Claudius Ptolemy

Claudius Ptolemy was born around 85AD in Egypt. He lived during a time of Hellenized Egypt but very little is known of his personal life.

His reputation rests on his unprecedented contributions to astronomy and geography and to the controversy that these works later caused as astronomers reacted to the famous work by Copernicus. In Ptolemy's universe, the earth sits at the center.

Ptolemy lived in Alexandria, Egypt. We know very little of his education. He mentions the astronomic data of Theon the mathematician who was probably Theon of Smyrna, one of his teachers. Many of his early works are dedicated to Syrus who may have also been one of his teachers.

Many of Ptolemy's major works still exist. His magnus opus is the Almagest which is Arabic for "the greatest". Its original name was the Mathematical Compilations. The Greeks soon started calling it The Greatest Compilations. Later on, the Arabic version became the basis of the Latin translation which is why Ptolemy's work is largely known as the Almagest.

The Almagest was one of Ptolemy's earliest works. It presents the Greek theory about the motions of the Sun, the moon, and the planets. This stood as the standard text on astronomy until Copernicus released his heliocentric theory in 1543.

It is a very ambitious work. Ptolemy writes (taken from MacTutor Biography):

We shall try to note down everything which we think we have discovered up to the present time; we shall do this as concisely as possible and in a manner which can be followed by those who have already made some progress in the field. For the sake of completeness in our treatment we shall set out everything useful for the theory of the heavens in the proper order, but to avoid undue length we shall merely recount what has been adequately established by the ancients. However, those topics which have not been dealt with by our predecessors at all, or not as usefully as they might have been, will be discussed at length to the best of our ability.

His geocentric astronomic model was based on Aristotle even though Aristarchus of Samos had argued for a heliocentric model. After presenting this model, he then introduces trigonometric models to predict the motions of the orbits. He presents tables of chords (which are similar to the sine function) and uses the concept of "epicycles" (circles-within-circles) to model the planetary movements. It is his theory of planets which is his greatest contribution. In his day, there were only 5 planets recognized and he presented detailed mathematical models for explaining their motions.

Although Hipparchus gets credit for much of the theory that Ptolemy presents, it is Ptolemy who was able to apply that theory to the planetary data.

The historian Toomer wrote (from MacTutor web site):

As a didactic work the "Almagest" is a masterpiece of clarity and method, superior to any ancient scientific textbook and with few peers from any period. But it is much more than that. Far from being a mere 'systemisation' of earlier Greek astronomy, as it is sometimes described, it is in many respects an original work.

There have been many who questioned whether Ptolemy deserves his reputation for being the greatest astronomer of his day. Tycho Brahe noticed that all of Ptolemy's star catalogue data was consistently off by 1 degree longitude. Although Ptolemy claimed that he had gathered the data. Brahe believed that Ptolemy had copied the data from another source. Some scholars have argued that Hipparchus is the true genius and Ptolemy merely copied Hipparchus's work. It is very hard to assess the truth of any of these claims since almost of all of Hipparchus's work is lost.

Regardless of these criticisms, it is undeniable that Ptolemy played the lead role in the geocentric theory of the solar system that ruled the day until the publication of Copernicus's famous work. Hipparchus may be the father of trigonometry but it is Ptolemy who became the lead voice in the application of trigonometry to the motions of the planets, sun, and moon.

References

• Biography of Ptolemy, MacTutor
  

Hipparchus of Rhodes

Hipparchus is considered by many to be the father of trigonometry because he was the first to organize measurements in relation to angles in a trigonometric table. Unfortunately, all but one of his math writings have been lost and the one text that remains is a minor work. Most of what we know today comes from the Greek astronomer Ptolemy.

Hipparchus was born around 190 B.C. in the town of Nicaea and died around 120 B.C. in Rhodes. He may have studied geometry in Alexandria. Unfortunately, not much more is known about his life.

His influence on astronomy and mathematics has been significant. He produced the first trigonometric table based on chords (the table lists the length of the chord that corresponds to a given angle). He discovered the precession of the equinoxes and accurately determined the length of a year to within 6.5 minutes. It is believed that Hipparchus created the first star catalogue which may have consisted of 650 stars. It is believed that he created this around 134 B.C. when a new star was said to have burst across the sky.

He was able to calculate the distance of the moon from the earth and developed a theory of the moon's motions based on epicycles. He also used epicycles to model the Sun's motions. He was perhaps the first to be able to predict solar and lunar eclipses. Hipparchus may have invented the astrolabe and used an equatorial ring to observe the sun's equinoxes.

Most of what is known about Hipparchus is found in Ptolemy's magnus opus Almagest. It was not Ptolemy's intentions to preserve the memory of Hipparchus. In fact, Ptolemy seems to assume that his reader has access to Hipparchus's original works.

References

• Hipparchus, Wikipedia
• Hipparchus, MacTutor
  

Archimedes

Archimedes was born around 287 B.C. in Syracuse. His father was an astronomer and it is believed that Archimedes was a relative of the King of Syracuse.

As a young man, Archimedes traveled to Egypt for it is said that while there, he invented what is today known as the Archimedes' screw.

