4 The route to calculus Flashcards
What is calculus
not suddenly invented by Newton & Leibniz in the seventeenth century, but had Greek origins.
Archimedes work on parabola?
??
Early history of curves
study of areas led to integration
study of tangents led to differentiation
Two studies developed separately until 1666:
Fundamental Theorem of
Calculus
integration & differentiation: inverses
Greek Origins of Calculus
Eudoxus: Method of Exhaustion
Euclid: Elements XII, Proposition 2: Areas of circles are to one another as the squares on their diameters
Archimedes:
Area of a parabolic segment
Volume of a sphere
*after Archimedes died nobody could carry on his calculus work
Eudoxus: Method of Exhaustion
QUESTION IN EXAM:
comment on the method
*Draw figure and say that the area is bigger is bigger than the area of a polygon drawn inside of it
*rigourous and AVOIDS infinitesimals and limits and approximations
*drawbacks to method: difficult to apply in practise and use two contractions to show a=b, no general method and have to know result before you begin
(show you can get a polygon as close as you like in equal area)
Curvilinear areas compared by exhausting them with polygons. Based on Eudoxus’ Principle :
Elements X, I If from a magnitude be taken more than its half, from remainder more that its half and so on, at length
there will remain a magnitude less than any given magnitude.
(keep on taking more than half away from a large number can eventually give any small number)
Natural; rigorous; avoids infinitesimals, Natural; rigorous; avoids infinitesimals,
limits & approximations
Difficult to apply: quantities Difficult a & b proved equal
by showing neither neither a< b nor a < b holds; no
general methods, each problem needs own
solution; result must be known before begins
Proofs by Exhaustion
IN EXAM
*must show which proofs can be proved by exhaustion in exam don’t need to know the proofs just results
Euclid
Elements XII, Proposition 2: Areas of circles are to
one another as the squares on their diameters .
Elements XII: Applied to circles, cones, pyramids, spheres.
(HAVE TO DRAW DIAGRAM)
- draw different polygons inside the shape and make them swell out to approximate the area
- if we have any large number we can take away 1/2 of it to get any small number smaller than anhy given
Archimedes
Quadrature of the Parabola, Proposition 24:
Parabolic segment has area four-thirds that of inscribed
triangle having the same base and equal height.
Sphere and Cylinder,
*Archimedes give at least three proofs of the area of a parabola
twice in the book and then in another
Proposition 34: Sphere has
volume four times that of the cone with base a great circle
of the sphere and height equal to its radius.
*do we need proof? by Archimedes non-rigorous method
Exhaustion of a circle Exhaustion of a circle c by inscribed regular polygons
PROOF EXAMPLE
Let a be any area. Then there is a regular polygon inscribed in c whose area differs from that of c by less than a
Proof Let p be regular polygon inscribed in c. Let p* be the regular polygon with vertices those of p, together with the midpoints of arcs joining adjacent vertices of p. Area between p* and c is less than half that between p and c. Eudoxus’ Principle shows that by doubling the number of vertices of p in this way sufficiently often, an inscribed regular polygon will be reached whose area differs from that of c by less than a. Q.E.D.
*example a triangle drawing inside of a circle. Adding extra points at the midpoints on the circle to form hexagon. and this is clearly a closer approximation as the errors are much smaller, taking away more than a half and get as close as you like. Half the error by adding midpoints.
Elements XII, Proposition 2
Let circles c, C
have diameters d, D & areas a, A. Then
a /A = d^2 / D^2
Proof We show that both inequalities below are untenable:
(i) a /A > d^2 / D^2
& (ii) a /A < d^2 / D^ 2
Suppose (i) holds: Then for some area a* < a such that
a* / A = d^2/ D^2
Let q be a regular polygon (area p) inscribed in c such that
a - p < a - a, whence a < p.
Let Q be a polygon (area P) similar to q inscribed in C.
By Elements XII, Proposition 1, p / P = d^2 /D^2 . Red-underlined equations show that a* / A = p / P, which is impossible, since a* < p & A > P (as Q lies in C). Thus (i) is untenable. A similar argument shows that (ii) is untenable. Thus, as desired, a /A = d^2 / D^2
*structure of all existing proofs:
to prove two things are equal we need to use two contradictions to show it is not bigger than than or less than so it must be equal to
- involves contradiction salad with an equality
- must understand how to start the proof by contradiction don’t have to carry on
Archimedes
c. 287 BC -212 BC
* for sure death date as killed by soldier
*Archimedes birthday is not accurate as it has been estimated from the age he looked when he died
Greatest mathematician of Antiquity.
