The value of sqrt{e} is approximately: | vedclass

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Question

(A) The exponential series is given by $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$To find $\sqrt{e}$,we set $x = \frac{1}{2}$:$\sqrt{e} =

Vedclass · Accepted Answer

$1.648$

Vedclass · Answer

(A) The exponential series is given by $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots$To find $\sqrt{e}$,we set $x = \frac{1}{2}$:$\sqrt{e} = e^{1/2} = 1 + \frac{1}{2} + \frac{(1/2)^2}{2!} + \frac{(1/2)^3}{3!} + \frac{(1/2)^4}{4!} + \dots$$= 1 + 0.5 + \frac{0.25}{2} + \frac{0.125}{6} + \frac{0.0625}{24} + \dots$$= 1 + 0.5 + 0.125 + 0.02083 + 0.00260 + \dots$$= 1.64843 \dots$Rounding to three decimal places,we get $1.648$.

The value of $\sqrt{e}$ is approximately:

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