Question #90ee2

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Answer 1 · 2016-03-02T03:08:38

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

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

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

$=(0+1)x$

$=x$