Question

How do you use the binomial series to expand `sqrt(z^2-1)`?

Answer 1

`sqrt(z^2-1) =i[1-1/2z^2 - 1/8z^4 - 1/16z^6+...]`

Explanation:

I'd quite like a double check because as a physics student I rarely get beyond `(1+x)^n ~~ 1+nx` for small x so I'm a bit rusty. The binomial series is a specialised case of the binomial theorem which states that

`(1+x)^n = sum_(k=0)^(oo) ((n),(k)) x^k`

With `((n),(k)) = (n(n-1)(n-2)...(n-k+1))/(k!)`

What we have is `(z^2-1)^(1/2)`, this is not the correct form. To rectify this, recall that `i^2 = -1` so we have:

`(i^2(1-z^2))^(1/2) = i(1-z^2)^(1/2)`

This is now in the correct form with `x = -z^2`

Therefore, the expansion will be:

`i[1 -1/2z^2 + (1/2(-1/2))/2z^4 - (1/2(-1/2)(-3/2))/6z^6 + ...]`

`i[1-1/2z^2 - 1/8z^4 - 1/16z^6+...]`