Bernoulli number for the Maclaurin series of $tan(x)$

Asked Nov 03 '19 at 15:39

Active Nov 03 '19 at 15:39

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I know that the series representation of $\tan(x)$ is:

$x+\frac{x^3}{3}+\frac{2x^5}{15}+\frac{17x^7}{315}...$

I can derive this by using the fact that $\tan(x)=\dfrac{\sin(x)}{\cos(x)}$. But I don't know how to find the general sigma formula for this function.

I have seen the general formula here: Maclaurin polynomial of tan(x)

But I don't know how the Bernoulli number is involved. Can you explain this further so I can understand?

asked Nov 03 '19 at 15:39

James Warthington

3,187

1

cf. the answer to this question – J. W. Tanner Nov 03 '19 at 15:43
The answer is not there, the link to ProofWiki is invalid – Amad Oct 06 '20 at 14:58
Here is the real answer: https://math.stackexchange.com/questions/2098941/bernoulli-numbers-taylor-series-expansion-of-tan-x – Amad Oct 06 '20 at 15:01

Bernoulli number for the Maclaurin series of $tan(x)$

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