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I've been working through these listed Maclaurin series and deriving all of them. For the listed trigonometric functions, I'm wondering if there are ways to do things:

  1. not have Bernoulli or Euler numbers in the series
  2. find the series without having to do polynomial division
  • $\sec x$
  • $\operatorname{sech} x$
  • $\tan x$
  • $\tanh x$

Additionally, on the list of mathematical series for wikipedia, the following also have Bernoulli numbers:

  • $\cot z$
  • $\csc z$
  • $\coth z$
  • $\operatorname{csch} z$

I'm sure I could find them by doing division, but is there a better way? (And by "better way", I mean a connection between other series, like how $\sinh x$ can be made up of variations of $e^x$, or how $\sin^{-1} x$ can be found by manipulating binomial series)

In an earlier question of mine (Two versions of Maclaurin series for $\arcsin(x)$), I learned that there's a way to not have the gamma function and so now I'm able to find that series -- but is there a way to not have these Bernoulli or Euler numbers (and is that way understandable at a single-variable calculus level?)

Ally
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  • Is having a recurrence for the coefficients fine? – Gary Jul 03 '23 at 01:19
  • @Gary, I think recurrence for the coefficients could be okay since that is likely within high school and first year university level. – Ally Jul 03 '23 at 01:23
  • See this for an example. – Gary Jul 03 '23 at 01:50
  • @Gary Okay, that makes sense, but I suppose my next question would be if there's a way to have the coefficients be expressed in a non-recurrence way? But, knowing my line of inquiry, even if there was, I'd wonder if there was an easier way. Basically, the heart of my question is "could I have a 'nice' series for these above functions like how there are 'nice' series for the others, where the only new variable is $n$ due to the series notation?" Is there something about these above series that make it so they can't be expressed in somewhat simple series? – Ally Jul 03 '23 at 03:05
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    Well, the coefficients of a power series of an analytic function are uniquely determined. You can replace the Bernoulli or Euler numbers by one of their explicit forms, but I am not sure if that makes things simpler. Indeed the coefficients of the Maclaurin series for, e.g., $\tan$ are not as nice as those for $\sin$ or $\cos$ but there is not much to do about it. They are what they are. – Gary Jul 03 '23 at 03:28
  • @Gary, that link is great but you're right that it's not particularly simpler. Every calculus textbook I've looked at has $\tan$ but none of them mention a "nice" series version. I suppose if there was, then it would have been mentioned. By chance with $\sin^{-1}$, after looking at textbooks and asking here, there was a lucky situation with the $\Gamma$ function being replaced with something simpler. I sorta hoped such replacement and simplification could be done with Bernoulli and Euler numbers if only I knew more math -- but it doesn't seem possible? Either way, thanks for the insights! – Ally Jul 03 '23 at 05:44
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    No, it is not possible. Otherwise we would not introduce those notations. – Gary Jul 03 '23 at 05:46
  • You can have nice series, so long as you aren't married to power series specifically. For example, $$\cot x = \sum_{n=-\infty}^\infty \frac{x}{x^2-n^2\pi^2}$$ – Ninad Munshi Jul 03 '23 at 05:54
  • To NinadMunshi, I'm basing my work off of Taylor series, so I'm only considering power series.

    @Gary (or anyone else who could answer), considering hypergeometric series (from equation 41 on the mathworld link), would there be a way to make hypergeometric version of those first four trig functions? Right now, for example, I could make $\sec(x) = \frac{1}{\cos(x)}$ and then substitute the hypergeometric series for $\cos(x)$ but the situation of non-simplification occurs. (https://math.stackexchange.com/a/4729642/454931 for series of $\cos(x)$) Explicitly, no "simple" hypergeometric formula.

     – Ally Jul 04 '23 at 03:26

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