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

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

• How does vanishing of those integrals imply the $\sup/\inf$ of the products are finite? – Wojowu Feb 1 '19 at 18:31
• Isn't $f_-(0) = -\frac{5}{4}$ negative? Similarly, what does $\log(\sin x - \tfrac{5}{4})$ mean? – Mateusz Kwaśnicki Feb 1 '19 at 19:28
• The original question is math.stackexchange.com/questions/2075374/… – Gerry Myerson Feb 1 '19 at 22:04
• Have you tried to replace 5/4 with any other positive real number in $f_{-}$ and $f_{+}$? – Sylvain JULIEN Feb 2 '19 at 0:25
• @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. – mick Feb 2 '19 at 8:44

(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 $$f_+(n)$$.