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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.

Phys
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  • Please describe what each term is: which terms are free, bound, variables, functions, etc.. – Jam Feb 04 '20 at 12:14
  • The sums in the denominator make no sense to me; how do they depend on $i$? They could ostensibly be factored out as constants. – Jam Feb 04 '20 at 12:17
  • Thanks for the suggestions, fixed – Phys Feb 04 '20 at 12:21
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    Can you show us the paper this came from? It still seems weird to me. If we take $N=T=1$ and $U_1(x)=U_2(x)=|x|+1$, the two expressions are clearly different. – Jam Feb 04 '20 at 12:27
  • Sure Jam, thanks. THe paper is https://arxiv.org/pdf/1706.02690.pdf. You can find the first formula in page 7, while the second bit is on page 8, "The effect of T". – Phys Feb 04 '20 at 12:30
  • This may be an instance of Taylor expansion.. – Trebor Feb 04 '20 at 12:34
  • Hi Trebor, if you mean the first line I wrote: yes, that is the result of a Taylor expansion. The lower branch is just an expanded exponential function. The problem is that they claim that type of proportionality. – Phys Feb 04 '20 at 12:39

2 Answers2

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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 $$\frac{1}{N}+\frac{1}{N^2}\left(U_1-\frac{U_2}{2T}\right)\frac 1T+\cdots$$

Claude Leibovici
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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 $A$.

lcv
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