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.
explanation: The document is a discussion about a mathematical proof, specifically Apostol's argument confirming the integral area of a parabolic segment. It contains a series of questions and partial answers, where the user is seeking clarification on the steps and reasoning in the proof. The document also includes responses from another user (CaptainBlack) providing explanations and further insights.
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distilabel_metadata.content: 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.
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion about a mathematical proof, specifically Apostol's argument confirming the integral area of a parabolic segment. It contains a series of questions and partial answers, where the user is seeking clarification on the steps and reasoning in the proof. The document also includes responses from another user (CaptainBlack) providing explanations and further insights.
  
  Label: multiple-question-and-answer