Thread: Why does this function act like this for large z?

I have been told that, for large z,

f(z) = [(z - 1) / (z + 1)]^(1/6)

= [(1-(1/z)) / (1 + (1/z))] by dividing through by z

= 1 - (1/3z) + (1/(18z^2)) + O(z^-3)

so write

f(z) = 1 - (1/3z) + g(z) where g(z) = O(z^-2) as z --> infinity

I have no idea how why the 3rd and 4th lines are as they are. What is the working behind them?

thanks

Do you know the Talor fomulla in Caculas?

Originally Posted by Shanks

Do you know the Talor fomulla in Caculas?

I can't see how to use it in this case

He is suggesting you expand in a Taylor's series about u= 0.