_{0}$ belongs to a set of points around $\displaystyle x_{0}$. In Your case $\displaystyle f(x)=x^{2}-x+\frac{1}{4}$ has only one zero [even if with multelicity two...] in $\displaystyle x_{0}= \frac{1}{2}$. In this case f(x) has in $\displaystyle x_{0}$ 'zero slope' so that we can define $\displaystyle x_{0}= \frac{1}{2}$ 'half- attracting fixed point' and that means that only initial values $\displaystyle a_{0} \le x_{0}$ can produce a sequence converging at $\displaystyle x_{0}$. According to the Theorem 4.1 of the post I indicated the values of $\displaystyle a_{0}$ that satisfy the condition $\displaystyle |f(x)| \le |x_{0}-x|$ will produce a sequence converging at $\displaystyle x_{0}$ and it is easy to verify that any value $\displaystyle -\frac{1}{2} \le a_{0} \le \frac{1}{2}$ will produce a sequence converging at $\displaystyle \frac{1}{2}$ and any other value of $\displaystyle a_{0}$ will produce a diverging sequence...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$