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I am looking up on the Bernoulli numbers and their connection to $\tan(x)$ power series and I see this thread:

Bernoulli numbers, taylor series expansion of tan x

In it, Rob John manipulate the series of

enter image description here

The part I don't understand is circled in red. What is the general formula for $-2\cot(2x)$? How do you combine the formula to produce the last one for $\tan(x)$. I don't understand that part of his manipulation. Can you do it in a few more steps so I can learn?

Bernard
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James Warthington
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  • this is just taking the two series for $cot(x)$ and $2cot(2x)$ and subtracting them, putting them inside the same series – fGDu94 Nov 23 '19 at 00:01
  • @George Dewhirst: But what is the sigma notation formula for $\2cot(2x)$. Of course I can manually do that with expanded form, but I wish to know how to manipulate it using the sigma notation formula. – James Warthington Nov 23 '19 at 00:08
  • just replace $x$ by $2x$ in the formula and take $2^{2n-1} $ out as a factor – fGDu94 Nov 23 '19 at 00:09
  • What about the subtracting part? How do you derive it, I mean the second red circle part. – James Warthington Nov 23 '19 at 00:11
  • the sum operator is linear – fGDu94 Nov 23 '19 at 00:12
  • @GeorgeDewhirst: Sorry I don't know what is a linear operator. I am still a noob haha! – James Warthington Nov 23 '19 at 00:13
  • haha ok basically it is fine to do this: $\sum a_n + \sum b_n = \sum (a_n + b_n)$ and likewise if it a subtraction – fGDu94 Nov 23 '19 at 00:14
  • @George Dewhirst I just don't know the second red circled part. – James Warthington Nov 23 '19 at 00:16
  • basically the $B_{2n}2^{2n}$ is common to both series, and the first term in the brackets is from $cot(x)$, the second is from $cot(2x)$ – fGDu94 Nov 23 '19 at 00:17
  • Let us continue this discussion in chat. – James Warthington Nov 23 '19 at 00:20
  • Basically, the part I don't understand is $x^{2n-1}-2 \cdot2^{2n-1}x^{2n-1}$ – James Warthington Nov 23 '19 at 00:22
  • That's just $x^{2n-1}-2(2x)^{2n-1}$, which is what you get when you do the things everyone has told you in the comments so far. – anon Nov 23 '19 at 00:24
  • $cot(x) - 2cot(2x) = \sum_{n=0}^{\infty}\frac{(-1)^n B_{2n} 2^{2n} x^{2n-1}}{(2n)!} - 2\sum_{n=0}^{\infty}\frac{(-1)^n B_{2n} 2^{2n} 2^{2n-1}x^{2n-1}}{(2n)!} = \sum_{n=0}^{\infty}\frac{(-1)^n B_{2n} (x^{2n-1} - 2*2^{2n-1} x^{2n-1})}{(2n)!}$ – fGDu94 Nov 23 '19 at 00:24
  • Ok, I see thank you! :) – James Warthington Nov 23 '19 at 00:27
  • One answer to this question can be found at https://math.stackexchange.com/a/4254671/945479. – qifeng618 Sep 19 '21 at 15:18
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