Question

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

Answer 1

Construct the Maclaurin series for `1/sqrt(1-t)` then substitute `t = x^2` to find:

`1/(sqrt(1-x^2)) = sum_(k=0)^oo (prod_(i=1)^k (2k-1))/(2^k k!) x^(2k)`

Explanation:

Let `f(t) = 1/sqrt(1-t)`

`f^((0))(t) = (1-t)^(-1/2)`

`f^((1))(t) = 1/2 (1-t)^(-3/2)`

`f^((2))(t) = 3/2^2 (1-t)^(-5/2)`

`f^((3))(t) = 15/2^3 (1-t)^(-7/2)`
...
`f^((k))(t) = (prod_(i=1)^k (2k-1))/2^k (1-t)^(-(2k+1)/2)`

So:

`f^((k))(0) = (prod_(i=1)^k (2k-1))/2^k`

Then the Maclaurin series for `f(t)` is:

`f(t) = sum_(k=0)^oo f^((k))(0)/(k!) t^k = sum_(k=0)^oo (prod_(i=1)^k (2k-1))/(2^k k!) t^k`

Substitute `t = x^2` to find:

`1/(sqrt(1-x^2)) = sum_(k=0)^oo (prod_(i=1)^k (2k-1))/(2^k k!) x^(2k)`