anil86
- 10
- 0
View Attachment to Understanding Taxes
- Context:
- Thread starter anil86
- Start date
-
- Tags
- Taxes
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
Discussion
chisigma
Gold Member
MHB
- 1,627
- 0
anil86 said:
Wellcome on MHB anil86!... You can start from the binomial sum...
$\displaystyle \frac{1}{\sqrt{1 - x^{2}}} = 1 + \frac{1}{2}\ x^{2} + \frac{3}{8}\ x^{4} + ...\ (1)$
... and from (1) ...
$\displaystyle \frac{x}{\sqrt{1 - x^{2}}} = x + \frac{1}{2}\ x^{3} + \frac{3}{8}\ x^{5} + ... \ (2)$
... so that the sum of Your series is... $\displaystyle S = \text{Im} \{\frac{e^{i\ \theta}}{\sqrt{1 - e^{2\ i\ \theta}}}\}\ (3)$
Are You able to proceed?...
Kind regards
$\chi$ $\sigma$
anil86
- 10
- 0
Attachments
chisigma
Gold Member
MHB
- 1,627
- 0
chisigma said:
Is...
$\displaystyle 1 - e^{2\ z} = - e^{z}\ (e^{z} - e^{- z}) -> \sqrt{1 - e^{2\ z}} = i\ e^{\frac{z}{2}}\ \sqrt {e^{z} - e^{- z}}\ (1)$
... so that for $\displaystyle z = i\ \theta $ is...
$\displaystyle \text{Im}\ \{\frac{e^{i\ \theta}}{\sqrt{1 - e^{2\ i\ \theta}}}\} = \text{Im}\ \{ \frac{e^{i\ \frac{\theta}{2}}}{i\ \sqrt{e^{i\ \theta} + e^{- i\ \theta}}} \} = \frac{\sin \frac{\theta}{2} + \cos {\frac{\theta}{2}}}{2\ \sqrt{\sin \theta}}\ (2)$
Kind regards
$\chi$ $\sigma$