composite function

vafakhlgh · **Joined:** Mon, 23 Aug 2010 15:52:35 UTC **Posts:** 42

man111 wrote:

$if \[f(x)=1+\frac{2}{x}\] and \\ $\ if f_{1}(x) = f(x) $\ and\\ $\ f_{2}(x) = f(f(x)) $\ and\\ $\ f_{n}(x) = f(f_{n-1}(x))\\ $\ Then find the no. of solution of the equation $\ f_{n}(x) = x.$

assuming x is a real number, consider number of solution of:
f_1=x

and a few others

From this can you find a pattern for the number of solutions of f_n

.

The pattern is: you get to solve x^2-x-2=0

fot each

and so there are two sloutions x=2

and

Sam Snyder · **Posted:** Fri, 8 Oct 2010 21:45:42 UTC

man111 wrote:

$\mbox{ if } \displaystyle{ \[f(x)=1+\frac{2}{x}\] } and \\ $\ \mbox{if } f_{1}(x) = f(x) $\ and\\ $\ f_{2}(x) = f(f(x)) $\ and\\ $\ f_{n}(x) = f(f_{n-1}(x))\\ $\ Then find the no. of solutions of the equation $\ f_{n}(x) = x.$

$\\ \mbox{Rewrite the function, } \displaystyle{f_1(x)=f(x)=\frac{x+2}{x}} \ . \\$

$\mbox{Then, } \displaystyle{f_2(x)=f(f_1(x))=\frac{f_1(x)+2}{f_1(x)}=\frac{\frac{x+2}{x}+2}{\frac{x+2}{x}}}=\frac{x+2+2x}{x+2}=\frac{3x+2}{x+2} \ . \\$

$\mbox{Similarly, } \displaystyle{f_3(x)=f(f_2(x)) =\frac{f_2(x)+2}{f_2(x)}=\frac{\frac{3x+2}{x+2}+2}{\frac{3x+2}{x+2}} =\frac{5x+6}{3x+2} } \ .$

$\mbox{ } \displaystyle{f_4(x)=f(f_3(x)) =\frac{11x+10}{5x+6} } \ .$

$\mbox{ } \displaystyle{f_5(x)=f(f_4(x)) =\frac{21x+22}{11x+10} \ . }$

$\mbox{ etc. }$

In general, f_n(x)

has the form $f_n(x) =\displaystyle{\frac{ax+b}{cx+d}} \ ,\ c\ne0$

Therefore, the equation f_n(x)=x

has two solutions in general. In fact, for this problem the two solutions are $x=2, \ -1$ for $n\ge1$ .

man111 · **Posted:** Wed, 20 Oct 2010 08:26:25 UTC

thanks........

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composite function

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