Question

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

Answer 1

The Maclaurin series is the same as the Taylor series, except it is expanded around `a = 0`.

So, you can start by assuming the Taylor series definition:

`sum_(n = 0)^N f^((n))(a)/(n!)(x-a)^n`

and modifying it to get:

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

Now, we can take the `n`th derivative. Let's say we go to `n = 3` only, because I know this is going to get a bit ridiculous to do.

`f^((0))(x) = color(green)(f(x)) = (1-x^2)^(-1) = color(green)(1/(1-x^2))`

`color(green)(f'(x)) = -(1-x^2)^(-2)(-2x) = (2x)(1-x^2)^(-2) = color(green)((2x)/(1-x^2)^2)`

`color(green)(f''(x)) = (2x)(-2(1-x^2)^(-3)(-2x)) + (1-x^2)^(-2)(2)`

`= 8x^2(1-x^2)^(-3) + 2(1-x^2)^(-2) = color(green)((8x^2)/(1-x^2)^3 + 2/(1-x^2)^2)`

`color(green)(f'''(x)) = [(8x^2)(-3(1-x^2)^(-4)(-2x)) + (1-x^2)^(-3)(16x)] + [2*(-2(1-x^2)^(-3)(-2x))]`

`= [(48x^3)(1-x^2)^(-4) + (1-x^2)^(-3)(16x)] + [8x(1-x^2)^(-3)]`

`= (48x^3)/(1-x^2)^(4) + (16x)/(1-x^2)^(3) + (8x)/(1-x^2)^(3)`

`= color(green)((48x^3)/(1-x^2)^(4) + (24x)/(1-x^2)^(3))`

So the Maclaurin series up to `n = 3` is:

`sum_(n = 0)^3 f^((n))(0)/(n!)x^n`

`= [1/(1-a^2)]/(0!)(x-a)^0 + [(2a)/(1-a^2)^2]/(1!)(x-a)^1 + [(8a^2)/(1-a^2)^3 + 2/(1-a^2)^2]/(2!)(x-a)^2 + [(48a^3)/(1-a^2)^(4) + (24a)/(1-a^2)^(3)]/(3!)(x-a)^3`

`= [1/(1-(0)^2)]/(0!)x^0 + [(2(0))/(1-(0)^2)^2]/(1!)x^1 + [(8(0)^2)/(1-(0)^2)^3 + 2/(1-(0)^2)^2]/(2!)x^2 + [(48(0)^3)/(1-(0)^2)^(4) + (24(0))/(1-(0)^2)^(3)]/(3!)x^3 + ...`

`= color(blue)(1 + x^2 + x^4 + ...)`

The odd terms just go away. How convenient!

Answer 2

Alternatively
`1/(1-x^2)= 1/((1-x)(1+x)) =1/2[ 1/(1-x) + 1/(1+x)]`
using `1/(1-x) = sum x^n` we get
`1/(1+x) = sum (-1)^n x^n`
and finally
`1/(1-x^2)= 1/2 sum( 1 + (-1)^n) x^n`
It's nice when you don't have to calculate derivatives :)

Answer 3

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

converging for `abs(x) < 1`

Explanation:

Note that:

`(1-t) sum_(k=0)^n t^k = sum_(k=0)^n t^k - t sum_(k=0)^n t^k`

`color(white)((1-t) sum_(k=0)^n t^k) = sum_(k=0)^n t^k - sum_(k=1)^(n+1) t^k`

`color(white)((1-t) sum_(k=0)^n t^k) = 1+color(red)(cancel(color(black)(sum_(k=1)^n t^k))) - color(red)(cancel(color(black)(sum_(k=1)^n t^k)))-t^(n+1)`

`color(white)((1-t) sum_(k=0)^n t^k) = 1-t^(n+1)`

Dividing both ends by `(1-t)`, we find:

`sum_(k=0)^n t^k = (1-t^(n+1))/(1-t)`

So if `abs(t) < 1`, we find:

`sum_(k=0)^oo t^k = lim_(n->oo) (1-t^(n+1))/(1-t) = 1/(1-t)`

Putting `t = x^2` we find:

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

converging for `abs(x) < 1`.

Answer 4

`1/(1-x^2) = 1 + x^2 + x^4 + x^6 + x^8 + x^10 + ...`

Explanation:

I will use a different approach because I'm a Mathematician (and therefore implicitly lazy), so I like to go for the shortest solution:

Firstly, and very importantly, power series are unique, so irrespective of what different methods are used to produce a series the end results must be the same.

On that basis, I will use the binomial theorem:

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

So that:

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

`1/(1-x^2) = (1+(-x^2))^(-1)`

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

`" " = 1 + (1)(x^2) + (1.2)/(1.2)(x^2)^2 +`
`" " (1.2.3)/(1.2.3)(x^2)^3 + (1.2.3.4)/(1.2.3.4)(x^2)^4 +`
`" " (1.2.3.4.5)/(1.2.3.4.5)(x^2)^5 + ...`

`" " = 1 + (x^2) + (x^2)^2 + (x^2)^3 + (x^2)^4 + (x^2)^5 + ...`

`" " = 1 + x^2 + x^4 + x^6 + x^8 + x^10 + ...`

Answer 5

`1/(1 - x^2) = 1 + x^2 + x^4 + ... + x^(2n) = sum_(n = 0)^oo x^(2n)`

Explanation:

Recall that the sum of an infinite geometric series is given by

`s_oo = a/(1 - r)`

Where the terms are `a, ar, ar^2, ..., ar^n`

Here, `a = 1`. If we let `r = x^2`, and plug it into the formula, we get:

`1/(1 - x^2) = 1 + 1(x^2) + 1(x^2)^2 + ...`

Which can be rewritten as

`1/(1 - x^2) = 1 + x^2 + x^4 = sum_(n = 0)^oo x^(2n)`

Hopefully this helps!