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

1. freckles

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

2. Loser66

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

3. Loser66

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

4. anonymous

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

5. anonymous

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

6. Loser66

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

7. anonymous

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...

8. IrishBoy123

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.