Question

How do you find `f^6(0)` where `f(x)=arctanx/x`?

Answer 1

`f^((6))(0) = -720/7`

Explanation:

We start by constructing the MacLaurin series for `arctan x`.
Consider the function:

`f(x) = 1/(1+x^2)`

This is the sum of a geometric series of ratio: `-x^2`, so that we have:

`1/(1+x^2) =sum_(n=0)^oo (-x^2)^n = sum_(n=0)^oo (-1)^nx^(2n)`

with radius of convergence `R=1`.

Within the interval `x in (-1,1)` we can therefore integrate term by term:

`int_0^x dt/(1+t^2) =sum_(n=0)^oo (-1)^n int t^(2n)dt`

`arctan x=sum_(n=0)^oo (-1)^n x^(2n+1)/(2n+1)`

and dividing by `x` both sides:

`arctanx/x = sum_(n=0)^oo (-1)^n x^(2n)/(2n+1)`

Now consider the standard expression of the MacLaurin series of `f(x)`

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

The two series are equal only if the coefficients of the same degree in `x` are equal, so that for `x^6` we have:

`f^((6))(0)/(6!) = -1/7`

and:

`f^((6))(0) = -720/7`