Infinite series as exponent

Calculus Level 2

e 1 2 − 1 4 + 1 6 − ⋯ = b \large e^{\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-{\cdots } }=\sqrt{b}

Enter ⌊ b ⌋ \lfloor b \rfloor .

Notation: ⌊ ⋅ ⌋ \lfloor \cdot \rfloor denotes the floor function .


The answer is 2.

Rafsan Rcc by Rafsan Rcc , 17, Bangladesh

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2 solutions

By Maclaurin's series ,

ln ⁡ ( 1 + x ) = x − x 2 2 + x 3 3 − ⋯ ⟹ ln ⁡ ( 1 + 1 ) = 1 − 1 2 + 1 3 − ⋯ e 1 2 − 1 4 + 1 6 − ⋯ = e 1 2 ( 1 − 1 2 + 1 3 − ⋯ ) = e 1 2 ln ⁡ 2 = 2 \begin{aligned} \ln (1+x) & = x - \frac {x^2}2 + \frac {x^3}3 - \cdots \\ \implies \ln (1 + 1) & = 1 - \frac 12 + \frac 13 - \cdots \\ e^{\frac 12 - \frac 14 + \frac 16 - \cdots} & = e^{\frac 12 \left(1-\frac 12 + \frac 13 - \cdots \right)} = e^{\frac 12 \ln 2} = \sqrt 2 \end{aligned}

Therefore ⌊ b ⌋ = ⌊ 2 ⌋ = 2 \lfloor b \rfloor = \lfloor 2 \rfloor = \boxed 2 .

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Rafsan Rcc
May 3, 2021

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You could do this instead

ln ⁡ 1 + x 2 = x 2 − x 2 4 … \frac{\ln{1+x}}{2}=\frac x2-\frac {x^2}{4} …

ln ⁡ 1 + 1 2 = Required exponent = ln ⁡ 2 \frac{\ln{1+1}}{2} = \text{Required exponent} = \ln{\sqrt 2}

Jason Gomez - 1 month, 1 week ago

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