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For $t\in(-1,1)$, let $$f(t):=\left(\frac{1+t}{1-t}\right)^{(1-t)/2}+\left(\frac{1-t}{1+t}\right)^{(1+t)/2}$$ and $$g(t):=\frac1{f(t)}.$$ Note that the functions $f$ and $g$ are even.

Question 1: Is it true that all the even-order derivatives $f^{(2k)}$ of $f$ at $0$ are negative, except for $k=0$ and $k=2$?

Question 2: Is it true that all the even-order derivatives $g^{(2k)}$ of $g$ at $0$ are positive?

Question 3: Is there a simple, explicit, and accurate upper bound on the even-order derivatives $g^{(2k)}$?

A correct and complete answer to any one of these three questions will be considered as a correct and complete answer to this entire post.

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  • $\begingroup$ In analyzing $f(t)$, it might suffice to study one part of $f$. That is, $f_1(t)=\left(\frac{1+t}{1-t}\right)^{(1-t)/2}$. I'm sure you knew that. $\endgroup$
    – T. Amdeberhan
    Commented Apr 29, 2022 at 18:28
  • $\begingroup$ @T.Amdeberhan : Yes, of course. Thank you for your comment. It may be interesting that apparently $f_1^{(2k)}(0)/(2k)!=f_1^{(2k+1)}(0)/(2k+1)!$ for all $k=0,1,\dots$. $\endgroup$
    – Iosif Pinelis
    Commented Apr 29, 2022 at 18:37
  • $\begingroup$ Also, what you put in the questions, such as #1, and restricting to $f_1(t)$, a similar manifestation works for the series of $F_1(t)=\frac{1-t}2\log\left(\frac{1+t}{1-t}\right)$. Inherited by $f_1(t)$. But, it is easier to see in $F_1(t)$ than $f_1(t)$. $\endgroup$
    – T. Amdeberhan
    Commented Apr 29, 2022 at 18:42
  • $\begingroup$ @T.Amdeberhan : Thank you for your further comment. Taking the log indeed helps. $\endgroup$
    – Iosif Pinelis
    Commented Apr 29, 2022 at 19:29

1 Answer 1

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Note that $$\ln(2g(t))=\frac{1}{2} \,\ln \left(1-t^2\right)+ t \tanh ^{-1}(t) =\sum_{k=1}^\infty\frac{t^{2k}}{2k(2k-1)}.$$ This immediately yields the positive answer to Question 2.

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  • $\begingroup$ @losif Pinelis By computing the sum of the two fractions defining $f(t)$, I find the same answer, excepted that I have $t \tanh^{-1}(t)$ instead of $2t \tanh^{-1}(t)$. But I get the same final formula. $\endgroup$
    – Christophe Leuridan
    Commented Apr 29, 2022 at 19:32
  • $\begingroup$ @ChristopheLeuridan : Thank you for your comment. That $2t$ was multiplied by $\frac12$, which you may not have noticed. I have now rewritten the expression in an equivalent, but slightly simpler, form. $\endgroup$
    – Iosif Pinelis
    Commented Apr 29, 2022 at 19:38
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    $\begingroup$ The coefficients of $2g(t)$, as an exponential generating function, are sequence A211393 of the OEIS (oeis.org/A211393) and this formula for $\ln(2g(t))$ can be found there. $\endgroup$
    – Ira Gessel
    Commented Apr 29, 2022 at 19:40
  • $\begingroup$ @IraGessel : Thank you for your comment. $\endgroup$
    – Iosif Pinelis
    Commented Apr 29, 2022 at 19:42

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