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SUMMARY
This discussion focuses on the mathematical derivation of a series using binomial sums and complex analysis. The key equations presented include the binomial series expansion for $\frac{1}{\sqrt{1 - x^{2}}}$ and its derivative $\frac{x}{\sqrt{1 - x^{2}}}$. The final result for the sum of the series is expressed as $S = \text{Im} \{\frac{e^{i\ \theta}}{\sqrt{1 - e^{2\ i\theta}}}\}$. Participants are encouraged to explore these mathematical concepts further to deepen their understanding.
PREREQUISITES- Understanding of binomial series expansions
- Familiarity with complex analysis and imaginary numbers
- Knowledge of calculus, particularly derivatives and integrals
- Ability to manipulate exponential functions and their properties
- Study the properties of binomial series and their applications
- Learn about complex functions and their derivatives
- Explore the concept of imaginary numbers in mathematical analysis
- Investigate the implications of the series sum in real-world applications
Mathematicians, students studying advanced calculus, and anyone interested in the applications of complex analysis in series summation.
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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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