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Using the three-point startpoint formula to find $f^{'}(7.4)$ where: $f(7.4)=-68.3193, f(7.6)=-71.6982, f(7.8)=-75.1576$

I got the approximated value: $f^{'}(7.4)=-16.69325$

then the actual error equals: $0.000367$ where $f(x)=ln(x+2)-(x+1)^{2}$

and the error bound equals: $0.000032$ where the error bound has the formula: $EB=max|(h^{2}/3)(f^{'''}(x)|$

My question is, how can the value of error BOUND is less than the value of actual error!.. I understood that the error bound has the greater value of the error may occure using this method of approximation.

soso sos
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1 Answers1

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Here the law I have found in the book!

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soso sos
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  • ah ok it's a not centered method – user Dec 03 '17 at 18:28
  • what do you mean by centered method please? – soso sos Dec 03 '17 at 18:30
  • "not centered"in the sense that you calculate the value of f'(x) at the left point $x_0$ – user Dec 03 '17 at 18:34
  • yes, by the startpoint formula, using $7.4, 7.6, 7.8$ what is the mistake? – soso sos Dec 03 '17 at 18:38
  • I really don't know, I obtain the same results! – user Dec 03 '17 at 19:07
  • OK thanks :( can anyone help please!!! – soso sos Dec 03 '17 at 19:18
  • Let's try with a shorter interval step h. – user Dec 03 '17 at 19:24
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    Finally I got the answer of my question :)

    The actual error is greater than the error bound because the accuracy of the error bound is greater than the accuracy of the data which are given in the original question! you can notice that the numbers in the question are with 4 decimals only!. I tried to solve with data with 9 decimals then I got more accurate answers so that the actual error is less than the error bound :)

    Thank you for your help!

     – soso sos Dec 03 '17 at 19:50
  • Well done! Thanks for the update.! – user Dec 03 '17 at 19:51