Question

How do you find the Maclaurin series for below function?

f(x)= `(x-cos x)/x^2` when x `!=` 0. And f(x)= `1/5` when x=0

Answer 1

You don't

Explanation:

Given:

`f(x) = { ((x-cos x)/x^2 " when " x != 0), (1/5 " when " x = 0) :}`

Note that:

`lim_(x->0) f(x) = lim_(x->0) (-1)/x^2 = -oo`

So `f(x)` is not even continuous at `x=0`, let alone differentiable.

Hence it has no Maclaurin series (i.e. Taylor expansion around `x=0`).