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 $\displaystyle a^n_2$ then it is true only for $\displaystyle n\geq 2$ , and inductively:
$\displaystyle a^2_{n+1}=\frac{1}{4}\left(a^2_n+14+\frac{49}{a_n^ 2}\right)>\frac{1}{4}\left(63+\frac{49}{a_n^2}\rig ht)$ .
Now this last expression is greater than 7, since $\displaystyle \displaystyle{\frac{1}{4}\left(63+\frac{49}{a_n^2} \right)>7\Longleftrightarrow \frac{49}{a_n^2}>28-63}$ , and since
this last inequality is trivial we're done.
Tonio
The difference equation defining the sequence is...
$\displaystyle \displaystyle \Delta_{n}= a_{n+1}-a_{n} = -\frac{a_{n}}{2} + \frac{7}{2 a_{n}} = \varphi(a_{n})$ (1)
The function $\displaystyle \varphi(x)$ for $\displaystyle x>0$ [for $\displaystyle x<0$ is $\displaystyle \varphi(-x)= - \varphi(x)$...] is illustrated here...
There is an 'attractive fixed point' at $\displaystyle x_{0} = \sqrt{7} = 2.645751311...$. Any $\displaystyle a_{1}>0$ will produce a sequence converging at $\displaystyle x_{0}$ but, because is [see 'red line'...] $\displaystyle |\varphi(x)|>|x_{0} - x|$ for $\displaystyle x< x_{0}$ and $\displaystyle |\varphi(x)|<|x_{0} - x|$ for $\displaystyle x> x_{0}$ , any $\displaystyle a_{1}< x_{0}$ will produce a sequence $\displaystyle a_{n}> x_{0}$ for $\displaystyle n>1$ ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$