5
$\begingroup$

After this post, I started wondering about possible approximations of the infinite product $$A_p=\prod _{k=p+1}^{\infty } \cos \left(\frac{p \,\pi}{2 k}\right)\tag 1$$ where $p$ is an integer. As far as I could see, there is no closed form expressions.

So, as I did in the linked question, I used Bhaskara I's approximation (for $-\frac \pi 2 \leq x\leq\frac \pi 2$) $$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\implies \cos\left(\frac{p\,\pi}{2k}\right)=\frac{4 \left(k^2-p^2\right)}{4 k^2+p^2}$$ and then computed $$B_p=\prod _{k=p+1}^{\infty }\frac{4 \left(k^2-p^2\right)}{4 k^2+p^2}=\frac{\Gamma \left( p+1-i\frac p 2\right)\,\, \Gamma \left( p+1+i\frac p 2\right)}{(2p)!}\tag 2$$ which does not seem to be very bad (see the table below).

Using Stirling approximation and Taylor series for supposed large values of $p$, I ended with $$\log(B_p)=\frac 12\log \left(\frac{5\pi}{4}\right)-\left(\log \left(\frac{16}{5}\right)+\cot ^{-1}(2)\right)p+\frac 12 \log(p)+\frac{11}{120 p}+O\left(\frac{1}{p^3}\right)\tag 3$$ What is interesting is to notice that $$\log \left(\frac{16}{5}\right)+\cot ^{-1}(2)\approx 1.62680$$ which is not so far from $\frac \pi 2$ ($3.56$% relative difference) and this already bring questions (at least, to me).

What is interesting is to see how close are the numbers in a logarithmic scale. $$\left( \begin{array}{cccc} p & (2) & (3) & (1) \\ 1 & -0.85190 & -0.85120 & -0.84448 \\ 2 & -2.17732 & -2.17725 & -2.16333 \\ 3 & -3.61661 & -3.61660 & -3.59727 \\ 4 & -5.10720 & -5.10719 & -5.08299 \\ 5 & -6.62701 & -6.62700 & -6.59815 \\ 6 & -8.16570 & -8.16570 & -8.13233 \\ 7 & -9.71760 & -9.71760 & -9.67979 \\ 8 & -11.2793 & -11.2793 & -11.2371 \\ 9 & -12.8485 & -12.8485 & -12.8019 \\ 10 & -14.4236 & -14.4236 & -14.3727 \\ 20 & -30.3496 & -30.3496 & -30.2557 \\ 30 & -46.4164 & -46.4164 & -46.2798 \\ 40 & -62.5413 & -62.5413 & -62.3621 \\ 50 & -78.6981 & -78.6981 & -78.4764 \end{array} \right)$$

Just out of curiosity, for $1 \leq p \leq 50$, I adjusted the parameters for the model $$\log(A_p)=a+b\,p+c\log(p)+\frac d p$$ and obtained a real good fit $$\begin{array}{clclclclc} \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval} \\ a & +0.69204 & 0.00006 & \{+0.69193,+0.69216\} \\ b & -1.62255 & 0.00000 & \{-1.62256,-1.62255\} \\ c & +0.50040 & 0.00003 & \{+0.50035,+0.50045\} \\ d & +0.08598 & 0.00007 & \{+0.08583,+0.08613\} \\ \end{array}$$ while is $(3)$, the coefficients are $(+0.68394,-1.62680,+0.50000,+0.09167)$ that is to say very very close.

For sure, all of this shows the high quality of the approximation of $\cos(x)$ but asks me questions about the value of $A_p$. If we truncate $(3)$ to $O\left(\frac{1}{p}\right)$,we should have $$A_p\sim \sqrt{\frac{5\pi p} 4 }\,e^{-\alpha \pi p}$$ with $\alpha \approx \frac 12$.

I wonder if we could find another better approximation of $A_p$ even at the price of more complex functions. Any idea would be welcome.

