Timeline for Higher order derivative of negative power of cosine function
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 14, 2017 at 23:33 | comment | added | Chee | Sorry that I misunderstood you. I want the first one, an exact algebraic formula. Your comments point out that asymptotics for large n would be useful too, which I would like to learn about. | |
| Dec 14, 2017 at 21:58 | comment | added | fedja | A polynomial in $\sigma$ with rational coefficients of the corresponding degree. What I mean is that there are at least three possible meanings of the request to "compute": looking for an exact algebraic formula (like the above) from which you cannot even figure out whether it is positive or negative without some deep thinking, looking for asymptotics for large $n$ in some regimes, which allows you to answer the sign and size questions, but is useless for algebraic properties, and looking for a stable efficient algorithm that allows you to compute the value. Which one is of interest for you? | |
| Dec 14, 2017 at 2:43 | comment | added | Chee | @fedja: say, what is it when n is 10, 100 or 100000? | |
| Dec 14, 2017 at 2:41 | comment | added | fedja | Now I need to compute $L^{(n)}(0)$. What is the exact meaning of the verb "to compute" in this sentence? | |
| Dec 11, 2017 at 21:15 | history | edited | Chee | CC BY-SA 3.0 |
added Morris's paper that provides recursion formula for this
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| Dec 11, 2017 at 13:02 | comment | added | Max Alekseyev | en.wikipedia.org/wiki/Faà_di_Bruno%27s_formula | |
| Dec 11, 2017 at 8:22 | comment | added | მამუკა ჯიბლაძე | Just look at the link - there seems to be a (more or less) explicit formula:$$L^{(2n)}(0)=\sum_{m=1}^nT_{n,m}\sigma^m$$with$$T_{n,m}=(-1)^{n+m}\sum_{k=1}^n(-1)^k\frac{S1(k, m)}{2^{k-1}k!}\sum_{j=1}^k(-1)^j\binom{2k}{k-j}j^{2n}$$where $S1(k,m)$ are the Stirling numbers of the first kind. | |
| Dec 11, 2017 at 7:32 | history | edited | Chee | CC BY-SA 3.0 |
added an update containing the determinantal formula
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| Dec 11, 2017 at 7:01 | comment | added | Chee | Thank you. If you know an analytic strategy instead of computing case by case, could you please point me to it? I do not get how triangular numbers come into play yet. | |
| Dec 11, 2017 at 6:40 | comment | added | მამუკა ჯიბლაძე | Some information about the triangle of numbers$$\begin{array}{rl}L(0)&=1\\L^{(2)}(0)/\sigma&=2+3\sigma,\\L^{(4)}(0)/\sigma&=16+30\sigma+15\sigma^2,\\L^{(6)}(0)/\sigma&=272+588\sigma+420\sigma^2+105\sigma^3,\\ \cdots \end{array}$$can be found at the OEIS A085734 page | |
| Dec 11, 2017 at 6:08 | history | asked | Chee | CC BY-SA 3.0 |