What is the Maclaurin series for? : $f(x) = x^2 - 3x$

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Answer 1 · 2017-08-04T01:14:18

As $f(x)$ is a polynomial, then the Maclaurin Series is itself. Thus the Maclaurin Series is:

$f(x) = -3x + x^2$

We have:

$f(x) = x^2 - 3x$

As $f(x)$ is a polynomial, then the Maclaurin Series is itself.

Not convinced? Then let us dewrive the series from first principles:

A Maclaurin series can be expressed in the following way:

$f(x) = f(0) + (f'(0))/(1!) x + (f''(0))/(2!) x^2 + (f'''(0))/(3!) x^3 + (f^((4))(0))/(4!) x^4 + ...$
$" "= sum_(n=0)^(∞) f^((n))(0)/(n!) x^n$

So, Let us find the derivatives, and compute the values at $x=0$

$f(x) =x^2-3x$
$f'(x) =2x-3$
$f''(x) =2$
$f'''(x) =0$ , and all higher derivatives are zero

And so when $x=0$ we have:

$f(0) =0$
$f'(0) =-3$
$f''(0) =2$
$f'''(0) =0$ , and all higher derivatives are zero
$...$

So the Maclaurin series is:

$f(x) = 0 + (-3)/(1!) x + (2)/(2!) x^2 + (0)/(3!) x^3 + (0)/(4!) x^4 + ...$
$" " = (-3)/(1) x + (2)/(1) x^2$
$" " = -3x + x^2 \ \ \$ QED

What is the Maclaurin series for? : f(x) = x^2 - 3x f(x) = x^2 - 3x