Question

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

Answer 1

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

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

Explanation:

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