有限差分の列挙とそれらの証明

Julia Advent Calendar 2024 シリーズ2 の9日目です🎄
https://qiita.com/advent-calendar/2024/julia

はじめに

FiniteDifferenceMatrices.jl と LaTeXStrings.jl を用いて中心差分と前進差分, 4次精度までの1階微分から4階微分に対応する 有限差分 (Finite Difference) を列挙し, それらを手で証明していきます.

有限差分の列挙

FiniteDifferenceMatrices.jl の fdcoefficient()で求めた係数を使い, LaTeXStrings.jl でうまく数式が表示されるようLaTeXにコンパイルできる文字列を生成します. コードはこちらにあります.
https://gist.github.com/ohno/bd52eb6461aaaeb5c2c3dec78017f33f

中心差分

2次精度

$$dfdx(x)=f(x+Δx)−f(x−Δx)2Δx+O(Δx2) \frac{\mathrm{d}f}{\mathrm{d}x}(x) = \frac{ f(x + \Delta x) - f(x - \Delta x) }{{2\Delta x}} + O(\Delta x^{2})$$

$$d2fdx2(x)=f(x+Δx)−2f(x)+f(x−Δx)Δx2+O(Δx2) \frac{\mathrm{d}^{2}f}{\mathrm{d}x^{2}}(x) = \frac{ f(x + \Delta x) - 2 f(x ) + f(x - \Delta x) }{{\Delta x}^{2}} + O(\Delta x^{2})$$

$$d3fdx3(x)=f(x+2Δx)−2f(x+Δx)+2f(x−Δx)−f(x−2Δx)2Δx3+O(Δx2) \frac{\mathrm{d}^{3}f}{\mathrm{d}x^{3}}(x) = \frac{ f(x + 2 \Delta x) - 2 f(x + \Delta x) + 2 f(x - \Delta x) - f(x - 2 \Delta x) }{{2\Delta x}^{3}} + O(\Delta x^{2})$$

$$d4fdx4(x)=f(x+2Δx)−4f(x+Δx)+6f(x)−4f(x−Δx)+f(x−2Δx)Δx4+O(Δx2) \frac{\mathrm{d}^{4}f}{\mathrm{d}x^{4}}(x) = \frac{ f(x + 2 \Delta x) - 4 f(x + \Delta x) + 6 f(x ) - 4 f(x - \Delta x) + f(x - 2 \Delta x) }{{\Delta x}^{4}} + O(\Delta x^{2})$$

4次精度

$$dfdx(x)=−f(x+2Δx)+8f(x+Δx)−8f(x−Δx)+f(x−2Δx)12Δx+O(Δx4) \frac{\mathrm{d}f}{\mathrm{d}x}(x) = \frac{ - f(x + 2 \Delta x) + 8 f(x + \Delta x) - 8 f(x - \Delta x) + f(x - 2 \Delta x) }{{12\Delta x}} + O(\Delta x^{4})$$

$$d2fdx2(x)=−f(x+2Δx)+16f(x+Δx)−30f(x)+16f(x−Δx)−f(x−2Δx)12Δx2+O(Δx4) \frac{\mathrm{d}^{2}f}{\mathrm{d}x^{2}}(x) = \frac{ - f(x + 2 \Delta x) + 16 f(x + \Delta x) - 30 f(x ) + 16 f(x - \Delta x) - f(x - 2 \Delta x) }{{12\Delta x}^{2}} + O(\Delta x^{4})$$

$$d3fdx3(x)=−f(x+3Δx)+8f(x+2Δx)−13f(x+Δx)+13f(x−Δx)−8f(x−2Δx)+f(x−3Δx)8Δx3+O(Δx4) \frac{\mathrm{d}^{3}f}{\mathrm{d}x^{3}}(x) = \frac{ - f(x + 3 \Delta x) + 8 f(x + 2 \Delta x) - 13 f(x + \Delta x) + 13 f(x - \Delta x) - 8 f(x - 2 \Delta x) + f(x - 3 \Delta x) }{{8\Delta x}^{3}} + O(\Delta x^{4})$$

$$d4fdx4(x)=−f(x+3Δx)+12f(x+2Δx)−39f(x+Δx)+56f(x)−39f(x−Δx)+12f(x−2Δx)−f(x−3Δx)6Δx4+O(Δx4) \frac{\mathrm{d}^{4}f}{\mathrm{d}x^{4}}(x) = \frac{ - f(x + 3 \Delta x) + 12 f(x + 2 \Delta x) - 39 f(x + \Delta x) + 56 f(x ) - 39 f(x - \Delta x) + 12 f(x - 2 \Delta x) - f(x - 3 \Delta x) }{{6\Delta x}^{4}} + O(\Delta x^{4})$$

