Question

How do you do the taylor series for expanding `1/(2+x^2)`, about a = 0 up to order n = 2?

Answer 1

Firstly we would find the Taylor series for the function centered at a=0, thus moreso finding is Maclaurin series.

we can make use of the common Maclaurin series of:

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

Now lets try and write our function to look similar to the common Maclaurin Series:

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

now we can just substitute `(-x^2/2)` in place of `x` in our common Maclaurin series, thus we get:

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

which can simplify a bit:

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

Now to get the expansion up to order n=2, we just follow the sum and get:

`(-1)^0 (x^2/2)^0 + (-1)^1 (x^2/2)^1 + (-1)^2 (x^2/2)^2`

which can simplify to:

`1 - x^2/2 + x^4/4`

note that this series is only convergent at `abs(x^2/2) < 1`

thus:

`-1< x^2/2 < 1`

`-2< x^2 < 2`

`-sqrt(2) < x < sqrt(2)`