It is believed that he studied with the students of Euclid in Alexandria because he mastered the ideas of Euclid and he references one Alexandrian mathematician, Conon of Samos, as a close friend. In his work On Spirals, he mentions a practice in Alexandria where mathematicians would give results but would not include proofs for those results. Archimedes disapproved of this practice and on one occasion sent false results to see how the
Alexandrian mathematicians responded.

Throughout his life, Archimedes made numerous engineering inventions. Despite his rising fame, Archimedes was most proud of his work in pure mathematics. Plutarch writes:

Archimedes possessed so high a spirit, so profound a soul, and such treasures of scientific knowledge, that though these inventions had now obtained him the renown of more than human sagacity, he yet would not deign to leave behind him any commentary or writing on such subjects; but, repudiating as sordid and ignoble the whole trade of engineering, and every sort of art that lends itself to mere use and profit, he placed his whole affection and ambition in those purer speculations where there can be no reference to the vulgar needs of life; studies, the superiority of which to all others is unquestioned, and in which the only doubt can be whether the beauty and grandeur of the subjects examined, of the precision and cogency of the methods and means of proof, most deserve our admiration.

Archimedes is responsible for contributing the term "eureka" into the English language. Eureka is Greek for "I found it." Archimedes was working on a problem for King Heiron when he figured out the solution while taking a bath. He was so excited, that he ran around Syracuse naked shouting "Eureka! Eureka!" which means "I have found it! I have found it!".

In his day, Archimedes was very well known. There are many references to him from contemporary writers. Interestingly, he was not famous for his mathematics but for his many mechanical devices that could be used in warfare. Plutarch writes about the use of these war machines when Marcellus led the Roman army against Syracuse in 212 BC:

... when Archimedes began to ply his engines, he at once shot against the land forces all sorts of missile weapons, and immense masses of stone that came down with incredible noise and violence; against which no man could stand; for they knocked down those upon whom they fell in heaps, breaking all their ranks and files. In the meantime huge poles thrust out from the walls over the ships and sunk some by great weights which they let down from on high upon them; others they lifted up into the air by an iron hand or beak like a crane's beak and, when they had drawn them up by the prow, and set them on end upon the poop, they plunged them to the bottom of the sea; or else the ships, drawn by engines within, and whirled about, were dashed against steep rocks that stood jutting out under the walls, with great destruction of the soldiers that were aboard them. A ship was frequently lifted up to a great height in the air (a dreadful thing to behold), and was rolled to and fro, and kept swinging, until the mariners were all thrown out, when at length it was dashed against the rocks, or let fall.

Ultimately, the Romans succeeded in taking Syracuse and Archimedes was killed during the invasion. Plutarch writes about three versions of the death of Archimedes. The most famous version is that he didn't notice the Roman soldier because he was intense on a math problem. In another version, he is confronted by the same soldier who tells him that he has come to kill him. Archimedes pleads for his life by saying that he has many mathematical works that have not yet been finished. The soldier is unmoved by this plea and kills him. In the third version, Archimedes carries his equipment to show Marcellus what can be done. The soldiers stop him and slay him because they believe that Archimedes has gold hidden inside the equipment.

Cicero writes that when Marcellus returned to Rome, he brought with him two inventions of Archimedes: a star globe, that mapped the sky onto a sphere, and the other was an orrery, an object that predicts the motions of the sun, moon, and planets.

Today, historians agree that Archimedes, Newton, and Gauss are the greatest mathematicians who have ever lived.

He made great advances in the method of integration that enabled him to find areas, volumes, and surface areas of geometric forms. He gave one of the first accurate approximations of π and found a method for estimating square roots. He invented a method for representing very large numbers. He invented the compound pulley, the law of the lever, he found a method for estimating the weight of an object immersed in a liquid, he came up with the concept of center of gravity. His favorite theorem was that the volume of a sphere is two-thirds the volume of a circumscribed cylinder.

Heath writes about Archimedes:

The treatises are, without exception, monuments of mathematical exposition; the gradual revelation of the plan of attack, the masterly ordering of the propositions, the stern elimination of everything not immediately relevant to the purpose, the finish of the whole, are so impressive in their perfection as to create a feeling akin to awe in the mind of the reader.

References

• Archimedes, Wikipedia
• Archimedes, MacTutor
• T. L. Heath, A history of Greek mathematics II (Oxford, 1931).
• Plutarch, Marcellus, MIT's on-line version

Euclid of Alexandria

Euclid of Alexandria is perhaps the greatest math teacher of all time. His textbook The Elements is the perhaps the most influential math book ever published.

Much of what we know about Euclid must be deduced from the various clues that are available in the writings of others. We know for example that he must have lived around the time of Ptolemy I in Alexandria from 325 - 265 BC from Proclus who writes:

He lived in the time of Ptolemy the First, for Archimedes, who lived after the time of the first Ptolemy, mentions Euclid. It is also reported that Ptolemy once asked Euclid if there was not a shorter road to geometry that through the Elements, and Euclid replied that there was no royal road to geometry.