前進差分

1次精度

$$dfdx(x)=f(x+Δx)−f(x)Δx+O(Δx) \frac{\mathrm{d}f}{\mathrm{d}x}(x) = \frac{ f(x + \Delta x) - f(x ) }{{\Delta x}} + O(\Delta x)$$

$$d2fdx2(x)=f(x+2Δx)−2f(x+Δx)+f(x)Δx2+O(Δx) \frac{\mathrm{d}^{2}f}{\mathrm{d}x^{2}}(x) = \frac{ f(x + 2 \Delta x) - 2 f(x + \Delta x) + f(x ) }{{\Delta x}^{2}} + O(\Delta x)$$

$$d3fdx3(x)=f(x+3Δx)−3f(x+2Δx)+3f(x+Δx)−f(x)Δx3+O(Δx) \frac{\mathrm{d}^{3}f}{\mathrm{d}x^{3}}(x) = \frac{ f(x + 3 \Delta x) - 3 f(x + 2 \Delta x) + 3 f(x + \Delta x) - f(x ) }{{\Delta x}^{3}} + O(\Delta x)$$

$$d4fdx4(x)=f(x+4Δx)−4f(x+3Δx)+6f(x+2Δx)−4f(x+Δx)+f(x)Δx4+O(Δx) \frac{\mathrm{d}^{4}f}{\mathrm{d}x^{4}}(x) = \frac{ f(x + 4 \Delta x) - 4 f(x + 3 \Delta x) + 6 f(x + 2 \Delta x) - 4 f(x + \Delta x) + f(x ) }{{\Delta x}^{4}} + O(\Delta x)$$

2次精度

$$dfdx(x)=−f(x+2Δx)+4f(x+Δx)−3f(x)2Δx+O(Δx2) \frac{\mathrm{d}f}{\mathrm{d}x}(x) = \frac{ - f(x + 2 \Delta x) + 4 f(x + \Delta x) - 3 f(x ) }{{2\Delta x}} + O(\Delta x^{2})$$

$$d2fdx2(x)=−f(x+3Δx)+4f(x+2Δx)−5f(x+Δx)+2f(x)Δx2+O(Δx2) \frac{\mathrm{d}^{2}f}{\mathrm{d}x^{2}}(x) = \frac{ - f(x + 3 \Delta x) + 4 f(x + 2 \Delta x) - 5 f(x + \Delta x) + 2 f(x ) }{{\Delta x}^{2}} + O(\Delta x^{2})$$

$$d3fdx3(x)=−3f(x+4Δx)+14f(x+3Δx)−24f(x+2Δx)+18f(x+Δx)−5f(x)2Δx3+O(Δx2) \frac{\mathrm{d}^{3}f}{\mathrm{d}x^{3}}(x) = \frac{ - 3 f(x + 4 \Delta x) + 14 f(x + 3 \Delta x) - 24 f(x + 2 \Delta x) + 18 f(x + \Delta x) - 5 f(x ) }{{2\Delta x}^{3}} + O(\Delta x^{2})$$

$$d4fdx4(x)=−2f(x+5Δx)+11f(x+4Δx)−24f(x+3Δx)+26f(x+2Δx)−14f(x+Δx)+3f(x)Δx4+O(Δx2) \frac{\mathrm{d}^{4}f}{\mathrm{d}x^{4}}(x) = \frac{ - 2 f(x + 5 \Delta x) + 11 f(x + 4 \Delta x) - 24 f(x + 3 \Delta x) + 26 f(x + 2 \Delta x) - 14 f(x + \Delta x) + 3 f(x ) }{{\Delta x}^{4}} + O(\Delta x^{2})$$

