Question

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?

Answer 1

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`