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 Post subject: composite function
PostPosted: Sun, 26 Sep 2010 09:09:33 UTC 
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Joined: Wed, 22 Sep 2010 07:26:43 UTC
Posts: 135
Location: (Pantnagar) Haldwani
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.


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 Post subject: Re: composite function
PostPosted: Sun, 26 Sep 2010 11:46:47 UTC 
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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
f_2=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 f_n=x and so there are two sloutions x=2 and x=-1

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 Post subject: Re: composite function
PostPosted: Fri, 8 Oct 2010 21:45:42 UTC 
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Joined: Fri, 10 Oct 2008 15:35:23 UTC
Posts: 179
Location: Clarksville, ARkansas
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.


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 Post subject:
PostPosted: Wed, 20 Oct 2010 08:26:25 UTC 
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Joined: Wed, 22 Sep 2010 07:26:43 UTC
Posts: 135
Location: (Pantnagar) Haldwani
thanks........


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