渐近展开下L值高精度估计组合式求解及推导问询

Let’s break down how to derive the high-precision estimates for both cases, starting with part (b) since you already know the result but want to understand the derivation.

Part (b): $ϕ(h) = L − c_1h − c_2h^2 − c_3h^3 − ···$ #

The core idea is to combine $ϕ(h)$ and $ϕ(h/2)$ to eliminate the leading error term ($O(h)$) and get a better estimate of $L$. Here’s the step-by-step breakdown:

1. First, write out the expansion for $ϕ(h/2)$:

   $ϕ(h/2) = L − c_1*(h/2) − c_2*(h/2)^2 − c_3*(h/2)^3 − ··· = L − \frac{c_1h}{2} − \frac{c_2h^2}{4} − \frac{c_3h^3}{8} − ···$

2. Multiply $ϕ(h/2)$ by 2 to align the coefficient of the $c_1h$ term with that in $ϕ(h)$:

   $2ϕ(h/2) = 2L − c_1h − \frac{c_2h^2}{2} − \frac{c_3h^3}{4} − ···$

3. Subtract $ϕ(h)$ from this result to cancel the $c_1h$ term:

   $2ϕ(h/2) − ϕ(h) = [2L − c_1h − \frac{c_2h^2}{2} − ···] − [L − c_1h − c_2h^2 − ···]$
   Simplifies to: $L + \frac{c_2h^2}{2} + \frac{3c_3h^3}{4} − ···$

The leading error term is now $O(h^2)$ instead of $O(h)$, so this combination gives a higher-precision estimate of $L$:
L ≈ 2ϕ(h/2) − ϕ(h)

Part (a): $ϕ(h) = L − c_1h^{1/2} − c_2h − c_3h^{3/2} − ···$ #

We use the exact same logic: eliminate the leading error term ($O(h^{1/2})$) by combining $ϕ(h)$ and $ϕ(h/2)$. Here’s how:

1. Write the expansion for $ϕ(h/2)$:

   $ϕ(h/2) = L − c_1\sqrt{h/2} − c_2*(h/2) − c_3*(h/2)^{3/2} − ··· = L − \frac{c_1\sqrt{h}}{\sqrt{2}} − \frac{c_2h}{2} − \frac{c_3h^{3/2}}{2\sqrt{2}} − ···$

2. We need coefficients $A$ and $B$ such that $Aϕ(h) + Bϕ(h/2)$ cancels the $O(\sqrt{h})$ term and leaves $L$ as the leading term. Set up these equations:

   • For the $L$ term: $A + B = 1$ (we want the coefficient of $L$ to be 1)
   • For the $\sqrt{h}$ term: $A + \frac{B}{\sqrt{2}} = 0$ (we want this coefficient to be 0)

3. Solve the system of equations:

   • From the second equation: $A = -\frac{B}{\sqrt{2}}$
   • Substitute into the first equation: $-\frac{B}{\sqrt{2}} + B = 1 → B(1 - \frac{1}{\sqrt{2}}) = 1 → B = 2 + \sqrt{2}$
   • Then $A = -(√2 + 1)$

4. The resulting combination is:

   $(2 + √2)ϕ(h/2) − (√2 + 1)ϕ(h)$

This cancels the leading $O(\sqrt{h})$ error term, leaving $L$ with a leading error of $O(h)$—a significant improvement in precision. So the high-precision estimate for $L$ is:
L ≈ (2 + √2)ϕ(h/2) − (√2 + 1)ϕ(h)

Key Takeaway #

Absolutely, the same logic applies to both parts: identify the leading error term, set up equations to cancel that term using the scaled version of $ϕ$ (here $ϕ(h/2)$), solve for the combination coefficients, and you get a higher-order estimate of $L$.

内容的提问来源于stack exchange，提问作者user530646