- #1

Forrest T

- 23

- 0

*Calculus Volume I*by Apostol, I didn't fully understand his description of the method of exhaustion. I noticed that Caltech and MIT don't even teach from this section of the book, and decided not to worry about it. I understand why everything in the proof is true, I just don't understand

*why*Apostol discarded parts of some equations. Please enlighten me.

Here is the part of the proof of the area under a parabola:

The author establishes the following equations:

1

^{2}+ 2

^{2}+ [itex]\cdots[/itex] + (

*n*- 1)

^{2}=

*n*

^{3}/3 -

*n*

^{2}/2 +

*n*/6

1

^{2}+ 2

^{2}+ [itex]\cdots[/itex] +

*n*

^{2}=

*n*

^{3}/3 +

*n*

^{2}/2 +

*n*/6

Then it says: For our purposes, we do not need the exact expressions given in the right-hand members of [these two equations]. All we need are the two

*inequalities*

1

^{2}+ 2

^{2}+ [itex]\cdots[/itex] + (

*n*- 1)

^{2}<

*n*

^{3}/3 < 1

^{2}+ 2

^{2}+ [itex]\cdots[/itex] +

*n*

^{2}which are valid for every integer

*n*[itex]\geq1[/itex]. These inequalities can be deduced easily as consequences of [the two above equations], or they can be proved directly by induction.

If we multiply both inequalities by

*b*

^{3}/

*n*

^{3}and make use of [the equations for the upper sum (

*S*) and lower sum (

_{n}*s*) we observe

_{n}*s*<

_{n}*b*

^{3}/3 <

*S*

_{n}for every

*n*. The author then proves that

*b*

^{3}/3 is the only number between

*s*and

_{n}*S*(proving the theorem).

_{n}It seems to me as though the author arbitrarily discards part of the first equations, and I don't understand the justification for this. Apostol led me to believe he was deriving the theorem, as Archimedes would have, through the method of exhaustion, rather than writing an unwieldy proof of a basic formula. In other words, I don't know why Archimedes would have discarded those parts of the first two equations while deriving the theorem.

Can someone please explain this to me? Thanks!