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Analysis by Its History
– p.1/57
Analysis by Its History
Traditional “Bourbaki style” instruction :sets,numbers,mappings
⇒limits,continuousfunctions
⇒ derivatives ⇒ integrals.
– p.1/57
Analysis by Its History
Traditional “Bourbaki style” instruction :sets,numbers,mappings
⇒limits,continuousfunctions
⇒ derivatives ⇒ integrals.
“ But the misunderstanding was that [Bourbaki] should be a textbook for everybody.That
was the big disaster.”
(Pierre Cartier, Math. Intell., 1998, p.25)
“Es ist ... diesen Meisterwerken [Laplace, Lagrange] kaum mehr ihre Werdegeschichte zu
entnehmen. Dadurch ist dem Leser die ... fuer einen selbständigen Geist grösste Freude
versagt, unter der Führung des Meisters die gefundenen Resultate selbsttätig gleichsam noch
einmal zu entdecken. In diesem Sinne mangelt den Werken der klassischen Zeit das
eigentlich erzieherische Moment.”
(Felix Klein, Vorl. Entw. der Mathematik 19. Jahrhundert, 1926)– p.1/57
Analysis by Its History
Traditional “Bourbaki style” instruction :sets,numbers,mappings
⇒limits,continuousfunctions
⇒ derivatives ⇒ integrals.
Historical development was the other direction :
Cantor 1872Dedekind ⇐ Cauchy 1821
Weierstrass ⇐ Newton 1665Leibniz 1675 ⇐
ArchimedesFermat 1638Cavalieri 1647
– p.2/57
Analysis by Its History
Traditional “Bourbaki style” instruction :sets,numbers,mappings
⇒limits,continuousfunctions
⇒ derivatives ⇒ integrals.
Historical development was the other direction :
Cantor 1872Dedekind ⇐ Cauchy 1821
Weierstrass ⇐ Newton 1665Leibniz 1675 ⇐
ArchimedesFermat 1638Cavalieri 1647
Integration :
Archimedes, parabola B. Cavalieri Fermat 1638, publ. 1679,y = x−2– p.2/57
Differentiation :
(Leibniz 1684)
(Euler 1748, Préface) (Newton 1671, publ. 1736)
Reduces integration to inversion of differentiation formulas ...The extent of this calculus is immense ... And it gives rise toaninfinity of surprising discoveries concerning curved or straighttangents, questionsDe maximis & minimis, inflexion points andcusps of curves, envelopes, caustics from reflexion or refraction,&c. as we shall see in this work.
(Joh. Bernoulli - Marquis de L’Hospital 1696)
Critics of “infinitely small differentials” by Lagrange (1797)...“differential quotient” dy
dx ⇒ “derivative” y′
”integral”∫
y dx ⇒ “primitive” – p.3/57
Limits : Cauchy 1821 crit. Lagrange (counter-ex.y = e−1
x2 ),places theory of limits on top of treatise :
(Cauchy 1821, p. 13)
and reintroduces “infinitely smalls” and derivatives as limits
(Cauchy 1821, p. 62) – p.4/57
Sets and Numbers :G. Cantor 1872 : first set theory notation
(first notation for a set, the setA of rational numbers (Math. Annalen 5, p. 123))
... for establishing a theory ofreal numbers (his setB) ...√
3 ist also nur ein Zeichen für eine Zahl, welche erst noch gefunden werden
soll, nicht aber deren Definition. Letztere wird jedoch in meiner Weise, etwa
durch
(1.7, 1.73, 1.732, . . .)
befriedigend gegeben.
(G. Cantor, Math. Annalen 33 (1889), p. 476)
... in a paper on convergence of trigonometric series.Hence :
Cantor 1872Dedekind ⇐ Cauchy 1821
Weierstrass ⇐ Newton 1665Leibniz 1675 ⇐
ArchimedesCavalieriFermat 1638
– p.5/57
However ! Euler : (the great pedagogue)Introductio1748:
(beginning of Euler’sIntroductio(1748, french transl. 1796))
I. Newton did the same: 1669:De Analysi; 1671:Fluxiones.
Contents :
Chapter I. Introductio in Analysin InfinitorumChapter II. Differential and Integral CalculusChapter III. Foundations of Classical AnalysisChapter IV. Calculus in Several Variables
– p.6/57
Chapter I. Introductio in Analysin InfinitorumThe Birth of Algebra: (Al-Khowârizmî, Bagdad 830)
Problem:“a square and ten roots of the same square are equal to39 in numbers”.