3次精度

$$dfdx(x)=2f(x+3Δx)−9f(x+2Δx)+18f(x+Δx)−11f(x)6Δx+O(Δx3) \frac{\mathrm{d}f}{\mathrm{d}x}(x) = \frac{ 2 f(x + 3 \Delta x) - 9 f(x + 2 \Delta x) + 18 f(x + \Delta x) - 11 f(x ) }{{6\Delta x}} + O(\Delta x^{3})$$

$$d2fdx2(x)=11f(x+4Δx)−56f(x+3Δx)+114f(x+2Δx)−104f(x+Δx)+35f(x)12Δx2+O(Δx3) \frac{\mathrm{d}^{2}f}{\mathrm{d}x^{2}}(x) = \frac{ 11 f(x + 4 \Delta x) - 56 f(x + 3 \Delta x) + 114 f(x + 2 \Delta x) - 104 f(x + \Delta x) + 35 f(x ) }{{12\Delta x}^{2}} + O(\Delta x^{3})$$

$$d3fdx3(x)=7f(x+5Δx)−41f(x+4Δx)+98f(x+3Δx)−118f(x+2Δx)+71f(x+Δx)−17f(x)4Δx3+O(Δx3) \frac{\mathrm{d}^{3}f}{\mathrm{d}x^{3}}(x) = \frac{ 7 f(x + 5 \Delta x) - 41 f(x + 4 \Delta x) + 98 f(x + 3 \Delta x) - 118 f(x + 2 \Delta x) + 71 f(x + \Delta x) - 17 f(x ) }{{4\Delta x}^{3}} + O(\Delta x^{3})$$

$$d4fdx4(x)=17f(x+6Δx)−114f(x+5Δx)+321f(x+4Δx)−484f(x+3Δx)+411f(x+2Δx)−186f(x+Δx)+35f(x)6Δx4+O(Δx3) \frac{\mathrm{d}^{4}f}{\mathrm{d}x^{4}}(x) = \frac{ 17 f(x + 6 \Delta x) - 114 f(x + 5 \Delta x) + 321 f(x + 4 \Delta x) - 484 f(x + 3 \Delta x) + 411 f(x + 2 \Delta x) - 186 f(x + \Delta x) + 35 f(x ) }{{6\Delta x}^{4}} + O(\Delta x^{3})$$

4次精度

$$dfdx(x)=−3f(x+4Δx)+16f(x+3Δx)−36f(x+2Δx)+48f(x+Δx)−25f(x)12Δx+O(Δx4) \frac{\mathrm{d}f}{\mathrm{d}x}(x) = \frac{ - 3 f(x + 4 \Delta x) + 16 f(x + 3 \Delta x) - 36 f(x + 2 \Delta x) + 48 f(x + \Delta x) - 25 f(x ) }{{12\Delta x}} + O(\Delta x^{4})$$

$$d2fdx2(x)=−10f(x+5Δx)+61f(x+4Δx)−156f(x+3Δx)+214f(x+2Δx)−154f(x+Δx)+45f(x)12Δx2+O(Δx4) \frac{\mathrm{d}^{2}f}{\mathrm{d}x^{2}}(x) = \frac{ - 10 f(x + 5 \Delta x) + 61 f(x + 4 \Delta x) - 156 f(x + 3 \Delta x) + 214 f(x + 2 \Delta x) - 154 f(x + \Delta x) + 45 f(x ) }{{12\Delta x}^{2}} + O(\Delta x^{4})$$

$$d3fdx3(x)=−15f(x+6Δx)+104f(x+5Δx)−307f(x+4Δx)+496f(x+3Δx)−461f(x+2Δx)+232f(x+Δx)−49f(x)8Δx3+O(Δx4) \frac{\mathrm{d}^{3}f}{\mathrm{d}x^{3}}(x) = \frac{ - 15 f(x + 6 \Delta x) + 104 f(x + 5 \Delta x) - 307 f(x + 4 \Delta x) + 496 f(x + 3 \Delta x) - 461 f(x + 2 \Delta x) + 232 f(x + \Delta x) - 49 f(x ) }{{8\Delta x}^{3}} + O(\Delta x^{4})$$

$$d4fdx4(x)=−21f(x+7Δx)+164f(x+6Δx)−555f(x+5Δx)+1056f(x+4Δx)−1219f(x+3Δx)+852f(x+2Δx)−333f(x+Δx)+56f(x)6Δx4+O(Δx4) \frac{\mathrm{d}^{4}f}{\mathrm{d}x^{4}}(x) = \frac{ - 21 f(x + 7 \Delta x) + 164 f(x + 6 \Delta x) - 555 f(x + 5 \Delta x) + 1056 f(x + 4 \Delta x) - 1219 f(x + 3 \Delta x) + 852 f(x + 2 \Delta x) - 333 f(x + \Delta x) + 56 f(x ) }{{6\Delta x}^{4}} + O(\Delta x^{4})$$