(from David Joyce's Web Site)

It is believed that Euclid must have attended Plato's Academy in Athens because of his deep knowledge about the works Eudoxus and Theaetetus. This is also clear since the organization of the Elements is an attempt to understand the Platonic ideal shapes. The Greek philosopher Proclus writes:

In his aim he [Euclid] was a Platonist, being in sympathy with this philosophy, whence he made the end of the whole "Elements" the construction of the so-called Platonic figures.

(from Euclid, MacTutor Biography)

It is probable that Euclid ran a school of mathematics in Alexandria. Pappus mentions that the Greek mathematician Apollonius learned geometry from the students of Euclid in Alexandria (Donald Lancon, here). There is a story told by Stobaeus in the fifth century:

... someone who had begun to learn geometry with Euclid, when he had learnt the first theorem, asked Euclid "What shall I get by learning these things?" Euclid called his slave and said "Give him threepence since he must make gain out of what he learns".

(from Heath, a History of Greek Mathematics)

Euclid had a reputation for being fair-minded, polite, and a serious scholar. Pappus writes that Euclid was:

... most fair and well disposed towards all who were able in any measure to advance mathematics, careful in no way to give offence, and although an exact scholar not vaunting himself.

(from Euclid, Mac Tutor Biography)

For some reason, Euclid did not write a preface to his great work as did many of the other Greek mathematicians. As a result of this, we are forced to rely on writings that came hundreds of years after his death.

It is believed that when Euclid was writing the Elements, he borrowed content from other textbooks that came earlier. For example, he presents definitions for oblong, rhombus, and rhomboid which are never used (Euclid, MacTutor).

The Elements consists of 13 books. The first two books focus on triangles, parallel lines, parallelograms, rectangles and squares. Included in these book is the parallel postulate, which is the starting point for non-Euclidean geometry, the famous Pythagorean Theorem, a construction for squaring a rectangle.

Books three and four deal with the properties of circles. Book five presents Eudoxus's theory of proportions. Book six deals with similar triangles.

Books seven to nine deal with number theory including Euclid's algorithm for greatest common divisor. Books ten deals with irrational numbers which is taken from Theaetetus's theory.

Books eleven through thirteen cover three-dimensional geometry. Book twelve ends with the proof that circles are to each other as the square of their diameters and spheres to each other as the cubes of their diameters. Book thirteen ends with the proof that there can be only 5 types of regular polyhedra (three-dimensional shapes with equal sides) which are the: tetrahedon (4-sided), cube (6-sided), octahedron (8-sided), dodecahedron (12-sided), and icosahedron (20-sided). These are of course the Platonic solids (see here for details):

1. Tetrahedron



2. Cube



3. Octahedron



4. Dodecahedron



5. Icosahedron



Euclid wrote other works besides the Elements including a work on perspective called Optics. Many of his works are lost including a work he did on conics which predates Apollonius.

The Elements was first published in book form in 1482. Since then, there have been over 1000 editions of this classic math book. It set the standard for rigor and clarity of mathematics and one of the chief themes in the history of mathematics is the effort to place other areas of mathematics on the same firm foundations at Euclid's Elements. Today, Euclid is known as the father of geometry.

References

• Euclid, MacTutor Biography
• Euclid, Wikipedia
• Euclid, David Joyce
  
• David Joyce, Euclid's Elements
  
• Sir Thomas Heath, History of Mathematics
• An Introduction to the Works of Euclid, Donald Lancon
  

More on Euler's Identity and Roots of Unity

When Leonhard Euler came up with his Formula and his Identity, he stood on the shoulders of many giants. In the next few blogs, I plan to focus a bit on some of the giants that Euler stood upon including: Euclid, Archimedes, Hipparchus, Ptolemy, Napier and Bernoulli.

Thinking about Euler's identity in the context of Fermat's Last Theorem raises some questions which I think need to be answered:

• Pi, Sine, and Cosine are based on Euclid's plane geometry. What validity can it have for number theory which is independent of Euclidean or non-Euclidean geometry?
• How is it possible for a number to be put to the power of an imaginary number? How can this construction possibly have any meaning?

It turns out that trigonometric functions can be defined independently of Euclid (see here). It also turns out that it is possible to use the Maclaurin Series to define a exponents so that they can include any complex power including i (see here)

One of the goals of this blog is to provide a complete set of proofs for each of the propositions that I present or to state those propositions as postulates. Implicit in the use of sine, cosine, and pi is a set of assumptions that are often not thought about:

• How can we be sure that all right triangles regardless of their size have the same ratio between their sides so that for a given angle, there is one and only one sine or cosine value? (See here for the answer)
  
• How can we be sure that pi is really constant? It may be obvious but where's the proof? How can we be sure that for all circles, the ratio of circumference-to-diameter is the same? (See here for the answer)
  

While these questions are very elementary, I think that it is still important to ask them.

In understanding Roots of Unity and Cyclotomic Integers, I think it is worthwhile to review the achievements of DeMoivre, Taylor, and Maclaurin.

I think it is important and valuable to understand all of these details before exploring Kummer's proof.

Lennart Carleson wins the Abel Prize!

Today, it was announced that Lennart Carleson has won the Abel Prize for 2006 for his work on harmonic analysis and the theory of smooth dynamical systems.

