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 March 12th, 2019, 01:58 PM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 591 Thanks: 87 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: 20,919 Thanks: 2203 One possibility is $e^x$. Thanks from idontknow
 March 12th, 2019, 02:41 PM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 591 Thanks: 87 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: 20,919 Thanks: 2203 Yes, just one: $f(x) = 0$. Thanks from idontknow
 March 13th, 2019, 04:52 AM #5 Senior Member   Joined: Dec 2015 From: somewhere Posts: 591 Thanks: 87 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
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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):

Quote:
 Originally Posted by idontknow (eq1) $\displaystyle c_1 e^{nx} +c_2 =c_{1}^{n} e^{nx} \; \;$
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: somewhere Posts: 591 Thanks: 87 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: 20,919 Thanks: 2203 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. Thanks from idontknow
 March 14th, 2019, 11:21 AM #9 Senior Member   Joined: Dec 2015 From: somewhere Posts: 591 Thanks: 87 A link about the method : https://math.stackexchange.com/quest...ifferentiation

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