証明

いくつかの有限差分を導出していきます. 証明にはTaylor展開

$$f(x+Δx)=f(x)+∂f(x)∂xΔx+12!∂2f(x)∂x2Δx2+13!∂3f(x)∂x3Δx3+14!∂4f(x)∂x4Δx4+⋯ ,f(x−Δx)=f(x)−∂f(x)∂xΔx+12!∂2f(x)∂x2Δx2−13!∂3f(x)∂x3Δx3+14!∂4f(x)∂x4Δx4+⋯ ,f(x+2Δx)=f(x)+2∂f(x)∂xΔx+222!∂2f(x)∂x2Δx2+233!∂3f(x)∂x3Δx3+244!∂4f(x)∂x4Δx4+⋯ ,f(x−2Δx)=f(x)−2∂f(x)∂xΔx+222!∂2f(x)∂x2Δx2−233!∂3f(x)∂x3Δx3+244!∂4f(x)∂x4Δx4+⋯ , \begin{aligned} f(x+\Delta x) &= f(x) + \frac{\partial f(x)}{\partial x} \Delta x + \frac{1}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{1}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + \frac{1}{4!} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + \cdots, \\ f(x-\Delta x) &= f(x) - \frac{\partial f(x)}{\partial x} \Delta x + \frac{1}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} - \frac{1}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + \frac{1}{4!} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + \cdots, \\ f(x+2\Delta x) &= f(x) + 2\frac{\partial f(x)}{\partial x} \Delta x + \frac{2^2}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{2^3}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + \frac{2^4}{4!} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + \cdots, \\ f(x-2\Delta x) &= f(x) - 2\frac{\partial f(x)}{\partial x} \Delta x + \frac{2^2}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} - \frac{2^3}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + \frac{2^4}{4!} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + \cdots, \end{aligned}$$

を用います.

中心差分, 2次精度, 1階微分

$$f(x+Δx)=f(x)+∂f(x)∂xΔx+12!∂2f(x)∂x2Δx2+O(Δx3)f(x−Δx)=f(x)−∂f(x)∂xΔx+12!∂2f(x)∂x2Δx2+O(Δx3) \begin{aligned} f(x+\Delta x) &= f(x) + \frac{\partial f(x)}{\partial x} \Delta x + \frac{1}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + O\left(\Delta x^{3}\right) \\ f(x-\Delta x) &= f(x) - \frac{\partial f(x)}{\partial x} \Delta x + \frac{1}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + O\left(\Delta x^{3}\right) \end{aligned}$$

の差を取ると,

$$f(x+Δx)−f(x−Δx)=2∂f(x)∂xΔx+O(Δx3)2∂f(x)∂xΔx=f(x+Δx)−f(x−Δx)+O(Δx3)∂f(x)∂x=f(x+Δx)−f(x−Δx)2Δx+O(Δx3)2Δx∂f(x)∂x=f(x+Δx)−f(x−Δx)2Δx+O(Δx2) \begin{aligned} f(x+\Delta x) - f(x-\Delta x) &= 2\frac{\partial f(x)}{\partial x} \Delta x + O\left(\Delta x^{3}\right) \\ 2\frac{\partial f(x)}{\partial x} \Delta x &= f(x+\Delta x) - f(x-\Delta x) + O\left(\Delta x^{3}\right) \\ \frac{\partial f(x)}{\partial x} &= \frac{f(x+\Delta x)- f(x-\Delta x)}{2\Delta x} + \frac{O\left(\Delta x^{3}\right)}{2\Delta x} \\ \frac{\partial f(x)}{\partial x} &= \frac{f(x+\Delta x)- f(x-\Delta x)}{2\Delta x} + O\left(\Delta x^{2}\right) \end{aligned}$$

