Question

How do you find the first two nonzero terms in Maclaurin's Formula and use it to approximate `f(1/3)` given `f(x)=sin(e^x)`?

Answer 1

The Maclaurin series for `sin(x)` is given by:

`sin(x)=sum_(n=0)^oo(-1)^n/((2n+1)!)x^(2n+1)=x-x^3/(3!)+x^5/(5!)+...`

Replacing `x` with `e^x`:

`sin(e^x)=sum_(n=0)^oo(-1)^n/((2n+1)!)(e^x)^(2n+1)=e^x-(e^x)^3/(3!)+(e^x)^5/(5!)+...`

So the first two nonzero terms are:

`f(x)=e^x-(e^x)^3/(3!)=e^x-e^(3x)/6`

So our approximation is:

`f(1/3)approxsin(e^(1/3))=e^(1/3)-e^(3(1/3))/6=e^(1/3)-e/6approx0.9426`

This compares to the actual value of `sin(e^(1/3))approx0.9847`.

Answer 2

`sin(e^(1/3)) ~~ 1`

Explanation:

The Maclaurin series for `f(x)` is given by:

`f(x) = sum_(n=0)^oo f^((n))(0)/(n!) x^n`

In our example:

`f^((0))(x) = sin(e^x)" "` so: `f^((0))(0) = sin(1)`

`f^((1))(x) = e^x cos(e^x)" "` so: `f^((1))(0) = cos(1)`

`f^((2))(x) = e^x cos(e^x) - e^(2x) sin(e^x)" "` so: `f^((2))(0) = cos(1)-sin(1)`

etc.

So the first few terms of the Maclaurin series for `sin(e^x)` are:

`(sin 1)+(cos 1)x+1/2(cos 1- sin 1)x^2+O(x^3)`

Using:

`sin 1 ~~ 0.8415`

`cos 1 ~~ 0.5403`

Then with `x=1/3` we have:

`sin(e^x) ~~ 0.8415+0.5403*1/3+1/2(0.5403-0.8415)*1/9+O(x^3)`

`color(white)(sin(e^x)) ~~ 0.8415+0.1801-0.0167+O(x^3)`

`color(white)(sin(e^x)) ~~ 1.0049`

Note that `sin` should be in the range `[-1, 1]` so we have not really used enough terms.

A calculator tells me:

`sin(e^(1/3)) ~~ 0.9846945`