# Math Help - Sequence Convergence

1. ## Sequence Convergence

Hello,

Suppose the sequence, a(n+1) = 1/2 (an + 7/an) with a1 = 2

Show that a^2n > 7

So i've said a^2(n+1) > 7

So then,

a^(n+1) - 7 = 1/4((an^2 + 7)^2/4a^2)

And then I hit a wall, any thoughts?

Thanks

2. Originally Posted by MattWT
Hello,

Suppose the sequence, a(n+1) = 1/2 (an + 7/an) with a1 = 2

Show that a^2n > 7

So i've said a^2(n+1) > 7

So then,

a^(n+1) - 7 = 1/4((an^2 + 7)^2/4a^2)

And then I hit a wall, any thoughts?

Thanks
This makes no sense. Can you write this in LaTeX or use better typsetting? Are you prove that $\displaystyle a_{n+1}=\frac{1}{2}\left(a_n+\frac{7}{a_n}\right)\ implies a_n^2>7$?

3. Yes, sorry for that, not to sure on how to use the Math tags, it dislikes my brackets.

4. Originally Posted by MattWT
Hello,

Suppose the sequence, a(n+1) = 1/2 (an + 7/an) with a1 = 2

Show that a^2n > 7

So i've said a^2(n+1) > 7

So then,

a^(n+1) - 7 = 1/4((an^2 + 7)^2/4a^2)

And then I hit a wall, any thoughts?

Thanks

If by "a^2n" you meant $a^n_2$ then it is true only for $n\geq 2$ , and inductively:

$a^2_{n+1}=\frac{1}{4}\left(a^2_n+14+\frac{49}{a_n^ 2}\right)>\frac{1}{4}\left(63+\frac{49}{a_n^2}\rig ht)$ .

Now this last expression is greater than 7, since $\displaystyle{\frac{1}{4}\left(63+\frac{49}{a_n^2} \right)>7\Longleftrightarrow \frac{49}{a_n^2}>28-63}$ , and since

this last inequality is trivial we're done.

Tonio

5. Originally Posted by MattWT
Hello,

Suppose the sequence, a(n+1) = 1/2 (an + 7/an) with a1 = 2

Show that a^2n > 7

So i've said a^2(n+1) > 7

So then,

a^(n+1) - 7 = 1/4((an^2 + 7)^2/4a^2)

And then I hit a wall, any thoughts?

Thanks
The difference equation defining the sequence is...

$\displaystyle \Delta_{n}= a_{n+1}-a_{n} = -\frac{a_{n}}{2} + \frac{7}{2 a_{n}} = \varphi(a_{n})$ (1)

The function $\varphi(x)$ for $x>0$ [for $x<0$ is $\varphi(-x)= - \varphi(x)$...] is illustrated here...

There is an 'attractive fixed point' at $x_{0} = \sqrt{7} = 2.645751311...$. Any $a_{1}>0$ will produce a sequence converging at $x_{0}$ but, because is [see 'red line'...] $|\varphi(x)|>|x_{0} - x|$ for $x< x_{0}$ and $|\varphi(x)|<|x_{0} - x|$ for $x> x_{0}$ , any $a_{1}< x_{0}$ will produce a sequence $a_{n}> x_{0}$ for $n>1$ ...

Kind regards

$\chi$ $\sigma$