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August 20th, 2017, 12:29 PM  #11 
Senior Member Joined: May 2016 From: USA Posts: 1,189 Thanks: 489 
Continuing on $f^1(x) = \dfrac{x}{\sqrt{1 + cx^2}} \implies$ $f^2(x) = \dfrac{\dfrac{x}{\sqrt{1 + cx^2}}}{\sqrt{1 + c \left ( \dfrac{x}{\sqrt{1 + cx^2}} \right )^2}} = \dfrac{\dfrac{x}{\sqrt{1 + cx^2}}}{\sqrt{1 + \dfrac{cx^2}{1 + cx^2}}} =$ $\dfrac{\dfrac{x}{\sqrt{1 + cx^2}}}{\sqrt{ \dfrac{1 + cx^2 + cx^2}{1 + cx^2}}} = \dfrac{\dfrac{x}{\cancel {\sqrt{1 + cx^2}}}}{\dfrac{\sqrt{1 + 2cx^2}}{\cancel {\sqrt{1 + cx^2}}}} = \dfrac{x}{\sqrt{1 + 2cx^2}}.$ Now let's try $f^3(x) = \dfrac{f^2(x)}{\sqrt{1 + c * \{f^2(x)\}^2}} = \dfrac{\dfrac{x}{\sqrt{1 + 2cx^2}}}{\sqrt{1 + c * \left \{\dfrac{x}{\sqrt{1 + 2cx^2}} \right \}^2}} =$ $\dfrac{\dfrac{x}{\sqrt{1 + 2cx^2}}}{\sqrt{1 + \dfrac{cx^2}{1 + 2cx^2}}} = \dfrac{\dfrac{x}{\sqrt{1 + 2cx^2}}}{\sqrt{ \dfrac{1 + 2cx^2 + cx^2}{1 + 2cx^2}}} =$ $\dfrac{\dfrac{x}{\cancel{\sqrt{1 + 2cx^2}}}}{\dfrac{\sqrt{1 + 3cx^2}}{\cancel{\sqrt{1 + 2cx^2}}}} = \dfrac{x}{\sqrt{1 + 3cx^2}}.$ I see a pattern namely $f^n(x) = \dfrac{x}{\sqrt{1 + cnx^2}}.$ Now all you have to do is to prove it. 
August 20th, 2017, 01:11 PM  #12  
Member Joined: Sep 2014 From: Morocco Posts: 43 Thanks: 0 
Hi Jeff, thanks for the semantic clarification. I have already posted: Quote:
 

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