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.
The text passage is a discussion about a mathematical proof, specifically about the area of a parabolic segment as described by Apostol and Archimedes. It includes a series of questions and answers related to the proof and the reasoning behind certain steps.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
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.
user
The text passage is a discussion about a mathematical proof, specifically about the area of a parabolic segment as described by Apostol and Archimedes. It includes a series of questions and answers related to the proof and the reasoning behind certain steps.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS