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The method of exhaustion for the area of a parabolic segment

  1. Jul 7, 2011 #1
    When I started reading Calculus Volume I by Apostol, I didn't fully understand his description of the method of exhaustion. I noticed that Caltech and MIT don't even teach from this section of the book, and decided not to worry about it. I understand why everything in the proof is true, I just don't understand why Apostol discarded parts of some equations. Please enlighten me.
    Here is the part of the proof that confuses me.

    The author establishes the following equations:

    12+22+[itex]\ldots[/itex]+[itex](n-1)2[/itex]=[itex]\frac{n3}{3}[/itex]-[itex]\frac{n2}{2}[/itex]+[itex]\frac{n}{6}[/itex]

    12+22+[itex]\ldots[/itex]+[itex](n)2[/itex]=[itex]\frac{n3}{3}[/itex]+[itex]\frac{n2}{2}[/itex]+[itex]\frac{n}{6}[/itex]

    Then it says: For our purposes, we do not need the exact expressions given in the right-hand members of [these two equations]. All we need are the two inequalities

    12+22+[itex]\ldots[/itex]+[itex](n-1)2[/itex]<[itex]\frac{n3}{3}[/itex]<12+22+[itex]\ldots[/itex]+[itex](n)2[/itex]

    which are valid for every integer n[itex]\geq1[/itex]. These inequalities can be deduced easily as consequences of [the two above equations], or they can be proved directly by induction.

    If we multiply both inequalities by [itex]\frac{b3}{n3}[/itex] and make use of [the equations for the upper sum, [itex]Sn[/itex], and lower sum, [itex]sn[/itex], we observe

    [itex]sn[/itex]<[itex]\frac{b3}{3}[/itex]<[itex]Sn[/itex]

    for every n. The author then proves that [itex]\frac{b3}{3}[/itex] is the only number between [itex]sn[/itex] and [itex]Sn[/itex], proving the theorem.

    It seems to me as though the author arbitrarily discards part of the first equations, and I don't understand the justification for this. In other words, I don't know why he would discard the last two terms in these equations. Doing this provides a more convenient proof of the theorem, although the author makes it seem as though he is deriving the result through the method of exhaustion, in the same way that Archimedes did. Archimedes would not have randomly discarded these terms unless he already knew how it worked out, in which case this is not really a derivation. Can someone please explain this to me?

    Thanks!

    Forrest T
     
  2. jcsd
  3. Jul 7, 2011 #2

    mathman

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    Fix your latex!
     
  4. Jul 7, 2011 #3
    Haha I did. I made a new one though, so ignore this thread.
     
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