x2x25x5x
5x5x25
x
x
5
5
(from Al-jabr w’al muqâbala; solution ofx2 + 10x = 39 by Al-Khowârizmî)
Solution:Attach two rectangles of 5 roots to the square and“complete the square” by adding the square52 = 25. Thissecond square is39 + 25 = 64, hence of root8. Thus our root is8 − 5 = 3. In modern notation (as in today’s high schools):
x2 + 10x = 39 ⇒ x2 + 10x + 25 = 64 ⇒ x + 5 = 8.– p.7/57
Cubic Equations. x3 + 6x = 20 (Scipio –Tartaglia – Cardano):
Cardano’sArs Magna1545
x xv
vv
uu
x xv
v
v
v
uu
Solution (same idea as Al-Khowârizmî’s first problem):Represent the term6x by 3 plates and3 columns, of totalvolume3uvx, attached to the cubex3. Hence
x = u − v, uv = 2 and u3 − v3 = 20.
Of the two quantitiesu3 and−v3 we thus know thesumand theproductu3(−v3) = −23 = −8. Thus they are the roots of
λ2 − 20λ − 8 = 0. – p.8/57
Cubic Equations. x3 + 6x = 20 (Scipio –Tartaglia – Cardano):
x2x25x5x
5x5x25
x
x
5
5
x xv
vv
uu
x xv
v
v
v
uu
Solution (same idea as Al-Khowârizmî’s first problem):Represent the term6x by 3 plates and3 columns, of totalvolume3uvx, attached to the cubex3. Hence
x = u − v, uv = 2 and u3 − v3 = 20.
Of the two quantitiesu3 and−v3 we thus know thesumand theproductu3(−v3) = −23 = −8. Thus they are the roots of
λ2 − 20λ − 8 = 0. – p.9/57
800 Years of Development of Algebraic Notation:
al-Khwarizmı (830)solution of
x2 + 21 = 10x
Cardano (1545)solution of
x3 + 6x = 20
Viète (1591)solution forA ofA2 + 2BA = Z
Descartes (1637)for probl. of Heraclitus
– p.10/57
Descartes’Geometrie. (Pappus, Book VII, Prop. 72)
Problem(Heraclitus):Given squareABDC
and given lengthBN ,find E on extendedAC
such thatEF = NB.
A
B
C
D
EN
F
(Descartes 1637, p. 387) – p.11/57
Descartes’ Solution.(Descartes 1637, p. 387-388)
A
B
C
D
EN
F⇒
a
u x
cc a−x
Pythagoras: u =√
a2 + x2 , Thales:x
u=
a − x
c,
x4 − 2ax3 + (2a2 − c2)x2 − 2a3x + a4 = 0
EulerE170:x
a+
a
x= y ⇒ y2 − 2y − c2
a2= 0 .
– p.12/57
Cartesian Coordinates
Problem.(Pappus, Introduction to Book VII)Given 3 (4,5,6,..) fixedlinesa, b, c, (d, ..), find set of pointsP such that
a
b c
O A
P
B C
a
b cd
O A
P
B C
D
PA · PB = (PC)2 PA · PB = PC · PD . – p.13/57
Descartes’ Solution.
a
b cy
xO A
P
B C
a
b cdy
xO A
P
B C
D
“Que le segment de la ligneOA, qui est entre les poinsO & A,soit nomméx. & queAP soit nomméy”
So were born the Cartesian Coordinates!– p.14/57
The Binomial Theorem.
Al-Karajı, Xth centuryPascal 1654
Example:n = 2: (Eucl. II.4);n = 3:(a+b)3 = a3+3a2b+3ab2+b3 or (1+x)3 = 1+3x+3x2+x3 .
n = 4 : (1 + x)4 = 1 + 4x + 6x2 + 4x3 + x4 .etc. – p.15/57
11 1
1 2 11 3 3 1
1 4 6 4 11 5 10 10 5 1
1 6 15 20 15 6 1
Pascal (1654, p. 7, “Consequence douziesme”):Ratios:11
21
12
31
22
13
41
32
23
14
51
42
33
24
15
61
52
43
34
25
16
– p.16/57
(1 + x)n = 1 +n
1x +
n(n−1)
1 · 2 x2 +n(n−1)(n−2)
1 · 2 · 3 x3 + . . .
