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Thread: [SOLVED] three-point formula for numerical differentiation

  1. #1
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    [SOLVED] three-point formula for numerical differentiation

    When using the three-point formulas for numerical differentiation, specifically for $\displaystyle f'$($\displaystyle x_0$)= (1/2h)[-3f($\displaystyle x_0$)+4f($\displaystyle x_0$+h)-f($\displaystyle x_0$+2h)]

    and $\displaystyle f'$($\displaystyle x_0$)=1/2h[f($\displaystyle x_0$+h)-f($\displaystyle x_0$-h]

    When you have a list of different x and their value at f(x) when you go to plug the formulas in, and you get $\displaystyle x_0$ then you go to $\displaystyle x_1$ how does the formula change? like do all the $\displaystyle x_0$'s get changed to $\displaystyle x_1$ and so forth? I got my first value done but now im getting stuck on the rest because I don't know what to do with the formula from here.
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  2. #2
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    haha never mind i figured it out..
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