You can check out the BBC story here. Here is the link to the Abel Prize web site.

The Abel Prize money is roughly $920,000. The prize is awarded by the Norwegian Academy of Sciences and Letters in recognition of "outstanding scientific work in the field of mathematics" (see here) . The prize will be presented in May by King Harald of Norway as part of a formal ceremony.

The prize is named in honor of the Norwegian mathematican Niels Henrik Abel.

Taylor Series

The Taylor Series is one of the most useful results in all of mathematics. The method in all its generality was first presented by Brook Taylor. To see an application for Taylor's Series, check out Euler's Formula.

The proof for Taylor's Series depends on basic calculus.

For anyone who does not feel comfortable with the notation f(x) or the concept of a mathematical function, start here.

For anyone who does not know about derivatives or about the notation f'(x), f''(x), or fⁿ(x), start here.

Theorem: Taylor Series
if
(a) f is a function with derivatives of all orders
(b) (lim n → inf)[f⁽ⁿ⁺¹⁾(z)]/(n+1)!(x-a)ⁿ⁺¹ = 0
then:
f(x) = f(a) + f'(a)(x-a) + [f''(a)/2!](x-a)² + .... + [f⁽ⁿ⁾(a)/n!](x-a)ⁿ + ....

Proof:

(1) Let n be an arbitrary positive integer.

(2) From Taylor's Formula, we have:
f(x) = P_n(x) + R_n(x)

where:

P_n(x) = f(a) + f'(a)(x-a) + f''(a)[(x-a)²]/2! + ... + f⁽ⁿ⁾[(x-a)ⁿ]/n!

and

R_n(x) = [f⁽ⁿ⁺¹⁾(z)]/(n+1)!(x-a)ⁿ⁺¹

where z is between a and x.

(3) From assumption b, we see that:
lim (n → inf) R_n(x) = 0.

(4) Then it follows that:
f(x) = lim (n → inf) P_n(x) =


= f(a) + f'(a)(x-a) + [f''(a)/2!](x-a)² + .... + [f⁽ⁿ⁾(a)/n!](x-a)ⁿ + ....

QED

References

• Edwards & Penny, Calculus and Analytic Geometry

Taylor's Formula

In today's blog, I will go over Taylor's Formula. This is a theorem that can be used to establish the Taylor Series and the MacLaurin Series which I use in my proof of Euler's Identity.

If you are not familiar with continuous functions or closed intervals, start here.

If you would like a review of derivatives, start here.

Theorem: Taylor's Formula

Let f(x) be a continuous function over the closed interval [a,b] that has an (n+1) derivative which is referenced by f⁽ⁿ⁺¹⁾(x) where n is a positive integer.

Then:

f(b) = f(a) + f'(a)(b-a) + ... + [f⁽ⁿ⁾(a)/n!](b-a)ⁿ + [f⁽ⁿ⁺¹⁾(z)/(n+1)!](b-a)ⁿ⁺¹)

for some number z which lies between a and b.

Proof:

(1) Let H = f(b) - f(a) - f'(a)(b-a) - [f''(a)/2!](b-a)² - ... - [f⁽ⁿ⁾(a)/n!](b-a)ⁿ

(2) Let K = H/(b-a)ⁿ⁺¹

(3) Let g(x) be a function such that:
g(x) = f(b) - f(x) - f'(x)(b-x) -[f''(x)/2!](b-x)² - ... - [f⁽ⁿ⁾(x)/n!](b-x)ⁿ - K(b-x)ⁿ⁺¹

(4) We can see that g(x) is a continuous function since:

(a) f(x) is continuous over [a,b] by the given.

(b) f⁽ⁿ⁾ is a continuous function since by the given we know that f(x) has an (n+1) derivative and since f(x) is differentiable at x, then it is continuous at x (see here)

(c) (b-x)ⁿ is continuous over [a,b] since f(x) is continuous since:

Let h(x)=b-x

h(x) is continuous since it is the addition of two continuous functions i(x)=b and j(x)=-x [See here for proof of the Addition Law]

(b-x)ⁿ is continuous by the Multiplication Law [See here for proof of the Multiplication Law]

(b-x)ⁿ = (b - x)*(b-x)*... since n is a positive integer.

(d) f(b) is continuous since it is a constant. (See here)

(e) Each f⁽ⁿ⁾(x)/n! is continuous since (-1/n!) can be thought of as a constant function h(x)=(-1/n!) and the multiplication of two continuous functions is itself continuous (see here)

(f) Finally, g(x) is continuous because the addition of a set of continuous functions is itself continuous (see here)

(5) We can see that g(a) = 0 since:

g(a) = H - H/(b-a)ⁿ⁺¹*(b-a)ⁿ⁺¹ = H - H = 0

(6) We can also see that g(b) = 0 since:

g(b) = f(b) - f(b) - f'(b)(b-b) - [f''(b)/2!](b-b)² - ... - [f⁽ⁿ⁾(b)/n!](b-b)ⁿ - K(b-b)ⁿ⁺¹

(7) By Rolle's Theorem, since g(x) is continuous on [a,b], we know that there exists a value z such that z ∈ [a,b] and g'(z) = 0. [See here for Rolle's Theorem]