$\endgroup$
  • $\begingroup$ I haven't worked out all the details yet, but using this inequality and because $0<\frac{p \pi}{2k}<\frac{\pi}{2}$ for $k\geq p+1$, we have $$\prod\limits_{k=p+1}\cos{\frac{p\pi}{2k}}\leq \frac{1}{e^{\frac{\pi^2 \cdot p^2}{8}\sum\limits_{k=p+1}\frac{1}{k^2}}}= \frac{1}{e^{\frac{\pi^2 \cdot p^2}{8}\left(\zeta(2)-\sum\limits_{k=1}^p\frac{1}{k^2}\right)}}=$$ and $$\sum\limits_{n\leq x}\frac{1}{n^s}=\frac{x^{1-s}}{1-s}+\zeta(s)+O(x^{-s})$$ for $s>0, s\ne 1$. $\endgroup$ – rtybase Apr 22 at 11:42
  • $\begingroup$ ... which leads to $\approx \frac{1}{e^{\alpha \pi p}}$ rather than $\approx \frac{\sqrt{p}}{e^{\alpha \pi p}}$ $\endgroup$ – rtybase Apr 22 at 11:42
  • $\begingroup$ @rtybase. This is interesting ! Thanks for that. But how could I explain the highly significant coefficient $c$ from the quick and dirty regression ? Cheers :-) $\endgroup$ – Claude Leibovici Apr 22 at 12:16
  • $\begingroup$ @rtybase. Based on my own answer, using the asymptotics of the gamma functions, I effectively do not see how $\log(p)$ could appear and then agree with you. The problem is that, using $1 \leq p \leq 200$, this $\log(p)$ term seems to be crucial. Any idea ? Cheers :-) $\endgroup$ – Claude Leibovici Apr 24 at 8:51
  • $\begingroup$ That's an interesting finding, indeed. Nope, no idea, but it's inline with Taylor's expansion of $\frac{1}{2}$ you mentioned, I believe for small $p$'s (or function wise, for $x$ around $0$)? $\endgroup$ – rtybase Apr 24 at 21:57
1
$\begingroup$

Considering that Bhaskara I's approximation $$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad \text{for} \qquad -\frac \pi 2 \leq x\leq\frac \pi 2$$ looks like a Padé approximant, I had the feeling that I could either pure Padé approximants built at $x=0$, that is to say $$f_n=\frac {1+\sum_{m=1}^n a_m x^{2m}}{1+\sum_{m=1}^n b_m x^{2m}}$$ or, may be better, $$g_n=\left(1-\frac{2x}{\pi }\right) \left(1+\frac{2x}{ \pi }\right)\frac {1+\sum_{m=1}^{n-1} c_m x^{2m}}{1+\sum_{m=1}^n d_m x^{2m}}$$ the second rational fraction being the Padé approximant of $\frac{\cos (x)}{\left(1-\frac{2x}{ \pi }\right) \left(1+\frac{2x}{ \pi }\right)}$; $g_n$ was chosen as potentially better than $f_n$ because the calculations will start closer and closer to $\frac \pi 2$. To confirm the validity of this choice were computed $$I_n=\int_0^{\frac \pi 2} \big(\cos(x)-f_n\big)^2\,dx \qquad \text{and}\qquad J_n=\int_0^{\frac \pi 2} \big(\cos(x)-g_n\big)^2\,dx$$ The table below shows clearly the superiority of $g_n$. $$\left( \begin{array}{ccc} n & I_n & J_n \\ 1 & 7.086 \times 10^{-5} & 7.962 \times 10^{-6} \\ 2 & 6.754 \times 10^{-11} & 1.909 \times 10^{-12} \\ 3 & 2.375 \times 10^{-18} & 3.014 \times 10^{-20} \end{array} \right)$$ while $$\int_0^{\frac \pi 2} \Big(\cos(x)-\frac{\pi ^2-4x^2}{\pi ^2+x^2}\Big)^2\,dx =1.489\times 10^{-6}$$

