text: 1. ## 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.
  
  2. Originally Posted by craigmain
  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
  
  CB
  
  3. Originally Posted by CaptainBlack
  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
  
  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?
  
  4. Originally Posted by craigmain
  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
  
  5. 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.