– p.17/57
(1 + x)n = 1 +n
1x +
n(n−1)
1 · 2 x2 +n(n−1)(n−2)
1 · 2 · 3 x3 + . . .
Newton 1665:“interpolate”formula forrational andnegativen.
manuscript Newton 1665Examples:1
1 + x= 1 − x + x2 − x3 + ...
1
1 − x= 1 + x + x2 + x3 + ...
(1 + x)1
2 = 1 + 12 x − 1·1
2·4 x2 + 1·1·32·4·6 x3 − ...
(1 + x)−1
2 = 1 − 12 x + 1·3
2·4 x2 − 1·3·52·4·6 x3 + ...
– p.17/57
Exponential Function :
1
P
VT
y
b
z
1 0
1y1
b
y2
b
y3
b b
y
x
(a) (b) (c)
Debaune 1638:Find curve with fixed subtangent (Fig. (a))?Descartes:“ . . . tentavit, sed non solvit”Leibniz 1684: z−y
b = y1 ⇒ z = (1 + b)y (Fig. (b))
Fig. (c): yi = (1 + b)i Euler 1748:
y = (1 + xN )
N= 1 + x +
x2(1− 1
N)
1·2 +x3(1− 1
N)(1− 2
N)
1·2·3 + . . .
= 1 + x + x2
1·2 + x3
1·2·3 + x4
1·2·3·4 + . . . = ex (N →∞)– p.18/57
Logarithms : (J. Napier 1614, J. Bürgi 1619, H. Briggs 1624)“Students usually find the concept of logarithms very difficult tounderstand.”
(B.L. van der Waerden 1957, p. 1)Logarithms transform
products→ sums and powers→ products.
– p.19/57
Logarithms : (J. Napier 1614, J. Bürgi 1619, H. Briggs 1624)“Students usually find the concept of logarithms very difficult tounderstand.”
(B.L. van der Waerden 1957, p. 1)Logarithms transform
products→ sums and powers→ products.Ancient computations by repeated square roots:
NumbersLogarithms10.0000 1.7.4989 0.875√√
103 = 5.6234 0.754.2170 0.625√
10 = 3.1623 0.52.3714 0.375√√
10 = 1.7783 0.251.3335 0.1251.0000 0.
.00
.25
.50
.75
1.00
1101/4101/2 103/4 10– p.19/57
More precise tables: incredible amount of work:
J. Napier, Edinburgi 1614 H. Briggs, Londini 1624 – p.20/57
Computation of Areas (Archimedes, Cavalieri, Fermat).Example:y = xa (Fermat 1638) :
B
Ba
θB
θaBa
θ2Bθ2aBa
y = xa
Idea:Choose grid points asgeometric sequenceB, θB, θ2B,. . . ,⇒ areas of rectangles also geometric sequence, hence
S = geom. series =Ba+1
a + 1if a > −1.
does not work for hyperbola (a = −1) . . .– p.21/57
Area of Hyperbola is a Logarithm :(Gregory of St. Vincent 1647, Mercator 1668) :
1 2 3 4 5 60
1
y = 1/x same areas
1 20
1
y = 1/x
0
1
areas
a
1
− x
+ x2
− x3
+ x4
.0
.5
a
− a2/2
+ a3/3
− a4/4
+ a5/5
11+x = 1 − x + x2...
ln(1 + x) = x − x2
2 + x3
3 ...
– p.22/57
Trigonometric Functions. “Traditional” (BI Mannheim 1962):
... and three pages later :
No word from where come these definitions ... – p.23/57
... no word about Archimedes ...
– p.24/57
... or Ptolemy :(150 A.D., printed 1813)
α60
60
cordα
Tables ofcordα for measures in Astronomy and Geography.– p.25/57
... or Regiomontanus ...
Bhaskara⇒ Arabs⇒ Regiomontanus(publ. 1533):Sinus.
0 1
1
cos α
sin α
tan α1
cot α
α
Around 1464, Regiomontanus computed a table (“SEQVITVRNVNC EIVSDEM IOANNIS Regiomontani tabula sinuum, persingula minuta extensa. . .”) giving the sine of all angles atintervals of 1 minute, with five decimals.
