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I have the problem for computing the j-derivative of a logarithm, with $j\gg1$ \begin{equation} c_j=\left.\frac{\partial^j}{\partial s^j}\log\left(1+Ae^s+Be^{2s}\right)\right|_{s=0}, \end{equation} being A and B real numbers ($0<A,B<1$). What I have done is to perform a Taylor expansion of the logarithm, finding \begin{equation} c_j=\left.\frac{\partial^j}{\partial s^j}\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}e^{ns}\left(A+Be^s\right)^n\right|_{s=0}. \end{equation} Finally, I can use the Newton binomia expression and derive inside the sum \begin{equation} c_j=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\sum_{k=0}^{n}{n\choose k}(2n-k)^jA^{n-k} B^{k} \end{equation} My question is to approximate this expression in the limit when $j\gg1$. I though about using some Stirling kind approximation, but I did not find any solution. Thanks for any help

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Just a rough idea. Let $\alpha, \beta$ be the zeros of $1+Ax+Bx^2$, then for $j\geq 1$ $$c_j = \left.\left(\frac{\partial}{\partial s}\right)^j \log( B(\alpha - e^s)(\beta-e^s) )\right|_{s=0} = \left.\left(\frac{\partial}{\partial s}\right)^j \log(\alpha-e^s) + \left(\frac{\partial}{\partial s}\right)^j \log(\beta-e^s)\right|_{s=0}.$$

Now, $$\log(\alpha-e^s) = \log \alpha + \log(1-\frac{e^s}{\alpha}) = \log \alpha - \sum_{k=1}^{\infty} \frac{e^{ks}}{k\alpha^k}$$ and hence $$ \left(\frac{\partial}{\partial s}\right)^j \log(\alpha-e^s) = - \sum_{k=1}^{\infty} \frac{k^{j-1}e^{ks}}{\alpha^k} = - \frac{\sum_{i=0}^{j-2} \left\langle {j-1\atop i}\right\rangle\left(\frac{e^s}{\alpha}\right)^{i+1}}{(1-\frac{e^s}{\alpha})^j},$$ and similarly for $\beta$. Here $\left\langle {\cdot\atop\cdot}\right\rangle$ are Eulerian numbers.

Therefore, $$c_j = - \frac{\sum_{i=0}^{j-2} \left\langle {j-1\atop i}\right\rangle\alpha^{j-1-i}}{(\alpha-1)^j} - \frac{\sum_{i=0}^{j-2} \left\langle {j-1\atop i}\right\rangle\beta^{j-1-i}}{(\beta-1)^j}.$$

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  • $\begingroup$ Thank you for the answer, I found it so useful. At the end I think that the series you get before defining the Eularian numbers can be sumed up by converting the sum into an integral:\begin{equation}\sum_{k=1}^\infty k^{j-1}/\alpha^k\to \int_{1}^{\infty}dx\;x^{j-1} e^{-\alpha x}\end{equation} $\endgroup$ – user3209698 Mar 15 '16 at 14:59
  • $\begingroup$ Do you think it would be a good approach to the nice exact result you got? $\endgroup$ – user3209698 Mar 15 '16 at 15:07
  • $\begingroup$ @user3209698: It's possible (under replacement of $\alpha$ with $\log \alpha$ in the integral). You know better what kind of result you need to get. A particular approach may also depend on actual values of $\alpha,\beta$. $\endgroup$ – Max Alekseyev Mar 15 '16 at 18:21
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As in M. Alekseyev’s post, let $1+Ax + Bx^2 = (a-x)(b-x).\,$ For $j>>1$
$$ c_j \sim -\Gamma(j) \bigl( \frac{1}{ \log^j{a} } + \frac{1}{ \log^j{b} } \bigr)$$ The key to a proof is to recognize that the infinite sum in M. Alekseyev’s post can be written as a polylogarithm, and then use the known asymptotic expansion $$ Li_{1-j}(x) \sim \frac{ \Gamma(j) }{\log^j(1/x)}$$

The asymptotic formula for $c_j$ should be interpreted in the following manner. Only sum the results if the two roots (a and b) have equal magnitude. Otherwise, you want the root that is closest to 1 in magnitude. (Incidentally this shows that if either a or b is 1, you have a problem!) The reason why this is mentioned is that for some (a, b), naively using both terms for the $c_j$ can lead to an imaginary part, which is nonsensical for real A and B. One can do the same kind of thing for cubics, quartics, etc. The ‘keep the largest term only, or sum the terms if several roots have the same magnitude and it’s the largest of the set ’ rule also applies. The convenience of the rule allows one to consider complex A, B, C… I haven’t proven the general applicability of the rule for arbitrary polynomials, however. I have written code to test up to the quartic case with various A, B, C, D, with magnitudes $\le 10$ and for j as small as 20, I get about 5 significant figures.

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  • $\begingroup$ Thank you for your answer. The first expression you get is the one that it is obtained by changing the sum by an integral in the solution given by M. Alekseyew. However, I didn't know about the existence of these polylogarithms, which are useful for simplifying the expressions. If you have some figures I would like to see them (if possible) $\endgroup$ – user3209698 Mar 16 '16 at 9:34

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