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This question is an old question from mathstackexchange.

Let $f_- (n) = \Pi_{i=0}^n ( \sin(i) - \frac{5}{4}) $

And let

$ f_+(m) = \Pi_{i=0}^m ( \sin(i) + \frac{5}{4} ) $

It appears that

$$\sup f_- (n) \inf f_+ (m) = \frac{5}{4} $$

Why is that so ?

Notice

$$\int_0^{2 \pi} \ln(\sin(x) + \frac{5}{4}) dx = Re \int_0^{2 \pi} \ln (\sin(x) - \frac{5}{4}) dx = \int_0^{2 \pi} \ln (\cos(x) + \frac{5}{4}) dx = Re \int_0^{2 \pi} \ln(\cos(x) - \frac{5}{4}) dx = 0 $$

$$ \int_0^{2 \pi} \ln (\sin(x) - \frac{5}{4}) dx = \int_0^{2 \pi} \ln (\cos(x) - \frac{5}{4}) dx = 2 \pi^2 i $$

That explains the finite values of $\sup $ and $ \inf $.. well almost. It can be proven that both are finite. But that does not explain the value of their product.


Update

This is probably not helpful at all , but it can be shown ( not easy ) that there exist a unique pair of functions $g_-(x) , g_+(x) $ , both entire and with period $2 \pi $ such that

$$ g_-(n) = f_-(n) , g_+(m) = f_+(m) $$

However i have no closed form for any of those ...

As for the numerical test i got about $ln(u) (2 \pi)^{-1}$ correct digits , where $u = m + n$ and the ratio $m/n$ is close to $1$.

Assuming no round-off errors i ended Up with $1.2499999999(?) $. That was enough to convince me.


I often get accused of " no context " or " no effort " but i have NOO idea how to even start here. I considered telescoping but failed and assumed it is not related. Since I also have no closed form for the product I AM STUCK.

I get upset when people assume this is homework. It clearly is not imho ! What kind of teacher or book contains this ?

——-

Example : Taking $m = n = 8000 $ we get

$$ max(f_-(1),f_-(2),...,f_-(8000)) = 1,308587092.. $$ $$ min(f_+(1),f_+(2),...,f_+(8000)) = 0,955226916.. $$

$$ 1.308587092.. X 0.955226916.. = 1.249997612208568.. $$

Supporting the claim.

I'm not sure if $sup f_+ = 7,93.. $ or the average of $f_+ $ ( $ 3,57..$ ) relate to the above $1,308.. $ and $0,955..$ or the truth of the claimed value $5/4$.

In principle we could write the values $1,308..$ and $0,955..$ as complicated integrals. By using the continuum product functions $f_-(v),f_+(w)$ where $v,w$ are positive reals.

This is by noticing $ \sum^t \sum_i a_i \exp(t \space i) = \sum_i a_i ( \exp((t+1)i) - 1)(\exp(i) - 1)^{-1} $ and noticing the functions $f_+,f_-$ are periodic with $2 \pi$.

Next with contour integration you can find min and max over that period $2 \pi$ for the continuum product functions.

Then the product of those 2 integrals should give you $\frac{5}{4}$.

—-

Maybe all of this is unnecessarily complicated and some simple theorems from trigonometry or calculus could easily explain the conjectured value $\frac{5}{4}$ .. but I do not see it.

——

—— Update This conjecture is part of a more general phenomenon.

For example the second conjecture :

Let $g(n) = \prod_{i=0}^n (\sin^2(i) + \frac{9}{16} ) $

$$ \sup g(n) \space \inf g(n) = \frac{9}{16} $$

It feels like this second conjecture could somehow follow from the first conjecture since

$$-(\cos(n) + \frac{5}{4})(\cos(n) - \frac{5}{4}) = - \cos^2(n) + \frac{25}{16} = \sin^2(n) + \frac{9}{16} $$

And perhaps the first conjecture could also follow from this second one ?

Since these are additional questions and I can only accept one answer , I started a new thread with these additional questions :

https://math.stackexchange.com/questions/3000441/why-is-inf-g-sup-g-frac916

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  • $\begingroup$ How does vanishing of those integrals imply the $\sup/\inf$ of the products are finite? $\endgroup$ – Wojowu Feb 1 '19 at 18:31
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    $\begingroup$ Isn't $f_-(0) = -\frac{5}{4}$ negative? Similarly, what does $\log(\sin x - \tfrac{5}{4})$ mean? $\endgroup$ – Mateusz Kwaśnicki Feb 1 '19 at 19:28
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    $\begingroup$ The original question is math.stackexchange.com/questions/2075374/… $\endgroup$ – Gerry Myerson Feb 1 '19 at 22:04
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    $\begingroup$ Have you tried to replace 5/4 with any other positive real number in $ f_{-} $ and $ f_{+} $? $\endgroup$ – Sylvain JULIEN Feb 2 '19 at 0:25
  • $\begingroup$ @Christian Rembling : they do not converge to 1. The shape is sine like. Even has the period 2 pi. If it converged to 1 then the sup or inf would be reached after a finite amount of step , and thus have a closed form. Then it would be a trivial problem. $\endgroup$ – mick Feb 2 '19 at 8:44
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(This is an extended comment, not a true answer. It provides a sort-of-closed-form expression for $f_+(n)$.)

There is a special function $S_2(\alpha; z)$, called the double sine function, which is meromorphic in $z \in \mathbb{C}$ and which satisfies $$ S_2(\alpha; z + 1) = \frac{S_2(\alpha; z)}{2 \sin(\tfrac{\pi}{\alpha} z)} \qquad \text{and} \qquad S_2(\alpha; z + \alpha) = \frac{S_2(\alpha; z)}{2 \sin(\pi z)} \, . $$ If we set $\alpha = 2 \pi$ and write simply $S_2(z) = S_2(2 \pi, z)$, we get $$ S_2(z + 1) = \frac{S_2(z)}{2 \sin(\tfrac{z}{2})} \, . $$ Choose $b > 0$ such that $\cosh b = \tfrac{5}{4}$, and write $\xi_\pm = \tfrac{\pi}{4} \pm b i$. Then $$ \sin(z) + \tfrac{5}{4} = 2 \sin(\tfrac{z}{2} + \xi_+) \sin(\tfrac{z}{2} + \xi_-) . $$ It follows that $$ \frac{2^n S_2(\xi_+) S_2(\xi_-)}{S_2(n + 1 + \xi_+) S_2(n + 1 + \xi_-)} = \prod_{k = 0}^n 2 \sin(\tfrac{k}{2} + \xi_+) \sin(\tfrac{k}{2} + \xi_-) = \prod_{k = 0}^n (\sin k + \tfrac{5}{4}) $$ is your sequence $f_+(n)$.

Now the double sine function is a special function that I know next to nothing about (I learned about it, as well as about many other fancy special functions, from Alexey Kuznetsov), but apparently it is reasonably well-understood. The refernences that I got from Alexey are:

  • S. Koyama and N. Kurokawa, Multiple sine functions, Forum Mathematicum 15 (2006), no. 6, 839–876;

  • S. Koyama and N. Kurokawa, Values of the double sine function, J. Number Theory 123 (2007), no. 1, 204–223.

I did not check them to see if they contain any information relevant to your questions.

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