Question

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

Answer 1

`x^2ln(3-x) = ln3 \ x^2 -1/3 x^3 + ...`

Explanation:

The 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`

Let use define the function,

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

First Term

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

So when `x=0` we have:

`f(0) = 0`

Second Term:

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

Using the Product Rule, in conjunction with the Chain Rule, we can differentiate wrt `x`:

`f'(x) = (x^2)(1/(3-x))(-1) + (2x)(ln(3-x)`
`\ \ \ \ \ \ \ \ \ = 2xln(3-x) - x^2/(3-x)`

And so:

`f'(0) = 0`

Third Term:

`f'(x) = 2xln(3-x) - x^2/(3-x)`

Using the Product Rule, and Quotient Rule, in conjunction with the Chain Rule, we can differentiate again wrt `x`:

`f''(x) = (2x)(1/(3-x))(-1) + (2)(ln(3-x)) - ( (3-x)(2x)-(x^2)(-1))/(3-x)^2`
`\ \ \ \ \ \ \ \ \ = 2ln(3-x) - (4x)/(3-x) - (x^2)/(3-x)^2`

And so:

`f''(0) = 2ln3`

Fourth Term:

`f''(x) = 2ln(3-x) - (4x)/(3-x) - (x^2)/(3-x)^2`

Using the Quotient Rule, in conjunction with the Chain Rule, we can differentiate again wrt `x`:

`f^((3))(x) = (2(x^2-9x+27))/(x-3)^3`
`:. f^((3))(0) = -2`

and so on ...

Using these calculations, wecan now construct the Maclaurin Series:

`f(x) = 0 + 0x + (2ln3)/(2!) x^2 + (2)/(3!) x^3 + ...`
`f(x) = ln3 \ x^2 -1/3 x^3 + ...`