Shaping Modern Mathematics

Raymond FloodGresham Professor of

Geometry

Lectures At the Museum of London

• Ghosts of Departed Quantities: Calculus and its Limits Tuesday 25 September 2012

• Polynomials and their Roots Tuesday 6 November 2012

• From One to Many Geometries Tuesday 11 December 2012

• The Queen of Mathematics Tuesday 22 January 2013

• Are Averages Typical? Tuesday 19 February 2013

• Modelling the World Tuesday 19 March 2013

Ghosts of Departed Quantities: Calculus and its Limits

Raymond FloodGresham Professor of

Geometry

What is the Calculus?

Integration is used to find areas of shapes in two-dimensional space or volumes in three dimensions.

Archimedes (c287 – 212BC)

the volume of a cylinder is 11/2 times that of the sphere it surrounds. Archimedes, by Georg

Andreas Böckler, 1661

What is the Calculus?

Differentiation is concerned with how fast things move or change. It is used to find speeds and the slopes of tangents to curves.

Apollonius’s Conics

David Gregory’s 1703 edition of Euclid’s Elements with Propositions from Euclid this time drawn on the sand

Halley’s 1710 edition of Apollonius with conic sections drawn in the sand

Torelli ‘s 1792 edition of Archimedes and this time with a spiral, in fact an Archimedean spiral, drawn in the sand.

Characterising the family of ideas called the calculus

• A systematic way of finding tangents

• A systematic way of finding areas

• Connecting tangents and areas

Isaac Newton 1642 - 1727

d = 4.9 t2

d = 4.9 t2

Average speed = distance travelled / time taken

Finding instantaneous speed from average speed

• Distance travelled in time t equals 4.9 t2

• At a later time, t + o, distance travelled is: 4.9 (t + o)2

Finding instantaneous speed from average speed

• Distance travelled in time t equals 4.9 t2

• At a later time, t + o, distance travelled is: 4.9 (t + o)2

• In the time interval t to t + 0 distance travelled is 4.9 (t + o)2 – 4.9 t2 = 9.8 t o + 4.9 o2

Finding instantaneous speed from average speed

• Distance travelled in time t equals 4.9 t2

• At a later time, t + o, distance travelled is: 4.9 (t + o)2

• In the time interval t to t + 0 distance travelled is 4.9 (t + o)2 – 4.9 t2 = 9.8 t o + 4.9 o2

• Divide by o to find average speed over the interval = 9.8 t + 4.9 o

Finding instantaneous speed from average speed

• Distance travelled in time t equals 4.9 t2

• At a later time, t + o, distance travelled is: 4.9 (t + o)2

• In the time interval t to t + 0 distance travelled is 4.9 (t + o)2 – 4.9 t2 = 9.8 t o + 4.9 o2

• Divide by o to find average speed over the interval = 9.8 t + 4.9 o• Shrink the interval i.e. allow o to approach zero • Then this average speed, 9.8 t + 4.9 o, approaches

the instantaneous speed 9.8 t

Average speeds are the slopes of the lines passing

through time t = 5

Instantaneous speed is the slope of the tangent at t = 5

Finding the speed from the distance

Distance = 4.9 t2

Finding the speed from the distance

Distance = 4.9 t2 Speed = 9.8 t

Finding the acceleration from the speed

Speed = 9.8 t

Finding the acceleration from the speed

Find the slopes of the tangents

Speed = 9.8 t Acceleration = 9.8

Find the slopes of the tangents

Find the slopes of the tangents

DIFFERENTIATIONAcceleration = 9.8Speed = 9.8 tDistance = 4.9 t2

INTEGRATION

Area equals 9.8 t

Find the areas

Acceleration = 9.8

INTEGRATION

Find the areas

Acceleration = 9.8Speed = 9.8 t

INTEGRATION

Find the areas

Area = 4.9 t 2

Speed = 9.8 t Acceleration = 9.8

INTEGRATION

Find the areas Find the areas

Speed = 9.8 t Acceleration = 9.8Distance = 4.9 t2

Find the areas Find the areas

Find the Slopes

Speed = 9.8 t Acceleration = 9.8Distance = 4.9 t2

DIFFERENTIATION

Differentiate t To get 1

Differentiate t2 To get 2t

Differentiate t3 To get 3t2

Differentiate t4 To get 4t3

Differentiate t5 To get 5t4

DIFFERENTIATIONDifferentiate tn to get ntn – 1

Differentiate t to get 1

Differentiate t2 to get 2t

Differentiate t3 to get 3t2

Differentiate t4 to get 4t3

Differentiate t5 to get 5t4

INTEGRATIONWe get tn On integrating ntn – 1

We get t On integrating 1

We get t2 On integrating 2t

We get t3 On integrating 3t2

We get t4 On integrating 4t3

We get t5 On integrating 5t4

Gottfried Leibniz 1646 - 1716

Binary Arithmetic

It is possible to use … a binary system, so that as soon as we have reached two we start again

from unity in this way:

(0) (1) (2) (3) (4) (5) (6) (7) (8) 0 1 10 11 100 101 110 111 1000

… what a wonderful way all numbers areexpressed by unity and nothing.

