My Math Forum  

Go Back   My Math Forum > College Math Forum > Real Analysis

Real Analysis Real Analysis Math Forum


Thanks Tree4Thanks
  • 1 Post By skipjack
  • 1 Post By skipjack
  • 1 Post By cjem
  • 1 Post By skipjack
Reply
 
LinkBack Thread Tools Display Modes
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.
idontknow is offline  
 
March 12th, 2019, 02:27 PM   #2
Global Moderator
 
Joined: Dec 2006

Posts: 20,373
Thanks: 2010

One possibility is $e^x$.
Thanks from idontknow
skipjack is offline  
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.
idontknow is offline  
March 12th, 2019, 08:40 PM   #4
Global Moderator
 
Joined: Dec 2006

Posts: 20,373
Thanks: 2010

Yes, just one: $f(x) = 0$.
Thanks from idontknow
skipjack is offline  
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) ?
idontknow is offline  
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):

Quote:
Originally Posted by idontknow View Post
(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.
Thanks from idontknow
cjem is online now  
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 .
idontknow is offline  
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.
Thanks from idontknow
skipjack is offline  
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
idontknow is offline  
Reply

  My Math Forum > College Math Forum > Real Analysis

Tags
equation, functional



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Functional Equation idontknow Elementary Math 4 July 30th, 2018 09:25 AM
Functional equation idontknow Elementary Math 2 March 22nd, 2018 12:31 PM
A functional equation Dacu Real Analysis 4 June 7th, 2015 05:41 AM
Functional Equation desum Elementary Math 0 May 10th, 2012 04:03 AM
Functional equation arnold Algebra 1 October 13th, 2011 04:23 AM





Copyright © 2019 My Math Forum. All rights reserved.