User Name Remember Me? Password

 Real Analysis Real Analysis Math Forum

 March 12th, 2019, 01:58 PM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 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: 21,034 Thanks: 2269 One possibility is $e^x$. Thanks from idontknow March 12th, 2019, 02:41 PM #3 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 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: 21,034 Thanks: 2269 Yes, just one: $f(x) = 0$. Thanks from idontknow March 13th, 2019, 04:52 AM #5 Senior Member   Joined: Dec 2015 From: somewhere Posts: 734 Thanks: 98 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: 313
Thanks: 112

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):

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: 734 Thanks: 98 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: 21,034 Thanks: 2269 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: 734 Thanks: 98 A link about the method : https://math.stackexchange.com/quest...ifferentiation Tags equation, functional Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post idontknow Elementary Math 4 July 30th, 2018 09:25 AM idontknow Elementary Math 2 March 22nd, 2018 12:31 PM Dacu Real Analysis 4 June 7th, 2015 05:41 AM desum Elementary Math 0 May 10th, 2012 04:03 AM arnold Algebra 1 October 13th, 2011 04:23 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      