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I posted a question on MSE about approximating Taylor series but Despite a bounty I did not receive any answers or comments.

Maybe you guys can help.

https://math.stackexchange.com/questions/1440931/proof-that-oint-r-dx-n-n-0

Let $f(x)$ be a real-entire function such that for all $x>0$ we have $f(x) > 0$, $f'(x) > 0$ , $f '' (x) > 0$.

And also $0 < D^M f(0) < D^{M-1} f(0)$.

Let $0<T<1$ and $n$ a positive integer.

Let $g(x,n) = \frac{f(x)}{x^n}$.

Let real $x_0(n)$ satisfy $x_0 (n)> 0$

$g ' (x_0(n),n) = 0$

Let $r(n)$ be a contour that contains the real interval $[T,x_0(n) + T]$ but not $0$ , not a negative number and no poles or zero's off the real line.

Conjecture A: there is Always an integer $N$ ( depending on $f$ ) such that for all $n > N$ $$ \frac {1} {2 \pi i} \oint_{r(n)} (x-T)^{n-1} f(x+T) - \sqrt n \ln(e+n) g(x,n) \frac{g '' (x,n) }{g ' (x,n) } < 0. $$

A similar conjecture that is probably not compatibel with Conjecture A (they probably cannot both be true , but I believe at least one is true).

Conjecture B: there is Always an integer $N$ ( depending on $f$ ) such that for all $n > N$ there exists a positive $w$ independent of $n$ such that $$ \frac {1} {2 \pi i} \oint_{r(n)} (x-T)^{n-1} f(x+T) - \frac{w g(x,n) g '' (x,n) \sqrt { f '' (x)} }{g ' (x,n) } < 0. $$

( the derivatives are with respect to $x$ , not $n$ )

How to prove/disprove them ?

In case these contour integrals look random/confusing notice that

$$ \frac {1} {2 \pi i} \oint_{r(n)} (x-T)^{n-1} f(x+T) $$

Is just the sequence of Taylor coëfficiënts for $f(x)$.

Also

$$ \frac {1} {2 \pi i} \oint_{r(n)} g(x,n) \frac{g '' (x,n) }{g ' (x,n) }$$

Is simply minimum( $f(x)/x^n $) for $x > 1.$

So basically we want to understand how good the intuïtive estimate min( $f(x)/x^n $) is compared to $D^n f(x) / n! $.

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  • $\begingroup$ Why the downvotes ? $\endgroup$ – mick Sep 28 '15 at 0:21
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    $\begingroup$ It would be best to copy the question (mentioning it is a copy) instead of providing a link. $\endgroup$ – Chris Ramsey Sep 28 '15 at 10:20
  • $\begingroup$ @Chris Ramsey : Done. $\endgroup$ – mick Sep 28 '15 at 19:50

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