Let $(x_n)$ and $(y_n)$ be two sequences dened recursively as follows: $$\left\{ \begin{array}{l} {x_1} = a > 0\\ {y_1} = b > 0\\ {x_{n + 1}} = \dfrac{{{x_n} + {y_n}}}{2}\\ {y_{n + 1}} = \sqrt {{x_n}{y_n}} \end{array} \right.$$ Find $\lim x_n$ and $\lim y_n$.
It's not so hard to improve that $\lim x_n=\lim y_n$ but I still can't find the limit of $(x_n)$ and $(y_n)$.
Do you have any idea? Please help me! Many thanks!
Let $a \le b$ and $k=b/a$ then limit we are looking for will be $L(a,b)=a\ f(k)$ where $f(k)=L(1,k)$ (not hard to prove, try it yourself)
So lets examine function f(x). $f(1)=1, f'(1)=1/2$ since f(x) is between $\sqrt{x}$ and $\frac{x+1}2$
Also (S): $f(x)=\sqrt xf(\frac{x+1}{2\sqrt x})$ since $L(x_1,y_1)=L(x_2,y_2)$
Lets say $x=(1+d)^2$ and write f in Taylor's series $f(1+x)=f(1)+f'(1)x/1!+f''(1)x^2/2!+...$
$$(S):f(1+2d+d^2)=(1+d)f(1+\frac{d^2}{2(1+d)})$$ $$f(1)+f'(1)(2d+d^2)/1!+f''(1)(2d+d^2)^2/2!+...=(1+d)(f(1)+f'(1)(\frac{d^2}{2(1+d)})/1!+f''(1)(\frac{d^2}{2(1+d)})^2/2!+...)$$ When you clean this mess little bit by substitution for f and f' $$d^2/4+\sum_{k=2}^∞ f^{(k)}(1)(2d+d^2)^k/k!=\sum_{k=2}^∞ f^{(k)}(1)\frac{d^{2k}}{2^k(1+d)^{k-1}}/k!$$ Ok, what now, we need such $f^{(k)}$ to work for any d, but you can't kill both $d^2$ and $d^3$ (as I tried) so when $d->0$ equation will not work...
So this is not an analytical function.