Timeline for Limit involving the fractional part and the Fibonacci numbers
Current License: CC BY-SA 4.0
31 events
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| Apr 1 at 23:51 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Apr 1 at 22:50 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Apr 1 at 21:32 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Apr 1 at 20:45 | comment | added | Babar | You're welcome thanks for your efforts! I tried other sequences of integers simpler than that of Fibonacci for which it seems that we have a repeating pattern and I think your proof could also adapt to these limits. This could be called a generalized Dirichlet divisor problem. | |
| Apr 1 at 18:57 | comment | added | GH from MO | @Babar Sorry if I sounded too harsh. In fact the problem is very nice. I will rewrite the proof in the next hour or so. The new version will be more streamlined, it will cover the case of even/odd jointly, and it will feature an error term $O(n^{1/2})$ as you expected. Stay tuned. | |
| Apr 1 at 16:05 | comment | added | Babar | Thank you very much. The appearance of Pi is lovely. Next time, I will be more careful in specifying what is conjectural. | |
| Apr 1 at 12:40 | comment | added | GH from MO | @SamHopkins Thank you. If you shift the indices by $1$, then the values of the original limit and the one in the "Added" section get flipped. At any rate, it is standard to assume that $F_0=0$, because we want $F_n$ to equal $(\varphi^n-\psi^n)(\varphi-\psi)$, where $\varphi=(1+\sqrt{5})/2$ and $\psi=(1-\sqrt{5})/2$. It serves harmony. Both Mathematica and SAGE think that $F_0=0$. | |
| Apr 1 at 12:02 | comment | added | Sam Hopkins | @GHfromMO: I think Fedor's point was that there are sometimes different conventions about how to index the Fibonacci numbers. Sometimes $F_0=0,F_1=1,F_2=1,\ldots$ but sometimes $F_0=1,F_1=1,F_2=2,\ldots$. See for example the discussion at the beginning of arxiv.org/abs/2209.08878. | |
| Apr 1 at 11:37 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Apr 1 at 11:36 | comment | added | GH from MO | @FedorPetrov If you permute the Fibonacci sequence, then the limit can change. For example, if you flip $F(2j-1)$ and $F(2j)$ for every $j$, then the limit changes to $\pi/8$. See my "Added" section. | |
| Apr 1 at 11:34 | comment | added | GH from MO | @Babar The "other limit" is not $2\log 2-1$ but $\pi/8$. See my "Added" section. BTW I don't know how you or your lecturer came up with the numbers $1-\log\left(\frac{1+\sqrt{5}}{2}\right)$ and $2\log 2-1$. Mathematics is not a guessing game. | |
| Apr 1 at 11:32 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Apr 1 at 10:35 | comment | added | Babar | I'm not sure I understand. The Fibonacci numbers here are $F(1)=F(2)=1$, then $F(n)=F(n-1)+F(n-2)$ so the limit is unique. | |
| Apr 1 at 10:24 | comment | added | Fedor Petrov | Can it be the case that the limit depends on how do you enumerate Fibonacci numbers? | |
| Apr 1 at 9:49 | comment | added | Babar | I have a very speculative idea. Let $\tau_{F}(n)$ be the number of divisors of $F(n)$ of the form $F(k)$. We have $\tau_{F}(1)=1$ then for $n\geq2$ we have $\tau_{F}(n)=\tau(n)+n\mod2$. Intuitively this leads me to think that the remainders $\sum_{k=1}^{n}\left\{ \frac{n}{k}\right\} -\left(1-\gamma\right)n$ and $\sum_ {k=1}^{2n}\left\{ \frac{F(2n)}{F(k)}\right\} -\frac{3\log2}{2}n$ behave the same way and so that we would optimally have$ \sum_{k=1}^{2n}\left\{ \frac{F(2n)}{F(k)}\right\} -\frac{3\log2}{2} n=O\left(n^{1/4+\varepsilon}\right)$. | |
| Mar 31 at 21:57 | comment | added | GH from MO | @Babar I think your last formula holds with a much better error term than $O(n^{1/2})$. The point is that the $k$-sum over each $I(r,n)$ can be determined with great precision, because the terms in it are exponentially close to $0$ or $1$. For example, for $n=10^6$, the sum is $\approx 1039707.22967$, while $\frac{3\log 2}{2}n\approx 1039720.77084$. So for $n=10^6$ the error is about $13.54117$, which is much smaller than $n^{1/2}$. | |
| Mar 31 at 21:40 | comment | added | Babar | Thanks a lot. My question about $M=O(n^{1/2})$ was: can we have something like the problem of Dirichlet divisors i.e. $$\sum_{k=1}^{2n}\left\{ \frac{F(2n)}{F(k)}\right\} =\frac{3\log2}{2}n+O\left(n^{1/2}\right)$$ I will try to see. | |
| Mar 31 at 21:33 | comment | added | GH from MO | @Babar Your last formula makes no sense as $n$ is a bound variable on the LHS, but a free variable on the RHS (see en.wikipedia.org/wiki/Free_variables_and_bound_variables). The limit in your first comment above makes sense, and I am sure it can be calculated by using the ideas in my post. I have no time to check if it equals $2\log 2-1$, but if you need confirmation, open a new question for it (featuring the limit in your first comment above). Thanks for accepting my answer! | |
| Mar 31 at 21:25 | vote | accept | Babar | ||
| Mar 31 at 21:25 | comment | added | Babar | Sorry for being unclear. If $f(x)=F(x)$ the Fibonacci numbers then we have two limits $$\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^{n}\left\{ \frac{f(n)}{f(k)}\right\} =\begin{cases} \frac{3\log2}{4} & n\ even\\ \ell & n\ odd \end{cases}$$ where I think $\ell=2\log2-1 $. | |
| Mar 31 at 21:13 | comment | added | GH from MO | @Babar It is not clear why you say "the other limit" as there is only one limit in your post. Of course there is an other limit in your recent comment, but by saying "the other limit" you seem to refer to earlier discussions (where this "other limit" was not present). I guess it is a language problem. Regarding $M$, it is only a technical quantity introduced in the proof (it is not present in the final result). Since the $k$-sum over each $I(r,n)$ converges rapidly, one can take $M$ to be a quantity growing with $n$ slowly, but why would you do that? I have not checked if $M=n^{1/2}$ is OK. | |
| Mar 31 at 21:02 | comment | added | Babar | Thank you very much for this proof which turns out to be quite technical. I suppose the other limit is $$\lim_{n\rightarrow\infty}\frac{1}{2n+1}\sum_{k=1}^{2n+1}\left\{ \frac{F(2n+1)}{F(k)}\right\} =2\log2-1$$ Do you think your $M$ can be $O(n^{1/2})$? | |
| Mar 31 at 0:59 | history | edited | GH from MO | CC BY-SA 4.0 |
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| Mar 30 at 19:30 | history | answered | GH from MO | CC BY-SA 4.0 |