Taylor Expansion for (a(1+z)^3 + b)^-1/2 around z=0 to First Order

[indie452]

Homework Statement

I have the function

(a(1+z)³ + b)^-1/2

and i need to taylor expand it around z=0 to the first order,
a and b are constants, there sum is equal to one.

I have the answer:
1 - (1+q)z
where q = a/2 - b

This is in my physics book but it does not explain the steps inbetween. I have tried to derive it myself but i can't get same answer (or close to it), and I don't want to just accept it cause i might need to do it in the exam in the summer.

 

[tiny-tim]

hi indie452!

a and b are constants, there sum is equal to one.

I have the answer:
1 - (1+q)z
where q = a/2 - b

are you sure you have the wrong answer?

(1+q) = 1 + a/2 - b = 1 + a/2 - 1 + a = 3a/2