Now, replacing $x$ by $\frac {p \pi}{2k}$ makes $$\cos \left(\frac{\pi p}{2 k}\right)=\frac {P_{m}(k^2)}{Q_{m}(k^2)}$$ where $P_m$ and $Q_m$ are homogeneous polynomials. $P_m$ shows only one real root and $Q_m$ only complex roots. By the end, limited to $n=2$, we can write $$\cos \left(\frac{\pi p}{2 k}\right)=\frac{(k-r_1 p)\,(k+r_1p)\,(k-r_2p)\,(k+r_2p)} {(k-s_1 p)\,(k+s_1p)\,(k-s_2p)\,(k+s_2 p) }$$ and if $$A_p=\prod _{k=p+1}^{\infty } \cos \left(\frac{p \,\pi}{2 k}\right)$$ then $$A_p\sim\frac{\Gamma [1+(1-s_1) p]\, \Gamma [1+(1+s_1) p]\, \Gamma [1+(1-s_2) p]\, \Gamma [1+(1+s_2) p]}{\Gamma [1+(1-r_1) p]\, \Gamma [1+(1+r_1) p]\,\Gamma [1+(1-r_2) p]\, \Gamma [1+(1+r_2) p]}$$

This leads to the following results (for the logarithms) $$\left( \begin{array}{cccc} p & \text{using } f_n & \text{using } g_n& \text{exact} \\ 1 & -0.844481 & -0.844481 & -0.844481 \\ 2 & -2.163326 & -2.163329 & -2.163327 \\ 3 & -3.597265 & -3.597275 & -3.597270 \\ 4 & -5.082976 & -5.082997 & -5.082989 \\ 5 & -6.598132 & -6.598168 & -6.598155 \\ 6 & -8.132292 & -8.132346 & -8.132328 \\ 7 & -9.679735 & -9.679810 & -9.679787 \\ 8 & -11.23699 & -11.23708 & -11.23706 \\ 9 & -12.80178 & -12.80190 & -12.80187 \\ 10 & -14.37255 & -14.37270 & -14.37266 \\ 20 & -30.25528 & -30.25578 & -30.25570 \\ 30 & -46.27897 & -46.27993 & -46.27979 \\ 40 & -62.36080 & -62.36228 & -62.36208 \\ 50 & -78.47457 & -78.47661 & -78.47636 \end{array} \right)$$ which looks much better than in my post.

$\endgroup$
1
$\begingroup$

As per the comments, but with references and links. Using this inequality and because $0<\frac{p \pi}{2k}<\frac{\pi}{2}$ for $k\geq p+1$, we have $$\color{red}{\prod\limits_{k=p+1}\cos{\frac{p\pi}{2k}}\leq} \frac{1}{e^{\frac{\pi^2 \cdot p^2}{8}\sum\limits_{k=p+1}\frac{1}{k^2}}}= \frac{1}{e^{\frac{\pi^2 \cdot p^2}{8}\left(\zeta(2)-\sum\limits_{k=1}^p\frac{1}{k^2}\right)}}=... \tag{1}$$ and (e.g. Apostol's "Introduction to Analytic Number Theory", page 55) $$\sum\limits_{n\leq x}\frac{1}{n^s}=\frac{x^{1-s}}{1-s}+\zeta(s)+O(x^{-s}) \tag{2}$$ for $s>0, s\ne 1$, leading to $$\sum\limits_{n\leq p}\frac{1}{n^2}=-\frac{1}{p}+\zeta(2)+O\left(\frac{1}{p^2}\right) \tag{3}$$ and, substituting in $(1)$ $$...=\frac{1}{e^{\frac{\pi^2 \cdot p^2}{8}\left(\frac{1}{p}-O\left(\frac{1}{p^2}\right)\right)}}= \frac{1}{e^{\frac{\pi^2 \cdot p}{8}-\frac{\pi^2 \cdot O(1)}{8}}}= \color{red}{\frac{e^{\frac{\pi^2}{8}\cdot O(1)}}{e^{\frac{\pi^2}{8}\cdot p}}}= \frac{O(1)}{e^{\frac{\pi^2}{8}\cdot p}}$$

Which leads to $\sim \frac{1}{e^{\alpha \pi p}}$ rather than $\sim \frac{\sqrt{p}}{e^{\alpha \pi p}}$.


