March 12th, 2019, 01:58 PM  #1 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68  Functional equation
$\displaystyle f(nx)=[f’(x) ]^{n} \; \;$ , $\displaystyle x\in \mathbb{R} , n>0$ . f(x)=? Last edited by idontknow; March 12th, 2019 at 02:01 PM. 
March 12th, 2019, 02:27 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
One possibility is $e^x$.

March 12th, 2019, 02:41 PM  #3 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
Since $\displaystyle f(x)$ can be an exponential function then $\displaystyle f(x)=a^x $, solving equation for a gives $\displaystyle \ln(a)=1 $ or $\displaystyle a=e$. Is there any other solution? Last edited by skipjack; March 12th, 2019 at 08:53 PM. 
March 12th, 2019, 08:40 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
Yes, just one: $f(x) = 0$.

March 13th, 2019, 04:52 AM  #5 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
My approach : the general solutions are of the form $\displaystyle y=c_1 e^x +c_2 \; $ after substitution to the general equation we must find the pairs $\displaystyle (c_1 , c_2 )$. (eq1) $\displaystyle c_1 e^{nx} +c_2 =c_{1}^{n} e^{nx} \; \;$ , $\displaystyle (c_1 , c_2 )$=? Now it is not easy to find the pairs but it can be seen that for the pair $\displaystyle (1,0)$ one of the solutions is $\displaystyle y=e^x$ . (eq1) has solutions only if $\displaystyle c_2=0$ but a proof is needed. After that the equation turns into $\displaystyle c_1 =c_{1}^{n} \; \; $ , $\displaystyle n>0$ . How to continue from (eq1) ? 
March 13th, 2019, 05:47 AM  #6 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry 
Once you've narrowed down the solutions to those of the form $f(x) = c_1 e^x + c_2$, it is elementary to show that $c_2 = 0$ and $c_1 = 0$ or $1$. Indeed, just plug in $x = 0$ and $x = 1$ into (eq1): to get an easy pair of simultaneous equations. The hard part of the problem is showing the general solution is indeed of the form $f(x) = c_1 e^x + c_2$. It's interesting that you were happy to assume this without proof, yet demanded proof for the trivial part of the problem. 
March 13th, 2019, 10:01 AM  #7 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
I used some methods similiar to this example : f(ax)=af(x) , (x+a)f’=f+af’ or xf’=f . The equation xf’=f has solution f(x)=Cx . 
March 13th, 2019, 03:23 PM  #8 
Global Moderator Joined: Dec 2006 Posts: 20,373 Thanks: 2010 
For $n = 1$, $f(x) = f’(x)$, which has solution $f(x) = \text{c}e^x$. It's easy to show that $\text{c}$ is 1 or 0.

March 14th, 2019, 11:21 AM  #9 
Senior Member Joined: Dec 2015 From: iPhone Posts: 436 Thanks: 68 
A link about the method : https://math.stackexchange.com/quest...ifferentiation 

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