关于构建三阶精度有限差分近似的权重求解问题

My professor gave us this question to solve, but I don't know have much familiarity with the topic.
Question
The forward difference is first order accurate and is defined to be $$ D_{+} = \frac{f(x+h) - f(x)}{h} $$ and the centered difference is $$ D_{0} = \frac{f(x+h)-f(x-h)}{2h}$$ Consider the finite difference approximation where $$ f'(x) = \frac{af(x+h) + bf(x) + cf(x-h)}{h}$$ where a,b,c are constants. The forward difference has constants $(a,b,c) = (1,-1,0)$ and the centered difference has constants $(a,b,c) = (\frac{1}{2}, 0, -\frac{1}{2})$. Are there any constants (a,b,c) that achieve 3rd order accuracy?

Work
So my approach was to take a look at the Taylor Series expansion of these terms.
\begin{align}
&af(x+h) = af(x)+ ahf'(x) + ah^2\frac{f''(x)}{2} + \dots \
&bf(x) = bf(x) \
&cf(x-h) = cf(x) - chf'(x) + ch^2\frac{f''(x)}{2} - \dots
\end{align}
We want to approximate the first order derivative, which means we need a,b, and c to eliminate the f(x), f''(x), f'''(x) and solve for f'(x). This gives us the matrix
$$
\begin{pmatrix}
1 & 1 & 1 \
1 & 0 & -1 \
1 & 0 & 1 \
1 & 0 & -1
\end{pmatrix} \times \begin{pmatrix} a \ b \ c\end{pmatrix} = \begin{pmatrix} 0 \ 1 \ 0 \ 0 \end{pmatrix}
$$
This setup must be wrong as the last line of my matrix has $a-c = 0$, but the second line has $a-c = 1$. Can someone help me with this?

I should add that when I look this up online I find that this does exist with weights $\frac{-11}{6}, \frac{-3}{2}, \frac{1}{3}$ Otherwise, I would write that it doesn't exist.

Thanks.

你的泰勒展开思路完全正确，但在构建方程组的时候出现了小问题，而且核心结论其实是：用三个点（$x-h, x, x+h$）的线性组合除以$h$来近似$f'(x)$，无法达到三阶精度，咱们一步步拆解原因：

第一步：正确展开泰勒级数并整理系数 #

先把$f(x+h)$和$f(x-h)$的泰勒展开写到三阶项（足够判断精度）：
$$
\begin{align*}
f(x+h) &= f(x) + h f'(x) + \frac{h^2}{2} f''(x) + \frac{h^3}{6} f'''(x) + O(h^4) \
f(x-h) &= f(x) - h f'(x) + \frac{h^2}{2} f''(x) - \frac{h^3}{6} f'''(x) + O(h^4)
\end{align*}
$$

将$af(x+h) + bf(x) + cf(x-h)$展开并合并同类项：
$$
\begin{align*}
af(x+h) + bf(x) + cf(x-h) &= (a+b+c)f(x) + (a - c)h f'(x) \
&\quad + \frac{a+c}{2}h^2 f''(x) + \frac{a - c}{6}h^3 f'''(x) + O(h^4)
\end{align*}
$$

第二步：明确三阶精度的要求 #

我们希望$\frac{af(x+h) + bf(x) + cf(x-h)}{h} = f'(x) + O(h^{3)$，也就是近似式的误差是$h}3$阶的。这意味着：

• 消去$f(x)$项：否则近似式里会出现$\frac{f(x)}{h}$的项，完全不符合导数的形式，因此要求：
  $$a + b + c = 0$$
• $f'(x)$的系数为1：这是我们要近似的目标项，因此：
  $$\frac{(a - c)h}{h} = a - c = 1$$
• 消去$f''(x)$项：如果保留这一项，误差会是$O(h)$或$O(h^2)$，无法达到三阶，因此要求：
  $$\frac{\frac{a+c}{2}h^2}{h} = \frac{a+c}{2}h = 0 \implies a + c = 0$$
• 消去$f'''(x)$项：要达到三阶精度，误差必须是$O(h^3)$，因此这一项的系数必须为0：
  $$\frac{\frac{a - c}{6}h^3}{h} = \frac{a - c}{6}h^2 = 0 \implies a - c = 0$$

第三步：发现矛盾 #

现在看上面的条件：

• 要求$a - c = 1$（保证$f'(x)$系数正确）
• 同时要求$a - c = 0$（消去$f'''(x)$项，达到三阶精度）

这两个方程是矛盾的，说明没有这样的常数$a,b,c$能同时满足所有四个条件。也就是说，仅用三个点的线性组合，最多只能达到二阶精度（就是你熟悉的中心差分，满足前三个条件，误差为$O(h^2)$）。

关于你查到的权重 #

你提到的权重$\frac{-11}{6}, \frac{-3}{2}, \frac{1}{3}$应该是对应包含更多点的高阶格式（比如加入$x+2h$或$x-2h$），或者是近似更高阶导数的格式，并非仅用$x-h, x, x+h$三个点的一阶导数近似。如果要实现三阶精度的一阶导数近似，通常需要用到四个点（比如$x-2h, x-h, x, x+h$），通过四个未知数来满足四个条件，从而消去$f(x), f''(x), f'''(x)$项，得到三阶精度的结果。

总结一下：仅用$(x-h, x, x+h)$三个点的线性组合来近似$f'(x)$，无法达到三阶精度，你的矩阵里出现的矛盾正好验证了这一点。

备注：内容来源于stack exchange，提问作者Calum