However, this doesn't seem to be very important for large $p$, simply because (I will use $-b$ to have $b=1.62255>0$): $$A_p=e^{\ln{A_p}}= e^{a-bp+c\ln{p}+\frac{d}{p}}= e^{a}\cdot e^{\frac{d}{p}} \cdot \frac{p^c}{e^{bp}}= O(1)\cdot \frac{1}{e^{(b-c)p}}\cdot \frac{p^c}{e^{cp}}\leq ...$$ because $\frac{p^c}{e^{cp}}\rightarrow 0, p\rightarrow\infty$ $$...\leq \frac{O(1)}{e^{(b-c)p}}$$ But $\frac{\pi^2}{8}> b-c$, thus, in a way $$\frac{O(1)}{e^{\frac{\pi^2}{8}\cdot p}}<\frac{O(1)}{e^{(b-c)p}}$$ from some large enough $p$ onwards.

$\endgroup$
  • 1
    $\begingroup$ Thanks for your answer ! $\endgroup$ – Claude Leibovici Apr 27 at 12:22
1
$\begingroup$

One can recognize a Riemann sum. The limit $$ A:=\color{blue}{\lim_{p\to\infty}\frac{\ln A_p}{p}}=\lim_{p\to\infty}\frac{1}{p}\sum_{k=p+1}^{\infty}\ln\cos\frac{p\pi}{2k}=\color{blue}{\int_{1}^{\infty}\ln\cos\frac{\pi}{2x}\,dx} $$ exists. Numerically, $A=-1.62254367352281126916452953\cdots$

The Euler-Maclaurin approach gives more details. For the first order, \begin{align} \ln A_p&=\left[\int_{p}^{\infty}-\int_{p}^{p+1}\right]\ln\cos\frac{p\pi}{2x}\,dx+\frac{1}{2}\ln\cos\frac{p\pi}{2(p+1)}+R_p \\&=Ap+\frac{\ln p}{2}-\int_{0}^{1}f_p(x)\,dx+\frac{1}{2}f_p(1)+R_p, \end{align} where $f_p(x)=\ln\left(p\cos\dfrac{p\pi}{2(p+x)}\right)\underset{p\to\infty}{\longrightarrow}\ln\dfrac{x\pi}{2}$, and \begin{align} R_p&=\frac{p\pi}{2}\int_{p+1}^{\infty}\left(\{x\}-\frac{1}{2}\right)\tan\frac{p\pi}{2x}\,\frac{dx}{x^2}&&\color{gray}{[x=n+(1-y)/2,\ -1\leqslant y\leqslant 1]} \\&=-\frac{p\pi}{2}\sum_{n=1}^{\infty}\int_{-1}^{1}yg_{p,n}(y)\,dy&&\color{gray}{\left[g_{p,n}(y)=\frac{\tan\dfrac{p\pi}{2(p+n)+1-y}}{\big(2(p+n)+1-y\big)^2}\right]} \\&=-\frac{p\pi}{2}\sum_{n=1}^{\infty}\int_{0}^{1}x\big(g_{p,n}(x)-g_{p,n}(-x)\big)\,dx&&\underset{p\to\infty}{\longrightarrow}\quad -R, \end{align} where (terms are nonnegative, so we can take the limit termwise) $$R=\int_{0}^{1}\sum_{n=1}^{\infty}\frac{x^2}{(2n+1)^2-x^2}\,dx=\int_{0}^{1}x\left(\frac{\pi}{4}\tan\frac{x\pi}{2}-\frac{x}{1-x^2}\right)\,dx.$$ Omitting the computations, $R=1-\dfrac{\ln 2\pi}{2}$, and finally $$\ln A_p=Ap+\dfrac{1}{2}\ln 4p+o(1).$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.