3
$\begingroup$

This is the first $40$ partial products on Desmos (Desmos gave me "undefined" for everything higher):

enter image description here

Looking at this, it doesn't appear to converge but then, while attempting to get Desmos to evaluate the summation with a higher upper bound $k$ using $\prod\limits_{n=1}^{k}\left(\int\limits_{-1}^{-10^{-7}}\left|\frac{\lfloor t^{1-n}\rfloor}{\lfloor t^{-n}\rfloor}\right|dt+\int\limits_{10^{-7}}^{1}\left|\frac{\lfloor t^{1-n}\rfloor}{\lfloor t^{-n}\rfloor}\right|dt\right)$, Desmos showed me the following:

enter image description here

This looks completely different and appears to converge to $1$. Once again attempting to get a higher upper bound, I used $\prod\limits_{n=1}^{k}\left(\int\limits_{-1}^{-10^{-2}}\left|\frac{\lfloor t^{1-n}\rfloor}{\lfloor t^{-n}\rfloor}\right|dt+\int\limits_{10^{-2}}^{1}\left|\frac{\lfloor t^{1-n}\rfloor}{\lfloor t^{-n}\rfloor}\right|dt\right)$. This allowed me a maximum upper bound of $155$ and the graph looked like this:

enter image description here

This, like the last, looks like it clearly converges to $1$. Because Desmos appears to be disagreeing with itself, I want to solve this problem analytically just to make sure. Here is my attempt:

$$\prod\limits_{n=1}^{\infty}\int\limits_{-1}^{1}\left|\frac{\lfloor t^{1-n}\rfloor}{\lfloor t^{-n}\rfloor}\right|dt$$ $$\prod\limits_{n=1}^{\infty}\left(\int\limits_{0}^{1}\frac{\lfloor t^{1-n}\rfloor}{\lfloor t^{-n}\rfloor}dt-\int\limits_{-1}^{0}\frac{\lfloor t^{1-n}\rfloor}{\lfloor t^{-n}\rfloor}dt\right)$$ $$\prod\limits_{n=1}^{\infty}\left(\sum_{i=1}^{\infty}\left(\left(i^{-\frac{1}{n}}-\left(i+1\right)^{-\frac{1}{n}}\right)\left(\frac{\lfloor i^{\frac{1}{n}}\rfloor}{i}\right)\right)+\sum_{i=1}^{\infty}\left(\left(i^{-\frac{1}{n}}-\left(i+1\right)^{-\frac{1}{n}}\right)\left(\frac{\lfloor (i+1)^{\frac{1}{n}}\rfloor}{i}\right)\right)\right)$$

Unfortunately, after this last step, I realized the summations only worked for when $n=2$ and I couldn't figure out how to fix them.

