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Generalised mean value theorem?

  1. Aug 18, 2007 #1
    Does anyone know the mean value theorem associated with the Taylor series. Representing the Taylor series a finite sum and an end term? I don't get how they get it to look that way.
  2. jcsd
  3. Aug 18, 2007 #2


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    can you be more specific about what you'Re asking?
  4. Aug 18, 2007 #3
    i have seen the taylor series derived that way..you consider the n-th antiderivative of the n derivative. when you apply the fundamental thm of calculus you will get a series of terms that define the taylor series, with a remainder term. you use the mean value theorem to show that the remainder will go to 0 as long as your function grows no faster than n!
  5. Aug 18, 2007 #4
    This could be it but I don't follow it very well.
  6. Aug 18, 2007 #5
    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.

    Arfken does exactly this derivation in his book (pg. 260 in the 2nd edition)
  7. Aug 19, 2007 #6

    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.
  8. Aug 19, 2007 #7
    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

    <happy 600 posts to me> :approve:
  9. Aug 20, 2007 #8
    THe problem is they still didn't mention how the mean value theorem works in the proof.
  10. Aug 20, 2007 #9
    you invoke the mean value thm in order to convert the integral into the remainder term. in other words,

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

    where y is an element of the interval (a,b)
  11. Aug 21, 2007 #10
    I've seen a proof of the genearlised mean value theorem in books and its long but it finally explained it fully in terms of Taylor series.
  12. Aug 21, 2007 #11
    i seem to recall that the MVT can be proven all by itself, and that the fundamental theorem of calculus can be derived from it
  13. Aug 22, 2007 #12
    But the generalised MVT is something else. It involves using induction.
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