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March 12th, 2019, 01:58 PM   #1
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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.
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March 12th, 2019, 02:27 PM   #2
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One possibility is $e^x$.
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March 12th, 2019, 02:41 PM   #3
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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.
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March 12th, 2019, 08:40 PM   #4
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Yes, just one: $f(x) = 0$.
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March 13th, 2019, 04:52 AM   #5
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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) ?
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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):

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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.
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March 13th, 2019, 10:01 AM   #7
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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 .
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March 13th, 2019, 03:23 PM   #8
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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.
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March 14th, 2019, 11:21 AM   #9
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A link about the method :
https://math.stackexchange.com/quest...ifferentiation
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