Interesting combo.

For 1ωn=k=1ncot2(kπ2n+1)(k=02ncot(x+kπ2n+1))\frac 1{\omega_n} =\prod_{k=1}^{n}\cot^{2}\left(\dfrac{k\pi}{2n+1}\right)\left(\sum_{k=0}^{2n}\cot\left(x+\frac{k\pi}{2n+1}\right)\right)

FindΩ=limn(limxπ4n+2(tanxtan(π4n+2))ωn)\Omega =\lim_{n\to\infty}\left(\lim_{x\to\frac{\pi}{4n+2}}\left(\dfrac{\tan x}{\tan\left(\frac{\pi}{4n+2}\right)}\right)^{\omega_n}\right) This is the proposal by teachers Marian Ursărescu and Florică Anastase, Romania and published by Romanian Mathematical Magazine.

Initially the answer of proposers agreed with me but later we had a slight different conclusions. As the per the proposers they arrived to 1e4π\frac{1}{\sqrt[\pi]{e^4}}.

If the answer of proposers is correct then probably I think I had done mistakes while evaluating the limits. I even tried on wolfram alpha it agrees with me ( doesn't provide the closed form but it approximates to my closed form) . Can anybody help me to correct my mistakes ? Thank you ! :)


Solution Here i wish to share my solution.

Firstly we recall Euler's formulaeimx=(cosmx+isinmx)=(cosx+isinx)me^{imx} = (\cos mx + i\sin mx)= (\cos x+ i\sin x)^m and further solving gives cosmx+isinmxsinmx=\frac{\cos mx +i\sin mx}{\sin^m x}=r=0m(mr)cotmrxir=r=0m2+l(1)r(m2r)cotm2rx+ir=1m2(m2r1)(1)rcotm2r+1x \displaystyle \sum_{r=0}^m { m\choose r} \cot^{m-r}x i^{r}= \displaystyle \sum_{r=0}^{\lceil \frac{m}{2}\rceil+ l }(-1)^{r}{ m\choose 2r} \cot^{m-2r}x+i\sum_{r=1}^{\lceil \frac{m}{2}\rceil}{ m\choose 2r-1}(-1)^{r} \cot^{m-2r+1}x where l=1,0l=1,0 if m m is even and odd respectively. We note that latter sum is the imaginary part and hence equating we get r=1m2(m2r1)cotm2r+1x=sinmxsinmx \sum_{r=1}^{\lceil \frac{m}{2}\rceil}{ m\choose 2r-1} \cot^{m-2r+1}x=\frac{\sin mx}{\sin^{m}x} on setting m=2n+1,  x=kπmm= 2n+1, \; x= \frac{k\pi}{m} and further expansion we get (2n+11)cot2nkπ2n+1(2n+13)cot2n2kπ2n+1++(2n+12n+1)=0 { 2n+1 \choose 1} \cot^{2n} \frac{k\pi}{2n+1} - { 2n+1 \choose 3} \cot^{2n-2}\frac{k\pi}{2n+1} +\cdots + { 2n+1 \choose 2n+1} =0 as the polynomial equation is of even degree and hence by Vieta's formula we have k=1ncot2(kπ2n+1)=(1)2n(2n+12n+1)(2n+11)=12n+1\prod_{k=1}^{n} \cot^2\left(\frac{k\pi}{2n+1}\right)=(-1)^{2n}\frac{{ 2n+1\choose 2n+1}}{{ 2n+1\choose 1}}=\frac{1}{2n+1} secondly we use the identity k=0n1sin(x+kπn)=sinnx2n1\displaystyle\prod_{k=0}^{n-1} \sin\left(x+\frac{k\pi}{n}\right) =\frac{\sin nx}{2^{n-1} } , taking log\log on both side and on differentiating with respect to xx, we get that, ie ddxlog(k=0n1sin(x+kπn))=ddxlog(sinnx2n1)k=0n1cot(x+kπn)=ncot(nx)\frac{d}{dx}\log\left(\displaystyle\prod_{k=0}^{n-1} \sin\left(x+\frac{k\pi}{n}\right)\right) =\frac{d}{dx}\log\left(\frac{\sin nx}{2^{n-1}}\right)\\ \Rightarrow \displaystyle \sum_{k=0}^{n-1} \cot\left(x+\frac{k\pi}{n}\right) =n\cot(nx). Replacing nn by 2n+12n+1 we yield k=02ncot(x+kπ2n+1)=(2n+1)cot((2n+1)x)\displaystyle \sum_{k=0}^{2n} \cot\left(x+\frac{k\pi}{2n+1}\right) =(2n+1)\cot((2n+1)x)and we deduce that ωn=((2n+1)cot((2n+1)x)2n+1)1=tan((2n+1)x)\omega_n = \left(\frac{(2n+1)\cot ((2n+1)x)}{2n+1}\right)^{-1} = \tan ((2n+1)x).

