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 Calculus Calculus Math Forum

 November 9th, 2017, 01:29 AM #1 Senior Member   Joined: Dec 2015 From: iPhone Posts: 447 Thanks: 69 Limit with sequence How to find this limit? $\displaystyle \lim _{x\rightarrow 0}x^{-1}(1-\prod \limits _{j=1}^{n}\sqrt[j]{\cos jx}) =$ $\displaystyle \lim _{x\rightarrow 0} \frac{1-\cos x \cdot \sqrt{\cos 2x}\cdot \sqrt{\cos 3x}\cdot ...\cdot \sqrt[n]{\cos nx}}{x}=\,?$ Last edited by skipjack; November 9th, 2017 at 02:34 AM. November 9th, 2017, 02:34 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,390 Thanks: 2015 What is the limit for n = 1? Can you find the limit for n = 2? Thanks from topsquark and idontknow November 9th, 2017, 04:31 AM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,623 Thanks: 2611 Math Focus: Mainly analysis and algebra I get a limit of zero using Taylor expansions. \begin{align*} \left. \begin{aligned} \cos{(x)} &= 1 - x^2 + \mathcal{O}(x^2) &\implies&& \cos{(nx)} &= 1 - (nx)^2 + \mathcal{O}(x^2) \\[8pt] &&&&\sqrt[n]{1 + x} &= 1 + \tfrac1nx + \mathcal{O}(x) \end{aligned} \right\} \implies \sqrt[n]{\cos{(nx)}} &= 1 - \tfrac1n(nx)^2 + \mathcal{O}(x^2) \\ &= 1 - nx^2 + \mathcal{O}(x^2) \end{align*} Thus the product \begin{align*} \prod_{n=1}^\infty \sqrt[n]{\cos{(nx)}} &= \prod_{n=1}^\infty \left(1 - nx^2 + \mathcal{O}(x^2) \right) \\ &= \left(1 - x^2 + \mathcal{O}(x^2) \right)\left(1 - 2x^2 + \mathcal{O}(x^2) \right)\left(1 - 3x^2 + \mathcal{O}(x^2) \right)\ldots \\ &= 1 + \mathcal{O}(x) & \text{(for some constant $a$)} \end{align*} by expanding the product term by term. Notice that, at this point we are assuming that the product has a finite value$^*$. We can motivate this by pointing out that for $x \lt \frac\pi2$, every term in the product is between $0$ and $1$, and thus the product must also be between $0$ and $1$. Now we can put this into the limit expression \begin{align*} \lim_{x \to 0} \frac{1 - \prod_{n=1}^\infty \sqrt[n]{\cos{(nt)}}}{x} &= \lim_{x \to 0} \frac{1 - \big(1 + \mathcal{O}(x)\big)}{x} \\ &= \lim_{x \to 0} \frac{ \mathcal{O}(x)}{x} \\ &= 0 \end{align*} https://www.desmos.com/calculator/b4rtdppucm $^*$ Note that the coefficient of $x^2$ is actually $\left(-\sum n\right)$ over the natural numbers. I assume that this leads to use of $\sum n = -\frac1{12}$. Thanks from topsquark and idontknow Last edited by v8archie; November 9th, 2017 at 05:19 AM. November 9th, 2017, 09:17 AM #4 Senior Member   Joined: Dec 2015 From: iPhone Posts: 447 Thanks: 69 $\displaystyle \lim_{x \to 0} \frac{1 - \prod_{n=1}^\infty \sqrt[n]{\cos{(nt)}}}{x}=-\lim_{x \to 0} \frac{ e^{\ln \prod_{n=1}^\infty \sqrt[n]{\cos{(nt)}}}-1}{x}=-\lim_{x \to 0}\sum _{n=1}^{\infty}(nx)^{-1}\ln cos(nt)=0$ November 9th, 2017, 09:50 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,623 Thanks: 2611 Math Focus: Mainly analysis and algebra I'm not convinced at all by that second equals sign. Tags limit, sequence Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Calculus 5 August 3rd, 2012 03:01 AM nappysnake Calculus 6 November 27th, 2011 09:30 AM vasukinv Calculus 7 April 2nd, 2011 12:21 AM everettjsj2 Calculus 1 February 26th, 2010 02:56 AM rhta Calculus 3 October 25th, 2009 10:08 AM

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