Results 1 to 5 of 5

Math Help - apostol and archimedes

  1. #1
    Junior Member
    Joined
    Jan 2009
    Posts
    57

    apostol and archimedes

    I am trying to understand apostol's argument confirming the integral area of a parabolic segment.

    He get's to:
    \frac{b^3}{3}-\frac{b^3}{n} < A < \frac{b^3}{3}+\frac{b^3}{n} for every n \ge 1

    There are three possibilities.

    A > \frac{b^3}{n} or A = \frac{b^3}{n} or A < \frac{b^3}{n}

    I am not really sure about this. I would get -\frac{b^3}{n} < A < \frac{b^3}{n}, so why does he get three possibilities. I understand any number is either less, equal or greater, but its still a little confusing.

    The next contradiction I think I sort of understand.

    Suppose A > \frac{b^3}{n} is true. From the second initial inequality we have:


    A-\frac{b^3}{3} < \frac{b^3}{n} for every n \ge 1

    Divide both sides and multiply to get:

    n < \frac{b^3}{A-\frac{b^3}{3}}

    But this is obviously false when

    (*) n \ge \frac{b^3}{A-\frac{b^3}{3}}

    Ok, clearly that is true, but why is it a contradiction that

    A > \frac{b^3}{n}

    Is it because (*) is always a fraction and can be at most 1 therefore n cannot be any less than that becuase n is bigger than one?

    Thanks
    Regards
    Craig.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by craigmain View Post
    I am trying to understand apostol's argument confirming the integral area of a parabolic segment.

    He get's to:
    \frac{b^3}{3}-\frac{b^3}{n} < A < \frac{b^3}{3}+\frac{b^3}{n} for every n \ge 1

    There are three possibilities.

    A > \frac{b^3}{n} or A = \frac{b^3}{n} or A < \frac{b^3}{n}
    Well the three possibilities trivially exist, that they can exist under the earlier inequality can be demonstrated if any value of n allows them.

    Try n=3, then we have from the top inequalities:

    0 < A < (2/3)b^3

    which permits all three cases A>b^3/3, A=b^3/3 and A<b^3/3

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2009
    Posts
    57
    Quote Originally Posted by CaptainBlack View Post
    Well the three possibilities trivially exist, that they can exist under the earlier inequality can be demonstrated if any value of n allows them.

    Try n=3, then we have from the top inequalities:

    0 < A < (2/3)b^3

    which permits all three cases A>b^3/3, A=b^3/3 and A<b^3/3

    CB
    I am not sure why that value is chosen though.
    If I were trying to solve the inequality i would eliminate \frac{b^3}{3} to get -\frac{b^3}{n} < A < \frac{b^3}{n}

    How did he know that the area was converging to that value?
    Last edited by craigmain; June 20th 2009 at 04:50 AM. Reason: MIstype
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by craigmain View Post
    I am not sure why that value is chosen though.
    If I were trying to solve the inequality i would eliminate \frac{b^3}{3} to get -\frac{b^3}{n} < A < \frac{b^3}{n}

    How did he know that the area was converging to that value?
    Apostol or Archimedes? Archimedes: by cutting the shape from a piece of material of known area density and weighing it. Apostol: by experimental calculation.

    and in either case by various huristic arguments and calculations.

    CB
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2009
    Posts
    57
    Thanks very much.
    I was worrying that I had missed something obvious. It's comforting to know they also struggled for a while before finding the final solution.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Apostol Section 10.20 #51
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 11th 2011, 06:50 PM
  2. Archimedes spiral?
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: April 8th 2011, 08:46 PM
  3. apostol calculus, trouble simplifying integral
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 12th 2010, 02:09 PM
  4. Proof of Green's theorem in Apostol's book
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: April 26th 2010, 11:06 PM
  5. Apostol - Analytic Number Theory
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: February 13th 2009, 11:01 AM

Search Tags


/mathhelpforum @mathhelpforum