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Using Taylor formulare How can I prove the following result? $ f''\left( x \right) \approx Af\left( x \right) + Bf\left( {x + h} \right) + Cf\left( {x + 2h} \right) + Df\left( {x + 3h} \right)\,\,h > 0 $

Obviously I have to consider the taylor expansion and "playing" with the terms, but that's my problem . Sorry for asking this stupid things

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    For reference, what you are looking at is "finite difference approximations" for the second derivative. In the numerical analysis literature finding optimal coefficients $A,B,C,D$ to minimise the error often uses the method of "underdetermined coefficients". One may wish to consult a textbook on "finite difference methods" for more information. For example, some discussion is given in Chapter 1 of Randall LeVeque's _Finite Different Methods for Ordinary and Partial Differential Equations_.2012-06-21

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As for $1 \le i \le 3$ \[ f(x+ih) = f(x) + ihf'(x) + \frac 12 {i^2}{h^2}f''(x) + \frac 16{i^3}{h^3}f'''(x) + O(h^4) \] So we have for your right hand side \begin{align*} f(x)\bigl(A + B + C + D)\\ {} + f'(x)\bigl(B + 2C + 3D\bigr)h\\ {} + f''(x)\frac 12\bigl(B + 4C + 9D\bigr)h^2\\ {} + f'''(x)\frac 16 \bigl(B + 8C + 27D)h^3\\ {} + O(h^4) \end{align*} So we want to have \begin{align*} A + B + C + D &= 0\\ B + 2C + 3D &= 0\\ B + 4C + 9D &= 0 \end{align*} Wlog we can let $A = 1$. Then our System gives \begin{align*} B + C + D &= -1\\ C + 2D &= 1\\ 3C + 8D &= 1 \end{align*} hence \begin{align*} B + C + D &= -1\\ C + 2D &= 1\\ 2D &= -2 \end{align*} so $D = -1$, $C = 3$, $B = -3$. Then $B + 8C + 27D = -3+24-27 = -6$. That gives \[ f'''(x) = \frac{-f(x) + 3f(x+h) - 3f(x+2h) + f(x+3h)}{h^3} + O(h). \]

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    @Daniel It's not worded in the best way. It's not really WLOG, but martini is choosing $A=1$ for concreteness, with the awareness that all other solutions will be given by rescaling everything by a constant. Notice in the end that $A = -1/h^3$, not $1$.2012-06-21
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Write \begin{align} f(x+h) & = f(x) + hf'(x) + \frac{h^2}2 f''(x) + \frac{h^3}{6}f'''(\zeta_1) \\\ f(x+2h) & = f(x) + 2hf'(x) + \frac{(2h)^2}2 f''(x) + \frac{(2h)^3}{6}f'''(\zeta_2) \\\ f(x+3h) & = f(x) + 3hf'(x) + \frac{(3h)^2}2 f''(x) + \frac{(3h)^3}{6}f'''(\zeta_3). \\\ \end{align} Then after you work out the equations, you get $ \begin{align} A f(x) + B f(x+h) & + C f(x+2h) + D f(x+3h) \\\ & = \\\ (A+B+C+D)f(x) + (B+2C + 3D)hf'(x) &+ \left( \frac B2 + 2C + \frac 92D \right)h^2 f''(x) + \text{error term} \end{align} $ where the error term contains the third derivatives. So just solve for $ \begin{align} A+B+C+D = 0 \\\ B+2C + 3D = 0 \\\ B +2C + 9D/2 = 1/h^2 \\\ \end{align} $ and you will find four functions of $h$ ($A(h)$, $B(h)$, $C(h)$ and $D(h)$) such that your approximation is as good as the upper bound you can put on the third derivative error term.

Hope that helps,

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Starting with $f'(x)h\approx f(x+h)-f(x),$ we get $\begin{align}f''(x)h^2&\approx f'(x+h)h-f'(x)h \\ &\approx f(x+2h)-f(x+h)-(f(x+h)-f(x))\\ &= f(x+2h)-2f(x+h)+f(x)\end{align}$ Repeat: $\begin{align}f'''(x)h^3&\approx f''(x+h)h-f''(x)h \\&\approx (f(x+3h)-2f(x+2h)+f(x+h))-(f(x+2h)-2f(x+h)+f(x))\\ &= f(x+3h)-3f(x+2h)+3f(x+h)-f(x)\end{align}$ You can go on as long as you like... The coefficients are binomial with alternating signs.

Perhaps this approach does not quality as using the Taylor expansion, but it seems less painful.