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

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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

---EDIT---

I want to explain better how to get a closed form for these numbers. I already mentioned that the periods of these functions are $$2 \pi$$ and how to use that.

But those few lines deserve more attention.

Basically this is what we do :

We use the fourier series

$$f(x) = \ln(\sin(x) + 5/4) = \sum_{k=1}^{\infty} \frac{-2 \cos(k(x + \pi/2))}{2^k k}$$

Now we use the inverse of the backward difference operator ( similar but distinct from the so-called "indefinite sum" which is defined as inverse of the forward difference operator )

In other words we solve for $$F(x)$$ such that

$$F(x) - F(x-1) = f(x)$$

We call this the " continuum sum " (CS) and write/define :

$$CS(f(x),x) := F(x)$$ $$CS(f(x),y) := F(y)$$

For clarity :

$$\sum_0^y f(x) = CS(f(x),y) - CS(f(x),-1)$$

$$\sum_0^0 f(x) = CS(f(x),0) - CS(f(x),-1) = f(0)$$

This operator is linear so we make use of that :

$$CS(2 \cos(k(x+\pi/2),x) = \csc(k/2) \sin(k(x+1/2 + \pi/2)) -1.$$

This implies that :

$$CS(f(x),x) = F(x) = - \sum_{k=1}^{\infty} \frac{\csc(k/2)\sin(k(x + 1/2 + \pi/2)) - 1}{2^k k}$$

and

$$G(x) = \frac{d F(x)}{dx} = - \sum_{k=1}^{\infty} \frac{\csc(k/2) \cos(k(x+1/2+\pi/2))}{2^k}$$

Now $$\sup f_+ = 7.93..$$ is the supremum of $$\exp(F(x) - F(-1)$$ and $$\inf f_+ = 0.95..$$ is the infimum of $$\exp(F(x) - F(-1)$$. And both these values are achieved at $$x$$ such that $$G(x) = 0$$.

The analogue for $$f_-$$ works.

So the numbers from the OP can ( more or less) be given by these infinite series and hence the whole conjectures can be stated by these infinite series.

We also know for instance

$$\sum_{k=1}^{\infty} \frac{1}{k 2^k} = \ln(2) , \sum_{k=1}^{\infty} \frac{(-1)^k}{k 2^k} = - \ln(3/2)$$

So that is hopefull.

Also the max and min of functions can be given by contour integrals but that might not make things easier ?

Many trig identities and symmetry are probably related. But I see no clear proof.

So that is how we compute the values and it might just help.

Also notice :

$$t(x) = \ln(\sin(x) - 5/4) = \ln(-1) + \ln(\sin(-x) + 5/4) = \ln(-1) + \sum_{k=1}^{\infty} \frac{-2 \cos(k(x + \pi/2))}{2^k k}$$

So the other case is no mystery.

And ofcourse the case $$\ln(\cos^2(x) - 9/16)$$ is also simply related. ( trig addition identities can be used )

• How does vanishing of those integrals imply the $\sup/\inf$ of the products are finite? Feb 1, 2019 at 18:31
• Isn't $f_-(0) = -\frac{5}{4}$ negative? Similarly, what does $\log(\sin x - \tfrac{5}{4})$ mean? Feb 1, 2019 at 19:28
• The original question is math.stackexchange.com/questions/2075374/… Feb 1, 2019 at 22:04
• Have you tried to replace 5/4 with any other positive real number in $f_{-}$ and $f_{+}$? Feb 2, 2019 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, 2019 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 $$f_+(n)$$.

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.

• See my edit perhaps where I explain better how these values can be computed.
– mick
Apr 18, 2021 at 22:36
• @mick: What values? Apr 18, 2021 at 22:43
• the supremums and infimums !
– mick
Apr 18, 2021 at 22:44