My Math Forum Generalizing with a function

 Pre-Calculus Pre-Calculus Math Forum

 August 20th, 2017, 12:29 PM #11 Senior Member   Joined: May 2016 From: USA Posts: 1,306 Thanks: 549 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
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Hi Jeff, thanks for the semantic clarification.

Quote:
 Originally Posted by Mifarni14 Hi again, sorry for the late reply. I simplified f(x) and f(f(x)) again, it turns out that I had to simplify more. What I then did was to conjecture the following: For each n in $\displaystyle \mathbb{N^*}$, (n iterations) $\displaystyle f \circ f \circ f \circ ... \circ f(x) = \frac{x}{\sqrt{1+ncx^2}}$ This was later demonstrated through induction.
Thanks for the help nonetheless.

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