中心差分, 2次精度, 2階微分

$$f(x+Δx)=f(x)+∂f(x)∂xΔx+12!∂2f(x)∂x2Δx2+13!∂3f(x)∂x3Δx3+O(Δx4)f(x−Δx)=f(x)−∂f(x)∂xΔx+12!∂2f(x)∂x2Δx2−13!∂3f(x)∂x3Δx3+O(Δx4) \begin{aligned} f(x+\Delta x) &= f(x) + \frac{\partial f(x)}{\partial x} \Delta x + \frac{1}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{1}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + O\left(\Delta x^{4}\right) \\ f(x-\Delta x) &= f(x) - \frac{\partial f(x)}{\partial x} \Delta x + \frac{1}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} - \frac{1}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + O\left(\Delta x^{4}\right) \end{aligned}$$

の和を取ると,

$$f(x+Δx)+f(x−Δx)=2f(x)+∂2f(x)∂x2Δx2+O(Δx4)∂2f(x)∂x2Δx2=f(x+Δx)−2f(x)+f(x−Δx)+O(Δx4)∂2f(x)∂x2=f(x+Δx)−2f(x)+f(x−Δx)Δx2+O(Δx4)Δx2∂2f(x)∂x2=f(x+Δx)−2f(x)+f(x−Δx)Δx2+O(Δx2) \begin{aligned} f(x+\Delta x) + f(x-\Delta x) &= 2f(x) + \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + O\left(\Delta x^{4}\right) \\ \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} &= f(x+\Delta x) - 2f(x) + f(x-\Delta x) + O\left(\Delta x^{4}\right) \\ \frac{\partial^{2} f(x)}{\partial x^{2}} &= \frac{f(x+\Delta x) - 2f(x) + f(x-\Delta x)}{\Delta x^{2}} + \frac{O\left(\Delta x^{4}\right)}{\Delta x^{2}} \\ \frac{\partial^{2} f(x)}{\partial x^{2}} &= \frac{f(x+\Delta x) - 2f(x) + f(x-\Delta x)}{\Delta x^{2}} + O\left(\Delta x^{2}\right) \end{aligned}$$

中心差分, 2次精度, 3階微分

$$f(x+Δx)=f(x)+∂f(x)∂xΔx+12!∂2f(x)∂x2Δx2+13!∂3f(x)∂x3Δx3+14!∂4f(x)∂x4Δx4+O(Δx5)f(x−Δx)=f(x)−∂f(x)∂xΔx+12!∂2f(x)∂x2Δx2−13!∂3f(x)∂x3Δx3+14!∂4f(x)∂x4Δx4+O(Δx5) \begin{aligned} f(x+\Delta x) &= f(x) + \frac{\partial f(x)}{\partial x} \Delta x + \frac{1}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{1}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + \frac{1}{4!} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + O\left(\Delta x^{5}\right) \\ f(x-\Delta x) &= f(x) - \frac{\partial f(x)}{\partial x} \Delta x + \frac{1}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} - \frac{1}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + \frac{1}{4!} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + O\left(\Delta x^{5}\right) \end{aligned}$$

の差を取って2倍すると,

$$f(x+Δx)−f(x−Δx)=2∂f(x)∂xΔx+13∂3f(x)∂x3Δx3+O(Δx5)2f(x+Δx)−2f(x−Δx)=4∂f(x)∂xΔx+23∂3f(x)∂x3Δx3+O(Δx5) \begin{aligned} f(x+\Delta x) - f(x-\Delta x) &= 2\frac{\partial f(x)}{\partial x} \Delta x + \frac{1}{3} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + O\left(\Delta x^{5}\right) \\ 2f(x+\Delta x) - 2f(x-\Delta x) &= 4\frac{\partial f(x)}{\partial x} \Delta x + \frac{2}{3} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + O\left(\Delta x^{5}\right) \end{aligned}$$

を得る. 次に

$$f(x+2Δx)=f(x)+2∂f(x)∂xΔx+222!∂2f(x)∂x2Δx2+233!∂3f(x)∂x3Δx3+244!∂4f(x)∂x4Δx4+⋯f(x−2Δx)=f(x)−2∂f(x)∂xΔx+222!∂2f(x)∂x2Δx2−233!∂3f(x)∂x3Δx3+244!∂4f(x)∂x4Δx4+⋯ \begin{aligned} f(x+2\Delta x) &= f(x) + 2\frac{\partial f(x)}{\partial x} \Delta x + \frac{2^2}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{2^3}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + \frac{2^4}{4!} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + \cdots \\ f(x-2\Delta x) &= f(x) - 2\frac{\partial f(x)}{\partial x} \Delta x + \frac{2^2}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} - \frac{2^3}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + \frac{2^4}{4!} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + \cdots \end{aligned}$$

