# apostol and archimedes

• Jun 20th 2009, 12:55 AM
craigmain
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.
• Jun 20th 2009, 02:08 AM
CaptainBlack
Quote:

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
• Jun 20th 2009, 03:50 AM
craigmain
Quote:

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?
• Jun 20th 2009, 09:18 AM
CaptainBlack
Quote:

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
• Jun 21st 2009, 09:34 PM
craigmain
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.