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 thatf(2x)=f′(x)⋅f″(x),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 thenke2ax=kaeax⋅ka2eax.Now by canceling eax, we havek=k2a3.This shows k=1,a=1 and finally f(x)=ex. But I want to solve the problem analytically.

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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 thatf(2x)=f′(x)⋅f″(x),where f&#039;, f&quot; are the first and second derivatives, respectively. It is not an ordinary differential equation.My last try was to put f(x)=keax, and thenke2ax=kaeax⋅ka2eax.Now by canceling eax, we havek=k2a3.This shows k=1,a=1 and finally f(x)=ex. But I want to solve the problem analytically.

glanzerjbdo · Accepted Answer

Step 1Too long for a comment.Assuming&amp;nbsp;f(x)&amp;nbsp;is polynomial of degree&amp;nbsp;n, we have&amp;nbsp;n=(n−1)+(n−2)&amp;nbsp;or&amp;nbsp;n=3. Therefore we can look for&amp;nbsp;f&amp;nbsp;in the following form:f(x)=ax3+bx2+cx+df&#039;(x)=3ax2+2bx+cf&#039;&#039;(x)=6ax+2bStep 2Then,&amp;nbsp;f(2x)=f&#039;(x)⋅f&#039;&#039;(x)&amp;nbsp;⟹&amp;nbsp;{8a=18a24b=18ab2c=4b2+6acd=2bcwhich gives the solutions&amp;nbsp;(0,0,0,0)&amp;nbsp;or&amp;nbsp;(4/9,0,0,0). So the only polynomial solution is&amp;nbsp;f(x)=49x3.

seskew192atp · Accepted Answer

Explanation:As was observed in answers and comments, the only polynomial solution is f(x)=49x3.Your own work leads to the solutions f(x)=1a3eaxAdding 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)=2a3sin⁡h(ax).Without the initial conditions f(0)=0 and using the Taylor series it&#039;s clear that there&#039;s more solutions, but probably there&#039;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.

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

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