– p.26/57
... or Euler ...
J.J. Stampioen(Leyden 1632)
cosinum anguli adA fore =rq − Cc
Ssr .
sinu crurisAB = S, cosinus eiusdem= C,sinu crurisAC = s et cosinu= c,cosinu baseosBC = q, et radio= r;
F.C. Maier( St. Pet., 1727)
cos :anguliA =cos :BC−cos :AB · cos :AC
sAB · sAC,
posito radio vel sinu toto1.
L. Euler(E14,1729)
A B
C
c
abcos A =
cos a − cos b · cos c
sin b · sin c
L. Euler(E214,1753)
– p.27/57
Formulas and Series (Ptolemy, Regiomontanus, Viète) :
0 1xy
x
1
cos ycos y
sin ysin y
cos y cos x
cos y sin xcos y sin x
sin y sin xsin y sin x
sin y cos xsin y cos x
sin(x + y) = sin x cos y + cos x sin y
cos(x + y) = cos x cos y − sin x sin y
Iterate (de Moivre 1730)(appear binom. coeff.):
cos nx = cosn x − n(n−1)1·2 sin2 x cosn−2 x + n(n−1)(n−2)(n−3)
1·2·3·4 sin4 x cosn−
Idea (Euler 1748, §134):x → 0, n → ∞, with nx fixed 7→ x,thencos x → 1, sin x → x, n sin x → nx 7→ x. Hence
cos x = 1 − x2
1 · 2 +x4
1 · 2 · 3 · 4 − ... sin x = x − x3
1 · 2 · 3 + ...– p.28/57
Picture:
5 10 15
−3
−2
−1
0
1
2
3 1
3
5
7
9
11
13
15
17
19
21
23
25
27
29
31
33
35
37
39
41
43
45y = sinx
5 10 15
−3
−2
−1
0
1
2
3
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
32
34
36
38
40
42
44y = cosx
– p.29/57
Series for tan. (how Newton did it,De Analysi1669):
y = tan x =sin x
cos x= a1x + a3x
3 + a5x5 + a7x
7 + . . . .
divide
x−x3
6+
x5
120−. . . = (a1x+a3x
3+a5x5+. . .)(1−x2
2+
x4
24−. . .).
compare coefficientsx, x3, x5 ... :
1 = a1, −1
6= −a1
2+ a3,
1
120=
a1
24− a3
2+ a5,
tan x = x +x3
3+
2x5
15+
17x7
315+
62x9
2835+
1382x11
155925+
21844x13
6081075...
mysterious series ... regularity?⇒ Euler-Maclaurin1736/42/55.– p.30/57
Arcus functions : Giventangentx, find arc??
D
E
x
arc
O
1
C
A B
C
F
x ∆x
1
1√1+x2
√1+x2
Proof of Leibniz 1676: Archimedes:arc= 2 · areaOEC.Thales, Pyth. and Eucl. I.41:2 · areaABC = AB · CF =
∆x
1 + x2= (1 − x2 + x4 − x6...) · ∆x.
(Fermat-Cavalieri)⇒ arctan x = x − x3
3+
x5
5− x7
7...
– p.31/57
Arcus sinus : Givensinus, find arcy ??
−1 0 1
1
x
xx
1
y∆y
∆x∆x√1 − x2
Newton,de Analysi1669
Original proof of Newton 1669: Similar triangles⇒∆y
∆x=
1√1 − x2
= 1 +1
2x2 +
1 · 32 · 4 x4 +
1 · 3 · 52 · 4 · 6 x6 + ...
as above by Fermat-Cavalieri ...
y = arcsin x = x +1
2
x3
3+
1 · 32 · 4
x5
5+
1 · 3 · 52 · 4 · 6
x7
7+ ...
Newton then:sin (invers.),cos(√...), tan(div.), arctan(invers.).– p.32/57
Chapter Iterminates with Section I.5 onComplex Numbers ...
G. Cardano (1545),(5 +√−15)(5 −
√−15) = 25 − (−15) = 40
“Complex” Debeaune curve:
0 1
1
0 1
1
0 1
1
1 + iϕ2
iϕ2
(1 + iϕ2 )2
N = 2
1 + iϕ4
(1 + iϕ4 )2
(1 + iϕ4 )3
(1 + iϕ4 )4
N = 4 N = 16
cos ϕ
sin ϕ ϕ
eiϕ
Beautiful relation eiϕ = cos ϕ + i sin ϕ (Euler 1743)– p.33/57
... and Section I.6 onContinued Fractions :
tan x =x
1 − x2
3 − x2
5 − x2
7 − x2
9 − . . .
tanm
n=
m
n − m2
3n − m2
5n − m2
7n − . . .Theorem (Lambert 1768): Ifx is rational,tan x is irrational.
