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 we have

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

Why is that so ?


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

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

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.


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 :



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 :

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


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

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 )


I wanted to make as little conjectures as possible when I posted this. Just one question/conjecture per post is the usual rule. But many related conjectures exist. Many might turn out to be equivalent or have already been shown to be equivalent. (such as the analogue cosine cases with $\frac{3}{4}$)

I thought it would be best not to flood with related conjectures and post the most important one. Which I did. But one comment of Richard can not be ignored.

Richard Stanley wrote

$$ \sup f_-(n)^2 - \inf f_+(n)^2 = \frac{4}{5} $$

He might not be the first to notice this, even if I exclude my mentor who came up with the main conjecture and related ones. (I think it was also mentioned on chat and comments on MSE)

Anyways it is also a nice conjecture.

Is it equivalent ? No.

Lets use shorthand notations : $x = \sup f_-(n), y = \inf f_+$.

Then combining both conjectures ( the original and the "Richard" variant) we get

$$x y = \frac{5}{4} , x^2 - y^2 = (x-y)(x+y) = \frac{4}{5}$$

Now this implies we can compute the value of $x$ (or $y$).

We get an interesting situation here.

If 2 conjectures are true than so is the third :

$$x y = \frac{5}{4}$$

$$ x^2 - y^2 = (x-y)(x+y) = \frac{4}{5} $$

$x$ is the positive real solution to $5 * 4^2 x^4 - 4^3 x^2 - 5^3 $ or $80 x^4 - 64 x^2 - 125$. NOTICE the 5's and 4's all over again.

(or closed form for $x = \frac{1}{2} \sqrt {\frac{8 + \sqrt {689}}{5}}$)

($689 = 13*53$ if that matters to anyone)

This was all known in the $90's$ by my mentor.

I call it the Raes-Stanley conjecture.

Update !

I had a talk with my mentor why he did not mention the

$$ \sup f_-(n)^2 - \inf f_+(n)^2 = \frac{4}{5} $$

part of the Raes-Stanley conjecture.

Although he noticed the apparant identity, he does not actually believe that last part.

He said that the value converges fast for $n,m > 210000$ to

$$ \sup f_-(n)^2 - \inf f_+(n)^2 = \frac{3.999789007..}{5} $$

In fact increasing $n,m$ from $210000$ to $314314$ both gives

$$ \sup f_-(n)^2 - \inf f_+(n)^2 = \frac{3.999789007..}{5} $$

so barely any noticeable change, while

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

seems already heuristically confirmed.

Testing for $n,m$ smaller than $100000$ might give the wrong impression and might be the cause of the mistake.

Numerical coincidence might then lead to wrong conclusions.

Roundoff errors might create an effect in the computations but he does not believe the "Stanley part "

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

can be saved by those errors.

If anyone can prove or argue or compute a higher value than $\frac{3.999789007..}{5}$ , please inform me.

Finally the Full Raes conjecture is the slight generalization :

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

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


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

Why is that so ?

$|*|$ means the absolute value here.


$$\inf |f_- (n)| = 0.157559...$$ $$\sup f_- (n) = 1.308592...$$

$$\inf f_+ (n) = 0.955225...$$ $$\sup f_+ (n) = 7.933553...$$

$$0.955225... * 1.308592... = 1.25 = \frac{5}{4}$$

$$0.157559... * 7.933553... = 1.25 = \frac{5}{4}$$

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

And let

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

It appears that

$$\sup g_- (n) \inf g_+ (m) = \frac{3}{4} $$


$$\inf |g_- (n)| \sup g_+ (m) = \frac{3}{4} $$

  • 1
    $\begingroup$ Isn't $f_-(0) = -\frac{5}{4}$ negative? Similarly, what does $\log(\sin x - \tfrac{5}{4})$ mean? $\endgroup$ Commented Feb 1, 2019 at 19:28
  • 3
    $\begingroup$ The original question is math.stackexchange.com/questions/2075374/… $\endgroup$ Commented Feb 1, 2019 at 22:04
  • 1
    $\begingroup$ Have you tried to replace 5/4 with any other positive real number in $ f_{-} $ and $ f_{+} $? $\endgroup$ Commented Feb 2, 2019 at 0:25
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    $\begingroup$ I find the OP's question quite entertaining. Immediate generalization: for $a\ge1$ (perhaps also more generally), replace $\sin(i)\pm5/4$ by $\sin(i)/a\pm(4a^2+1)/(4a^2)$. Once again, the corresponding integral of log vanishes, and the conjecture seems to be valid for any $a$. Also note that $\sin^2(i)+9/16=(17/16-\cos(2i)/2)$, so this essentially corresponds to $a=2$. $\endgroup$ Commented Feb 3, 2019 at 17:46
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    $\begingroup$ Very speculative, but could it be true that $\sup f_{-}(n)^2-\inf f_{+}(m)^2=4/5$? $\endgroup$ Commented Jan 27 at 19:52

