f1(x)=x/2+10
fn(x)=f1(fn-1(x)). n=2
Then evaluate. Lim fn(x). n tends to infinity.

Question

f1(x)=x/2+10
 fn(x)=f1(fn-1(x)).         n>=2
     Then evaluate.   Lim fn(x). n tends to infinity.

freckles · Answer

is this what you said?
$$f_1(x)=\frac{x}{2}+10 \ f_n(x)=f_1(f_{n-1}(x)) , n \ge 2  $$

I would find the first few terms and see if I can find a pattern for an explicit form for f_n

loser66 · Answer

$f_\color{red}{2}(x) = f_1*f_{2-1}= f_1^\color{red}{2}$
$f_\color{red}{3}(x) = f_1*f_{3-1}= f_1^\color{red}{3}$
$f_\color{red}{4}(x) = f_1*f_{4-1}= f_1^\color{red}{4}$
--------------------------------
$f_\color{red}{n}(x) = f_1*f_{2-1}= f_1^\color{red}{n }=(\dfrac{x+20}{2})^n$

loser66 · Answer

the middle term of the last line is wrong, it should be $f_1f_{n-1} $

anonymous · Answer

@Loser66 I disagree:
$$f_1(x)=\frac{x+20}{2}~~\implies~~f_2(x)=\frac{\dfrac{x+20}{2}+20}{2}=\frac{x+60}{4}\neq\left(\frac{x+20}{2}\right)^2$$

anonymous · Answer

I believe you're mistaking composition for multiplication:
$$f(f(x))\neq f(x)\times f(x)$$

loser66 · Answer

oh yeah!! I misread the problem. :) 
Thanks for pointing it out.

anonymous · Answer

I'm wondering if there's a way to do this with generating functions... Here's what I have so far.

Denote $a_1=f_1(x)$ and $a_n=f_n(x)$, so we have the recurrence relation
$$\begin{cases}a_1=\dfrac{x}{2}+10\\
a_n=\dfrac{a_{n-1}}{2}+10&\text{for }n\ge2\end{cases}$$
Then denote the generating function by $F(y)=\displaystyle\sum_{n=1}^\infty a_ny^n$.

We have
$$\begin{align*}
a_n&=\frac{a_{n-1}}{2}+10\\
\sum_{n=2}^\infty a_ny^n&=\frac{1}{2}\sum_{n=2}^\infty a_{n-1}y^n+10\sum_{n=2}^\infty y^n\\
F(y)-a_1y&=\frac{y}{2}\sum_{n=2}^\infty a_{n-1}y^{n-1}+\frac{10y^2}{1-y}\\
\left(1-\frac{y}{2}\right)F(y)&=\frac{10y^2}{1-y}+a_1y\\
F(y)&=20-2a_1+\frac{20}{1-y}+\frac{20-4a_1}{2-y}\\
&=20-2a_1+20\sum_{n=0}^\infty y^n+(10-2a_1)\sum_{n=0}^\infty\left(\frac{y}{2}\right)^n
\end{align*}$$
but I'm just not seeing where to go from here...

irishboy123 · Answer

the first 5 terms are:
$f_1 = \frac{x + 20}{2}$
$f_2 = \frac{x + 60}{4}$
$f_1 = \frac{x + 140}{8}$
$f_1 = \frac{x + 300}{16}$
$f_1 = \frac{x + 620}{32}$

this demands: $f_n(x) = \frac{x + (2^n - 1)20}{2^n}$ which you can stuff back into the recursion to get $f_{n+1}(x)$.

and so for the limit we use $f_n = \frac{\frac{x}{2^n} + (1 - \frac{1}{2^n})20}{1}$.

20.