# Functional equation

#### idontknow

$$\displaystyle f(nx)=[fâ€™(x) ]^{n} \; \;$$ , $$\displaystyle x\in \mathbb{R} , n>0$$ .
f(x)=?

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#### skipjack

Forum Staff
One possibility is $e^x$.

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#### idontknow

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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#### skipjack

Forum Staff
Yes, just one: $f(x) = 0$.

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#### idontknow

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) ?

#### cjem

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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#### idontknow

I used some methods similiar to this example :
f(ax)=af(x) , (x+a)fâ€™=f+afâ€™ or xfâ€™=f .
The equation xfâ€™=f has solution f(x)=Cx .

#### skipjack

Forum Staff
For $n = 1$, $f(x) = fâ€™(x)$, 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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