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I recently took part of user https://math.stackexchange.com/users/72031/paramanand-singh 's proof for the error term in the trapezoidal rule. However, it doesn't seem to agree with other literature error terms since it doesn't have $n^2$ in the denominator. My immediate reaction to this was to be that this could be for a single-interval and thus not for the composite. Is this true? If so, how could turn this into the composite error bound?

user7722867
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When you use trapezoidal rule you split the interval $[a, b] $ into $n$ subintervals of length $h=(b-a) /n$ and thus the error involved is sum of all errors in each subinterval. And clearly this is equal to $$n\cdot\frac{h^{3}}{12}f''(\xi)=\frac{(b-a)^{3}}{12n^{2}}f''(\xi)$$ and you get the $n^{2}$ in denominator.

Note: The above proof assumes that $f''$ is continuous on $[a, b] $.

Paramanand Singh
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  • Hello again! I saw that the proof you provided was based on Hardy's "A course of pure mathemathics". I want to cite it in a highschool research paper and have gotten a copy of the book from the Project Gutenberg Ebooks. Could you provide me with a page number for the exercise? – user7722867 Sep 14 '17 at 13:58
  • @user7722867: you should search for Simpson's rule in the index. The exercise just after this topic is what you need. I am not giving exact page number as I have 10th edition and Gutenberg is 3rd edition. – Paramanand Singh Sep 14 '17 at 15:37
  • Is this the page you are referencing? https://i.imgur.com/UqKsZqD.png . There doesn't look like there is much to go after here. Perhaps I should cite both your proof and the book – user7722867 Sep 15 '17 at 06:43
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    It appears you need 10th edition, where Hardy proves the Simpson's rule (which is bit difficult) and leaves trapezoidal for an exercise. I used the same proof technique for Trapezoidal rule. You can give a reference to my proof as well as cite Hardy's book 10the edition page 329. – Paramanand Singh Sep 15 '17 at 07:48
  • Hello one last time. I just looked at your proof and realized that the step where you differentiate to get g'(t) looks strange to me. You are differentiating with respect to t, but the result you provide is the exact same one as if you differentiated with respect to the x. The antiderivative with respect to x when differentiated with respect to t won't surely be the same thing as f(x), or am I wrong? – user7722867 Oct 22 '17 at 11:04
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    @user7722867: don't worry you can go even after the last time. There is no variable $x$ involved so one can't use derivative with $x$. The $x$ in $f(x) , dx$ in integral sign is a dummy variable. It does not serve any purpose. – Paramanand Singh Oct 22 '17 at 11:14
  • Nevermind. I realized that since it is a definite integral, it is treated as a constant with respect to t. Thanks again! – user7722867 Oct 23 '17 at 11:47