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
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