2 Answers 2


Taking logarithms lets us write $$\ln(\sup f_-(n)\inf f_+(m))=\sup\ln f_-(n)+\inf\ln f_+(m).$$ (We will assume that $n$ is odd so that $f_-(n)>0$). Let's begin by computing $\ln f_+(m)$. The Fourier expansion $$\ln(\sin(x)+5/4)=\sum_{k=1}^\infty\frac{-2\cos(k(x+\pi/2))}{2^kk}$$ let us write $\ln f_+(m)$ as \begin{align*} \ln f_+(m)&=\ln\prod_{j=0}^m(\sin(j)+5/4)\\ &=\sum_{j=0}^m\ln(\sin(j)+5/4)\\ &=\sum_{j=0}^m\sum_{k=1}^\infty\frac{-2\cos(k(j+\pi/2))}{2^kk}\\ &=-\sum_{k=1}^\infty\frac{1}{2^kk}\sum_{j=0}^m2\cos(k(j+\pi/2))\\ &=-\sum_{k=1}^\infty\frac{1}{2^kk}\sum_{j=0}^me^{ik(j+\pi/2)}+e^{-ik(j+\pi/2)}\\ &=-\sum_{k=1}^\infty\frac{1}{2^kk}\left[e^{ik\pi/2}\frac{1-e^{ik(m+1)}}{1-e^{ik}}+e^{-ik\pi/2}\frac{1-e^{-ik(m+1)}}{1-e^{-ik}}\right]\\ &=-\sum_{k=1}^\infty\frac{1}{2^kk}\left[e^{ik\pi/2}\frac{e^{-ik/2}-e^{ik(m+\frac{1}{2})}}{e^{-ik/2}-e^{ik/2}}+e^{-ik\pi/2}\frac{e^{ik/2}-e^{-ik(m+\frac{1}{2})}}{e^{ik/2}-e^{-ik/2}}\right]\\ &=\sum_{k=1}^\infty\frac{e^{ik\pi/2}(e^{-ik/2}-e^{ik(m+\frac{1}{2})})-e^{-ik\pi/2}(e^{ik/2}-e^{-ik(m+\frac{1}{2})})}{2^kk(e^{ik/2}-e^{-ik/2})}\\ &=\sum_{k=1}^\infty\frac{e^{ik\pi/2}e^{-ik/2}-e^{-ik\pi/2}e^{ik/2}}{2^kk(e^{ik/2}-e^{-ik/2})}-\sum_{k=1}^\infty\frac{e^{ik\pi/2}e^{ik(m+\frac{1}{2})}-e^{-ik\pi/2}e^{-ik(m+\frac{1}{2})}}{2^kk(e^{ik/2}-e^{-ik/2})}\\ &=C_+-h_+(m+\tfrac{1}{2}). \end{align*} Note that this is a continuous function of $m$ with period $2\pi$. Similarly, \begin{align*} \ln f_-(n)&=\ln\prod_{j=0}^n(\sin(j)-5/4)\\ &=\ln\prod_{j=0}^n(\sin(j-\pi)+5/4)\\ &=\sum_{j=0}^n\ln(\sin(j-\pi)+5/4)\\ &=\sum_{j=0}^n\sum_{k=1}^\infty\frac{-2\cos(k(j-\pi/2))}{2^kk}\\ &=[\text{the same computation as for $f_+$ but with all of the $\pi$'s negated}]\\ &=\sum_{k=1}^\infty\frac{e^{-ik\pi/2}e^{-ik/2}-e^{ik\pi/2}e^{ik/2}}{2^kk(e^{ik/2}-e^{-ik/2})}-\sum_{k=1}^\infty\frac{e^{-ik\pi/2}e^{ik(n+\frac{1}{2})}-e^{ik\pi/2}e^{-ik(n+\frac{1}{2})}}{2^kk(e^{ik/2}-e^{-ik/2})}\\ &=C_--h_-(n-\tfrac{1}{2}). \end{align*} That was a bit of a mouthful, but we are very close. We have managed to write \begin{align*} \ln(\sup f_-(n)\inf f_+(m))&=\sup(C_--h_-(n+\tfrac{1}{2}))+\inf(C_+-h_+(m+\tfrac{1}{2}))\\ &=C_-+C_++\inf h_-(x)+\sup h_+(x). \end{align*} The key observation is that $h_+(-x)=-h_-(x)$ (or vice versa), so that $\inf h_-(x)+\sup h_+(x)=0$. Finally, \begin{align*} C_-+C_+&=\sum_{k=1}^\infty\frac{e^{-ik\pi/2}e^{-ik/2}-e^{ik\pi/2}e^{ik/2}+e^{ik\pi/2}e^{-ik/2}-e^{-ik\pi/2}e^{ik/2}}{2^kk(e^{ik/2}-e^{-ik/2})}\\ &=\sum_{k=1}^\infty\frac{-(e^{-ik\pi/2}+e^{ik\pi/2})}{2^kk}\\ &=\sum_{k=1}^\infty\frac{-(e^{-i(2k)\pi/2}+e^{i(2k)\pi/2})}{2^{2k}(2k)}\\ &=\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k}\left(\frac{1}{4}\right)^k\\ &=\ln\left(\frac{1}{4}+1\right)=\ln\left(\frac{5}{4}\right). \end{align*} Thus, $\sup f_-(n)\inf f_+(m)=\tfrac{5}{4}$.

