验证|e^z-∑ₖ=0^10T zᵏ/k!|≤e^(-9T)及斯特林近似相关问题

Alright, let's break down your question step by step:

First, let's address the core question: does the inequality (|e^z - \sum_{0 \leq k \leq 10T} \frac{z^k}{k!}| \leq e^{-9T}) hold when (z \in \mathbb{C}), (T \in \mathbb{R}), and (|z| \leq T)?

1. Verifying the Remainder Bound #

You already correctly noted that (|\sum_{k > 10T} \frac{z^k}{k!}| \leq \sum_{k > 10T} \frac{|z|^k}{k!} \leq \sum_{k > 10T} \frac{T^k}{k!}) (by the triangle inequality and (|z| \leq T)). Now we just need to confirm this real-valued sum is bounded by (e^{-9T}):

• For small (T): The Taylor series of (e^T) converges extremely quickly. For example, when (T=0.5), (10T=5), the remainder (e^{0.5} - \sum_{k=0}^5 \frac{0.5^k}{k!} \approx 3 \times 10^{-6}), which is way smaller than (e^{-4.5} \approx 0.011). For (T=1), the remainder is ~(2.7 \times 10^{-8}), compared to (e^{-9} \approx 1.2 \times 10^{-4}).
• For large (T): We can use large deviation theory for Poisson random variables. The sum (\sum_{k > 10T} \frac{T^k}{k!} = e^T \cdot P(X > 10T)), where (X \sim \text{Poisson}(T)). The large deviation bound gives (P(X > 10T) \leq e^{-T(10\ln10 - 9)} \approx e^{-14.03T}), so the remainder is bounded by (e^T \cdot e^{-14.03T} = e^{-13.03T}), which is strictly smaller than (e^{-9T}) (since (13.03 > 9)).

In both cases, the inequality holds.

2. Precision of Stirling's Approximation #

Stirling's formula for factorials has a well-known asymptotic expansion with explicit error terms:
[
n! = \sqrt{2\pi n} \left(\frac{n}{e}\right)^n \left(1 + \frac{1}{12n} + \frac{1}{288n^2} - \frac{139}{51840n^3} - \cdots\right)
]

• For positive integers (n): The approximation gets exponentially better as (n) increases. For (n=10), the basic Stirling estimate has a relative error of ~0.8%; for (n=100), it drops to ~0.08%; for (n=1000), it's ~0.008%.
• The error term for the basic approximation ((n! \approx \sqrt{2\pi n}(n/e)^n)) is bounded by (e^{1/(12n)} - 1) (from the remainder in the expansion), which is negligible for large (n).

3. Stirling's Approximation for Complex Numbers #

Yes, Stirling's formula extends to the complex plane! For any complex number (z) with (|\arg z| < \pi) (i.e., (z) does not lie on the negative real axis), we have the asymptotic expansion:
[
\Gamma(z+1) = z! \approx \sqrt{2\pi z} \left(\frac{z}{e}\right)^z \left(1 + \frac{1}{12z} + \frac{1}{288z^2} - \cdots\right)
]
This holds as (|z| \to \infty) with (|\arg z| < \pi). The precision is analogous to the real case: the larger (|z|) is, the better the approximation, and the error terms follow the same asymptotic behavior.

内容的提问来源于stack exchange，提问作者user257