Call L(n)=limxπ4n+2(tanxtanβ)ωn L(n)= \displaystyle\lim_{x\to\frac{\pi}{4n+2}}\left(\frac{\tan x}{\tan \beta} \right)^{\omega_n} . As we can observe that we have 11^{\infty} limit form. Since as soon as x14n+2x\to\frac{1}{4n+2},function wnw_n\to\infty and tanxtanβ1\frac{\tan x}{\tan\beta}\to 1 with β=π4n+2\beta =\frac{\pi}{4n+2}. So we can either make the direct use the formula for 11^{\infty} or without of it too. ie

limxβ(tanxtanβ)ωn=limxβ(1+tanxtanβ1)ωn=limxβ(1+1tanxtanβ1)ωn=exp(limxβ(tanxtanβ1)tan((2n+1)x))=exp(limxβ(tanxtanβtanβcot((2n+1)x)))\displaystyle\lim_{x\to\beta}\left(\frac{\tan x}{\tan \beta}\right)^{\omega_n}=\displaystyle\lim_{x\to\beta}\left(1+\frac{\tan x}{\tan \beta}-1\right)^{\omega_n}= \displaystyle\lim_{x\to\beta}\left(1+\frac{1}{\frac{\tan x}{\tan \beta}-1}\right)^{\omega_n} \\= \exp\left(\displaystyle\lim_{x\to\beta }\left(\frac{\tan x}{\tan \beta }-1\right)\tan((2n+1)x)\right)=\exp\left(\displaystyle\lim_{x\to\beta }\left(\frac{\tan x-\tan\beta}{\tan\beta \cot ((2n+1)x)}\right)\right) . Here we have the limit of the form 00\frac{0}{0} so we use L-hopital's rule to get exp(limxβsec2x(2n+1)csc2((2n+1)x)tanβ)=exp(secβsinβ(2n+1))\exp\left(\displaystyle \lim_{x\to\beta}\frac{\sec^2 x}{-(2n+1)\csc^2((2n+1)x)\tan\beta}\right)=\exp\left(\frac{\sec \beta}{-\sin\beta (2n+1)}\right) . And therefore, limnL(n)=exp(limn2ππ(4n+2)tanπ4n+2)=exp(2πlimnπ(4n+2)tanπ4n+2)=e2π=1e2π\lim_{n\to\infty} L(n)=\exp\left(-\displaystyle \lim_{n\to\infty} \frac{\frac{2\pi}{\pi(4n+2)}}{\tan \frac{\pi}{4n+2}}\right)=\exp\left(-\frac{2}{\pi}\displaystyle \lim_{n\to\infty} \frac{\frac{\pi}{(4n+2)}}{\tan \frac{\pi}{4n+2}}\right) =e^{-\frac{2}{\pi}}=\frac{1}{\sqrt[\pi]{e^{2}}}

Note by Naren Bhandari
1 month ago

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@Mark Hennings Sir, can you help me?
Thank you !! :)

Naren Bhandari - 4 weeks, 1 day ago

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I tend to agree with you The inner limit is limxβ(tanxtanβ)tan(πx2β)  =  limy0(tan(βy)tanβ)cot(πy2β)  =  limy0F(y) \lim_{x \to \beta} \left(\frac{\tan x}{\tan \beta}\right)^{\tan\big(\tfrac{\pi x}{2\beta}\big)} \; = \; \lim_{y \to 0}\left(\frac{\tan(\beta-y)}{\tan\beta}\right)^{\cot\big(\tfrac{\pi y}{2\beta}\big)} \; = \; \lim_{y \to 0}F(y) where β=π4n+2\beta = \tfrac{\pi}{4n+2}, so that limy0lnF(y)=  limy0ln(tan(βy)tanβ)tanπy2β  =  limy0sec2ytanβtanytanβsec2y1+tanβtanyπ2βsec2πy2β=  2βπ(tanβ+cotβ)  =  4βπsin2β\begin{aligned} \lim_{y \to 0}\ln F(y) & = \; \lim_{y \to0}\frac{\ln\left(\frac{\tan(\beta-y)}{\tan\beta}\right)}{\tan \tfrac{\pi y}{2\beta}} \; = \; \lim_{y \to 0} \frac{-\frac{\sec^2y}{\tan \beta - \tan y} - \frac{\tan\beta \sec^2y}{1 + \tan\beta\tan y}}{\frac{\pi}{2\beta}\sec^2\tfrac{\pi y}{2\beta}} \\ & = \; -\frac{2\beta}{\pi}(\tan\beta + \cot\beta) \; = \; - \frac{4\beta}{\pi \sin2\beta} \end{aligned} which in turn tends to 2π-\tfrac{2}{\pi} as nn \to \infty, so that β0\beta \to 0. Exponentiating gives the limit 1e2π\frac{1}{\sqrt[\pi]{e^2}}.

I have not looked at the original question. You have transcribed it correctly?

Mark Hennings - 4 weeks ago

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Yes, I have it correctly, Sir. I don't know how to attach the photo of it here in comment section so I'm here providing the link of it Original problem as I got this problem on Facebook wall of proposer. Please do check once. Thank your very much for your response. :)

Naren Bhandari - 3 weeks, 6 days ago

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