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We have:

$f(x+2h) = f(x) +2hf'(x) + 2h^2f''(x) + \frac{4h^3}{3}f'''(\zeta_1)$, where $\zeta_1 \in (x, x+2h) $

$f(x+h) = f(x) +hf'(x) + \frac{h^2}{2}f''(x) + \frac{h^3}{6}f'''(\zeta_2)$, where $\zeta_2 \in (x, x+h) $

Substituting into the formula gives :

$$\frac{f(x) +2hf'(x) + 2h^2f''(x) + \frac{4h^3}{3}f'''(\zeta_1) - 2(f(x) +hf'(x) + \frac{h^2}{2}f''(x) + \frac{h^3}{6}f'''(\zeta_2))}{h^2}$$

$$= f''(x) + \frac{4h}{3}f'''(\zeta_1) - \frac{h}{3}f'''(\zeta_2) $$

$$=f''(x) + \frac{h}{3}(4f'''(\zeta_1) - f'''(\zeta_2))$$

$$=f''(x) + \frac{h}{3}(3f'''(\zeta)), ~\zeta \in (x,x+2h)$$

$$= f''(x) + hf'''(\zeta))$$

Does this seem correct? I am unsure about combining the error terms at the end

Btzzzz
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1 Answers1

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No, it is not correct, and you are missing a $h$ in the last formula. However, to solve the problem you do not need to combine the remainder terms.

Lutz Lehmann
  • 126,666
  • Okay i've edited the h in. Then what should i do because i've seen examples elsewhere where terms are combined? – Btzzzz Mar 30 '16 at 23:19
  • You can not apply the intermediate value theorem if there is no intermediate value. You can find $c\in(a,b)$ so that $g(a)+g(b)=2g(c)$, but $\frac13(4g(a)-g(b))$ is outside the segment $[g(a),g(b)]$. Take one more term in the Taylor series to remove any doubt on the error order. – Lutz Lehmann Mar 30 '16 at 23:28
  • Okay thanks, so $f''(x) + \frac{2h^2}{3}f''''(\zeta_1) - \frac{2h^2}{24}f''''(\zeta_1)$ = $f''(x) + \frac{7h^2}{12}f''''(\zeta)$ should work for the segment? – Btzzzz Mar 30 '16 at 23:45
  • That still is not a convex combination. It will be true for many functions, but I think one could construct counter examples for that step. -- You also missed the $hf'''(x)$ term. – Lutz Lehmann Mar 31 '16 at 01:32