Born Syracuse Sicily. Slain in Roman sack of city.
Studied at Alexandria? Mathematical friends there.
Seminal research in pure and applied mathematics.
Fanciful stories told about him.
More known of him than any other Greek mathematician.
*attacked romans by catapults, invented pulleys, mirrors to burn sails etc
Quality, quantity, originality, and diversity of output helped survival of works.
Some of his books have letters to friends, providing
invaluable background information.
Inventions: Archimedean screw, multiple pulleys
Anecdotes: King’s crown with its Eureka, defence of city, dramatic death, tombstone fuel his legend.
Plutarch’s Life of Marcellus (75 AD), describes siege of Syracuse. Gives firm date of death as 212 BC. The c.287 BC birth date less secure - based on an unproven assumption that he died aged seventy five.
- Archimedes Tombstone:
- everything was definitely Archimedes as work unlike euclid’s elements
Archimedes’ Resume
Equilibrium of Plane Figures I, II
Quadrature of the Parabola
(quadrature means area)
Sphere and Cylinder I, II
Spirals
Conoids and Spheroids
Floating Bodies I, II
Measurement of the Circle
(inequality for pi)
Sandreckoner
THE METHOD
Archimedes principle floating bodies:
Quadrature of the Parabola Quadrature of the Parabola
Proposition 24: Area of parabolic segment
LEARN THIS***
all you need to know is on this sheet?
Proposition 24: Area of parabolic segment is four-thirds that of inscribed triangle with same base and equal height
triangle drawn in parabola with vertices ABC midpoints? marked D E
S is parabolic segment cut off by chord AC.
B is point on arc
AC furthest from AC.
Distance of B from AC is height of S.
Triangle ABC has same base and height as S.
Of all triangles inscribed in S with base AC, triangle ABC has greatest area.
ABC is first polygon in exhausting S. Points D and E on arcs AB and BC, respectively, are furthest from AB and BC.
Sum of areas of triangles ABD and BCE is quarter that of
triangle ABC.
ADBEC is second polygon in exhausting S. Etc.
*the height of The Triangle is the same as the parabolic segment and the base
PROVED the area of the triangle is 4/3 of/..
- use this formula to find the integral from -a to a of x^2 dx
- first approximation from series of triangles inside the segment
start off with the triangle ABC then add 2 extra triangles according to…
original + (1/4) +(1/16)+….
=4/3
geometric progression
- FIRST EXHAUSTION is ABC triangle
- NEXT PENTAGON ADBEC
- NEXT ONE has extra point at each chord
- the extra area, shaded triangles, is 1/4 of the triangle
- and the sides furthest away
The New York Times The New York Times NEW YORK, TUESDAY, JULY 16, 1907 BIG LITERARY FIND IN CONSTANTINOPLE
Savant Discovers Books by Archimedes, Copied about 900 A D COPENHAGEN, July 15 - Professor J. L. Heiberg of
the University of Copenhagen has discovered a number of
palimpsests which, in addition to prayers and psalms of
the twelfth century, include works by Archimedes that
had hitherto been thought lost.
The Turkish authorities would not allow the professor to remove the manuscript, but he was allowed to copy it
*but quite a lot as had been written over to make for prayers
Sale of the Century
Auction: Christie’s New York, 29 October 1998
The Method letter
______________Syracuse
______________Sicily
Eratosthenes
Librarian
University of Alexandria
Egypt
Greetings
On a former occasion, I sent you some theorems discovered by me, but withheld their proofs. In
my new book, The Method, that I now send to you, I write out the proofs in full and explain the
peculiarity of a certain method which will enable you to investigate problems in mathematics by means of mechanics
Archimedes
The Method
In The Method Archimedes describes the preliminary mechanical investigations that paved the way for many of his mathematical discoveries.
In his discovery method, parallel cross sections of , a figure of unknown content are balanced against those of another of known content and centroid, using his principle of moments.
We sketch Archimedes’ discovery proof of the volume of a sphere, in which a sphere and a cone are balanced against a cylinder, the volumes of a cone and a cylinder being known.
Proposition 2: Volume of a sphere of radius r is (4/3)pir^3
method of exhaustion
this method is rigorous, avoids infinitesimal and limits by using indirect methods, proves a = b by contradiction argument, by showing that neither is less than or more than is tenable hard to apply answer must be known and guest in advance, there is no general methods and it’s used in the element book 12
