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

1 Answer
Jul 13, 2016

Answer:

#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+...]#