基于泰勒定理推导数值微分二阶导数近似公式遇阻求助

Hey there! Let's work through this problem step by step to untangle where you might be getting stuck deriving the second-order numerical differentiation formula from Taylor series.

First, let's recap the one-sided Taylor expansion you started with (truncated to the second-order term):

$f(a+h) = f(a) + h f'(a) + \frac{h^2}{2} f''(a) + O(h^3)$

If you're trying to solve for $f''(a)$ directly from this single equation, you'll hit a roadblock: it still contains the unknown first derivative $f'(a)$. That's probably where your confusion is coming in—one-sided expansions alone can't eliminate the first-order term, so we need to use a central difference approach by expanding both $f(a+h)$ and $f(a-h)$.

Step 1: Write Taylor Expansions for Both Directions #

First, expand $f(a+h)$ out to the third-order term (we'll see why in a second):
$ f(a+h) = f(a) + h f'(a) + \frac{h^2}{2} f''(a) + \frac{h^3}{6} f'''(a) + O(h^4) $

Now expand $f(a-h)$ by substituting $-h$ into the Taylor formula—note the sign changes for odd-powered terms:
$ f(a-h) = f(a) - h f'(a) + \frac{h^2}{2} f''(a) - \frac{h^3}{6} f'''(a) + O(h^4) $

Step 2: Add the Two Expansions to Eliminate First-Order Terms #

If we add the left-hand sides and right-hand sides of these two equations together, the first-derivative terms and third-derivative terms cancel out perfectly:
$ f(a+h) + f(a-h) = 2f(a) + h^2 f''(a) + O(h^4) $

This leaves us with an equation that only includes $f''(a)$ as the unknown derivative—exactly what we need!

Step 3: Rearrange to Solve for $f''(a)$ #

Now we can rearrange the equation to isolate $f''(a)$:

1. Subtract $2f(a)$ from both sides:
   $ f(a+h) + f(a-h) - 2f(a) = h^2 f''(a) + O(h^4) $
2. Divide both sides by $h^2$:
   $ f''(a) = \frac{f(a+h) + f(a-h) - 2f(a)}{h^2} + O(h^2) $

This is the standard second-order central difference approximation for the second derivative, with a truncation error of $O(h^2)$.

Why Your Initial One-Sided Expansion Wasn't Working #

If you tried to solve for $f''(a)$ using only the one-sided expansion $f(a+h)$, you'd end up with:
$ f''(a) = \frac{2\left[f(a+h) - f(a) - h f'(a)\right]}{h^2} $

The problem here is that $f'(a)$ is still unknown. You could substitute a first-derivative approximation (like the forward difference $f'(a) \approx \frac{f(a+h)-f(a)}{h}$), but this would result in a lower-accuracy approximation (with error $O(h)$) instead of the higher-precision central difference we derived.

Common Algebra Mistakes to Check #

If you were making an error during rearrangement, double-check these details:

• Taylor series coefficients: The second-order term's coefficient is $\frac{1}{2!} = \frac{1}{2}$—it's easy to mix up factorial denominators.
• Sign changes for $f(a-h)$: Any odd-powered term (like $h f'(a)$ or $\frac{h^3}{6} f'''(a)$) will flip signs when substituting $-h$.
• Combining terms when adding expansions: Don't forget that $f(a) + f(a) = 2f(a)$—it's a common oversight that throws off the entire rearrangement.

Hope this breakdown clears up your confusion and helps you get past the stuck point!

内容的提问来源于stack exchange，提问作者apostato34