Now, “les esprits” are sufficiently prepared for ...– p.34/57
Chapter II. Differential and Integral CalculusEt j’ose dire que c’est cecy le problésme le plus utile, & le plusgeneral non seulement que ie sçache, mais mesme que i’ayeiamais desiré de sçauoir en Geometrie. . .
(Descartes 1637, p. 342)
Problem. Let y = f(x) be a given curve. At each pointx wewish to know theslope, thetangentor thenormal.Motivations.– Calculation of the angles under which two curves intersect(Descartes);– construction of telescopes (Galilei), of clocks (Huygens1673);– search for the maxima, minima of a function (Fermat 1638);– velocity and acceleration of a movement (Galilei 1638,Newton 1687);– astronomy, verification of the Law of Gravitation (Kepler,Newton). – p.35/57
Example : Parabola. (x + ∆x)2 = x2 + 2x ∆x + ∆x2
0
2x ∆x + ∆x 2
x
∆xy = x2
Drawing by Joh. Bernoulli 1691
∆y = (x + ∆x)2 − x2 = 2x ∆x + ∆x2 ⇒ dy
dx= 2x
(known by Apollonius 250 B.C.). Similar:
y = xn ⇒ dy
dx= nxn−1.
– p.36/57
Exponential function and logarithm:
y = ex
x = ln y y
∆x
x0
∆y
1
1
Thales:
∆y
∆x=
y
1⇒
dy
dx= y = ex,
dx
dy=
1
y.
Sinus and cosinus.Newton’s picture,Leibniz’ symbols,Thales:
0 1
1
∆s
−∆c−∆c
∆x
c
s1 x
s = sin x,
c = cos x,
ds
dx= c = cos x,
dc
dx= −s = − sin x.
– p.37/57
“His positi calculi regulae erunt tales:”(Leibniz 1684)
Productrule: u
∆u
v ∆v
uv
v∆u
u∆v⇒ d(uv) = u dv + v du
Fractions:
v
∆v
w = 1/v−∆w
1
∗
∗ ⇒w =
1
v⇒ − v dw = w dv
dw = −w
vdv = −dv
v2
Quotient rule. (second access by algebra):
u + ∆u
v + ∆v− u
v=
v∆u − u∆v
v2 + v∆v⇒ d(
u
v) =
v du − u dv
v2 – p.38/57
Inverse function.
0 1
∆x
∆y
slope = 1/2
0
1
∆x
∆y
slope = 2
y =√
x x = y2
dy
dx=
1dxdy
Composition (chain rule).
10
1
∆x
∆y
x
y
10
1
∆x
∆z
x
z
10
1
∆z
∆y
z
yy = sin 2x z = 2x y = sin z
dy
dx=
dy
dz· dz
dx
– p.39/57
Problemes “de maximis et minimis”Example : Fermat’s Principle
. . . et trouver la raison de la réfraction dans notre principecommun, qui est que la nature agit toujours par les voies les pluscourtes et les plus aisées.
(Fermat to De La Chambre, août 1657,Œuvres2, p. 354)
α1α1
α2α2
xx l − xl − x
a
A
b
B
v1v1
v2v2
Drawing of Joh. Bernoulli
“cujus accuratissimam demonstrationem a principio nostroderivatam exhibet superoir analysis”
– p.40/57
Leibniz’ solution of Fermat’s principle.Leibniz (1684) proudly solves Fermat’s problem “in tribuslineis” :
T =
√a2 + x2
v1+
√
b2 + (ℓ − x)2
v2= min !
by differentiating with respect tox
T ′ =1
v1
2x
2√
a2 + x2︸ ︷︷ ︸
sin α1
− 2(ℓ − x)
2√
b2 + (ℓ − x)2
︸ ︷︷ ︸
sin α2
1
v2= 0.
in France:Descartes’ law; in Denmark:Snellius’ law.
