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- Thread starter pivoxa15
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quasar987

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can you be more specific about what you'Re asking?

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This could be it but I don't follow it very well.

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i know, but i dont feel like typing out all of the latex. just write out taking N definite integrals over some interval for a function that has been differentiated N times.This could be it but I don't follow it very well.

Arfken does exactly this derivation in his book (pg. 260 in the 2nd edition)

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Quite good explanations. This stuff was left unclear to me when I was sleeping in the calculus classes, but now I feel like understanding the integral trickery that is done in the Wikipedia's article.

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that is exactly the method that i was referring to. as long as the function doesn't grow faster than n! you will be fine and the remainder term will go to zero

Quite good explanations. This stuff was left unclear to me when I was sleeping in the calculus classes, but now I feel like understanding the integral trickery that is done in the Wikipedia's article.

<happy 600 posts to me>

- #8

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THe problem is they still didn't mention how the mean value theorem works in the proof.

Quite good explanations. This stuff was left unclear to me when I was sleeping in the calculus classes, but now I feel like understanding the integral trickery that is done in the Wikipedia's article.

- #9

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you invoke the mean value thm in order to convert the integral into the remainder term. in other words,THe problem is they still didn't mention how the mean value theorem works in the proof.

[tex]

\int_a^b f(x) dx = (b - a)f(y)

[/tex]

where y is an element of the interval (a,b)

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But the generalised MVT is something else. It involves using induction.

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