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I feel a bit lost with this problem. I've looked into Romberg integration a bit and tried to apply it here, but haven't been able to conclude anything. I also read somewhere that there is a correlation between the difference, $T(2h) - T(h)$, and the step size $h$, but I'm not sure what that correlation is. For example, the difference is $0.0008 = 8*10^{-4}$ and I know the error for the trapezoidal rule is $\mathcal{O}{(h^2)}$. My question is can this information be used in order to answer the question and if so, how? If not, I'd appreciate any guidance.

$\textbf{The problem}$

The integral $\int_{0}^{1}\sqrt{\frac{1}{2}+2e^{-x}sin(2x^2)}dx$ has been approximated using the step sizes $0.2$ and $0.1$. The result was $T(0.2)=1.6426, T(0.1)=1.6418$. What step size, $h$ (approximately), should be used in order to get an error less than $8*10^{-8}$ with the least amount of work?

  1. $10^{-2}$
  2. $10^{-3}$
  3. $10^{-4}$
  4. $10^{-5}$
  5. $10^{-6}$
Moo
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1 Answers1

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We know that the error is (dropping higher-order terms)

$$E(r) = Ch^2$$

for some constant C. We also know that $E(0.2) - E(0.1) = 0.0008$, thus $C=\frac{8}{3}\cdot 10^{-2}$.

Now, we need h such that $E(h) < 8 \cdot 10^{-8}$, hence $h ^2 < 3 \cdot 10^{-6}$.

lisyarus
  • 15,517
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    I've accepted your answer, but I noticed an error when I worked through it. At the very last step it should be $h^2 < 3 * 10^{-6}$. – Moo Feb 11 '20 at 13:30
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    @Moo Indeed, thank you! Very good for you to work it out yourself. – lisyarus Feb 11 '20 at 13:47