Question

find the maclaurin series for f(x) using the definition of a maclaurin series. [assume that f has a power series expansion. do not show that rn(x) → 0.] f(x) = 9(1 − x)−2

Answer 1

It looks like you have

`f(x) = \frac9{(1-x)^2}`

Compute the first several derivatives and evaluate them at
`x=0`.

`f'(x) = (9\cdot2)/((1-x)^3) \implies f'(0) = 18`

`f''(x) = (9\cdot2\cdot3)/((1-x)^4) \implies f''(0) = 54`

`f'''(x) = (9\cdot2\cdot3\cdot4)/((1-x)^5) \implies f'''(0) = 216`

and so on, so that the
`n`-th derivative is

`f^((n))(x) = (9(n+1)!)/((1-x)^(n+2)) \implies f^((n))(0) = 9(n+1)!`

Then the Maclaurin series of
`f(x)` is

`\displaystyle f(x) = \sum_(n=0)^\infty (f^((n))(0))/(n!)x^n`

`\displaystyle \frac9{(1-x)^2} = \boxed{\sum_(n=0)^\infty 9(n+1)x^n}`

Alternatively, we can start from the Maclaurin series

`\displaystyle \frac1{1-x} = \sum_(n=0)^\infty x^n`

Differentiate both sides and multiply by 9.

`\displaystyle \frac1{(1-x)^2} = \sum_(n=0)^\infty n x^(n-1)`

`\displaystyle \frac1{(1-x)^2} = \sum_(n=1)^\infty n x^(n-1)`

`\displaystyle \frac1{(1-x)^2} = \sum_(n=0)^\infty (n+1) x^n`

`\displaystyle \frac9{(1-x)^2} = \sum_(n=0)^\infty 9 (n+1) x^n`