$\endgroup$

1 Answer 1

1
$\begingroup$

This is not a complete answer (too big for a comment) but this will help you get started. Define $h_0 = 1$, $h_n : [-1,1]\setminus\{0\} \to \mathbb{N}=\{1,2,\cdots\}$ by $h_n(x) = \text{Abs}(\lfloor x^{-n} \rfloor)$ and take $\epsilon = -\frac{1}{n}$, for $n \in \mathbb{N}$. The definition of $h_n$ can be simplified as following: $$h_n(x) = \begin{cases}k , \,\, n \text{ even and } x \in [-k^{\epsilon},-(k+1)^{\epsilon}) \cup ((k+1)^{\epsilon},k^{\epsilon}], \, k \in \mathbb{N} \\ k , \,\, n \text{ odd and } x \in (-(k-1)^{\epsilon},-k^{\epsilon}] \cup ((k+1)^{\epsilon},k^{\epsilon}], \, k \ge 2 \\ 1, \,\, n \text{ odd and } x \in \{-1\} \cup (2^{\epsilon},1] \end{cases}.$$ In terms of $h_n$, the terms of the product can be written as $a_n = \int_{-1}^{1} h_{n-1}(t)/h_n(t) dt $. With the above simplified definition of $h_n$, we can already compute $a_1$, which is, $$a_1 = \int_{-1}^{1} \frac{dt}{h_1(t)} = \int_{1/2}^{1} dt + \sum_{k=2}^{\infty} \frac{1}{k}\left(\int_{\frac{1}{k+1}}^{\frac{1}{k}}+\int_{\frac{-1}{k-1}}^{\frac{-1}{k}}dt\right) = \frac{1}{2}+\sum_{k=2}^{\infty}\frac{2}{k(k^2-1)}=1.$$ However, other terms can't be expressed in closed form, this can be shown by computing $a_2$, $$\begin{gather*}a_2 = \int_{-1}^{1} \frac{h_1(t)}{h_2(t)}dt = \sum_{k=1}^{\infty} \frac{1}{k}\left(\int_{-\frac{1}{\sqrt{k}}}^{-\frac{1}{\sqrt{k+1}}}h_1(t) dt+\int_{\frac{1}{\sqrt{k+1}}}^{\frac{1}{\sqrt{k}}} h_1(t) dt\right) \\ = \sum_{l=1}^{\infty} \sum_{k=l^2}^{(l+1)^2-1} \frac{1}{k}\left((l+1)\int_{-\frac{1}{\sqrt{k}}}^{-\frac{1}{\sqrt{k+1}}} dt + l \int_{\frac{1}{\sqrt{k+1}}}^{\frac{1}{\sqrt{k}}} dt\right) \\ = \sum_{l=1}^{\infty} \sum_{k=l^2}^{(l+1)^2-1} \frac{2l+1}{k}\left( \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}} \right) \approx 1.42722.\end{gather*}$$ To get an estimate of $a_n$, observe that for $n$ even $\frac{1}{x^n}-1 \le h_n(x) \le \frac{1}{x^n}$ and for $n$ odd, $\frac{1}{x^n}-1 \le h_n(x) \le \frac{1}{x^n}, x \in (0,1]$, $-\frac{1}{x^n} \le h_n(x) \le 1-\frac{1}{x^n}, x \in [-1,0)$. For $n$ odd, we have $$ \int_{-1}^{1} \frac{1}{h_n(t) t^{n-1}} dt- \int_{-1}^{1} \frac{1}{h_n(t)} dt \le a_n \le \int_{-1}^{1} \frac{1}{h_n(t) t^{n-1}} dt$$ where, $$\begin{align*} \int_{-1}^{1} \frac{1}{h_n(t) t^{n-1}} dt &= \int_{2^{\epsilon}}^{1} \frac{1}{t^{n-1}}dt + \sum_{k=2}^{\infty} \frac{1}{k}\left( \int_{-(k-1)^{\epsilon}}^{-k^{\epsilon}} + \int_{(k+1)^{\epsilon}}^{k^{\epsilon}} \frac{1}{t^{n-1}}dt\right) \\ &= \frac{1}{n-2}\left[(2^{1+2\epsilon}-1)+\sum_{k=2}^{\infty} \frac{1}{k}\left( (k+1)^{1+2\epsilon}-(k-1)^{1+2\epsilon}\right) \right]\end{align*}$$ and, $$\int_{-1}^{1} \frac{dt}{h_n(t)} = \int_{2^{\epsilon}}^{1}1 + \sum_{k=2}^{\infty} \frac{1}{k}\left( \int_{-(k-1)^{\epsilon}}^{-k^{\epsilon}} + \int_{(k+1)^{\epsilon}}^{k^{\epsilon}} dt\right) = (1-2^{\epsilon})+\sum_{k=2}^{\infty} \frac{1}{k}\left( (k-1)^{\epsilon}-(k+1)^{\epsilon}\right). $$ The sum in the above expression converges because $(k+1)^{1+2\epsilon}-(k-1)^{1+2\epsilon} \sim (2+4\epsilon)k^{2\epsilon}$ as $k \to \infty$.

Edit 1: Just realized that $\lim_{n\to\infty}\int_{-1}^{1} \frac{1}{h_n(t)} dt = 0$. For $n$ odd case, $$ \begin{align*}0 \le \int_{-1}^{1} \frac{1}{h_n(t)} dt &\le -\int_{-1}^{0} t^n dt + \int_0^{2^{\epsilon}} \frac{t^n}{1-t^n}dt + \int_{2^{\epsilon}}^{1} 1 dt \\ &= \frac{1}{n+1}+\frac{1}{n}\int_{0}^{\frac{1}{2}}\frac{t^{1/n}}{1-t}dt+(1-2^{-\frac{1}{n}}) \end{align*} $$ therefore, ($n$ odd), $$ a_n -\int_{-1}^{1} \frac{1}{h_n(t) t^{n-1}} dt = O\left(\frac{1}{n}\right).$$

Edit 2: If (big if) we can directly substitute the difference $(k+1)^{1+2\epsilon}-(k-1)^{1+2\epsilon} \sim (2+4\epsilon)k^{2\epsilon}$ in the expression for $\int_{-1}^{1} \frac{1}{h_n(t) t^{n-1}} dt$, then the summation is $\zeta\left(1+\frac{2}{n}\right)$ (Reimann zeta function). Taylor expanding this, we obtain that $$ a_n = 1 + O\left(\frac{1}{n}\right).$$ This will imply that the product diverges from this result.

$\endgroup$
3
  • 1
    $\begingroup$ Thanks for the incredible contribution! $\endgroup$ Commented Feb 4 at 4:15
  • 1
    $\begingroup$ No problem! I am still unsure if the product converges because the sum is getting larger as $\epsilon$ gets small. However, the denominator is getting large as well due to presence of $n$. $\endgroup$
    – Sam
    Commented Feb 4 at 4:19
  • $\begingroup$ Unless there is something wrong in the calculations, I believe that the product diverges. @DylanLevine $\endgroup$
    – Sam
    Commented Feb 4 at 10:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .