表达式的大O记号渐近分析：求分离主导项及误差项的方法

Alright, let's walk through how to find the leading term and Big-O error for your expression as $x \to \infty$. I'll break this down into straightforward steps so you can follow along:

First, note that as $x$ approaches infinity, $\ln x$ also goes to infinity. That means the constant $-1$ inside the square root becomes totally negligible compared to $\ln x$. We can leverage this to simplify the square root term using a binomial expansion (since $1/\ln x$ will tend to 0, making it a small correction term).

Start by factoring out the dominant part of the square root's argument:
$$\sqrt{2(\ln x - 1)} = \sqrt{2\ln x \left(1 - \frac{1}{\ln x}\right)} = \sqrt{2\ln x} \cdot \sqrt{1 - \frac{1}{\ln x}}$$

Now use the binomial approximation for $(1 + \epsilon)^{1/2}$ where $\epsilon = -1/\ln x$ (since $\epsilon \to 0$ as $x \to \infty$):
$$\sqrt{1 - \frac{1}{\ln x}} = 1 - \frac{1}{2\ln x} - \frac{1}{8(\ln x)^2} + O\left(\frac{1}{(\ln x)^3}\right)$$

Multiply this by $\sqrt{2\ln x}$ to expand the full square root term:
$$\sqrt{2(\ln x - 1)} = \sqrt{2\ln x} - \frac{\sqrt{2\ln x}}{2\ln x} - \frac{\sqrt{2\ln x}}{8(\ln x)^2} + O\left(\frac{\sqrt{\ln x}}{(\ln x)^3}\right)$$

Simplify each term to see their growth rates:

• The second term simplifies to $\frac{1}{\sqrt{2\ln x}}$
• The third term becomes $\frac{\sqrt{2}}{8(\ln x)^{3/2}}$
• The error term reduces to $O\left(\frac{1}{(\ln x)^{5/2}}\right)$ (since $\frac{\sqrt{\ln x}}{(\ln x)^3} = \frac{1}{(\ln x)^{5/2}}$)

So now we have:
$$\sqrt{2(\ln x - 1)} = \sqrt{2\ln x} - \frac{1}{\sqrt{2\ln x}} - \frac{\sqrt{2}}{8(\ln x)^{3/2}} + O\left(\frac{1}{(\ln x)^{5/2}}\right)$$

Substitute this expanded square root back into your original expression:
$$\frac{1}{A}\left(\ln x + \sqrt{2(\ln x - 1)}\right) = \frac{1}{A}\left[\ln x + \sqrt{2\ln x} - \frac{1}{\sqrt{2\ln x}} - \frac{\sqrt{2}}{8(\ln x)^{3/2}} + O\left(\frac{1}{(\ln x)^{5/2}}\right)\right]$$

Now we just need to rank the terms by their growth rate as $x \to \infty$:

• $\ln x$ grows linearly with $\ln x$ (this is the leading term, since it's the fastest-growing)
• $\sqrt{2\ln x}$ grows like $(\ln x)^{1/2}$ (slower than $\ln x$)
• All remaining terms tend to 0 as $x \to \infty$

Depending on how precise you need to be, you can write the expression in a few ways:

1. Basic leading term with error:
   $$\frac{1}{A}\left(\ln x + \sqrt{2(\ln x - 1)}\right) = \frac{\ln x}{A} + O\left(\sqrt{\ln x}\right)$$
   This captures that all other terms grow slower than $\ln x$.

2. Leading term plus next subdominant term:
   $$\frac{1}{A}\left(\ln x + \sqrt{2(\ln x - 1)}\right) = \frac{\ln x}{A} + \frac{\sqrt{2\ln x}}{A} + O\left(\frac{1}{\sqrt{\ln x}}\right)$$
   This includes the next fastest-growing term and bounds the error to terms that decay like $1/\sqrt{\ln x}$.

3. Even tighter expansion:
   $$\frac{1}{A}\left(\ln x + \sqrt{2(\ln x - 1)}\right) = \frac{\ln x}{A} + \frac{\sqrt{2\ln x}}{A} - \frac{1}{A\sqrt{2\ln x}} + O\left(\frac{1}{(\ln x)^{3/2}}\right)$$
   This adds the first decaying term and tightens the error bound further.

• Always start by identifying which terms dominate as the variable approaches the limit—this tells you where to focus.
• Use Taylor or binomial expansions for terms with small correction factors (like $1 - 1/\ln x$ here) to break down subdominant terms.
• Simplify each term to compare their growth rates; Big-O notation is all about capturing the rate of growth/decay of the remaining terms.

内容的提问来源于stack exchange，提问作者N. Younger