Hmm... No solutions... (1) First, show that we only have to see when all $ n $ : $ f(n)>0 $. (2) Let's say $ g(n)=ln(2017f(n+325)) $, then $ g(m+m) \leq g(m)+g(n) $. (3) If there is function $ g:N \to R $ that $ g(m+n) \leq g(m)+g(n) $ and there exist $ a $ that $ \frac{g(a)}{a}<C $ then, there exist big $ N $ that all $ m \geq N $ get $ \frac{g(m)}{m} < C $. (4) get big $ l $ that $ \frac{g(2017^{l}-325)}{2017^{l}-325} < \frac{g(2017^k-325)}{2017^k} $ and use (3) Done

toto1234567890 wrote: (1) First, show that all $ n $ : $ f(n)>0 $. How can one do it easily? I tried to show it but failed. So my solution detours by not taking the $\log$ but just somehow thinking everything in a multiplicative fashion, which is similar to yours anyway.

jaydoubleuel wrote: toto1234567890 wrote: (1) First, show that all $ n $ : $ f(n)>0 $. How can one do it easily? I tried to show it but failed. So my solution detours by not taking the $\log$ but just somehow thinking everything in a multiplicative fashion, which is similar to yours anyway. I don't think I did like this in the exam...(There WAS a easier way....) (1)It's easy to see $ f(325)>0 $. (2) We can see that $ 2017f(m)f(2017) \geq f(m+1692) $ so, $ f(2017^2-2 \times 1692)>0 $. So, we get $ 2017^2-2 \times 1692 $ in $ m $ and use $ 1692 a + (2017^2-2017-1692) b $ gets all big even numbers... So, we get odd numbers $ f(n)>0 $.

toto1234567890 wrote: jaydoubleuel wrote: toto1234567890 wrote: (1) First, show that all $ n $ : $ f(n)>0 $. How can one do it easily? I tried to show it but failed. So my solution detours by not taking the $\log$ but just somehow thinking everything in a multiplicative fashion, which is similar to yours anyway. I don't think I did like this in the exam...(There WAS a easier way....) (1)It's easy to see $ f(325)>0 $. (2) We can see that $ 2017f(m)f(2017) \geq f(m+1692) $ so, $ f(2017^2-2 \times 1692)>0 $. So, we get $ 2017^2-2 \times 1692 $ in $ m $ and use $ 1692 a + (2017^2-2017-1692) b $ gets all big numbers... Doesn't $$ f(n) = (-1)^{n+1}2^{(325-n)/325} $$satisfy all conditions?

any other solution

jaydoubleuel wrote: toto1234567890 wrote: I don't think I did like this in the exam...(There WAS a easier way....) (1)It's easy to see $ f(325)>0 $. (2) We can see that $ 2017f(m)f(2017) \geq f(m+1692) $ so, $ f(2017^2-2 \times 1692)>0 $. So, we get $ 2017^2-2 \times 1692 $ in $ m $ and use $ 1692 a + (2017^2-2017-1692) b $ gets all big numbers... Doesn't $$ f(n) = (-1)^{n+1}2^{(325-n)/325} $$satisfy all conditions? Sorry. It should have been $$ f(n) = 2(-1)^{n+1}4032^{-n/325} $$

Any other solution

Sorry for my inconvience solution

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toto1234567890 wrote: Hmm... No solutions... (1) First, show that we only have to see when all $ n $ : $ f(n)>0 $. (2) Let's say $ g(n)=ln(2017f(n+325)) $, then $ g(m+m) \leq g(m)+g(n) $. (3) If there is function $ g:N \to R $ that $ g(m+n) \leq g(m)+g(n) $ and there exist $ a $ that $ \frac{g(a)}{a}<C $ then, there exist big $ N $ that all $ m \geq N $ get $ \frac{g(m)}{m} < C $. (4) get big $ l $ that $ \frac{g(2017^{l}-325)}{2017^{l}-325} < \frac{g(2017^k-325)}{2017^k} $ and use (3) Done Can you give full solution ? Thank you so much , your idea is good

I have a solution which I can't sure if it is right,so please check it. It suffices to prove for any $k \in \mathbb{N}$,$f(n)^{c_k} < f(c_k)^n$ holds for sufficiently large $n$. Let $$n=a_t (2017^t-235)+a_{t-1} (2017^{t-1}-235) + \cdots a_k(2017^k-235)+R_n$$where $t$ is the largest positive integer such that $2017^t-235 < n$ and $a_t,\cdots,a_k,R_n $ are non-negative integer taken in order,with every number as large as possible(but keep $a_t$ and $R_n$ positive).Then $R_n \leq 2017^k-235$. We only need to consider the case when $f(n)>0$ We know by condition 1 that $$f(n) \leq \prod_{i=k}^t (2017f(2017^i))^{a_i} \cdot f(R_n)$$(Note that there aren't any neggative problems).Fix $k$ ,then $R_n$ is bounded (say, by $M$) and so is $f(2017^k)^{R_n}$. It suffices to show that $$M \prod_{i=k}^t (2017f(2017^i))^{a_i} < f(2017^k)^{\frac{n}{2017^k}}$$Which is (after some calculation)equivlant to $$ \prod_{i=k}^t \bigl( \frac{f(2017^k)^{2017^i}}{f(2017^i)^{2017^k}C(k)} \bigr)^{a_i} > \frac{M^{2017^k}}{f(2017^k)^{R_n}} \cdots (*) $$where $C(k) = 2017^{2017^k} \cdot f(2017^k)^{325}$ is fixed,and the RHS of (*) is bounded. We know from condition 2 that $$d := \frac{f(2017^k)^{2017}}{f(2017^{k+1})} > 1$$and for any $j > k+1$,$$f(2017^{k+1})>f(2017^{j})^{2017^{k+1-j}}$$So$$ \frac{(f(2017^k))^{2017}}{f(2017^{j})^{2017^{k+1-j}}}>d$$and $$\frac{(f(2017^k))^{2017^j}}{f(2017^{j})^{2017^{k}}}>d^{j-1}$$The RHS goes to $+\infty$ as $j \to \infty$,so we can deduce that (*) holds for sufficiently large $n$.

I'm trying to understand why $$M \prod_{i=k}^t (2017f(2017^i))^{a_i} < f(2017^k)^{\frac{n}{2017^k}} \implies \prod_{i=k}^t \bigl( \frac{f(2017^k)^{2017^i}}{f(2017^i)^{2017^k}C(k)} \bigr)^{a_i} > \frac{M^{2017^k}}{f(2017^k)^{R_n}} \cdots (*)$$in J_bucca's post I've spent some time trying to get the second inequality from the first and have so far got: $$\frac{M^{2017^k}}{f(2017^k)^{R_n}}<\frac{\prod_{i=k}^{t}(f(2017^i)^{a_i(2017^i-235)})}{\prod_{i=k}^{t} (2017f(2017^i)^{a_i})}$$