Leibniz’s Calculating machine

The machine’s crucial innovation was a stepped gearing wheel with a variable number of teeth along its length, which allowed multiplication on turning a handle.

Leibniz notationd (or dy/dx) notation for differentiation:referring to the change in y divided by the change in x

∫ notation for the integration:finding areas under curves by summing lines.He defined omnia l (all the ls), which he then represented by an elongated S for sum, the integral sign, ∫.

First appearance of the Integral sign, ∫on October 29th 1675

Leibniz’s 1684 account of his Differential Calculus

Leibniz’s rules for differentiation

• For any constant a: d(a) = 0, d(ay) = a dy

• d(v + y) = dv + dy

• d(vy) = v dy + y dv

• d(v/y) = (y dv − v dy) / y2

The Priority dispute

Developed Calculus1665 – 1667Published1704 -1736

Developed Calculus1673– 1676Published 1684 - 1686

Brachistochrone problemSuppose that you roll a ball down a ramp from a point A to another point B. Which curve should the ramp be if the ball is to reach B in the shortest possible time?

Johann Bernoulli1667 – 1748

A cycloid is the curve traced by a fixed point on a circle rolling along a straight line; one can think of a cycloid as the curve traced out by a piece of

mud on a bicycle tyre when the bicycle is wheeled along.

Model to illustrate that the cycloid gives the path of quickest

descent

Bishop Berkeley 1685 - 1753

If to be is to be perceived?

If to be is to be perceived? There was a young man who said God, Must find it exceedingly odd When he finds that the tree Continues to beWhen no one's about in the Quad.

Dear Sir, your astonishment's odd I'm always about in the Quad And that's why the tree Continues to be Since observed by, yours faithfully, God

Ronald Knox

Bishop Berkeley’s Queries

•Query 64 Whether mathematicians, who are so delicate in religious points, are strictly scrupulous in their own science? Whether they do not submit to authority, take things upon trust, and believe points inconceivable’? Whether they have not their mysteries, and what is more, their repugnancies and contradictions?

And what are these same evanescent Increments? They are neither finite Quantities nor Quantities infinitely small, nor yet nothing. May we not call them the ghosts of departed quantities?

Finding instantaneous speed from average speed

• Distance travelled in time t equals 4.9 t2

• At a later time, t + o, distance travelled is: 4.9 (t + o)2

• In the time interval t to t + 0 distance travelled is 4.9 (t + o)2 – 4.9 t2 = 9.8 t o + 4.9 o2

• Divide by o to find average speed over the interval = 9.8 t + 4.9 o• Shrink the interval i.e. allow o to approach zero Shrink the interval i.e. allow o to approach zero • Then this average speed, 9.8 t + 4.9 o, approaches Then this average speed, 9.8 t + 4.9 o, approaches

the instantaneous speed 9.8 tthe instantaneous speed 9.8 t

Dividing by ZeroTo Prove that 5 = 8

0 x 5 = 0 x 8 as they are both 0.If we are able to divide by 0 and do so we get 5 = 8

Finding instantaneous speed from average speed

• Distance travelled in time t equals 4.9 t2

• At a later time, t + o, distance travelled is: 4.9 (t + o)2

• In the time interval t to t + 0 distance travelled is 4.9 (t + o)2 – 4.9 t2 = 9.8 t o + 4.9 o2

• Divide by o to find average speed over the interval = 9.8 t + 4.9 o• Shrink the interval i.e. allow o to approach zero Shrink the interval i.e. allow o to approach zero • Then this average speed, 9.8 t + 4.9 o, approaches Then this average speed, 9.8 t + 4.9 o, approaches

the instantaneous speed 9.8 tthe instantaneous speed 9.8 t

Jean-Baptiste le Rond

d'Alembert, 1717 – 1783

Bernard Bolzano 1781 – 1849

Augustin-Louis Cauchy (1789 – 1857)

Limit

Earthrise Apollo 8, December 24th 1968

Polynomials and their RootsTuesday 6th November 2012

X2 - 8 X + 15 = 0 has two real solutions.

X2 - 8 X + 16 = 0 has one real solution.

X2 - 8 X + 17 = 0 has no real solutions.

19th Century Mathematical Physics

Wednesday, 31 October 2012 - 4:00pmBarnard’s Inn Hall

A series of talks on Lord Kelvin, Peter Guthrie Tait and James Clerk Maxwell

ByMark McCartney, Julia Collins and Raymond Flood