Approximate formula for $f''(x_0)$ of a smooth function $f(x)$ using an interpolating polynomial











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Given three points $x_{-1}$, $x_{0} = x_{-1} + h_0$, and $x_{1} = x_{0} + h_1$ where $h_1 neq h_0$ use the second degree interpolating polynomial in Newton form, differentiated twice to obtain



$$f''(x_0) approx 2f[x_{-1},x_0,x_1]$$



Show that this approximation is, in general, first order accurate.



My attempt so far has been to use the polynomial interpolation error formula
$$f(x) = p_2(x) + f[x_{-1},x_0,x_1,x](x-x_{-1})(x-x_0)(x-x_1)$$
differentiated twice and evaluated at $x_0$. So far, I have not seen anything that would give me a first order result. Any solutions/hints would be greatly appreciated.










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    Given three points $x_{-1}$, $x_{0} = x_{-1} + h_0$, and $x_{1} = x_{0} + h_1$ where $h_1 neq h_0$ use the second degree interpolating polynomial in Newton form, differentiated twice to obtain



    $$f''(x_0) approx 2f[x_{-1},x_0,x_1]$$



    Show that this approximation is, in general, first order accurate.



    My attempt so far has been to use the polynomial interpolation error formula
    $$f(x) = p_2(x) + f[x_{-1},x_0,x_1,x](x-x_{-1})(x-x_0)(x-x_1)$$
    differentiated twice and evaluated at $x_0$. So far, I have not seen anything that would give me a first order result. Any solutions/hints would be greatly appreciated.










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Given three points $x_{-1}$, $x_{0} = x_{-1} + h_0$, and $x_{1} = x_{0} + h_1$ where $h_1 neq h_0$ use the second degree interpolating polynomial in Newton form, differentiated twice to obtain



      $$f''(x_0) approx 2f[x_{-1},x_0,x_1]$$



      Show that this approximation is, in general, first order accurate.



      My attempt so far has been to use the polynomial interpolation error formula
      $$f(x) = p_2(x) + f[x_{-1},x_0,x_1,x](x-x_{-1})(x-x_0)(x-x_1)$$
      differentiated twice and evaluated at $x_0$. So far, I have not seen anything that would give me a first order result. Any solutions/hints would be greatly appreciated.










      share|cite|improve this question













      Given three points $x_{-1}$, $x_{0} = x_{-1} + h_0$, and $x_{1} = x_{0} + h_1$ where $h_1 neq h_0$ use the second degree interpolating polynomial in Newton form, differentiated twice to obtain



      $$f''(x_0) approx 2f[x_{-1},x_0,x_1]$$



      Show that this approximation is, in general, first order accurate.



      My attempt so far has been to use the polynomial interpolation error formula
      $$f(x) = p_2(x) + f[x_{-1},x_0,x_1,x](x-x_{-1})(x-x_0)(x-x_1)$$
      differentiated twice and evaluated at $x_0$. So far, I have not seen anything that would give me a first order result. Any solutions/hints would be greatly appreciated.







      numerical-methods






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      asked Nov 16 at 15:21









      geo17

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