(8) If we differentiate on g(x), we get:
g'(x) = -f'(x) + f'(x) -f⁽²⁾(x)(b-x) + f⁽²⁾(x)(b-x) - (1/2!)f⁽³⁾(x)(b-x)² + (1/2!)f⁽³⁾(x)(b-x)² - (1/3!)f⁽⁴⁾(x)(b-x)³ + ... + [1/(n-1)!]f⁽ⁿ⁾(x)(b-x)^n-1 - (1/n!)f⁽ⁿ⁺¹⁾(x)(b-x)ⁿ + (n+1)K(b-x)ⁿ since:

(a) g(x) = f(b) - f(x) - f'(x)(b-x) -[f''(x)/2!](b-x)² - ... - [f⁽ⁿ⁾(x)/n!](b-x)ⁿ - K(b-x)ⁿ⁺¹

(b) By Lemma 3 here, we can differentiate each individual product in the sum.

(c) f(b) is a constant so d/dx(f(b)) = 0 [See here for Constant Rule]

(d) d/dx(-f(x)) = -f'(x) [See here for details]

(e) d/dx[-f'(x)(b-x)] = -f''(x)(b-x) - f'(x)(-1) = f'(x) -f''(x)(b-x) [See here for the Product Rule]

(f) d/dx[(-f''(x)/2!)(b-x)²] = (-f⁽³⁾(x)/2!)(b-x)² + [-f''(x)/2!](2)(b-x)(-1)] =
= (-f⁽³⁾/2!)(b-x)² + f''(x)(b-x) [See here for Generalized Power Rule]

(g) d/dx([-f⁽ⁿ⁾(x)/n!](b-x)ⁿ) = (-f⁽ⁿ⁺¹⁾(x)/n!)(b-x)ⁿ + [-f⁽ⁿ⁾(n)/n!](n)(b-x)^n-1*(-1) =
= (-f⁽ⁿ⁺¹⁾(x)/n!)(b-x)ⁿ +[ f⁽ⁿ⁾/(n-1)!](b-x)^n-1

(h) Since K is a constant
d/dx(-K(b-x)ⁿ⁺¹) = (n+1)(-K)(b-x)ⁿ(-1) = (n+1)K(b-x)ⁿ

(9) We see that most of the terms cancel out so that we get:
g'(x) = (n+1)K(b-x)ⁿ - (1/n!)f⁽ⁿ⁺¹⁾(x)(b-x)ⁿ

(10) Applying the fact that g'(z) = 0 from step #7 gives us:
g'(z) = (n+1)K(b-z)ⁿ - (1/n!)f⁽ⁿ⁺¹⁾(z)(b-z)ⁿ = 0

(11) We can divide both sides by (b-z)ⁿ to get:
(n+1)K - (1/n!)f⁽ⁿ⁺¹⁾(z) = 0

(12) Now we can rearrange (#11) to get:
K = [(1/n!)f⁽ⁿ⁺¹⁾(z)]/(n+1) = [f⁽ⁿ⁺¹⁾(z)]/(n+1)!

(13) Now, using (#12) and (#3) with x=a, we get:
g(a) = 0 = f(b) - f(a) - f'(a)(b-a) - [f''(a)/2!](b-a)² - ... - [f⁽ⁿ⁾(a)/n!](b-a)ⁿ - [(b-a)ⁿ⁺¹][f⁽ⁿ⁺¹⁾(z)/(n+1)!]

(14) Now, if we move over all the elements after f(b), we get:
f(b) = f(a) + f'(a)(b-a) + [f''(a)/2!](b-a)² + ... + [f⁽ⁿ⁾(a)/n!](b-a)ⁿ + [(b-a)ⁿ⁺¹][f⁽ⁿ⁺¹⁾(z)/(n+1)!]

QED

References

• Edwards & Penny, Calculus and Analytic Geometry

Euler's Formula

Today's proof for Euler's Formula is based on the Taylor's Series. Euler's Formula is the equation:

e^ix = cosx + isinx

In a previous blog, I spoke about Euler's Identity which is derived from Euler's Formula. Richard Cotes was the first person to provide a proof but the great popularizer of this result was Leonhard Euler. Euler's Identity is used in the construction of cyclotomic integers which are used in Kummer's proof of Fermat's Last Theorem for regular primes.

Lemma 1: Maclaurin's Series

f(x) = f(0) + (x/1!)f'(0) + (x²/2!)f''(0) + (x³/3!)f'''(0) + ... + (xⁿ/n!)fⁿ(0)

(1) Taylor's series gives us:
f(x) = f(a) + f'(a)(x-a) + [f''(a)/2!](x-a)² + .... + [f⁽ⁿ⁾(a)/n!](x-a)ⁿ + ....
[See here for its proof]

(2) Now, if a=0, then we have:
f(x) = f(0) + (x/1!)f'(0) + (x²/2!)f''(0) + (x³/3!)f'''(0) + ... + (xⁿ/n!)fⁿ(0).

QED

Lemma 2: e^x = 1 + (x/1) + (x²/2!) + (x³/3!) + ....

