Approximating a derivative: How to complete this proof of $f'(x_2) = \frac{f_0 - 8f_1 + 8f_3 - f_4}{12 h} + \frac{h^4}{30}f^\mathrm{V}(\xi)$?

Question

Fix five equally spaced nodes as $x_i = x_0 + ih$ where $h > 0$, $x_0\in\mathbb{R}$, and $i = 0, 1, 2, 3, 4$. Let us also denote $f_i := f(x_i)$.

Exercise. Assume that $f\in \operatorname{C^5}[x_0, x_4].$ Show that there exists some $\xi(x_2)=:\xi\in[x_0, x_4]$ such that $$f'(x_2) = \dfrac{f_0 - 8f_1 + 8f_3 - f_4}{12 h} + \dfrac{h^4}{30}f^\mathrm{V}(\xi).\label{E}\tag{E}$$

Solution. Using a method of undetermined coefficients and approximation by Taylor polynomials with Lagrangian remainders, I believe to have shown that

$$f'(x) = \dfrac{f_0 - 8f_1 + 8f_3 - f_4}{12 h} + \frac{h^4}{30} \frac{16\, f^\mathrm{V}(\xi_2) - 4\, f^\mathrm{V}(\xi_1)}{12} \tag{1}$$

where $\xi_1, \xi_2 \in[x_0, x_4]$, and $x:=x_2.$

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Here is the more detailed explanation. (Skip forward to section named Question if you wish). First for $k = 1, 2$ using Taylor polynomials and Lagrangian remainders

$$f(x\pm kh) = f(x) \pm f'(x)\, kh + f''(x)\, \frac{k^2 h^2}{2} \pm f'''(x)\, \frac{k^3 h^3}{6} + f^\mathrm{IV}(x)\,\frac{k^4 h^4}{24} \pm f^\mathrm{V}(\xi_\pm^k)\,\frac{k^5 h^5}{120} \label{A1}\tag{A1}$$

where $\xi_\pm^k$ is between $x$ and $x \pm k h$. Note also that $x_1 = x - h$, $x_3 = x + h$ and so on.

Let us view the expression $Af_0 + Bf_1 + Cf_3 + Df_4$ where $A, B, C, D$ are to be determined. After substituting $f_0, f_1, f_3, f_4$ from the earlier Taylor expansion $\eqref{A1}$ into this expression, one gets after further dividing both sides by $h$ that

\begin{align*}\frac{Af_0 + Bf_1 + Cf_3 + Df_4}{h} = \, (&A + B + C + D)\,\frac{f(x)}{h} + (-2A - B + C + 2D)\, f'(x)\\ +&(4A + B + C + 4 D)\,f''(x)\, \frac{h}{2} \\ +&\, (-8A -B + C + 8D)\,f'''(x)\, \frac{h^2}{6} + (16A + B + C + 16D)\, f^\mathrm{IV}(x)\,\frac{ h^3}{24}\\ +& \left[-32A\, f^\mathrm{V}(\xi_-^2) - B\, f^\mathrm{V}(\xi_-^1) + C\, f^\mathrm{V}(\xi_+^1) + 32D\, f^\mathrm{V}(\xi_+^2)\right]\,\frac{h^4}{120}. \label{A2}\tag{A2}\end{align*}

Next we attempt to determine the coefficients $A, B, C, D$ in such a way that we are left with $f'(x)$ and $h^4$ terms on the RHS of $\eqref{A2}$. This gives us the system

$$ \begin{cases} A + B + C + D = 0,\\ -2A - B + C + 2D = 1, \\ 4A + B + C + 4 D = 0, \\ -8A -B + C + 8D = 0,\\ 16A + B + C + 16D = 0. \end{cases}\label{A3}\tag{A3} $$

The unique solution is $A = - D = \dfrac{1}{12}$, $-B = C = \dfrac{2}{3}.$ If we denote the $h^4$ term by $-\mathcal R(x)$, then substituting the values of the coefficients back into $\eqref{A2}$, we get

$$\dfrac{f_0 - 8f_1 + 8f_3 - f_4}{12 h} + \mathcal R(x) = f'(x).\label{A4}\tag{A4}$$

