How do you derive the maclaurin series for #f(x)=e^(1/x^2)#?

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Answer 1 · 2016-11-24T21:41:54

Really you can't because #e^(1/x^2)# does not have continuous derivatives in #x=0#

The general formula of the McLaurin series is:

#sum_0^oo (f^"(n)"(0))/(n!#

but it requires that #f(x)# is continuous in #x=0# together with its derivatives of all orders, which #e^(1/x^2)# is not.

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

#f^'(x) = -2/x^3e^(1/x^2)#

#f^"(2)"(x) = 6/x^4*e^(1/x^2)+4/x^6e^(1/x^2) = (6x^2+4)/x^6e^(1/x^2)#

Answer 2 · 2016-11-24T23:05:55

The series would involve negative powers and so no such formal Taylor or Maclaurin series exists.

The function #f(x)=e^(1/x^2)# is not defined for #x=0# so technically #f(x)# is not a real analytical function.

However we can form a Laurent Series if we extend the function #f(z)# into the Complex Plane #CC#, and remove the singularity at #z=0#, and we get:

# e^(1/z^2) = sum_(k-0)^oo (1/z^2)^k/(k!) " "(z!=0)#

# e^(1/z^2) = 1 + 1/z^2 + 1/(2z^4) + 1/(6z^6) + 1/(24z^8) + ... " "(z!=0)#