If 0

Question

(A) Given $x(y^3 - 1) = 1$,we have $x = \frac{1}{y^3 - 1}$.Let $k = \frac{1}{x} = y^3 - 1$. Note that $x = \frac{1}{k}$.The series is $S = 2(\frac{1}{

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$\log \left( \frac{y^3}{2 - y^3} \right)$

Vedclass · Answer

(A) Given $x(y^3 - 1) = 1$,we have $x = \frac{1}{y^3 - 1}$.Let $k = \frac{1}{x} = y^3 - 1$. Note that $x = \frac{1}{k}$.The series is $S = 2(\frac{1}{x}) + \frac{2}{3}(\frac{1}{x})^3 + \frac{2}{5}(\frac{1}{x})^5 + \dots = 2k + \frac{2}{3}k^3 + \frac{2}{5}k^5 + \dots$We know the logarithmic expansion $\log \left( \frac{1+k}{1-k} \right) = 2(k + \frac{k^3}{3} + \frac{k^5}{5} + \dots) = 2k + \frac{2}{3}k^3 + \frac{2}{5}k^5 + \dots$Substituting $k = y^3 - 1$:$S = \log \left( \frac{1 + (y^3 - 1)}{1 - (y^3 - 1)} \right) = \log \left( \frac{y^3}{1 - y^3 + 1} \right) = \log \left( \frac{y^3}{2 - y^3} \right)$.

If $0 < y < 2^{1/3}$ and $x(y^3 - 1) = 1$,then $\frac{2}{x} + \frac{2}{3x^3} + \frac{2}{5x^5} + \dots$ is equal to:

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