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I know that $$\cos(z^4)-1=-\frac{z^8}{2!}+\frac{z^{16}}{4!}+...$$ but how do I take the reciprocal of this series (please do not use little-o notation)? Or are there better methods to obtain the required series?

ryang
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    A follow-up? Still working on this function? Is this homework? – Julien Mar 26 '13 at 04:46
  • @julien Hi Julien! I decided to create a new question because I feel it deserves its own answer. No, it's not homework. If you don't mind to elaborate on the little-O notation usage in the original question, I'd love to read it. – ryang Mar 26 '13 at 05:14
  • Hi Ryan. Definitely, the full Laurent series is a different question which deserves its own answer. O and o are useful in many situations. To begin with limit determination. It allows you to keep the relevant part in a Taylor expansion and to throw away the remainder. Unlike equivalents, O's and o's can be safely added, subtracted, etc... – Julien Mar 26 '13 at 05:24
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    The reciprocal doesn't help you a lot and I don't see a nice system here – Dominic Michaelis Mar 26 '13 at 05:40
  • @DominicMichaelis How do you mean by not seeing a nice system? And do you mean that the way to obtain the Laurent series isn't by taking the reciprocal as I suggested? If so, then how can I produce the Laurent series? – ryang Mar 27 '13 at 01:18
  • Long division... – GEdgar Mar 27 '13 at 01:47
  • @Ryan I think the reciprocal doesn't help you to find the full taylor series. My first idea would be leipnizformula to get all derivatives but to be honest I guess you won't find an explicit formula (to be more honest I didn't even try after calculating the first 10 derivatives with mathematica the terms got about 100 times to long for me) – Dominic Michaelis Mar 27 '13 at 07:07
  • @DominicMichaelis Hey, did you mean for finding the residue at 0 instead? If so, yes, it's crazy to take derivatives. For this particular problem, to find the resiude at 0 (which is an order-8 pole), long division (though I'm still not comfortable with long division involving infinite series) to first obtain the Laurent series seems to be the best method. – ryang Mar 27 '13 at 10:58
  • @Ryan no i mean for the Laurentseries $$\sum_{n=-8}^\infty a_n z^n $$ the long division won't work well – Dominic Michaelis Mar 27 '13 at 10:59
  • Isn't the series explicit from the formula? Why need to perform division? – ryang Mar 27 '13 at 11:02

4 Answers4

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Long Division ... still worth learning

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To get more terms in the quotient, use more terms in the divisor.

GEdgar
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  • Thank you. While I understand the mechanics of the above division, I am unsure why it works, because for polynomial long division I'm accustomed to listing the polynomials in decreasing instead of increasing powers. I also don't know how the radius of convergence of a Laurent series obtained by division is related to that of the original Taylor series (in this case, the series in the denominator). – ryang Mar 27 '13 at 03:55
  • That doesn't help you to get the full laurent series, and when you wan't another term you have to calculate all, I see no advantage to making the taylor series by taking derivates often – Dominic Michaelis Mar 27 '13 at 06:21
  • @DominicMichaelis What derivatives? – ryang Mar 27 '13 at 11:01
  • @Ryan, you get the coefficients of the laurent series for $n>$ with $$a_n=\frac{f^{(n)}}{n!}$$ just like the taylor series. – Dominic Michaelis Mar 27 '13 at 11:05
  • And you think computing the 8th derivative of $f(z)$ is easier than the long-division calculation above? – GEdgar Mar 27 '13 at 15:48
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Of course the easy way is to ask a CAS. This is from Maple:

$$ (-2) z^{(-8)} - \frac{1}{6} - \frac{1}{120} z^{8} - \frac{1}{3024} z^{16} - \frac{1}{86400} z^{24} - \frac{1}{2661120} z^{32} - \\ \quad{}\quad{}\frac{691}{59439744000} z^{40} - \frac{1}{2874009600} z^{48} - \frac{3617}{355687428096000} z^{56} - \\ \quad{}\quad{}\frac{43867}{150267476975616000} z^{64} - \frac{174611}{21127833228902400000} z^{72} - \\ \quad{}\quad{}\frac{77683}{335740477128376320000} z^{80} - \frac{236364091}{36822263841994427596800000} z^{88} \\ \quad{}\quad{} - \frac{657931}{3722690410399436636160000} z^{96} + \operatorname{O} \bigl(z^{104}\bigr) $$

