Finding non-zero function such that \(\displaystyle{f{{\left({2}{x}\right)}}}={f}'{\left({x}\right)}\cdot{f}{''}{\left({x}\right)}\). I am

Yaritza Estrada

Yaritza Estrada

Answered question


Finding non-zero function such that f(2x)=f(x)f(x).
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(x)0 such that
where f', f" are the first and second derivatives, respectively. It is not an ordinary differential equation.
My last try was to put f(x)=keax, and then
Now by canceling eax, we have
This shows k=1,a=1 and finally f(x)=ex. But I want to solve the problem analytically.

Answer & Explanation



Beginner2022-04-08Added 13 answers

Step 1
Too long for a comment.
Assuming f(x) is polynomial of degree n, we have n=(n1)+(n2) or n=3. Therefore we can look for f in the following form:
Step 2

 f(2x)=f'(x)f''(x)  {8a=18a24b=18ab2c=4b2+6acd=2bc
which gives the solutions (0,0,0,0) or (4/9,0,0,0). So the only polynomial solution is f(x)=49x3.



Beginner2022-04-09Added 12 answers

As was observed in answers and comments, the only polynomial solution is f(x)=49x3.
Your own work leads to the solutions f(x)=1a3eax
Adding the initial conditions f(0)=0,f(0)0 and assuming that f is analytic, looking at the coefficients of the Taylor series we get f(x)=2a3sinh(ax).
Without the initial conditions f(0)=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.

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