Quadrature of the parabola with the square pyramidal number

Size: px

Start display at page:

Download "Quadrature of the parabola with the square pyramidal number"

Sybil Irma Barker
5 years ago
Views:

1 Quadrature of the parabola with the square pyramidal umber By Luciao Acora We perform here a ew proof of the Archimedes theorem o the quadrature of the parabolic segmet, executed without the aid of itegral calculus, but usig oly the square pyramidal umber ad covergece criteria umerical sequeces. The traslatio of the discussio i the umerical field will happe usig as uit of measuremet of the areas ivolved i the proof, equivalet triagles, suitably idetified i the grid of costructio of the parabolic segmet. Itroductio The Quadrature of the parabola is oe of the first works composed by Archimedes. It has as subject the quadrature of the parabolic segmet, amely the costructio (with ruler ad compass) of a polygo equivalet to it. For parabolic segmet Archimedes meas the area betwee a straight lie ad a parabola, coceived as a sectio of a right coe. The work opes with a itroductio to the basic properties of the parabola; the move to perform the quadrature of the parabola i a mechaical way, with cosideratios o the lever equilibrium; fially we get to the geometric proof of the quadrature, performed applyig the exhaustio method. Our proof revisits i a moder key the work of Archimedes, usig the same figure that he uses i the propositio 16, where it is proved the fudametal result that the triagle ABC is triple of the parabolic segmet. Archimedes uses a triagle ABC rectagle i B, havig show, i the previous propositio 15, that the result for such a situatio geeralizes to a parabolic segmet with base o-perpedicular to the axis (ote 1). I the ext prop. 17 Archimedes ifers from this result the other, more kow, that the parabolic segmet is 4/ of the iscribed triagle. Propositio 16 Let AB the base of a parabolic segmet, ad draw through B the straight lie BC parallel to the axis of the parabola to meet the taget at A i C. I say that the area of the parabolic segmet is oe-third of the ABC triagle area. Fig. 1 1

2 Proof Split the segmets AB ad BC ito six equal parts ad lead, from split poits o AB parallels to BC, ad from poits o BC lies joiig with A. The parabola passes through the poits of itersectio of the grid, as draw, because, for oe of its properties, it cuts the vertical lies of the grid i the same ratio i which the vertical lies cut the segmet AB. Cosider the saw-tooth figure that circumscribes the parabolic segmet. The area of this figure exceeds the area of the segmet of a quatity that is equal to the overall area of the teeth. If we icrease the umber of divisios o AB ad BC, the excess area teds to zero as teds to ifiity. I other words: the area of the saw-tooth figure coverges to the area of the parabolic segmet, as teds to ifiity. I the graph, the saw-tooth figure is divided ito six vertical stripes composed: the first of 6 equivalet triagles, ad the other strips, respectively, of 5, 4,,, 1 trapezoids, equivalet to each other i each strip. Now cosider the triagle (show i gree) with a vertex at the poit A. We will use this triagle as the measuremet uit of the areas i the couts that follow: The triagle ABM cotais: (sum of the first 6 odd umbers) 6 gree triagles (the sum of the first odd umbers is ). The triagle ABC cotais: gree triagles. I geeral, for ay umber of divisios of AB ad BC, the triagle ABC cotais gree triagles. The circumscribed saw-tooth figure A(cir.) cotais (for the equivalece of trapezes view above): A ( cir.) gree triagles. (1) We see that i (1) the successio of the addeds is formed by products i which: the first factors give the sequece of the first six odd umbers ad the secod the successio, backward, of the first six atural umbers. We represet ow the (1) with the followig scheme (that also plays the dislocatio of the trapezoids i the figure ad their cotets, makig it more itelligible the cout): Fig.

3 The schematized coutig shows that the total umber of gree triagles i the A(cir.) figure is give by the sum of the squares of the first 6 atural umbers (there is also a diagoal path i the scheme, which leads to the same coclusio. It therefore appears that, the area of the saw-tooth figure, expressed i gree triagles, is give by the square pyramidal umber P 6 : 6 A( cir.) P6 k The geeralizatio to ay umber of divisios, it follows from the possibility to exted the scheme of Figure to the umber, addig rows cotaiig successive sequeces of odd umbers, util the -th. The result is, i geeral, that: A ( cir.) 1 + ( 1) + 5 ( ) ( 1) 1 Ad the area of the umpteeth saw-tooth figure, which circumscribes the parabolic segmet, will be expressed by the square pyramidal umber P : ( cir.) A P k This circumstace, together with the geeral result obtaied for the area of triagle ABC (which is equal to ), we ca reduce the proof to the simple check of the followig relatioship: lim k where the umerator to the first member is the -th square pyramidal umber P. You kow that the sum i the umerator of () is: 1 P after which the limit () follows from the fact that the ratio of two polyomials of the same degree i the variable teds (as teds to ifiity) to the ratio betwee respective leadig coefficiets (coefficiets of terms of maximum degree). But () states that: the area (measured i gree triagles) of the circumscribed figure is oe-third the area of the triagle ABC, as teds to ifiity. Follows the statemet i the propositio 16. ()

4 The proof "from below" So that a proof could be called "complete" requires two estimates, oe from above ad oe from below, ie, with a figure out ad a iside from the parabolic segmet. Fig. Iscribig a saw-tooth figure A(is.) i the parabolic segmet, as i the figure, oe ca see that it cotais: A ( is.) gree triagles. But the umber 55 appears to be the 5th square pyramidal umber; therefore, followig the same reasoig made above, we ca write: A ( is.) 1 ( 1) + ( ) ( ) 1 ad, for the umpteeth area of the iscribed figure: ( is.) 1 1 A P k Thus, the proof "from below" follows from the equality: lim 1 k which is also true, sice the sum i the umerator of () is: P which is still a third-degree polyomial i with leadig coefficiet equal to 1/. () 4

5 Propositio 17 The immediate cosequece of the propositio 16 is the propositio 17 with which Archimedes proves the fudametal theorem o the area of the parabolic segmet (but usig a mechaical method): The area of the parabolic segmet is 4/ of the triagle havig the same base ad equal height. Fig. 4 For a property of the parabola: LM MN, the: A PS ABC / 4/ ABM The same thig whe you cosider the half ALM of the parabolic segmet ABM. I fact we have: ALM AMN, ad the parabolic segmet with base AM has area equal to 1/ of AMN (prop. 15) ad therefore of ALM. Doublig follows the statemet. Sice the propositio 17 is a corollary of 16, havig demostrated geometrically 16, the it is also geometrically proved the fudametal theorem. Notes 1 - The model chose for our proof provides a opportuity to show i a differet way as stated by Archimedes i the propositio 15, i which its proof is geeralized to a parabolic segmet with base ot perpedicular to the axis. Our proof ca refer (without chagig aythig i the text) to a figure more geeral obtaied by shiftig arbitrarily the segmet BC o its straight lie, like this: 5

6 Fig. 6 I fact, the trasformatio does ot affect ay of equivalece relatios betwee trapezoids ad triagles used i the proof, which are the essece of the proof itself. Refereces L. Acora, Quadratura della parabola co il umero piramidale quadrato, i Archimede (Le Moier), 4/014 Liks 1 - See aimatio of the proof at

CALCULUS BASIC SUMMER REVIEW

CALCULUS BASIC SUMMER REVIEW NAME rise y y y Slope of a o vertical lie: m ru Poit Slope Equatio: y y m( ) The slope is m ad a poit o your lie is, ). ( y Slope-Itercept Equatio: y m b slope= m y-itercept=