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 November 8th, 2016, 07:00 AM #1 Newbie   Joined: Nov 2016 From: Spain Posts: 1 Thanks: 0 Limit involving derivative of zeta function Hello. I would like to know what is the value of the derivative of 1/ (zeta(x)) as x tends to 1: $\displaystyle \displaystyle \lim_{x \to 1} \frac{d\frac{1}{\zeta(x)}}{dx}$ As you get an indetermination of the type $\displaystyle 0\infty$, what I have tried so far is to apply L'Hôpital's rule, but with no success. Thank you. November 8th, 2016, 08:40 AM #2 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 What is your definition of the zeta function? Can't find it. November 9th, 2016, 06:47 AM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 $\displaystyle \zeta(p)=\sum_{n=1}^{\infty}\frac{1}{n^{p}}=\sum_{ n=1}^{\infty}(\frac{1}{n})^p$, p>1 $\displaystyle \frac{\mathrm{d} }{\mathrm{d} p}\zeta(p)=\sum_{n=1}^{\infty}\log(\frac{1}{n}) \frac{1}{n^{p}}=-\sum_{n=1}^{\infty}\log(n)\frac{1}{n^{p}}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} p}\frac{1}{\zeta(p))}= -\frac{1}{\zeta(p)^{2}}\frac{\mathrm{d} }{\mathrm{d} p}\zeta(p)=\frac{1}{\zeta(p)^{2}} \sum_{n=1}^{\infty}\log(n)\frac{1}{n^{p}}$ $\displaystyle \lim_{p\rightarrow 1}\frac{\mathrm{d} }{\mathrm{d} p}\frac{1}{\zeta(p))}=0\lim_{p\rightarrow 1}\sum_{n=1}^{\infty}\log(n)\frac{1}{n^{p}}=?$ Ref $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}a^x=(\log a)a^x$ Thanks from topsquark Last edited by skipjack; November 9th, 2016 at 11:37 PM. November 9th, 2016, 11:18 AM #4 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Ref previous post. $\displaystyle \frac{\mathrm{d} }{\mathrm{d} p}\frac{1}{\zeta(p))}= -\frac{1}{\zeta(p)^{2}}\frac{\mathrm{d} }{\mathrm{d} p}\zeta(p)=\frac{1}{\zeta(p)^{2}} \sum_{n=1}^{\infty} \log(n)\frac{1}{n^{p}}$ $\displaystyle \lim_{p\rightarrow 1}\frac{\mathrm{d} }{\mathrm{d} p}\frac{1}{\zeta(p))}= \lim_{p\rightarrow 1}\sum_{n=1}^{\infty}\frac{\log(n)\frac{1}{n^{p}}} {\zeta(p)^{2}}= 0$ $\displaystyle \lim_{p\rightarrow 1}\zeta(p)=\infty$ Last edited by skipjack; November 9th, 2016 at 11:45 PM. November 9th, 2016, 10:09 PM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Previous post incorrect. Limit is indeterminate. Situation is similar to $\displaystyle \lim_{k\rightarrow \infty}\frac{1}{k} \sum_{n=1}^{\infty}\frac{1}{n}$ L'Hôpitals rule, differentiating with respect to p, doesn't help. Last edited by skipjack; November 9th, 2016 at 11:47 PM. November 9th, 2016, 11:49 PM #6 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 $\displaystyle \lim_{x \to 1} \frac{d\frac{1}{\zeta(x)}}{dx} = 1$ November 10th, 2016, 06:45 AM   #7
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Quote:
 Originally Posted by skipjack $\displaystyle \lim_{x \to 1} \frac{d\frac{1}{\zeta(x)}}{dx} = 1$
Thanks. How'd you get it? November 10th, 2016, 11:44 AM #8 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 It's easily deduced from this expansion. November 10th, 2016, 03:49 PM #9 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 How? Thanks from topsquark November 14th, 2016, 09:59 AM #10 Global Moderator   Joined: Dec 2006 Posts: 21,026 Thanks: 2257 For x close enough to 1, it gives 1/ζ(x) as being approximately x - 1, which has derivative 1. Thanks from topsquark Tags derivative, function, infinity, involving, limit, zeta, zeta function 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 Adam Ledger Number Theory 0 April 14th, 2016 05:15 PM Dougy Number Theory 1 October 11th, 2012 06:53 PM mathbalarka Complex Analysis 12 September 21st, 2012 08:10 AM AFireInside Calculus 4 August 31st, 2011 10:24 PM Ziaris Real Analysis 0 February 18th, 2009 07:47 PM

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