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Could you please give me formulas for the fourth order of accuracy $O(h^4)$ of the $``$forward and backward$"$ difference methods to find the approximation of $f''(x_0)$?

Ahmed
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2 Answers2

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Add the two Taylor series $f(x\pm h) =f(x)\pm hf'(x)+h^2f''(x)/2\pm h^3f'''(x)/6 +O(h^4) $.

marty cohen
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  • Please, I need the fourth order of accuracy : Forward and Backward difference methods for the secondderivative. What you wrote is the second order finite difference method for the second derivative. – Ahmed Jul 17 '17 at 15:23
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Thanks a lot for your efforts. I answer the question:

I used the coefficients in this table: https://en.wikipedia.org/wiki/Finite_difference_coefficient

The forward is \begin{align} f''(x_0) =& \frac{1}{h^2} \Big(\frac{15}{4}f(x_0) - \frac{77}{6}f(x_0+h) + \frac{107}{6}f(x_0+2h) - 13 f(x_0+3h) \\&+ \frac{61}{12} f(x_0+4h) - \frac{5}{6} f(x_0+5h)\Big) \end{align} The bachward is \begin{align} f''(x_0) =& \frac{1}{h^2} \Big(-\frac{15}{4}f(x_0) + \frac{77}{6}f(x_0-h) - \frac{107}{6}f(x_0-2h) + 13 f(x_0-3h) \nonumber\\ &- \frac{61}{12} f(x_0-4h) + \frac{5}{6} f(x_0-5h)\Big). \end{align}

Ahmed
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