Yaritza Estrada

2022-04-07

Finding non-zero function such that $f\left(2x\right)={f}^{\prime}\left(x\right)\cdot f{}^{\u2033}\left(x\right)$ .

I am thankful if someone can help me or show me the clue. As honestly as possible, I got stuck on this problem.

I need some help in finding a$f\left(x\right)\ne 0$ such that

$f\left(2x\right)={f}^{\prime}\left(x\right)\cdot f{}^{\u2033}\left(x\right),$

where f', f" are the first and second derivatives, respectively. It is not an ordinary differential equation.

My last try was to put$f\left(x\right)=k{e}^{ax}$ , and then

$k{e}^{2ax}=ka{e}^{ax}\cdot k{a}^{2}{e}^{ax}.$

Now by canceling$e}^{ax$ , we have

$k={k}^{2}{a}^{3}.$

This shows$k=1,a=1$ and finally $f\left(x\right)={e}^{x}$ . But I want to solve the problem analytically.

I am thankful if someone can help me or show me the clue. As honestly as possible, I got stuck on this problem.

I need some help in finding a

where f', f" are the first and second derivatives, respectively. It is not an ordinary differential equation.

My last try was to put

Now by canceling

This shows

glanzerjbdo

Beginner2022-04-08Added 13 answers

Step 1

Too long for a comment.

Assuming $f\left(x\right)$ is polynomial of degree $n$, we have $n=(n-1)+(n-2)$ or $n=3$. Therefore we can look for $f$ in the following form:

$\begin{array}{rl}f\left(x\right)& =a{x}^{3}+b{x}^{2}+cx+d\\ {f}^{\text{'}}\left(x\right)& =3a{x}^{2}+2bx+c\\ {f}^{\text{'}\text{'}}\left(x\right)& =6ax+2b\end{array}$

Step 2

Then,

$f\left(2x\right)={f}^{\text{'}}\left(x\right)\cdot {f}^{\text{'}\text{'}}\left(x\right)\phantom{\rule{0ex}{0ex}}\u27f9\phantom{\rule{0ex}{0ex}}\{\begin{array}{l}8a=18{a}^{2}\\ 4b=18ab\\ 2c=4{b}^{2}+6ac\\ d=2bc\end{array}$

which gives the solutions $(0,0,0,0)$ or $(4/9,0,0,0)$. So the only polynomial solution is $f\left(x\right)=\frac{4}{9}{x}^{3}$.

seskew192atp

Beginner2022-04-09Added 12 answers

Explanation:

As was observed in answers and comments, the only polynomial solution is$f\left(x\right)=\frac{4}{9}{x}^{3}$ .

Your own work leads to the solutions$f\left(x\right)=\frac{1}{{a}^{3}}{e}^{ax}$

Adding the initial conditions$f\left(0\right)=0,{f}^{\prime}\left(0\right)\ne 0$ and assuming that f is analytic, looking at the coefficients of the Taylor series we get $f\left(x\right)=\frac{2}{{a}^{3}}\mathrm{sin}h\left(ax\right)$ .

Without the initial conditions$f\left(0\right)=0$ and using the Taylor series it's clear that there's more solutions, but probably there's no simple formula for the coefficients of the series.

Without the assumption of the function been analytic, I have no idea how the problem could be attacked.

As was observed in answers and comments, the only polynomial solution is

Your own work leads to the solutions

Adding the initial conditions

Without the initial conditions

Without the assumption of the function been analytic, I have no idea how the problem could be attacked.

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