大B下(1-x₁/B)⁻¹的一阶二项式近似有效性技术问询

Absolutely, your approximation is completely valid here! Let’s walk through why this works step by step to confirm:

Breaking down the binomial approximation application #

You referenced the rule that when (|a x| \ll 1), the binomial approximation gives:
$$(1+x)^a \approx 1 + a x$$

In your case, we’re looking at the term (\left(1 - \frac{x_1}{B}\right)^{-1}). Let’s map this to the approximation formula:

• Here, (x = -\frac{x_1}{B})
• The exponent (a = -1)

Now, check the key condition: (|a x| = \left| (-1) \times \left(-\frac{x_1}{B}\right) \right| = \frac{x_1}{B}). Since (x_1) is a small positive number relative to the large (B), (\frac{x_1}{B} \ll 1)—this satisfies the approximation’s requirement perfectly.

Substituting into the formula gives:
$$\left(1 - \frac{x_1}{B}\right)^{-1} \approx 1 + (-1) \times \left(-\frac{x_1}{B}\right) = 1 + \frac{x_1}{B}$$

Applying this to the original expression #

Plugging this approximation back into your original term:
$$-B \left(1 - \frac{x_1}{B}\right)^{-1} \approx -B \left(1 + \frac{x_1}{B}\right)$$

If you simplify this further, you get (-B - x_1), which is a clean first-order approximation of the original expression. For most practical purposes where (x_1 \ll B), this is more than accurate enough—higher-order terms (like (\left(\frac{x_1}{B}\right)^2)) would be negligible here.

内容的提问来源于stack exchange，提问作者pafnuti