1 - x + frac{x^2}{2!} - frac{x^3}{3!} + dots | vedclass

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Question

(B) The expansion of the exponential function $e^y$ is given by $e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \dots \infty$. Substituting $y = -x$

Vedclass · Accepted Answer

$e^{-x}$

Vedclass · Answer

(B) The expansion of the exponential function $e^y$ is given by $e^y = 1 + y + \frac{y^2}{2!} + \frac{y^3}{3!} + \dots \infty$. Substituting $y = -x$ into the series,we get: $e^{-x} = 1 + (-x) + \frac{(-x)^2}{2!} + \frac{(-x)^3}{3!} + \dots \infty$ $e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots \infty$. Thus,the correct option is $B$.

$1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots \infty = $

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