The method of exhaustion for the area of a parabolic segment

In summary, the conversation is about the confusion surrounding the method of exhaustion in Calculus Volume I by Apostol. The author establishes equations and inequalities and then discards parts of the equations to provide a more convenient proof of the theorem. This is causing confusion as it seems like the author is arbitrarily discarding terms and not fully explaining the reasoning behind it. The conversation ends with one person asking for clarification on this issue.
  • #1
Forrest T
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0
When I started reading 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 that confuses me.

The author establishes the following equations:

12+22+[itex]\ldots[/itex]+[itex](n-1)2[/itex]=[itex]\frac{n3}{3}[/itex]-[itex]\frac{n2}{2}[/itex]+[itex]\frac{n}{6}[/itex]

12+22+[itex]\ldots[/itex]+[itex](n)2[/itex]=[itex]\frac{n3}{3}[/itex]+[itex]\frac{n2}{2}[/itex]+[itex]\frac{n}{6}[/itex]

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

12+22+[itex]\ldots[/itex]+[itex](n-1)2[/itex]<[itex]\frac{n3}{3}[/itex]<12+22+[itex]\ldots[/itex]+[itex](n)2[/itex]

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 [itex]\frac{b3}{n3}[/itex] and make use of [the equations for the upper sum, [itex]Sn[/itex], and lower sum, [itex]sn[/itex], we observe

[itex]sn[/itex]<[itex]\frac{b3}{3}[/itex]<[itex]Sn[/itex]

for every n. The author then proves that [itex]\frac{b3}{3}[/itex] is the only number between [itex]sn[/itex] and [itex]Sn[/itex], proving the theorem.

It seems to me as though the author arbitrarily discards part of the first equations, and I don't understand the justification for this. In other words, I don't know why he would discard the last two terms in these equations. Doing this provides a more convenient proof of the theorem, although the author makes it seem as though he is deriving the result through the method of exhaustion, in the same way that Archimedes did. Archimedes would not have randomly discarded these terms unless he already knew how it worked out, in which case this is not really a derivation. Can someone please explain this to me?

Thanks!

Forrest T
 
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  • #2
Fix your latex!
 
  • #3
Haha I did. I made a new one though, so ignore this thread.
 

Related to The method of exhaustion for the area of a parabolic segment

What is the method of exhaustion for finding the area of a parabolic segment?

The method of exhaustion is a mathematical technique used to approximate the area of a shape by dividing it into smaller, simpler shapes and summing their areas.

How does the method of exhaustion apply to a parabolic segment?

In the case of a parabolic segment, the shape is divided into an infinite number of triangles that each have the same area. By summing the areas of these triangles, an approximation of the total area of the segment can be found.

What is the history of the method of exhaustion?

The method of exhaustion was first developed by ancient Greek mathematicians, such as Eudoxus and Archimedes, to solve problems related to areas and volumes. It was later refined and used extensively by European mathematicians during the 16th and 17th centuries.

Are there any limitations to using the method of exhaustion for finding the area of a parabolic segment?

Yes, the method of exhaustion is not a precise method and relies on approximations. The more triangles used, the closer the approximation will be, but it will never be exact. Additionally, the method can become increasingly complex and time-consuming when applied to more complex shapes.

How is the method of exhaustion used in modern mathematics?

The method of exhaustion is still used in modern mathematics, particularly in the fields of calculus and analysis. It serves as the basis for more advanced techniques, such as integration, and is used to prove important theorems, such as the area under a curve.

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