Question

How do you find the Maclaurin Series of `f(x) = 8x^3 - 3x^2 - 4x + 3`?

Answer 1

`f(x) = sum_(n=0)^oo (f^((n))(0))/(n!)x^n =3-4x-3x^2+8x^3`

Explanation:

In a sense you already have it in the question, though a Maclaurin series is usually written in ascending powers of `x`.

Using the general formula for a Maclaurin series, we have:

`f(x) = sum_(n=0)^oo (f^((n))(0))/(n!)x^n`

In our example:

`f^((0))(x) = 8x^3-3x^2-4x+3 color(white)(X)rArrcolor(white)(X) f^((0))(0) = 3`

`f^((1))(x) = 24x^2-6x-4 color(white)(XXX)rArrcolor(white)(X) f^((1))(0) = -4`

`f^((2))(x) = 48x-6 color(white)(XXXXXX)rArrcolor(white)(X) f^((2))(0) = -6`

`f^((3))(x) = 48 color(white)(XXXXXXXXX)rArrcolor(white)(X) f^((3))(0) = 48`

`f^((n))(x) = 0` for `n >= 4`

So:

`f(x) = sum_(n=0)^oo (f^((n))(0))/(n!)x^n`

`=(3/(0!))x^0 + ((-4)/(1!))x^1 + ((-6)/(2!))x^2 + (48/(3!))x^3`

`=3-4x-3x^2+8x^3`