の差を取ると,

$$f(x+2Δx)−f(x−2Δx)=4∂f(x)∂xΔx+83∂3f(x)∂x3Δx3+O(Δx5) \begin{aligned} f(x+2\Delta x) - f(x-2\Delta x) &= 4\frac{\partial f(x)}{\partial x} \Delta x + \frac{8}{3} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + O\left(\Delta x^{5}\right) \end{aligned}$$

を得る. さらに, 得られた2式

$$2f(x+Δx)−2f(x−Δx)=4∂f(x)∂xΔx+23∂3f(x)∂x3Δx3+O(Δx5)f(x+2Δx)−f(x−2Δx)=4∂f(x)∂xΔx+83∂3f(x)∂x3Δx3+O(Δx5) \begin{aligned} 2f(x+\Delta x) - 2f(x-\Delta x) &= 4\frac{\partial f(x)}{\partial x} \Delta x + \frac{2}{3} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + O\left(\Delta x^{5}\right) \\ f(x+2\Delta x) - f(x-2\Delta x) &= 4\frac{\partial f(x)}{\partial x} \Delta x + \frac{8}{3} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + O\left(\Delta x^{5}\right) \end{aligned}$$

の差を取って左右を入れ替えると

$$−63∂3f(x)∂x3Δx3+O(Δx5)=2f(x+Δx)−2f(x−Δx)−f(x+2Δx)+f(x−2Δx)2∂3f(x)∂x3Δx3=f(x+2Δx)−2f(x+Δx)+2f(x−Δx)−f(x−2Δx)+O(Δx5)∂3f(x)∂x3=f(x+2Δx)−2f(x+Δx)+2f(x−Δx)−f(x−2Δx)2Δx3+O(Δx5)2Δx3∂3f(x)∂x3=f(x+2Δx)−2f(x+Δx)+2f(x−Δx)−f(x−2Δx)2Δx3+O(Δx2) \begin{aligned} - \frac{6}{3} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + O\left(\Delta x^{5}\right) &= 2f(x+\Delta x) - 2f(x-\Delta x) - f(x+2\Delta x) + f(x-2\Delta x) \\ 2 \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} &= f(x+2\Delta x) - 2f(x+\Delta x) + 2f(x-\Delta x) - f(x-2\Delta x) + O\left(\Delta x^{5}\right) \\ \frac{\partial^{3} f(x)}{\partial x^{3}} &= \frac{ f(x+2\Delta x) - 2f(x+\Delta x) + 2f(x-\Delta x) - f(x-2\Delta x) }{2\Delta x^{3}} + \frac{O\left(\Delta x^{5}\right)}{2\Delta x^{3}} \\ \frac{\partial^{3} f(x)}{\partial x^{3}} &= \frac{ f(x+2\Delta x) - 2f(x+\Delta x) + 2f(x-\Delta x) - f(x-2\Delta x) }{2\Delta x^{3}} + O\left(\Delta x^{2}\right) \end{aligned}$$

中心差分, 2次精度, 4階微分

$$f(x+Δx)=f(x)+∂f(x)∂xΔx+12!∂2f(x)∂x2Δx2+13!∂3f(x)∂x3Δx3+14!∂4f(x)∂x4Δx4+15!∂5f(x)∂x5Δx5+O(Δx6)f(x−Δx)=f(x)−∂f(x)∂xΔx+12!∂2f(x)∂x2Δx2−13!∂3f(x)∂x3Δx3+14!∂4f(x)∂x4Δx4−15!∂5f(x)∂x5Δx5+O(Δx6) \begin{aligned} f(x+\Delta x) &= f(x) + \frac{\partial f(x)}{\partial x} \Delta x + \frac{1}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{1}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + \frac{1}{4!} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + \frac{1}{5!} \frac{\partial^{5} f(x)}{\partial x^{5}} \Delta x^{5} + O\left(\Delta x^{6}\right) \\ f(x-\Delta x) &= f(x) - \frac{\partial f(x)}{\partial x} \Delta x + \frac{1}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} - \frac{1}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + \frac{1}{4!} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} - \frac{1}{5!} \frac{\partial^{5} f(x)}{\partial x^{5}} \Delta x^{5} + O\left(\Delta x^{6}\right) \end{aligned}$$

