k is real constant $\gt = 1$. Is $a_n$ for $f(x)$ positive, increasing, and $\lt 1$, where $n\lt= e^{k-1}$?

$$f(x) = \sum_{n=0}^{\infty} a_n x^n = \exp(-\exp(k-ex))$$

$f(x)$ is the double exponential as described. Prove the conjecture that the Taylor series coefficients for f(x) are positive, increasing, and all less than 1, up to $a_n$, where $n\lt= e^{k-1}$, and also probably for ceiling as well, $n \lt= \lceil e^{k-1}\rceil$. The first two or three terms above n are still increasing in magnitude and positive, but the Taylor series coefficients become chaotic, and soon have both negative and positive values.

For example, consider k=3, where $f(0)=\exp(-\exp(3))$, where $f(x)$ has the following Taylor series. The first negative term is $a_{12}$. $\lceil e^{k-1}\rceil = 8$. The first term>=1 is $a_9$.

k3=     0.00000000189217869484
+x^ 1*  0.000000103309456270
+x^ 2*  0.00000267984048780
+x^ 3*  0.0000437878433105
+x^ 4*  0.000503376203252
+x^ 5*  0.00430132652897
+x^ 6*  0.0280820601079
+x^ 7*  0.141245147887
+x^ 8*  0.540609219094
+x^ 9*  1.49907379722
+x^10*  2.54732645227
+x^11*  0.276584318427
+x^12* -11.9671868638
+x^13* -30.4747754691
+x^14* -10.2766978768

I derived my conjecture using non-rigorous methods, by finding optimal radii to do a Cauchy circle integral for the various coefficients of a_n, by using the inverse of $ e x \exp(k-ex)$ as a "frequency approximation" guide for what radius to use for the Cauchy integral. This inverse function has a branch singularity at x=1/e corresponding to where the Taylor series coefficients become chaotic. I'm hoping someone else can give a more rigorous approach.

I am interested in using these approximations to help rigorously answer a math stack question about a conjectured nowhere analytic function. The next step is to get an approximation for the "radius of convergence" for each Taylor series coefficient, $\exp(-\ln(a_n)/n)$, for all Taylor series coefficients in the well behaved region, $n\lt= e^{k-1}$.

I have also looked at the chaotic Taylor series coefficients. This is a closely related but possibly more difficult problem. Consider the Taylor series for $$f2(x) = \exp(-\exp(k - \frac{\pi x}{2}))$$ Here, we are interested in where the |a_n| reaches its maximum magnitude, but this is more difficult because the Taylor series coefficients are chaotic for n>exp(k-1), from the result above. This function scales the Taylor series coefficients from $f(x)$ by $(\frac{\pi}{2e})^n$. The second conjecture is that |a_m| reaches a maximum near $m \approx 0.5\pi e^k$, where $a_m \approx r^m$, where r approaches 1 as k increases. For the k=3 example, the maximum magnitude of $a_m$ occurs near n=31. This is also the minimum for the radius of convergence approximation, $\exp(-\ln(a_n)/n)$.

If we consider k=10 for $f(x)$ than $n=\lceil e^9 \rceil = 8104$. The first conjecture would be that all Taylor series coefficients are positive up that point. This would also hold for $f2(x)$. For $f2(x)$ for k=10, we would expect the Taylor series coefficients to peak in magnitude near $m \approx 0.5\pi e^{10} \approx 34599$. These numbers quickly become too large to actually compute the Taylor series coefficients by brute force, which is what makes these approximations useful. In the problem I mentioned, I have used approximations for the radius of convergence for k in the millions, far beyond what can be computed.


$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} \expo{-\expo{k - \expo{}x}}&= \sum_{\ell = 0}^{\infty}{\pars{-\expo{k - \expo{}x}}^{\ell} \over \ell!} = \sum_{\ell = 0}^{\infty}{\pars{-\expo{k}}^{\ell} \over \ell!}\, \expo{-\ell\expo{}x} = \sum_{\ell = 0}^{\infty}{\pars{-\expo{k}}^{\ell} \over \ell!}\, \sum_{\ell'=0}^{\infty}{\pars{-\ell\expo{}x}^{\ell'} \over \ell'!} \\[3mm]&= \sum_{\ell = 0}^{\infty}{\pars{-\expo{k}}^{\ell} \over \ell!}\, \sum_{\ell'=0}^{\infty}{\pars{-\ell\expo{}}^{\ell'}x^{\ell'} \over \ell'!} =\sum_{\ell'=0}^{\infty}x^{\ell'}\bracks{{\pars{-1}^{\ell'} \over \ell'!}% \sum_{\ell = 0}^{\infty}{\pars{-\expo{k}}^{\ell}\ell^{\ell'} \over \ell!}\,} \end{align}

$$ \expo{-\expo{k - \expo{}x}}=\sum_{n = 0}^{\infty}a_{n}x^{n}\,,\qquad a_{n}\equiv {\pars{-1}^{n} \over n!}\,\expo{-\expo{k}}B_{n}\pars{-\expo{k}} $$ where $B_{n}\pars{x}$ is the Bell Polynomial. Also See this link.

  • $\begingroup$ Thanks Felix! I thought about pursuing Bell polyomials, but I hadn't done it yet. I got your Bell polynomials implemented; and I think your formula has a small typo, missing a factor of $e^n$, but if I map $a_n \mapsto a_n\times e^n$, then I get the same polynomial I posted for k=3 (which I generated with pari-gp). I don't know how to use the Bell polynomials to prove the pattern holds, for increasing values of k, that the Taylor series coefficients increase, and then become chaotic, but I will investigate. $\endgroup$ – Sheldon L Mar 13 '14 at 3:01

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