(1) Let f(x) = e^x

(2) From the properties of e^x [See here for details]:
f(x) = e^x → f(0) = e⁰ = 1
f'(x) = e^x → f'(0) =e⁰ = 1
fⁿ(x) = e^x → fⁿ(0) = e⁰ = 1

(3) We know that e^x is continuous since it has a derivative at each point. [See here for details of why this is true]

(4) By Lemma 1 above, we have:
e^x = 1 + (x/1!)(1) + (x²/2!)(1) + ... (xⁿ/n!)(1)

QED

Lemma 3: sinx = x - (x³/3!) + (x⁵/5!) - (x⁷/7!) + ...

(1) Let f(x) = sin x

(2) From the properties of sin, we know:

f(0) = sin(0) = 0 [See here for details if needed]

f'(x) = cos x → f'(0) = 1 [See here for proof if needed]

f''(x) = -sin(x) → f'(0) = 0 [See here for proof if needed]

f'''(x) = -cos x → f'(0) = -1

(3) From this, we see that:

fⁿ(0) = 0 if n is even.
fⁿ(0) = 1 if (n-1)/2 is even
fⁿ(0) = -1 if (n-1)/2 is odd


(4) Putting this all together gives us:
sin x = (x/1)(1) + (x²/2!)(0) + (x³/3!)(-1) + ...

QED

Lemma 4: cos x = 1 - (x²)/2! + (x⁴/4!) - (x⁶/6!) + ...

(1) Let f(x) = cos x

(2) From the properties of cos, we know:

f(0) = cos(0) = 1 [Details if needed are found here]

f'(x) = -sin x → f'(0) = 0 [Details if needed are found here]

f''(x) = -cos(x) → f'(0) = -1

f'''(x) = sin x → f'(0) = 0

(3) From this, we see that:

fⁿ(0) = 0 if n is odd.
fⁿ(0) = 1 if (n/2) is even
fⁿ(0) = -1 if (n/2) is odd

(4) Putting this all together gives us:
cos x = 1 + (x²/2!)(-1) + (x³/3!)(0) + (x⁴/4!)(1) + ...

QED

Theorem: Euler's Formula

e^ix = cos x + isin x

(1) From Lemma 2, we have:
e^ix = 1 + ix + (ix)²/2! + (ix)³/3! + ...

(2) Since i² = -1 and i⁴ = 1, this gives us: (for details on i, see here)
e^ix = (1 - x²/2! + x⁴/4! + ...) + i(x - x³/3! + x⁵/5! + ...)

(3) From Lemma 4 above, we see that:
cos(x) = (1 - x²/2! + x⁴/4! + ...)

(4) From Lemma 3 above, we see that:
isin(x) = i(x - x³/3! + x⁵/5! + ...)

(5) Combining step #2 with step #3 and step #4 gives us:
e^ix = cos x + i sin x.

QED

Corollary: De Moivre's Formula

(cos x + isin x)ⁿ = cos(nx) + isin(nx)

Proof:

(1) (e^ix)ⁿ = e^inx

(2) (cos x + i sin x)ⁿ = cos(nx) + isin(nx) [Applying Euler's formula above]

QED

Euler's Identity

One of the simplest and most elegant equations is Euler's Identity:
e^πi + 1 = 0.

It states that Euler's number e (which is equal to roughly 2.718...) to the power of i*π (taken at roughly 3.14...) is equal to -1. This equation is used to derive cyclotomic integers which are used in Kummer's proof for Fermat's Last Theorem for regular primes.

To be fair, for many people, xⁱ does not have a clear value. What does it mean to put an exponent to an imaginary power? In mathematics, it is ok if the details are not intuitive as long as they are logically consistent. The Maclaurin Series, for example, can be used to define exponents to a complex power (see here). Newton did something similar when he generalized the Binomial Theorem to include complex powers (see here for details)

Theorem: Euler's Identity is the equation: e^πi = -1

This equation derives directly from Euler's Formula:
e^ix = cos x - isin x [See here for proof]

We get:
e^πi = cos(π) + isin(π) = -1 + i(0) = -1. [Review of e (see here), sin and cos (see here), i (see here), and π (see here)]

QED

Corollary: e^πi + 1 = 0

This directly follows from above.

QED

Cyclotomic Integers

In Ernst Kummer's groundbreaking partial proof on Fermat's Last Theorem, he used cyclotomic integers.

Cyclotomic integers are based the equation xⁿ = 1 (they get their name from this equation which is known as the cyclotomic equation). The solutions to this equation are known as the Roots of Unity. The solutions to n=1 and n=2 are well known (1, -1). What is less well known is that as n increases, there are complex solutions that also solve this equation (these are also known as DeMoivre Numbers).

For some, it may seem strange that complex numbers emerge out of x³ = 1. The insight to why this is so comes from considering Euler's Identity:

e^iπ=-1

In my next blog, I will go into the details behind this equation. Still, using this equation, we can see that three solutions to x³ = 1 are:
1, e^(2iπ)/3, e^(4iπ)/3

Since (e^(2iπ)/3)³ = e^2iπ = (-1)² = 1.

