Fractional identity?

Question

Is there any way to re-write the following

$$p_{\hat y} \simeq\frac{1}{N-\frac{1}{T}U_1(x)+\frac{1}{2T^2}U_2(x)}$$

such that

$$p_{\hat y}\propto \left(U_1-\frac{U_2}{2T}\right)/T $$

$N$ is a positive integer number, $U_1$ and $U_2$ are functions of $x$ but they are always positive, $T$ is a parameter ranging in $[0,+\infty]$.

I found this on a paper, but I cannot understand why the signs of $U_1$ and $U_2$ change from minus to plus and vice-versa. Maybe there is some fractional identity that I forgot about. Thanks.

Answer 1

You can write your expression as

$$ p = (N - A)^{-1} $$

When $N$ is large this is

\begin{align} p &= N^{-1} (1- A/N)^{-1} \\ &= N^{-1} (1+ A/N +O(1/N^2)) \\ &=1/N + A/N^2 +O(1/N^3)) \end{align}

If you omit the first term, meaning that you are not interested in an offset, the remaining one is indeed proportional to $A$.

Answer 2

What it looks to me is that this is the result of a Taylor expansion for large values on $N$. $$\frac{1}{N-\frac{1}{T}U_1+\frac{1}{2T^2}U_2}=\frac{1}{N}+\frac{\frac{U_1}{T}-\frac{U_2}{2 T^2}}{N^2}+O\left(\frac{1}{N^3}\right)$$ that is to say $$\frac{1}{N}+\frac{1}{N^2}\left(U_1-\frac{U_2}{2T}\right)\frac 1T+\cdots$$