Functional equation

Dec 2015
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Earth
\(\displaystyle f(nx)=[f’(x) ]^{n} \; \;\) , \(\displaystyle x\in \mathbb{R} , n>0\) .
f(x)=?
 
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Dec 2015
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169
Earth
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?
 
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Dec 2015
1,082
169
Earth
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) ?
 
Aug 2017
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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):

(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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Dec 2015
1,082
169
Earth
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 .
 

skipjack

Forum Staff
Dec 2006
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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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