The solution then for any value of n is e^(2iπ)/n and this solution is by convention represented by the Greek symbol zeta ζ. The set of integers then is Z[ζ] where each integer can be constructed from the rational integers a,b using a + bζ. This construction follows the same pattern as what Euler did for Z[√-3] (see here) and what Gauss did for Z[i] (see here).

For any value n, there are n different roots of unity which can be generated by e^(2iπk)/n where k is any integer 0, 1, 2, ..., up to n-1. For this reason, one speaks about the value ζ (equal to e^2iπ/n) as a primitive root of unity.

One might reasonably ask if all cyclotomic integers created in this way are characterized by unique factorization. In using quadratic integers, Euler made a mistake by assuming that Z[√-3] has unique factorization which it does not. Gabriel Lame assumed that all cyclotomic integers where characterized by unique factorization when he presented his proof of Fermat's Last Theorem. This assumption turns out to be the major flaw in Lame's proof.

This then becomes the major question which Kummer sought to resolve. Under what circumstances are cyclotomic integers characterized by unique factorization? The answer to this question led Kummer to his important partial proof of Fermat's Last Theorem. He showed that Fermat's Last Theorem holds for a certain set or primes which he called "regular primes." I talk more about the properties of cyclotomic integers here.

References

• Roots of Unity, Wikipedia
  
• Cyclotomic Equation, MathWorld
  

Quadratic Reciprocity: Last Step

In today's blog I continue the proof for the Principle of Quadratic Reciprocity. If you are not familiar with the concepts of quadratic reciprocity, quadratic residues, or the Legendre Symbol start here.

If you would like to start this proof at the beginning, start here.

If you would like to start with Gauss's Lemma, start here.

Today's proof is taken from Gareth A. Jones and J. Mary Jones' Elementary Number Theory.

Lemma 1:
(a) Let x' = (p+1)/2 - x
(b) Let y' = (q+1)/2 - y
Then
qx - py is greater than -(p/2) if and only if qx' - py' is less than q/2.

(1) Assume that qx - py is greater than (-p/2)

(2) (p/2) + qx - py is greater than 0

(3) (p +q-q+pq-pq)/2 + qx - py is greater than 0

(4) (p - q + pq - pq)/2 + qx - py is greater than (-q/2)

(5) -q(p+1)/2 + p(q+1)/2 + qx - py is greater than (-q/2)

(6) q(p+1)/2 - p(q+1)/2 - qx + py is less than (q/2)

(7) q[(p+1)/2 - x] - p[(q+1)/2 - y] is less than (q/2)

(8) qx' - py' is less than (q/2)

(9) We can use the same reasoning to show that:
if qx' - py is less than (q/2), then qx - py is greater than (-p)/2.

QED

Lemma 2: Given:
(a) Let P = { 1, 2, ... (p-1)/2 }
(b) Let Q = {1, 2, ... (q-1)/2 }
(c) Let A be the set of points PxQ such that qx-py is greater than (-p)/2. [See diagram in Lemma 3]
(d) Let B be the set of points PxQ such that qx-py is less than (q/2) [See diagram in Lemma 3]
(e) Let α be the number of points in A
(f) Let β be the number of points in B
Then:
There is a one-to-one mapping between α and β [So that, they both have the same number of points]

Proof:

(1) Let's define the following function:
p(x,y) = (x',y') = [ (p+1)/2 - x, (q+1)/2 - y ]

(2) p(A) = p(B) since qx - py is greater than (-p)/2 implies that qx' - py is less than (q/2). [See Lemma 1 above]

(3) p(B) = p(A) since qx - py is less than (q/2) implies that qx' - py' is greater than (-p)/2. [See Lemma 1 above]

QED

Lemma 3: Given
(a) p,q are distinct, odd primes
(b) P is a set = { 1, 2, ... (p-1)/2 }
(c) Q is a set = { 1, 2, ... (q-1)/2 }
(d) x,y are integers such that x ∈ P and y ∈ Q
(e) (-p/2) is less than qx - py which is less than (q/2).
(f) μ = the number of elements (x,y) ∈ P x Q where (-p/2) is less than qx - py which is less than -1.
(g) ν = the number of elements (y,x) ∈ Q x P where (-q/2) is less than py - qx which is less than -1.

Then:
(-1)^{μ + ν} = (-1)^{[(p-1)(q-1)]/4}

Proof:

(1) The set of assumptions can be represented as a graph:

We see that:
(a) R is the set of all points PxQ = [(p-1)/2][(q-1)/2] = (p-1)(q-1)/4

(b) μ + ν = R ∩ S

(c) A = the area of PxQ that falls above S

(d) B = the area of PxQ that falls below S

(2) We can further define α, β such that:

α = the number of points ∈ A
β = the number of points ∈ B

(3) From this, we see that the number of points R ∩ S is:
(p-1)(q-1)/4 - (α + β)

(4) Now we can show that α = β or in other words that there is a one-to-one mapping between elements in α with elements in β by the following:

(a) We can do this using the following rotation p(x,y):

p(x,y) = (x',y') = ( [p+1]/2 - x, [q+1]/2 - y )

(b) So, for example, the corners of R before the rotation are: (1,[q-1]/2), ([p-1]/2,[q-1]/2), (1,1), and ([p-1]/2, 1).

