
PreCalculus PreCalculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
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  
Member Joined: Sep 2014 From: Morocco Posts: 43 Thanks: 0 
Hi Jeff, thanks for the semantic clarification. I have already posted: Quote:
 

Tags 
function, generalizing 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
guessing the base function of a real function that meets certain requirements  vlekje5  PreCalculus  11  March 27th, 2017 01:58 PM 
Limit generalizing  MisaKr  Calculus  2  October 24th, 2016 12:13 PM 
Generalizing the prime number theorem. Sorta.  standardmalpractice  Math  1  March 21st, 2016 09:47 AM 
Derivation of tau function, sigma, euler and mobius function  msgelyn  Number Theory  2  January 12th, 2014 04:13 AM 
Generalizing a recursive series  boyo  Applied Math  1  November 20th, 2009 09:39 AM 