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November 15th, 2016, 08:40 AM   #11
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Quote:
Originally Posted by skipjack View Post
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
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November 15th, 2016, 09:37 AM   #12
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That is incorrect. The limit is 1.

$$\begin{align*}
f(x) &= \frac1x + e^x = \frac{1+xe^{x}}{x} \\
g(x) = \frac1{f(x)} &= \frac{x}{1+xe^{x}} \\
&\approx x\big(1-x(1-x)\big) = x-x^2+x^3 \\
g'(x) & \approx 1 - 2x \\
\lim_{x \to 0} g'(x) = 1
\end{align*}$$
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November 15th, 2016, 11:03 AM   #13
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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 11:05 AM.
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November 15th, 2016, 02:49 PM   #14
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Quote:
Originally Posted by zylo View Post
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 View Post
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.
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November 15th, 2016, 06:19 PM   #15
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Quote:
Originally Posted by skipjack View Post
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 12:04 AM.
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