 My Math Forum Limit involving derivative of zeta function

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November 15th, 2016, 09:40 AM   #11
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Quote:
 Originally Posted by skipjack For x close enough to 1, it gives 1/ζ(x) as being approximately x - 1, which has derivative 1.
Gross oversimplification.

Suppose:
f(x)=1/x+e$\displaystyle ^{x}$
Then
$\displaystyle \lim_{x\rightarrow 0}\frac{\mathrm{d} \frac{1}{f(x))}}{\mathrm{d} x}$=0 November 15th, 2016, 10:37 AM #12 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 Math Focus: Mainly analysis and algebra That is incorrect. The limit is 1. \begin{align*} f(x) &= \frac1x + e^x = \frac{1+xe^{x}}{x} \\ g(x) = \frac1{f(x)} &= \frac{x}{1+xe^{x}} \\ &\approx x\big(1-x(1-x)\big) = x-x^2+x^3 \\ g'(x) & \approx 1 - 2x \\ \lim_{x \to 0} g'(x) = 1 \end{align*} November 15th, 2016, 12:03 PM #13 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 Thanks. I made an algebraic mistake. Suppose: f(x)=1/x+g(x) Then $\displaystyle \lim_{x\rightarrow 0}\frac{\mathrm{d} \frac{1}{f(x))}}{\mathrm{d} x}= \lim_{x\rightarrow 0}\frac{1-x^{2}g'(x)}{(1+xg(x))^{2}}$ Which depends on behavior of g(x) From https://en.wikipedia.org/wiki/Riemann_zeta_function $\displaystyle \zeta (s)={\frac {1}{s-1}}+\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{n!}}\gamma _{n}\;(s-1)^{n}$. ---------------------------------------------- $\displaystyle \zeta$ (s) = $\displaystyle {\frac {1}{s-1}}$+g(s-1) In this case Lim is indeed 1. skipjack is right. I still question the argument that if f(x) $\displaystyle \approx$ h(x). f'(x) $\displaystyle \approx$ h'(x) Last edited by zylo; November 15th, 2016 at 12:05 PM. November 15th, 2016, 03:49 PM   #14
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Quote:
 Originally Posted by zylo Which depends on behavior of g(x)
Notice that the approach I took demonstrates that the derivative is 1 for every $g(x)$ having a Maclaurin expansion. Also (I think) for every $g(x)$ that is continuous at zero.

Quote:
 Originally Posted by zylo I still question the argument that if f(x) $\displaystyle \approx$ h(x). f'(x) $\displaystyle \approx$ h'(x)
Again, if you consider series expansions, the result is pretty obvious, at least in the case that $h(x)$ is a polynomial. November 15th, 2016, 07:19 PM   #15
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Quote:
 Originally Posted by skipjack For x close enough to 1, it gives 1/ζ(x) as being approximately x - 1, which has derivative 1.
OP uses $\displaystyle \zeta$(x), presumably x real.
Replace x with s complex to get $\displaystyle \zeta$(s)

Wiki article says $\displaystyle \zeta$(s) is analytic and therefore can be expressed as a Laurent series.
$\displaystyle \zeta$(s)=$\displaystyle \sum a_{n}(s-1)^{n}$
Therefore $\displaystyle \zeta$(x) = $\displaystyle \sum a_{n}(x-1)^{n}$ and above quote works.

The point is that the analyticity of $\displaystyle \zeta$(s) allows you to find a power series for $\displaystyle \zeta$(x).

Last edited by skipjack; November 16th, 2016 at 01:04 AM. 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 06:15 PM Dougy Number Theory 1 October 11th, 2012 07:53 PM mathbalarka Complex Analysis 12 September 21st, 2012 09:10 AM AFireInside Calculus 4 August 31st, 2011 11:24 PM Ziaris Real Analysis 0 February 18th, 2009 08:47 PM

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