# Why does this function act like this for large z?

• Nov 26th 2009, 03:45 AM
Siknature
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
• Nov 26th 2009, 05:40 AM
Shanks
Do you know the Talor fomulla in Caculas?
• Nov 26th 2009, 07:41 AM
Siknature
Quote:

Originally Posted by Shanks
Do you know the Talor fomulla in Caculas?

I can't see how to use it in this case
• Nov 26th 2009, 05:24 PM
HallsofIvy
He is suggesting you expand $\frac{1- u}{1+ u}$ in a Taylor's series about u= 0.