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Want to get this: $$\tan z = \frac{1}{i}\frac{{{e^{iz}} - {e^{ - iz}}}}{{{e^{iz}} + {e^{ - iz}}}} = \frac{1}{i}\left( {1 - \frac{2}{{{e^{2iz}} - 1}} + \frac{4}{{{e^{4iz}} - 1}}} \right)$$ as a start. Getting lost in the algebra

$$\frac{{{e^{iz}} - {e^{ - iz}}}}{{{e^{iz}} + {e^{ - iz}}}} = \frac{{{e^{iz}} + {e^{ - iz}} - 2{e^{ - iz}}}}{{{e^{ - iz}} + {e^{iz}}}} = \frac{{{e^{iz}} + {e^{ - iz}} - 2{e^{ - iz}}}}{{{e^{ - iz}} + {e^{iz}}}} = 1 - \frac{{2{e^{ - iz}}}}{{{e^{ - iz}} + {e^{iz}}}}$$

and also $$\frac{{2{e^{ - iz}}}}{{{e^{ - iz}} + {e^{iz}}}} = \frac{2}{{{e^{2iz}} + 1}} = \frac{{ - 2 + 4}}{{{e^{2iz}} + 1}} = \frac{{ - 2}}{{{e^{2iz}} + 1}} + \frac{4}{{{e^{2iz}} + 1}}$$

Please someone, show me the right way.

Robert Z
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Megamatics
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  • See https://math.stackexchange.com/questions/2098941/bernoulli-numbers-taylor-series-expansion-of-tan-x – Robert Z Aug 10 '19 at 08:59
  • I have already seen this. I am asking specifically how to get the equation that I need to get started. Where is that in your proposed answer? – Megamatics Aug 10 '19 at 09:10

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Work backwards! We have that $$\begin{align} {1 - \frac{2}{{{e^{2iz}} - 1}} + \frac{4}{{{e^{4iz}} - 1}}}&=\frac{e^{4iz} - 1-2({e^{2iz}} + 1)+4}{{{e^{4iz}} - 1}}\\&=\frac{e^{4iz}-2e^{2iz}+1}{{e^{4iz}} - 1}= \frac{(e^{2iz}-1)^2}{{e^{4iz}} - 1}\\&=\frac{e^{2iz}-1}{{e^{2iz}} + 1}=\frac{{{e^{iz}} - {e^{ - iz}}}}{{{e^{iz}} + {e^{ - iz}}}}.\end{align}$$ Supplement: since Bernoulli numbers are related to the expansion of $x/(e^x-1)$, we try to write the LHS in terms of functions like $1/(w^n-1)$ where $w=e^{iz}$, $$\frac{{{e^{iz}} - {e^{ - iz}}}}{{{e^{iz}} + {e^{ - iz}}}}=\frac{w^2-1}{w^2 + 1}=1-\frac{2}{w^2 + 1}=1-\frac{2(w^2-1)}{w^4 - 1}\\=1-\frac{2(w^2+1)-4}{w^4 - 1}=1-\frac{2}{w^2 - 1}+\frac{4}{w^4 - 1}.$$

Robert Z
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  • Thanks very helpful. Working backwards was ideal for verification and you have successfully answered the question, but as a supplemental question that follows how do you come up with the equation in the first place? Is it just a question of intuition and making an informed guess at a formula? – Megamatics Aug 10 '19 at 09:34
  • That is inspirational. Don't know why I didn't try the obvious substitution. – Megamatics Aug 10 '19 at 09:50