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From: charlescalculus_robertobaggio on 28 Feb 2010 17:58
Hi, I have a maths problem which asks to show that if x_0 > 3, then the sequence x_(n+1) = ((x_n)^2 + 6)/5 is strictly increasing. I do know the difference between a sequence that is said to be increasing or strictly increasing. However, the only way I can see why this is the case is by using a cobweb plot.

Plot y = x and y = (x^2 + 6)/5 on a graph, then choose your initial value and with a software like mathematica I see that there are no stable fixed points (as in Dynamical Systems) for x_0 > 3.

Anyone know of an analytical proof?
From: W. Dale Hall on 1 Mar 2010 04:49
charlescalculus_robertobaggio(a)hotmail.com wrote:
> Hi, I have a maths problem which asks to show that if x_0> 3, then
> the sequence x_(n+1) = ((x_n)^2 + 6)/5 is strictly increasing. I do
> know the difference between a sequence that is said to be increasing
> or strictly increasing. However, the only way I can see why this is
> the case is by using a cobweb plot.
>
> Plot y = x and y = (x^2 + 6)/5 on a graph, then choose your initial
> value and with a software like mathematica I see that there are no
> stable fixed points (as in Dynamical Systems) for x_0> 3.
>
> Anyone know of an analytical proof?

How about just showing that, given your assumption on x_0,
the difference x_(n+1) - x_n is greater than zero? It seems
odd you don't seem to have considered this approach.

Let x = x_n, and take the expression for x_(n+1) - x_n:

(x^2 + 6)/5 - x

and clear the denominator to get

x^2 + 6 - 5x

which you rearrange to this form:

x^2 - 5x + 6

Then, show that if x > 3, the above quadratic is greater than zero.

No offense intended, but for a person who chose a nickname
"charlescalculus" you don't seem to use much algebra.

From: charlescalculus_robertobaggio on 28 Feb 2010 19:11
W. Dale Hall, thanks for the reply. However, i am only able to show if x^2 - 5x + 6 > 0, then x > 3 (Completing the square). The question asks for the implication to be the other way around i.e. if x>3, then x^2 - 5x + 6 > 0.

Do I have a situation where both ways the implication is true?
From: Henry on 1 Mar 2010 07:35
On 1 Mar, 10:11, "charlescalculus_robertobag...(a)hotmail.com"
<charlescalculus_robertobag...(a)hotmail.com> wrote:
> W. Dale Hall, thanks for the reply. However, i am only able to show if x^2 - 5x + 6 > 0, then x > 3 (Completing the square). The question asks for the implication to be the other way around i.e. if x>3, then x^2 - 5x + 6 > 0.
>
> Do I have a situation where both ways the implication is true?

The implication is the reverse of the way you have it,
since if x^2 - 5x + 6 > 0
then it is possible that x < 2.
E.g. for x=1, x^2 - 5x + 6 = 2 > 0

To prove "if x>3, then x^2 - 5x + 6 > 0"
you should start by factoring x^2 - 5x + 6

From: charlescalculus_robertobaggio on 28 Feb 2010 22:25
Thanks Henry, I see what you mean. Factoring x^2 - 5x + 6, I have (x-3)(x-2). If x > 3, then I have (positive number)(positive number) > 0. Thanks.
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