– p.41/57
Leibniz’ solution of Fermat’s principle.Leibniz (1684) proudly solves Fermat’s problem “in tribuslineis” :
T =
√a2 + x2
v1+
√
b2 + (ℓ − x)2
v2= min !
by differentiating with respect tox
T ′ =1
v1
2x
2√
a2 + x2︸ ︷︷ ︸
sin α1
− 2(ℓ − x)
2√
b2 + (ℓ − x)2
︸ ︷︷ ︸
sin α2
1
v2= 0.
in France:Descartes’ law; in Denmark:Snellius’ law.
However: Min or Max ??Die Prüfung, ob ein Maximum oder Minimum vorhanden ist, macht am meisten
Schwierigkeiten ...(Weierstrass,Differentialrechnung, Vorl. an dem königl. Gewerbeinstitut, 1861)
... need to consult higher derivative ... – p.41/57
Higher derivatives.But the velocities of the velocities, the second, third, fourth, andfifth velocities, &c., exceed, if I mistake not, all humanunderstanding. The further the mind analyseth and pursueththese fugitive ideas the more it is lost and bewildered;. . .
(Bishop Berkeley 1734,The Analyst)
Joh. Bernoulli 1691/92
y0
y1
x0 x1
y0
y1
x0 x1
y′′ > 0
y′′ < 0
– p.42/57
Taylor (1715) Interpolation polynomial∆x → 0, n → ∞:
Hierin liegt nun tatsächlich einGrenzubergang von unerhorterKuhnheit. (F. Klein 1908, Zweite Aufl., p. 509)
x0 x1 x2
f(x)
p(x)
∆x = 0.4
x0x1x2
f(x)
p(x)
∆x = 0.2
x0x2
f(x)
p(x)∆x = 0.1
Polyn.: p(x) = y0 +x − x0
1
∆y0
∆x+
(x − x0)(x − x1)
1 · 2∆2y0
∆x2,
Series : f(x) = f(x0) + (x − x0)f′(x0) +
(x − x0)2
2!f ′′(x0) + . . .
ALL above series are examples !!– p.43/57
Envelopes of families of curves.Ex.: Canon ball trajectories:
Mon Frere, Professeur àBale, a pris de là occasion de rechercherplusieurs courbes que la Nature nous met tous les jours devantles yeux. . .
(Joh. Bernoulli 1692)
.5 1.0
.2
.4
y = ax − x2(1 + a2)
2– p.44/57
Envelopes of families of curves.Ex.: Canon ball trajectories:
Mon Frere, Professeur àBale, a pris de là occasion de rechercherplusieurs courbes que la Nature nous met tous les jours devantles yeux. . .
(Joh. Bernoulli 1692)
.5 1.0
.2
.4
.5 1.0
.2
.4
y = ax − x2(1 + a2)
2
∂y
∂a= 0
– p.44/57
Envelopes of families of curves.Ex.: Canon ball trajectories:
Mon Frere, Professeur àBale, a pris de là occasion de rechercherplusieurs courbes que la Nature nous met tous les jours devantles yeux. . .
(Joh. Bernoulli 1692)
.5 1.0
.2
.4
.5 1.0
.2
.4
.5 1.0
.2
.4
y = ax − x2(1 + a2)
2
∂y
∂a= 0 y = (1 − x2)/2
– p.44/57
Further examples : (Jac. und Joh. Bernoulli 1692)
−1 0 1
5 10 15
5
10
15
xx
yy
Caustic of a circle Lines (Dürer 1525)
y = −√
1 − x2/3(1
2+ x2/3) y = x − 2
√13x + 13
– p.45/57
Curvature (Newton 1671: inters. of two neighboring normals)
There are few Problems concerning Curves more elegant thanthis, or that give a greater Insight into their nature.
(Newton 1671, Engl. pub. 1736, p. 59)
a
f(a)
(x0,y0)
y0 − f(a)y0 − f(a)
a − x0a − x0 a + ∆a
(Newton,Fluxions1671, publ. 1736)
y = f(a) − 1
f ′(a)(x − a) r =
(1 + (f ′(a))2)3/2
|f ′′(a)|– p.46/57
Integral Calculus.. . . notam
∫pro summis, ut adhibetur notad pro differentiis. . .
(Letter of Leibniz to Joh. Bernoulli, March 8/18, 1696)
. . . vocabulum i n t e g r a l i s etiamnum usurpaverim. . .
