How do you find the Maclaurin series for #f(t) =t^3(e^(-t^2))# centered at 0?

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Answer 1 · 2017-01-13T00:05:54

#t^3e^(-t^2) = sum_(n=0)^oo (-1)^n t^(2n+3)/(n!)#

Start with the MacLaurin series of the exponential function:

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

Substitute #x =-t^2#

#e^(-t^2) = sum_(n=0)^oo (-t^2)^n/(n!) = sum_(n=0)^oo (-1)^nt^(2n)/(n!) #

Then multiply by #t^3# term by term:

#t^3e^(-t^2) =sum_(n=0)^oo (-1)^n t^3t^(2n)/(n!) = sum_(n=0)^oo (-1)^n t^(2n+3)/(n!)#

I would also note that a MacLaurin series is always centered at #x=0#, otherwise it is called a Taylor series.