limit

**Dinkydoe** · June 29th 2011, 06:10 AM

Hoi guys,

suppose we have $p_{k+1}\geq p_k^9+9p_k^8(1-p_k)$

I'd like to show that $p_k\to 1$ as $k\to\infty$ if $p_0>0.968...$

Now, this seems quite obvious maybe...as $p_{k+1}>p_k$ if $p_0>0.968$
But how can we see $p_k\to 1$ ?

**CaptainBlack** · June 29th 2011, 06:43 AM

Originally Posted by Dinkydoe

Hoi guys,

suppose we have $p_{k+1}\geq p_k^9+9p_k^8(1-p_k)$

I'd like to show that $p_k\to 1$ as $k\to\infty$ if $p_0>0.968...$

Now, this seems quite obvious maybe...as $p_{k+1}>p_k$ if $p_0>0.968$
But how can we see $p_k\to 1$ ?

Suppose:

$p_{k+1}=100p_k^9+1000$

It clearly satisfies the inequality but $p_k \to \infty$ as $k \to \infty$

So what am I missing?

(I may be completely misunderstanding the question as I am having to read raw LaTeX as this machine is not rendering any MHF LaTeX correctly)

CB

**Dinkydoe** · June 29th 2011, 07:04 AM

sorry, I need to add $p_0 \in (0.968, 1]$

I wonder how we can see $p_k\to 1$ ...as $p_k$ clearly is growing with such a starting value. But, why can it not converge
to some specific value a in (0.968,1)

edit: furthermore, is it even obvious that $p_k$ can not exceed the value 1??

**chisigma** · June 29th 2011, 08:01 PM

Originally Posted by Dinkydoe

Hoi guys,

suppose we have $p_{k+1}\geq p_k^9+9p_k^8(1-p_k)$

I'd like to show that $p_k\to 1$ as $k\to\infty$ if $p_0>0.968...$

Now, this seems quite obvious maybe...as $p_{k+1}>p_k$ if $p_0>0.968$
But how can we see $p_k\to 1$ ?

Let's suppose to have the sigh $=$ instead of sigh $\ge$ so that the 'recursive relation' is...

$p_{k+1}= 9\ p_{k}^{8} - 8\ p_{k}^{9}$ (1)

The (1) can be written as...

$\Delta_{k}= p_{k+1}-p_{k}= -p_{k} + 9\ p_{k}^{8} - 8\ p_{k}^{9}= f(p_{k})$ (2)

The $f(x)$ is represented here...

There are three zeroes in $x_{0}=0$ , $x_{1}= .967689763902...$ and $x_{2}=1$ . $x_{0}$ and $x_{2}$ are both 'attractive fixed points' and $x_{1}$ is a 'repulsive fixed point'. Any $p_{0}< x_{1}$ will produce a sequence converging at $x_{0}$ and any $x_{1}< p_{0} \le x_{2}$ will produce a sequence converging at $x_{2}$ . The case $p_{0} > x_{2}$ is left to You...

Kind regards

$\chi$ $\sigma$

**Dinkydoe** · July 1st 2011, 09:04 AM

Thanks, I have another question. Dunno If I should make a new thread for this...this was actually part of an article I was reading. (and didn't quite understand all the details).

Suppose we have $p_{k+1}=f(p_k)=p_k^9+9p_k^8(1-p_k)$ then we have $p_k\to 1$ as $k\to\infty$ if $p_0\in (0.968,1]$

Then the following observation is made: $\sum_k(1-p_k)<\infty$ . Now this very unclear to me...Can
we see from the recurrence relation $p_k=f(p_k)$ that $p_k$ goes to 1 fast enough?

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