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I'm trying to compute the Taylor series for $\sec(x)$ but the derivatives are getting way too complicated. So how do I go about this without having to calculate all the derivatives? I tried to build some kind of relationship with the series for $\cos(x)$ but I didn't get anything meaningful.

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    $\begingroup$ how many terms are you meant to expand to? $\endgroup$
    – Chinny84
    Jan 22, 2018 at 18:54
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    $\begingroup$ it is $$1+\frac{x^2}{2}+\frac{5x^4}{24}+\frac{61x^6}{720}+O(x^7)$$ $\endgroup$ Jan 22, 2018 at 18:56
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    $\begingroup$ if you KNOW the answer then post it $\endgroup$ Jan 22, 2018 at 19:00
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    $\begingroup$ @Trey That's very unlikely. $\endgroup$ Jan 22, 2018 at 19:02
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    $\begingroup$ $$\frac{1}{\cos\left(x\right)}=\frac{1}{1+\displaystyle \sum_{n=1}^{+\infty}\frac{\left(-1\right)^{n}}{(2n)!}x^{2n}}=\sum_{p=0}^{+\infty}\left(\sum_{n=1}^{+\infty}\frac{\left(-1\right)^{n}}{(2n)!}x^{2n}\right)^p $$ That's all i can get $\endgroup$
    – Atmos
    Jan 22, 2018 at 19:04

2 Answers 2

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Look at the Boustrophedon table: $$\matrix{1\\0&1\\1&1&0\\0&1&2&2\\5&5&4&2&0\\0&5&10&14&16&16\\61&61&56&46&32&16&0\\0&61&122&178&224&256&272&272\\1385&1385&1324&1202&1024&800&544&272&0}$$ etc. Each row is the series of partial sums of the previous row, but at each stage one reverses the order we add up and enter the partial sums. Any, from the first column we read off $$\sec x=1+\frac{x^2}{2!}+\frac{5x^4}{4!}+\frac{61x^6}{6!}+\frac{1385x^8}{8!}+\cdots.$$ The right-most elements also give $$\tan x=x+\frac{2x^3}{3!}+\frac{16x^5}{5!}+\frac{272x^7}{7!}+\cdots.$$

There's a good discussion on this in Concrete Mathematics by Graham, Knuth and Patashnik.

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    $\begingroup$ It's clear by going through the table that it works, but my question is, why does it work? $\endgroup$
    – imranfat
    Nov 27, 2020 at 21:00
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    $\begingroup$ @imranfat This is like one of those Ramanujan answers. $\endgroup$
    – Mark C
    Dec 13, 2021 at 6:59
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There is the expression $$\sec z = 1 + {z^2\over 2} + {5z^4\over 24} + \cdots + {(-1)^n E_{2n}\over (2n)!} z^{2n} + \cdots$$ for the Taylor series, where $E_{2n}$ is an Euler number; see Abramowitz and Stegun, {\sl Handbook of Mathematical Functions}, p.~75, Equation 4.3.69; and the discussion of Euler numbers on pp.~804--805, and the table of values for Euler numbers on p.~810. I don't know if there is an efficient way of computing Euler numbers for a large index, but it is at least a different place to start. (I stumbled on this looking for a proof that all the Taylor coefficients of~$\sec(z)$ are positive.)

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