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

##### 2 Answers

**Maclaurin Series**

The Maclaurin series for

#e^x=sum_(n=0)^oox^n/(n!)=1+x+x^2/(2!)+x^3/(3!)+x^4/(4!)+...#

Replacing

#e^(-1/x)=sum_(n=0)^oo(-1/x)^n/(n!)=sum_(n=0)^oo(-1)^n/(n!)x^-n#

#color(white)(e^(-1/x))=1-1/x+1/((2!)x^2)-1/((3!)x^3)+1/((4!)x^4)+...#

See description:

The Taylor Series about

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

#### Explanation:

Let

The Taylor Series about the pivot point

# f(x) = f(a) + f'(a)(x-a) + (f''(a))/(2!)(x-a)^2 + (f^((3))(a))/(3!)(x-a)^3 + ... + (f^((n))(a))/(n!)(x-a)^n + ... #

As no pivot point for the Taylor Expansion Series has been provided it would be usual practice to assume that

# f(x) = f(0) + f'(0)x + (f''(0))/(2!)x^2 + (f^((3))(0))/(3!)x^3 + ... + (f^((n))(0))/(n!)x^n + ... #

However,

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

Using the well know series for

# e^(-1/z) = 1-1/z+1/(2!z^2)-1/(3!z^3)+1/(4!z^4)+...#

We can however form a Taylor Series about another pivot point so lets do so about

Firstly, we have:

# f(1)=e^(-1) =1/e #

We need the first derivative:

# f'(x)=e^(-1/x)/x^2 #

# :. f'(1)=e^(-1)/1 =1/e #

And the second derivative (using quotient rule):

# f''(x) = ( (x^2)(e^(-1/x)/x^2)-(e^(-1/x))(2x) ) / (x^2)^2 #

# " " = ( e^(-1/x)(1-2x) ) / (x^4) #

# :. f''(1) = -1/e #

#vdots#

And so the Taylor Series about

# f(x) = 1/e + 1/e(x-1) + (-1/e)/(2!)(x-1)^2 + (f^((3))(a))/(3!)(x-1)^3 + ... #

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