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I've been going through some past exam papers and came across a question which asks you to compute the Taylor expansion of $\sec(z)$ given that $$\sec(z)=\frac{1}{\cos(z)},$$ and $$\cos(z)=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{(2n)!}.$$

The problem is I dont know how to compute the reciprocal, and I have found it difficult to find how to do so online. Could someone point me in the direction of some webpage that goes through such problems( I can't afford a book), or perhaps explain a general method for computing the reciprocal in such cases ?

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  • $\begingroup$ I don't think it's possible to get an explicit general formula for the coefficients in this case, because there's no pattern on the coefficients. However, it's possible to calculate the first few terms. $\endgroup$ – Frank Lu May 10 '18 at 21:50
  • $\begingroup$ @FrankLu Do you know anywhere that goes through examples so I could learn that way instead , if no general method exists ? $\endgroup$ – excalibirr May 10 '18 at 21:53
  • $\begingroup$ You may want to take a look into this link math.stackexchange.com/questions/1577978/… $\endgroup$ – caverac May 10 '18 at 22:00
  • $\begingroup$ Take a look also at this Wikipedia page en.wikipedia.org/wiki/Taylor_series#Trigonometric_functions $\endgroup$ – Orest Bucicovschi May 10 '18 at 22:13
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If$$\sec z=a_0+a_1z+a_2z^2+\cdots,$$then, since $\sec$ is an even function, $a_k=0$ when $k$ is odd. So, in fact,$$\sec z=a_0+a_2z^2+a_4z^4+\cdots$$On the other hand$$\cos(z)\sec(z)=1\iff\left(1-\frac{z^2}{2!}+\frac{z^4}{4!}+\cdots\right)\left(a_0+a_2z^2+a_4z^4+\cdots\right)=1$$and therefore$$\left\{\begin{array}{l}a_0=1\\a_2-\frac{a_0}{2!}=0\\a_4-\frac{a_2}{2!}+\frac{a_0}{4!}=0\\\vdots\end{array}\right.$$From these equalities, you can abtain te first coefficients of the Taylor series of $\sec$.

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    $\begingroup$ I remember you showed me a similar trick before, hadn't thought to use it here but it's such a nice way to answer these type of questions I really must dedicate it to memory from now on to have in my back pocket :) $\endgroup$ – excalibirr May 10 '18 at 22:05
  • $\begingroup$ see here math.stackexchange.com/questions/2616561/taylor-series-for-secx $\endgroup$ – user547564 May 11 '18 at 11:18
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Hint. One may recall the generating function of the Bernoulli numbers, $$ \frac{u}{e^u-1}=\sum_{n=1}^{\infty} B_n \frac{u^n}{n!}, \quad |u|<2\pi,\tag1 $$ then one may apply it by writing $$ \begin{align} \frac{1}{\cos z}&=\frac{2}{e^{iz}+e^{-iz}} \\\\&=\frac{2e^{iz}}{e^{2iz}+1} \\\\&=\frac{2e^{iz}}{e^{2iz}-1}-\frac{4e^{iz}}{e^{4iz}-1} \\\\&=\frac{1}{iz}\left(\frac{2iz}{e^{2iz}-1}-\frac{4iz}{e^{4iz}-1}\right)e^{iz} \end{align} $$ and by recalling that $$ e^{iz}=\sum_{n=0}^\infty\frac{(iz)^n}{n!}. $$

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Hint:

Just perform the division of $1$ by the expansion of $\cos z$ by increasing powers, up to the order you want.

You should find in a few lines that it begins with $$\sec z=1+\frac{z^2}2+\frac{5z^4}{24}+\frac{61z^6}{720}+\dotsm,$$ if I'm not mistaken.

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