Question

Question #90ee2

Answer 1

The first nonzero term of the Maclaurin series for the given function is `x`

Explanation:

The general form of the Maclaurin series for a function `f(x)` is

`sum_(n=0)^oof^((n))(0)/(n!)x^n`

For the given function, at `n=0` we have

`f(0)/(0!)x^0 = f(0) = 0`

As this is not nonzero, we move on to `n=1`

`(f'(0))/(1!)x^1 = f'(0)*x`

The first derivative of `f` is

`f'(x) = -2x^2e^(-x^2)+e^(-x^2)`

Thus we have our first nonzero term as

`f'(0)*x = (-2*0^2e^(-0^2)+e^(-0^2))x`

`=(0+1)x`

`=x`