How do you find the first three terms of a Maclaurin series for f(t) = (e^t - 1)/t using the Maclaurin series of e^x?

1 Answer
NickTheTurtle
Apr 20, 2018

We know that the Maclaurin series of #e^x# is

#sum_(n=0)^oox^n/(n!)#

We can also derive this series by using the Maclaurin expansion of #f(x)=sum_(n=0)^oof^((n))(0)x^n/(n!)# and the fact that all derivatives of #e^x# is still #e^x# and #e^0=1#.

Now, just substitute the above series into
#(e^x-1)/x#

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

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

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

#=sum_(n=1)^oox^(n-1)/(n!)#

If you want the index to start at #i=0#, simply substitute #n=i+1#:

#=sum_(i=0)^oox^i/((i+1)!)#

Now, just evaluate the first three terms to get

#~~1+x/2+x^2/6#

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