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Question

Given $f : \mathbb{R}$ $\rightarrow$ $\mathbb{R}$ a continuous function satisfying the functional equation : $$ f(x+y) - 3^y f(x) = 3^x f(y)$$ for all $x,y$ $\in \mathbb{R}$.

Let $$L = \lim_{x\to \infty} \left(\frac{f(x)}{f'(x)} + \frac{f'(x)}{f''(x)}+\cdots+ \frac{f^{n-1}(x)}{f^n(x)}\right)$$

Find L

My approach:

I could not find $f(x)$ itself. Usually, my approach is partially differentiating and setting one variable to zero, but it does not work in this case.

$$f'(x+y) - 3^y f(x) =(\ln3) 3^x f(y) $$

Setting $y=0$ simply gives $0=0$.

Any hints would help a lot. Thank you

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    $\begingroup$ What is $f'(x)$? You are only assuming that $f(x)$ is continuous. Did you mean to assume it was $C^{\infty}$? Smooth? Analytic? Or are you claiming that those properties follow from the functional equation? $\endgroup$
    – lulu
    Commented 11 hours ago
  • $\begingroup$ Do you have initial conditions for $f(x)$? The constant function $f(x)=0$ satisfies the functional equation, but $L$ is undefined for that example. $\endgroup$
    – lulu
    Commented 11 hours ago
  • $\begingroup$ @lulu I don't know, I guess it's not a zero function, as limit exists. And I don't know some terms as I'm in high school, could you elaborate? $\endgroup$
    – Someone
    Commented 11 hours ago
  • $\begingroup$ All I am saying is that you need more assumptions before you can even start work on the question. I don't think it's appropriate for us to try to guess at the intended assumptions. $\endgroup$
    – lulu
    Commented 11 hours ago
  • $\begingroup$ I suggest: search for another solution, or try to prove that $0$ is the only solution. $\endgroup$
    – lulu
    Commented 11 hours ago

1 Answer 1

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For a start.

If $f$ is continuous then so is $g(x) = f(x)/3^x$ and we have $$g(x+y)=g(x)+g(y)$$ so , since $g$ is continuous $g(x)= ax$ for some real $a$. Then $f(x) = ax3^x$ and now it should be easy?

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