の和を取って4倍すると,

$$f(x+Δx)+f(x−Δx)=2f(x)+22!∂2f(x)∂x2Δx2+24!∂4f(x)∂x4Δx4+O(Δx6)4f(x+Δx)+4f(x−Δx)=8f(x)+4∂2f(x)∂x2Δx2+13∂4f(x)∂x4Δx4+O(Δx6) \begin{aligned} f(x+\Delta x) + f(x-\Delta x) &= 2 f(x) + \frac{2}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{2}{4!} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + O\left(\Delta x^{6}\right) \\ 4f(x+\Delta x) + 4f(x-\Delta x) &= 8 f(x) + 4 \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{1}{3} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + O\left(\Delta x^{6}\right) \end{aligned}$$

を得る. 次に

$$f(x+2Δx)=f(x)+2∂f(x)∂xΔx+222!∂2f(x)∂x2Δx2+233!∂3f(x)∂x3Δx3+244!∂4f(x)∂x4Δx4+255!∂5f(x)∂x5Δx5+O(Δx6)f(x−2Δx)=f(x)−2∂f(x)∂xΔx+222!∂2f(x)∂x2Δx2−233!∂3f(x)∂x3Δx3+244!∂4f(x)∂x4Δx4−255!∂5f(x)∂x5Δx5+O(Δx6) \begin{aligned} f(x+2\Delta x) &= f(x) + 2\frac{\partial f(x)}{\partial x} \Delta x + \frac{2^2}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{2^3}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + \frac{2^4}{4!} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + \frac{2^5}{5!} \frac{\partial^{5} f(x)}{\partial x^{5}} \Delta x^{5} + O\left(\Delta x^{6}\right) \\ f(x-2\Delta x) &= f(x) - 2\frac{\partial f(x)}{\partial x} \Delta x + \frac{2^2}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} - \frac{2^3}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + \frac{2^4}{4!} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} - \frac{2^5}{5!} \frac{\partial^{5} f(x)}{\partial x^{5}} \Delta x^{5} + O\left(\Delta x^{6}\right) \end{aligned}$$

の和を取ると,

$$f(x+2Δx)+f(x−2Δx)=2f(x)+4∂2f(x)∂x2Δx2+43∂4f(x)∂x4Δx4+O(Δx6) \begin{aligned} f(x+2\Delta x) + f(x-2\Delta x) &= 2f(x) + 4 \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{4}{3} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + O\left(\Delta x^{6}\right) \end{aligned}$$

を得る. さらに, 得られた2式

$$4f(x+Δx)+4f(x−Δx)=8f(x)+4∂2f(x)∂x2Δx2+13∂4f(x)∂x4Δx4+O(Δx6)f(x+2Δx)+f(x−2Δx)=2f(x)+4∂2f(x)∂x2Δx2+43∂4f(x)∂x4Δx4+O(Δx6) \begin{aligned} 4f(x+\Delta x) + 4f(x-\Delta x) &= 8 f(x) + 4 \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{1}{3} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + O\left(\Delta x^{6}\right) \\ f(x+2\Delta x) + f(x-2\Delta x) &= 2f(x) + 4 \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{4}{3} \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + O\left(\Delta x^{6}\right) \end{aligned}$$

の差を取ると

$$6f(x)−∂4f(x)∂x4Δx4+O(Δx6)=4f(x+Δx)+4f(x−Δx)−f(x+2Δx)−f(x−2Δx)∂4f(x)∂x4Δx4=f(x+2Δx)−4f(x+Δx)+6f(x)−4f(x−Δx)+f(x−2Δx)+O(Δx6)∂4f(x)∂x4=f(x+2Δx)−4f(x+Δx)+6f(x)−4f(x−Δx)+f(x−2Δx)Δx4+O(Δx2) \begin{aligned} 6 f(x) - \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} + O\left(\Delta x^{6}\right) &= 4f(x+\Delta x) + 4f(x-\Delta x) - f(x+2\Delta x) - f(x-2\Delta x) \\ \frac{\partial^{4} f(x)}{\partial x^{4}} \Delta x^{4} &= f(x+2\Delta x) - 4f(x+\Delta x) + 6 f(x) - 4f(x-\Delta x) + f(x-2\Delta x) + O\left(\Delta x^{6}\right) \\ \frac{\partial^{4} f(x)}{\partial x^{4}} &= \frac{f(x+2\Delta x) - 4f(x+\Delta x) + 6 f(x) - 4f(x-\Delta x) + f(x-2\Delta x) }{\Delta x^{4}} + O\left(\Delta x^{2}\right) \end{aligned}$$

