Taylor approximation of $R=1-\frac{2y}{N^2}$

Question

Can somebody please explain to me how $R=1-\frac{2y}{N^2}$, (for large N) by Taylor approximation gives $R^N \approx 1 + \frac{2y}{N}$?

Answer 1

$$\begin{align}R^N&=\left(1-\frac{2y}{N^2}\right)^N \\&=1-\frac{2y}{N^2}\cdot N+\left(\frac{2y}{N^2}\right)^2\cdot \frac{N(N-1)}{2!}-\left(\frac{2y}{N^2}\right)^3\cdot\frac{N(N-1)(N-2)}{3!}+\cdots \\&=1-\frac{2y}{N}+O(N^{-4})\cdot O(N^2)-O(N^{-6})\cdot O(N^3)+\cdots\\&=1-\frac{2y}{N}+o\left(\frac1N\right)\end{align}$$ So for large $N$, the first two terms contribute the most.

Answer 2

Let $f(x)=(1-x)^N$. Then using $$ f(x)=f(0)+f'(0)x+\frac{1}{2}f''(\theta)x^2, \theta\in(0,x)$$ one has $$(1-x)^N=1-Nx+\frac{1}{2}N(N-1)(1-\theta)^{N-2}x^2. $$ Letting $x=\frac{2y}{N^2}$, one has $$ R^N=1-\frac{2y}{N}+\frac12N(N-1)(1-\theta)^{N-2}\frac{4y^2}{N^4}=1-\frac{2y}{N}+O(N^{-2}).$$

Answer 3

$$R=1-\frac{2y}{N^2}\implies \log(R)=\log\left(1-\frac{2y}{N^2} \right)\implies\log(R^N)=N\log\left(1-\frac{2y}{N^2} \right)$$ Now, use Taylor $$\log\left(1-\frac{2y}{N^2} \right)=-\frac{2 y}{N^2}-\frac{2 y^2}{N^4}+O\left(\frac{1}{N^6}\right)$$ $$\log(R^N)=N\log\left(1-\frac{2y}{N^2} \right)=-\frac{2 y}{N}-\frac{2 y^2}{N^3}+O\left(\frac{1}{N^5}\right)$$ $$R^N=e^{\log(R^N)}=1-\frac{2 y}{N}+\frac{2 y^2}{N^2}+O\left(\frac{1}{N^3}\right)=1-\frac{2 y}{N}+O\left(\frac{1}{N^2}\right)$$