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Bonaventura Cavalieri (1598-1647)

Bonaventura Cavalieri (1598-1647). Joined the religious order Jesuati in Milan in 1615 while he was still a boy. In 1616 he transferred to the Jesuati monastery in Pisa. After meeting Galileo, considered himself a disciple of his.

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Bonaventura Cavalieri (1598-1647)

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  1. Bonaventura Cavalieri (1598-1647) • Joined the religious order Jesuati in Milan in 1615 while he was still a boy. • In 1616 he transferred to the Jesuati monastery in Pisa. After meeting Galileo, considered himself a disciple of his. • In 1629 Cavalieri was appointed to the chair of mathematics at Bologna • Cavalieri's theory of indivisibles, presented in his Geometria indivisibilis continuorum nova of 1635. After criticism Cavalieri improved his exposition publishing Exercitationes geometricae sex which became the main source for 17th Century mathematicians. • Galileo wrote: “few, if any, since Archimedes, have delved as far and as deep into the science of geometry.”

  2. Bonaventura Cavalieri (1598-1647) Cavalieri’s principle: If two plane figures cut by a set of parallel lines intersect, on each of these straight lines, equal chords, the two figures are equivalent [i.e. of equivalent area]. Cavalieri’s triangle paradox:

  3. Bonaventura Cavalieri (1598-1647) • Cavalieri thinks of a triangle or any plane figure as being made up of all lines (a.l.) of the figure. The area is then given by a.l.. • When he writes all squares (a.s.) of a plane figure he means the all the squares obtained by producing a square from each line of the plane figure. • When he writes all cubes (a.c.) of a plane figure he means the all the cubes obtained by producing a cube from each line of the plane figure. • We should think of the word “all” as a kind of integration sign.

  4. Bonaventura Cavalieri (1598-1647) Proposition 21: ac(AD)=4ac(ACF)=4ac(FDC). Notice that “together with” means plus and “with” means that the collection of indivisibles is over the product. With abuse of notation we will write just the product. Going sentence by sentence and giving comments in [ ], + denotes “together with” and juxtaposition denotes “with” : • ac(AD)=ac(ACF)+ac(FDC)+3al(ACF)as(FDC)+3as(ACF)al(FDC). [(NE)3=(NH+HE)3=NE3+HE3+3NH(HE)2+3(NH)2HE]. • ac(AD)=al(AD)as(AD): ac(AD)/[al(AD)as(FDC)]=al(AD)as(AD)/[al(AD)as(FDC)] =as(AD)/as(FDC)=3 [For this one needs that each line of AD has the same length and thus the term al(AD) yields a constant factor. Also: as(AD)/as(FDC)=3 from the quadrature of the parabola.] • ac(AD)=3al(AD)as(FDC), al(AD)as(FDC)=al(ACF)as(FDC)+al(FDC)as(FDC), al(FDC)as(FDC)=ac(FDC). [NE(HE)2=(NH+HE)(HE)2=NH(HE)2+(HE)3.] • ac(AD)=3 (ac(FDC)+al(ACF)as(FDC)) • ac(AD)=ac(ACF)+ac(FDC)+3 al(FDC)as(ACF)+3al(ACF)as(FDC). • 3al(ACF)as(FDC)= 3al(ACF)as(FDC) • Cryptic: Subtract 6 from RHSs of 4 and 5. • ac(ACF)+ac(FDC)+3as(ACF)al(FDC)=3ac(FDC). • ac(ACF)+ac(FDC)=2ac(ACF), since ac(ACF)=ac(FDC). [The triangles are congruent.] • 3al(ACF)as(FDC)+3as(ACF)al(FDC)+ ac(ACF)+ac(FDC)=ac(AD)=4ac(FDC)=4ac(FAC). [From 1,8,9,11]. • al(ACF)as(FDC)=as(ACF)al(FDC), by symmetry. • 3al(ACF)as(FDC)=3as(ACF)al(FDC) [This is not needed but clarifies the proof. Each summand of 1 contributes the same.]

  5. Bonaventura Cavalieri (1598-1647) • Proposition 21: ac(AD)=4ac(ACF)=4ac(FDC) • ac(AD)=ac(ACF)+ac(FDC)+3al(ACF)as(FDC)+3as(ACF)al(FDC) • ac(AD)/al(AD)as(FDC)=al(AD)as(AD)/al(AD)as(FAD) =as(AD)/as(FAD)=3 • Now al(AD)=al(ACF)+al(FDC) and al(FDC)as(FDC)=ac(FDC) so ac(AD)=3al(AD)as(FDC)=3al(ACF)as(FDC)+3ac(FDC) • By symmetry: • ac(ACF)=ac(FDC) • al(ACF)as(FDC)=as(ACF)al(FDC) • So (1) ac(AD)=2ac(FDC)+6al(ACF)as(FDC) (2) ac(AD)=3ac(FDC)+3al(ACF)as(FDC) thus 3al(ACF)as(FDC)=ac(FDC) and: ac(AD)=2(ac(FDC)+3al(ACF)as(FDC))=4ac(FDC)

  6. Bonaventura Cavalieri (1598-1647) Effectively Cavalieri integrates of x=y3 from 0 to 1 without the fundamental theorem by decomposing the square [0,1]X[0,1] into the two triangles above and below x=y, and using that Cavalieri does slightly more than this, since he integrates u=v3 between 0 and 1 in any linear coordinates system. It is enough to consider x=au+bv, y=cv. (why?)

  7. Bonaventura Cavalieri (1598-1647) Proposition 23. Any parallelogram has twice the area of its first diagonal space, three times the area of its second diagonal space, four times the area of its third diagonal space etc. Rephrased: In any (not necessarily Cartesian) linear coordinate system the area under the curve x=yn between a and b is 1/n+1 times the area of the parallelogram [0,a]x[0,b].

  8. Bonaventura Cavalieri (1598-1647) • The link between proposition 23 and 21 is given by noting that for third diagonal space al are given by the FI. • Now EF:FI=DA3:AF3 • all DA3=ac(AC) • all AF3= ac(ABC)=ac(CDA), • all EF= al(AC)=area(AC) • all FI=al(3rd diagonal space)=area(3rd diagonal space). • So the area of the parallelogram is to the area of the third diagonal space as ac(AC):ac(ABC)=1/4 (by proposition 21).

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