前進差分, 1次精度, 1階微分

$$f(x+Δx)=f(x)+∂f(x)∂xΔx+O(Δx2) f(x+\Delta x) = f(x) + \frac{\partial f(x)}{\partial x} \Delta x + O\left(\Delta x^{2}\right)$$

から直ちに

$$f(x+Δx)=f(x)+∂f(x)∂xΔx+O(Δx2)∂f(x)∂xΔx=f(x+Δx)−f(x)+O(Δx2)∂f(x)∂x=f(x+Δx)−f(x)Δx−O(Δx2)Δx∂f(x)∂x=f(x+Δx)−f(x)Δx+O(Δx) \begin{aligned} f(x+\Delta x) &= f(x) + \frac{\partial f(x)}{\partial x} \Delta x + O\left(\Delta x^{2}\right) \\ \frac{\partial f(x)}{\partial x} \Delta x &= f(x+\Delta x) - f(x) + O\left(\Delta x^{2}\right) \\ \frac{\partial f(x)}{\partial x} &= \frac{f(x+\Delta x) - f(x)}{\Delta x} - \frac{O\left(\Delta x^{2}\right)}{\Delta x} \\ \frac{\partial f(x)}{\partial x} &= \frac{f(x+\Delta x) - f(x)}{\Delta x} + O\left(\Delta x\right) \end{aligned}$$

前身差分, 1次精度, 2階微分

まず

$$f(x+Δx)=f(x)+∂f(x)∂xΔx+12!∂2f(x)∂x2Δx2+13!∂3f(x)∂x3Δx3+O(Δx4) \begin{aligned} f(x+\Delta x) &= f(x) + \frac{\partial f(x)}{\partial x} \Delta x + \frac{1}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{1}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + O\left(\Delta x^{4}\right) \end{aligned}$$

を2倍して,

$$2f(x+Δx)=2f(x)+2∂f(x)∂xΔx+22!∂2f(x)∂x2Δx2+23!∂3f(x)∂x3Δx3+O(Δx4) \begin{aligned} 2 f(x+\Delta x) &= 2f(x) + 2\frac{\partial f(x)}{\partial x} \Delta x + \frac{2}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{2}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + O\left(\Delta x^{4}\right) \end{aligned}$$

となる. これと

$$f(x+2Δx)=f(x)+2∂f(x)∂xΔx+42!∂2f(x)∂x2Δx2+83!∂3f(x)∂x3Δx3+O(Δx4) \begin{aligned} f(x+2\Delta x) &= f(x) + 2 \frac{\partial f(x)}{\partial x} \Delta x + \frac{4}{2!} \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + \frac{8}{3!} \frac{\partial^{3} f(x)}{\partial x^{3}} \Delta x^{3} + O\left(\Delta x^{4}\right) \end{aligned}$$

の差を取ると,

$$f(x+2Δx)−2f(x+Δx)=−f(x)+∂2f(x)∂x2Δx2+O(Δx3) \begin{aligned} f(x+2\Delta x) - 2 f(x+\Delta x) = - f(x) + \frac{\partial^{2} f(x)}{\partial x^{2}} \Delta x^{2} + O\left(\Delta x^{3}\right) \end{aligned}$$

となり, これを整理して次式を得る.

$$∂2f(x)∂x2=f(x+2Δx)−2f(x+Δx)+f(x)Δx2+O(Δx) \begin{aligned} \frac{\partial^{2} f(x)}{\partial x^{2}} &= \frac{f(x+2\Delta x) - 2f(x+\Delta x) + f(x)}{\Delta x^{2}} + O\left(\Delta x\right) \end{aligned}$$

まとめ

有限差分法 (Finite Difference Method, FDM) で用いる公式を列挙・証明しました.