If |x|

Question

(A) Given the series expansion: $y = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$ This is the standard logarithmic series expansion for

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$y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots$

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(A) Given the series expansion: $y = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$ This is the standard logarithmic series expansion for $\log(1+x)$ where $|x| So,$y = \log(1+x)$. Taking the exponential of both sides: $e^y = 1+x$. Therefore,$x = e^y - 1$. The Taylor series expansion for $e^y$ is $e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots$. Substituting this into the expression for $x$: $x = (1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots) - 1$. $x = y + \frac{y^2}{2!} + \frac{y^3}{3!} + \ldots$.

If $|x| < 1$ and $y = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots$,then $x$ is equal to :

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