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I found the following formula here: Taylor Series of $\tan x$.

Taylor series of $\tan x$:

$$\tan x = \sum_{n\,=\,1}^\infty \frac {(-1)^{n-1}2^{2n} (2^{2n}-1) B_{2n}} {(2n)!} x^{2n - 1} $$.

I do not understand what the Bernoulli number means in this formula. My problems are the following ones:

I do not know anything about Bernoulli numbers. I do not know how to expand the Bernoulli number in the formula you see further up. What expansion will be the final result? How did you arrive there? Why did you arrive there? How did you know that you should use the specific Bernoulli-related terms that you used in your answer? What is the purpose of the Bernoulli number in this formula?

A great deal of thanks.

J. W. Tanner
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Andreas
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  • The Bernoulli numbers appear from Riemann zeta function. Do you about it? – Nosrati Jan 15 '17 at 16:35
  • @MyGlasses Nope. – Andreas Jan 15 '17 at 16:42
  • @MyGlasses But you could make a explanation with it - I can make some research on the Riemann zeta func. – Andreas Jan 15 '17 at 16:45
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    The best book, that I know, about Bernoulli numbers, Riemann zeta function is Apostol Tom Mike, Introduction to Analytic Number Theory, Springer-Verlag, New York, (1976)(352s) , pp266-270. – Nosrati Jan 15 '17 at 16:51
  • @MyGlasses https://i.stack.imgur.com/ATuev.png Is this the book that you mean? Is it the first or the second volume that you are referring to? – Andreas Jan 15 '17 at 16:58
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    Yes. Volume one I mean. Volume two speaks about pure number theory. It has another name i don't remember – Nosrati Jan 15 '17 at 17:01
  • @MyGlasses That book is to far beyond my current level of mathematical understanding. Am I screwed in understanding the example of the taylor series of tan x, that you see further up, if so? It should not be the case. – Andreas Jan 15 '17 at 18:03
  • Where will you use $\tan $ series, if possible to say. – Nosrati Jan 15 '17 at 18:50
  • @MyGlasses I don't know really, I just found the series particularly interesting. – Andreas Jan 15 '17 at 18:53
  • Oh, sorry. I see that you did present some context. – robjohn Jan 17 '17 at 04:38

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The exponential generating function for the Bernoulli Numbers is $$ \sum_{n=0}^\infty\frac{B_nx^n}{n!}=\frac{x}{e^x-1}\tag{1} $$ The even part of $(1)$ is $$ \sum_{n=0}^\infty\frac{B_{2n}x^{2n}}{(2n)!}=\frac x2\coth\left(\frac x2\right)\tag{2} $$ Since $\cot(x)=i\coth(ix)$, by substituting $x\mapsto ix$, we get $$ \begin{align} \sum_{n=0}^\infty\frac{(-1)^nB_{2n}x^{2n}}{(2n)!} &=\frac{ix}2\coth\left(\frac{ix}2\right)\\ &=\frac x2\cot\left(\frac x2\right)\tag{3} \end{align} $$ Therefore, $$ \cot(x)=\sum_{n=0}^\infty\frac{(-1)^nB_{2n}2^{2n}x^{2n-1}}{(2n)!}\tag{4} $$ Since $\tan(x)=\cot(x)-2\cot(2x)$, we get $$ \begin{align} \tan(x) &=\sum_{n=0}^\infty\frac{(-1)^nB_{2n}2^{2n}\left(x^{2n-1}-2\cdot2^{2n-1}x^{2n-1}\right)}{(2n)!}\\ &=\sum_{n=1}^\infty\frac{(-1)^{n-1}B_{2n}\left(2^{2n}-1\right)2^{2n}x^{2n-1}}{(2n)!}\tag{5} \end{align} $$

robjohn
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