(Letter of Joh. Bernoulli to Leibniz, April 7, 1696)
(Newton, publ. 1736)BB
D
y
dxz
xa
z
y
dx
dz = y dx
z =∫
y dx
“... sic nobis summæ & differentiæ seu∫
& d, reciproquæ sunt.” – p.47/57
ALL above examples are inversions of simple diff. rules:∫
xn dx =xn+1
n + 1+ C (n 6= −1)
∫1
xdx = ln x + C
∫1
1 + x2dx = arctan x+C
∫1√
1 − x2dx = arcsin x+C
Many new applications:Surfaces, arc lengths, gravity centers,total energy,...
x = −ax = −a
x = a
dxdx
(a2 − x2)1/2
0 10
1
dx
dx
dy dyds
– p.48/57
Differential equations. Example:Leibniz’ Isochrone:(ισoς =equal,χρoνoς =time)
B
O
A
C
Body falling fromO;
Galilei 1638:Velocity: (dy
dt)2
= −2gy
Leibniz, Sept. 1687:Given constantb,
find curveABC such thatdy
dt= −b.
Vir CeleberrimusChristianus Hugenius(Oct. ’87)gives solution “sed suppressa demonstratione & explicatione”;
Leibniz 1689:“celeb. Auctoris Demonstratio Synthetica”;
Jak. Bernoulli 1690:Solution with “modern” calculus⇒ ...
– p.49/57
Jakob’s solution of Leibniz’ isochrone. (A.E. 1690, p. 218)
−1
1
dy
dx
y
x
1
√
−1 − 2gyb2
S1
S2
A
B
C
Galilei: (ds
dt)2
= −2gy
want: (dy
dt)2
= b2
divide:1 · dx = −√
−1 − 2gy
b2· dy.
x =b2
3g(−1 − 2gy
b2)3/2
.
“Solutio sit linea paraboloeides quadrato cubica. . .” (Leibniz)
first use of term“integral”, first differential equation solved“by separation of variables”.
– p.50/57
Galilei. Discorsi 1638:
Giornata seconda:“ ...questa catenella si piega in figuraparabolica,...” p.54Giornata terza:“Da quanto si è dimostrato sembra si possaricavare che il movimento più veloce da estremo ad estremo nonavviene lungo la linea più breve, cioè la retta, ma lungo un arcodi cerchio.”
AB
C
D
E
F
G
“... Mr. Leibnits remarque en Galiléedeux fautes considerables:c’est quecet homme-là, qui étoit, sans contredit,le plus clairvoyant de son temsdans cette matiére, vouloit conjecturerque la courbe de la chainetteétoit une Parabole, que celle de laplus vite descente étoit un Cercle.."(Joh. Bernoulli, 1697)
– p.51/57
Other differential equations treated:Catenary
(Leibniz, Joh.Bernoulli 1691)
Brachystochrone
(Joh. and Jak. Bernoulli 1696)
Tractrix
(Leibniz 1676, 1693)
Isochronous pendulum
(Huygens 1673) – p.52/57
Now, “les esprits” are prepared for the search of more rigor ...
3. Foundations of Classical Analysis– What is a derivative really?Answer: a limit.– What is an integral really?Answer: a limit.– What is an infinite seriesa1 + a2 + a3 + . . . really?
Answer: a limit.This leads to– What is a limit?Answer: a number.And, finally, the last question:– What is a number?Answer very difficult : Euklid Book V, Bolzano, Weierstrass,Dedekind, Cantor, Méray, Heine (all 1872).
Here starts the “traditional” itinerarynumbers⇒ limits ofsequences⇒ series⇒ absolutely conv. ser.⇒ continuosfunctions⇒ uniform convergence⇒ integration⇒differentiation⇒ Taylor series. . . – p.53/57
The judgement of the students.
At the end of the year the students were given a questionary andasked to check the
easiestmost difficult
most interesting
sections of the course. With the results ...
– p.54/57
1990: easy difficult interest.
– p.55/57
1991: easy diffic. inter.
– p.56/57
the easiest at the beginning...the most difficult towards the end...and the most interesting in the middle ...
The results ...
– p.57/57
the easiest at the beginning...the most difficult towards the end...and the most interesting in the middle ...
The results ...
are convincing!!Thank you.
– p.57/57
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