If 0

Question

(A) Given $x(y^3 - 1) = 1$,we have $\frac{1}{x} = y^3 - 1$. Since $0 < y < 2^{1/3}$,we have $0 < y^3 < 2$,so $-1 < y^3 - 1 < 1$. Thus,$|1/x| < 1$. The

Vedclass · Accepted Answer

$\log \left[ \frac{y^3}{y^3 - 2} \right]$

Vedclass · Answer

(A) Given $x(y^3 - 1) = 1$,we have $\frac{1}{x} = y^3 - 1$. Since $0 Thus,$|1/x| The given series is $S = 2 \left[ \frac{1}{x} + \frac{1}{3x^3} + \frac{1}{5x^5} + \dots \right]$. Using the logarithmic series expansion $\ln \left( \frac{1+t}{1-t} \right) = 2 \left( t + \frac{t^3}{3} + \frac{t^5}{5} + \dots \right)$ for $|t| $S = \ln \left( \frac{1 + 1/x}{1 - 1/x} \right) = \ln \left( \frac{x+1}{x-1} \right)$. Substituting $x = \frac{1}{y^3 - 1}$: $S = \ln \left( \frac{\frac{1}{y^3 - 1} + 1}{\frac{1}{y^3 - 1} - 1} \right) = \ln \left( \frac{1 + y^3 - 1}{1 - (y^3 - 1)} \right) = \ln \left( \frac{y^3}{2 - y^3} \right)$. Note: The provided option $(A)$ is $\ln \left( \frac{y^3}{y^3 - 2} \right)$,which is equivalent to $\ln \left( -\frac{y^3}{2 - y^3} \right)$. Given the domain constraints,the expression simplifies to $\ln \left( \frac{y^3}{2 - y^3} \right)$.

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 = $

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