The cosine version can be proved in a similar way. We can write \begin{align*} \ln g_+(m)&=\ln\prod_{j=0}^m(\cos(j)+5/4)\\ &=\sum_{j=0}^m\ln(\sin(j+\pi/2)+5/4)\\ &=\sum_{j=0}^m\sum_{k=1}^\infty\frac{-2\cos(k(j+\pi))}{2^kk}\\ &=[\text{the same computation as for $f_+$ but with all of the $\pi/2$'s replaced by $\pi$'s}]\\ &=\sum_{k=1}^\infty\frac{e^{ik\pi}e^{-ik/2}-e^{-ik\pi}e^{ik/2}}{2^kk(e^{ik/2}-e^{-ik/2})}-\sum_{k=1}^\infty\frac{e^{ik\pi}e^{ik(m+\frac{1}{2})}-e^{-ik\pi}e^{-ik(m+\frac{1}{2})}}{2^kk(e^{ik/2}-e^{-ik/2})}\\ &=C_+-h_+(m+\tfrac{1}{2}) \end{align*} and \begin{align*} \ln g_-(n)&=\ln\prod_{j=0}^n(\cos(j)-5/4)\\ &=\ln\prod_{j=0}^n(\sin(j-\pi/2)+5/4)\\ &=\sum_{j=0}^n\ln(\sin(j-\pi/2)+5/4)\\ &=\sum_{j=0}^n\sum_{k=1}^\infty\frac{-2\cos(k(j+0))}{2^kk}\\ &=[\text{the same computation as for $f_+$ but with all of the $\pi/2$'s replaced by $0$'s}]\\ &=\sum_{k=1}^\infty\frac{e^{-ik/2}-e^{ik/2}}{2^kk(e^{ik/2}-e^{-ik/2})}-\sum_{k=1}^\infty\frac{e^{ik(n+\frac{1}{2})}-e^{-ik(n+\frac{1}{2})}}{2^kk(e^{ik/2}-e^{-ik/2})}\\ &=C_--h_-(n+\tfrac{1}{2}). \end{align*} This time we have $h_-(-x)=-h_-(x)$ and $h_-(x+\pi)=h_+(x)$, so that $\inf h_-(x)+\sup h_+(x)=0$. And we can compute $$C_-+C_+=\sum_{k=1}^\infty\frac{-1}{2^kk}+\sum_{k=1}^\infty\frac{(-1)^{k+1}}{2^kk}=\ln\left(\frac{1}{2}\right)+\ln\left(\frac{1}{2}+1\right)=\ln\left(\frac{3}{4}\right).$$ Thus, $\sup g_-(n)\inf g_+(m)=\tfrac{3}{4}$.

  • $\begingroup$ Nice ! Could you do the cos part too ? $\endgroup$
    – mick
    Commented Mar 23 at 12:03
  • 1
    $\begingroup$ @mick Yes, done. $\endgroup$ Commented Mar 23 at 18:36
  • $\begingroup$ I upvoted. I wait for accepting and giving the bounty because there might be other good answers. But you are close to getting the accepted answer :) $\endgroup$
    – mick
    Commented Mar 24 at 12:48
  • 1
    $\begingroup$ A similar way was also my attempt for solving this, and it's good to see it coming to a result (not having checked all, though). For the sake of clarity, two things might probably be worth mentioning: (1) The series defining $C_\pm$ and $h_\pm(x)$ above are absolutely convergent (exponentially decreasing), because $\pi$ has a finite irrationality measure (en.wikipedia.org/wiki/Liouville_number), thus $1/|e^{ik/2}-e^{-ik/2}|$ has an upper bound, polynomial in $k$. (2) The (odd) positive integers mod $2\pi$ are dense in $\mathbb{R}\mod2\pi$, hence sup and inf can be taken over the reals. $\endgroup$ Commented Mar 25 at 0:28

(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.

  • $\begingroup$ See my edit perhaps where I explain better how these values can be computed. $\endgroup$
    – mick
    Commented Apr 18, 2021 at 22:36
  • $\begingroup$ @mick: What values? $\endgroup$ Commented Apr 18, 2021 at 22:43
  • $\begingroup$ the supremums and infimums ! $\endgroup$
    – mick
    Commented Apr 18, 2021 at 22:44
  • $\begingroup$ My mentor came up with the idea around $1990$ so I wonder if there are earlier references ? Also if anyone has read these papers please inform me. $\endgroup$
    – mick
    Commented Jan 28 at 17:50
  • $\begingroup$ See here for the second paper mentioned by Mateusz : sciencedirect.com/journal/journal-of-number-theory/vol/123/… what should look like ; pdf.sciencedirectassets.com/272482.... $\endgroup$
    – mick
    Commented Mar 12 at 23:39

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