# What is the taylor expansion of #e^(-1/x)#?

Maclaurin Series

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The Taylor Series about

#e^(-1/x) = 1/e + (x-1)/e - (x-1)^2/(2e) + ... #

Technically this is the end of the question - There is no such series.

Firstly, we have:

We need the first derivative:

And the second derivative (using quotient rule):

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The Taylor expansion of ( e^{-1/x} ) about ( x = 0 ) is:

[ e^{-1/x} = \sum_{n=0}^{\infty} \frac{(-1)^n}{n!} \left(\frac{1}{x}\right)^n ]

This is the Maclaurin series for ( e^{-1/x} ), centered at ( x = 0 ).

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