Question

How do you find the maclaurin series expansion of `f(x)=(1-x)^-2`?

Answer 1

`1/(1-x)^2 = sum_(n=0)^oo (n+1)x^n`

converging for `absx < 1`

Explanation:

Start from the geometric series:

`sum_(n=0)^oo x^n = 1/(1-x)`

converging for `abs(x) < 1`.

Note now that:

`1/(1-x)^2 = d/dx (1/(1-x)) = d/dx( sum_(n=0)^oo x^n)`

and inside the interval of convergence we can differentiate the series term by term, so:

`1/(1-x)^2 = sum_(n=0)^oo d/dx (x^n) = sum_(n=1)^oo nx^(n-1)`

Changing the index to start from `0`:

`1/(1-x)^2 = sum_(n=0)^oo (n+1)x^n`

Answer 2

`f(x) = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + ...+ nx^(n-1) + ...`

Explanation:

We could alternatively derive a MacLaurin Series by using the Binomial Expansion: The binomial series tell us that:

`(1+x)^n = 1+nx + (n(n-1))/(2!)x^2 (n(n-1)(n-2))/(3!)x^3 + ...`

And so for the given function, we can replace "`x`" by `-x` and substitute `n=-2`:

`f(x) = (1-x)^(-2)`

`\ \ \ \ \ \ \ = 1+(-2)(-x) + ((-2)(-3))/(2!)(-x)^2 + ((-2)(-3)(-4))/(3!)(-x)^3 + ((-2)(-3)(-4)(-5))/(4!)(-x)^4 + ...`

`\ \ \ \ \ \ \ = 1 + (2)x + (2.3)/(1.2)x^2 + (2.3.4)/(1.2.3)x^3 + (2.3.4.5)/(1.2.3.4)x^4 + ...`

`\ \ \ \ \ \ \ = 1 + 2x + 3x^2 + 4x^3 + 5x^4 + ...+ nx^(n-1) + ...`