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Find the $f'(x=0)$ using Five-point endpoint formula. where $f$ is a long vector of length $n$, say $n=11$.

  x       f   
  0    0.001
 0.1   0.435  
 0.2   0.765   
 0.3   0.897  
 0.4   0.875
 0.5   0.786
 0.6   0.776
 0.7   0.994
 0.8   0.564
 0.9   0.987
  1    0.657

Five-point endpoint formula is

$f'(x_0)=[−25f(x_0) + 48f(x_0 + ℎ) − 36f(x_0 + 2ℎ) + 16f(x-0 + 3ℎ) − 3(x_0 + 4ℎ)]/12h$,

and at $x=0$

$f'(0)=[−25f(0) + 48f(0.1) − 36f(0.2) + 16f(0.3) − 3f(0.4)]/1.2$,

but the answer is not correct. Can someone help me?

Thank you in advance

kind regards,

Bibigul

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In your formula you missed an $f()$. It should be $$f'(x_0)=[−25f(x_0) + 48f(x_0 + ℎ) − 36f(x_0 + 2ℎ) + 16f(x_0 + 3ℎ) − 3f(x_0 + 4ℎ)]/12h.$$

With this change you get $$f'(0) = (-25\times 0.001+48\times 0.435 -36\times 0.765 +16\times 0.897 -3\times 0.875)/1.2 = 4.202$$

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  • $\begingroup$ thank you for the comment, it's a typo error here but I did the same procedure as you mention. I am little bit confuse here because it is five endpoint formula and we are not taking the end point. $\endgroup$
    – Bibigul
    Nov 25 '18 at 3:54
  • $\begingroup$ if we divide our data into four equal parts by taking h=4. Can we do like that? $\endgroup$
    – Bibigul
    Nov 25 '18 at 4:02

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