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Discussion Overview
The discussion revolves around mathematical series and their representations, particularly focusing on binomial sums and complex analysis. Participants explore the derivation and manipulation of series related to the function $\frac{1}{\sqrt{1 - x^{2}}$ and its implications in a broader mathematical context.
Discussion Character
- Technical explanation
- Mathematical reasoning
Main Points Raised
- One participant introduces the binomial sum representation of $\frac{1}{\sqrt{1 - x^{2}}}$ and its expansion, suggesting it as a starting point for further exploration.
- Another participant reiterates the same binomial sum and proposes a related series involving $\frac{x}{\sqrt{1 - x^{2}}}$, indicating a connection to the original series.
- There is a suggestion that the sum of the series can be expressed in terms of the imaginary part of a complex function involving $e^{i\theta}$.
- A subsequent post introduces a transformation involving $z = i\theta$ and provides a new expression for the imaginary part, indicating a potential pathway for further analysis.
Areas of Agreement / Disagreement
Participants appear to share a common interest in the mathematical series and their properties, but there is no explicit consensus on the implications or correctness of the derived expressions. The discussion remains exploratory with multiple approaches being presented.
Contextual Notes
Some assumptions regarding the convergence of the series and the conditions under which the transformations are valid are not fully addressed. The dependence on specific definitions of the functions involved may also influence the interpretations presented.
- #2
chisigma
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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
$\chi$ $\sigma$
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anil86
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chisigma
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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
$\chi$ $\sigma$
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