GEdgar
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The function $f(z)=\frac1{\cos(z^4)-1}$ can be rearranged as \begin{align*} f(z)&=-\frac1{1-\cos(z^4)}\\ &=-\frac{1}{2\sin^2\frac{z^4}2}\\ &=-\frac12\biggl(\frac{2}{z^4}\biggr)^2 \Biggl(\frac{\frac{z^4}{2}}{\sin\frac{z^4}{2}}\Biggr)^2\\ &=-\frac{2}{z^8} \Biggl\{1+\sum_{q=1}^{\infty}(-1)^q\Biggl[\sum_{k=1}^{2q}\frac{(2)_k}{k!} \sum_{j=1}^k(-1)^j\binom{k}{j} \frac{T(2q+j,j)}{\binom{2q+j}{j}}\Biggr]\frac{z^{8q}}{(2q)!}\Biggr\},\quad z^4<2\pi, \end{align*} where $(2)_k$ is the Pochhammer symbol or the rising factorial and $T(2q+j,j)$ denotes the central factorial numbers of the second kind. The last step comes from employing Theorem 4.1 in the paper

  1. Feng Qi and Peter Taylor, Series expansions for powers of sinc function and closed-form expressions for specific partial Bell polynomials, Applicable Analysis and Discrete Mathematics 18 (2024), no. 1, in press; available online at https://doi.org/10.48550/arXiv.2204.05612.
qifeng618
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From \begin{equation} \begin{aligned}\label{sin-square-recip-ser} \biggl(\frac{x}{\sin x}\biggr)^2&=1+2B_2x^2-4\sum_{k=2}^{\infty}(-1)^k\Biggl[\bigl(2^{2k-1}-1\bigr)B_{2k}\\ &\quad-\sum_{j=1}^{k-1}\binom{2k}{2j}\bigl(2^{2j-1}-1\bigr)\bigl(2^{2k-2j-1}-1\bigr)B_{2j}B_{2k-2j}\Biggr]\frac{x^{2k}}{(2k)!}\\ &=1+\frac{x^2}{3}+\frac{x^4}{15}+\frac{2 x^6}{189}+\frac{x^8}{675}+\frac{2 x^{10}}{10395}+\frac{1382 x^{12}}{58046625}+\dotsm \end{aligned} \end{equation} for $x\in(-\pi,\pi)$, we can arrive at \begin{align*} \frac1{\cos(z^4)-1}&=-\frac12\biggl(\frac2{z^4}\biggr)^2\Biggl(\frac{\frac{z^4}2}{\sin\frac{z^4}2}\Biggr)^2\\ &=-\frac2{z^8} \Biggl\{1+\frac{B_2}2z^8-4\sum_{k=2}^{\infty}(-1)^k\Biggl[\bigl(2^{2k-1}-1\bigr)B_{2k}\\ &\quad-\sum_{j=1}^{k-1}\binom{2k}{2j}\bigl(2^{2j-1}-1\bigr)\bigl(2^{2k-2j-1}-1\bigr)B_{2j}B_{2k-2j}\Biggr]\frac{z^{8k}}{2^{2k}(2k)!}\Biggr\} \end{align*} for $z^4\in\bigl(-2\pi,2\pi\bigr)$.

References

  1. Xue-Yan Chen, Lan Wu, Dongkyu Lim, and Feng Qi, Two identities and closed-form formulas for the Bernoulli numbers in terms of central factorial numbers of the second kind, Demonstratio Mathematica 55 (2022), no. 1, 822--830; available online at https://doi.org/10.1515/dema-2022-0166.
qifeng618
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