Comparing this to $\eqref{E}$, what remains to be shown is that the expression

$$\mathcal R(x) = \left[32A\, f^\mathrm{V}(\xi_-^2) + B\, f^\mathrm{V}(\xi_-^1) - C\, f^\mathrm{V}(\xi_+^1) - 32D\, f^\mathrm{V}(\xi_+^2)\right]\,\frac{h^4}{120}\label{A5}\tag{A5}$$

or, after substituting the solution coefficients and simplifying, that the expression

$$\mathcal R(x) = \frac{h^4}{30} \frac{8\, f^\mathrm{V}(\xi_-^2) - 2\, f^\mathrm{V}(\xi_-^1) - 2\, f^\mathrm{V}(\xi_+^1) + 8\, f^\mathrm{V}(\xi_+^2)}{12}\label{A6}\tag{A6}$$

is somehow equal to

$$\dfrac{h^4}{30}f^\mathrm{V}(\xi)\label{A7}\tag{A7}$$

for some $\xi\in[x_0, x_4]$. Because $f^\mathrm{V}$ is continuous, by the intermediate value theorem we get

\begin{align*} f^\mathrm{V}(\xi_-^1) + f^\mathrm{V}(\xi_+^1) = 2 f^\mathrm{V}(\xi_1),\label{A8}\tag{A8}\\ f^\mathrm{V}(\xi_-^2) + f^\mathrm{V}(\xi_+^2) = 2 f^\mathrm{V}(\xi_2),\label{A9}\tag{A9} \end{align*}

where $\xi_1 \in(x - h, x + h)$ and $\xi_2 \in(x - 2h, x + 2h)$. Therefore,

$$\mathcal R(x) = \frac{h^4}{30} \frac{16\, f^\mathrm{V}(\xi_2) - 4\, f^\mathrm{V}(\xi_1)}{12}.\label{A10}\tag{A10}$$

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Question.

• If I could show that for some $\xi\in [x_0, x_4]$ $$16f^\mathrm{V}(\xi_1) - 4f^\mathrm{V}(\xi_2) = 12f^\mathrm{V}(\xi),\label{Q}\tag{Q}$$ the proof would be complete. Is this achievable?

If it isn't always possible to do, there is probably a mistake somewhere...

Answer 1

Apply the extended mean value theorem four times \begin{align} \frac{a(h)}{b(h)}&=\frac{f(x-2h)-8f(x-h)-12hf'(x)+8f(x+h)-f(x+2h)}{h^5} \\ =\frac{a'(h_1)}{b'(h_1)}&=\frac{-2f'(x-2h_1)+8f'(x-h_1)-12f'(x)+8f'(x+h_1)-2f'(x+2h_1)}{5h_1^4} \\ =\frac{a''(h_2)}{b''(h_2)}&=\frac{f''(x-2h_2)-2f''(x-h_2)+2f''(x+h_2)-f''(x+2h_2)}{5h_2^3} \\ =\frac{a'''(h_3)}{b'''(h_3)}&=\frac{-2f'''(x-2h_3)+2f'''(x-h_3)+2f'''(x+h_3)-2f'''(x+2h_3)}{15h_3^2} \\ =\frac{a^{(4)}(h_4)}{b^{(4)}(h_4)}&=\frac{2f^{(4)}(x-2h_4)-f^{(4)}(x-h_4)+f^{(4)}(x+h_4)-2f^{(4)}(x+2h_4)}{15h_4} \end{align} with $h>h_1>h_2>h_3>h_4>0$, $b(h)=h^5$ and $a(0)=a'(0)=a''(0)=a'''(0)=a^{(4)}(0)=0$. Now apportion the last term as \begin{align} &=\frac{2(f^{(4)}(x-2h_4)-f^{(4)}(x-h_4))+(f^{(4)}(x-h_4)-f^{(4)}(x+h_4))+2(f^{(4)}(x+h_4)-f^{(4)}(x+2h_4))}{15h_4} \\ &=\frac2{15}(-f^{(5)}(\xi_1)-f^{(5)}(\xi_2)-f^{(5)}(\xi_3)) \end{align} where $x-2h_4<\xi_1<x-h_4$, $x-h_4<\xi_2<x+h_4$, $x+h_4<\xi_3<x+2h_4$ by the simple mean value theorem. By the intermediate value theorem there is some $\xi\in(x-2h_4,x+2h_4)$ with $3f^{(5)}(\xi)=f^{(5)}(\xi_1)+f^{(5)}(\xi_2)+f^{(5)}(\xi_3)$ so that in total \begin{align} &\frac{f(x-2h)-8f(x-h)-12hf'(x)+8f(x+h)-f(x+2h)}{h^5} \\ &=-\frac2{5}f^{(5)}(\xi), \end{align} the claim follows.