(c) After the rotation, the corners of R' are: ([p-1]/2,1), (1,1), ([p-1]/2,[q-1]/2), and (1,[q-1]/2).

(d) So, we see that R' has the same corners as R.

(e) By Lemma 2, we see that there is a one-to-one correspondence between A and B.

(5) So, this means that α + β ≡ 2*α ≡ 0 (mod 2).

(6) But this means that the α + β does not change the result since:
(-1)^{μ + ν} = (-1)^{μ + ν}*(-1)^2α

So:
(-1)^{μ + ν} = (-1)^{μ + ν + 2*α} = (-1)^(p-1)(q-1)/4

QED

Quadratic Reciprocity: Expanding on Gauss's Lemma

In today's blog I continue the proof for the Principle of Quadratic Reciprocity. If you are not familiar with the concepts of quadratic reciprocity, quadratic residues, or the Legendre Symbol start here.

If you would like to start this proof at the beginning, start here.

If you would like to start with Gauss's Lemma, start here.

Today's proof is taken from Gareth A. Jones and J. Mary Jones' Elementary Number Theory.

Lemma 1:
if
(a) p,q are odd primes
(b) P = the set of integers {1, 2, ..., (p-1)/2}
(c) N = the set of integers P = { -1, -2, ..., -(p-1)/2 }
(d) qx ≡ n (mod p) where n ∈ N, x ∈ P
then:
there exists a y such that (-p/2) is less than (qx - py) which is less than 0.

Proof:

(1) There exists y such that qx - n = py [Since qx ≡ n (mod p) -- see here for review of modular arithmetic]

(2) This then gives us: qx - py = n.

(3) qx - py is less than 0 since n is less than 0.

(4) qx - py is greater than (-p/2) since n ≥ -(p-1)/2 = (-p+1)/2 which is greater than -p/2.

QED

Lemma 2: if p,q,x,y are integers such that:
(a) p,q are positive, odd primes
(b) (-p/2) is less than qx - py
(c) qx - py is less than 0
(d) x,y are positive.
Then:
(a) there is only 1 value that y can have.
(b) 0 is less than qx/p which is less than y
(c) y is less than (qx/p) + 1/2.

Proof:

(1) Let y be any integer that satisfies the conditions (a),(b),(c)

(2) There is no value y' that y' is an integer and less than y which satisfies (a),(b),(c) since:

(a) qx - p(y-1) = qx - py + p

(b) We know that absolute(p) is greater than absolute(qx-py) since (-p/2) is less than (qx-py) which is less than 0.

(c) Since p is positive and qx-py is negative, we know that qx-py + p is greater than 0.

(3) There is no value y' such that y' is an integer and greater than y which satisfies (a),(b),(c) since:

(a) qx - p(y+1) = qx - py - p

(b) Since -p is greater than -p/2, we know that qx-py-p is less than (-p/2).

(4) From (a),(b),(c), we know that (p/2) is greater than (py-qx) which is greater than 0.

(5) Dividing (4) by p gives us:
(1/2) is greater than y - (qx/p) which is greater than 0.

(6) From (5), we know that y is greater than (qx/p).

(7) From (5), we also know that (1/2) + (qx/p) is greater than y.

(8) Since q,x,p are positive, we know that qx/p is greater than 0.

(9) By (#8),(#6), we know that y is greater than 0.

QED

Lemma 3: if q,p,x,y are integers such that:
(a) p,q are positive, odd primes
(b) (-p/2) is less than qx - py
(c) qx - py is less than 0
(d) x ∈ { 1, 2, ..., (p-1)/2 }
(e) y is a positive integer
Then:
y is less than (qx/p) + 1/2 ≤ [q(p-1)]/2p + 1/2 which is less than q+1/2

Proof:

(1) From Lemma 2, we know that:
0 is less than (qx/p) which is less than y which is less than (qx/p) + 1/2.

(2) From (d), we know that x ≤ (p-1)/2.

(3) Combining (1) and (2), we have:
(qx/p) + (1/2) ≤ [q(p-1)]/2p + (1/2)

(4) We can conclude that [q(p-1)]/2p + (1/2) is less than (q+1)/2 since:

(a) [q(p-1)]/2p + (1/2) = (qp - q + p)/2p = [(qp/p - q/p + p/p)]/2 =
= (q - q/p + 1)/2 = (q + 1 - q/p)/2

(b) q/p is greater than 0 since q,p are odd primes.

(c) From (a) and (b), we have:
(q + 1 - q/p)/2 is less than (q+1)/2 [ Since 1 - q/p is less than 1 ]

QED

Lemma 4: if q,p are odd primes, then there exist μ, ν where:

(a) Using the Legendre Symbol (see here):


(b) P = set of integer = {1, 2, ... (p-1)/2 }

(c) Q = set of integers = {1, 2, ... (q-1)/2 }

(d) μ + ν = the number of pairs (x,y) ∈ P x Q such that -p/2 is less than (qx-py) which is less than q/2.

Proof:

(1) We can start out with the observation that:

where μ is the number of pairs (x,y) ∈ PxQ such that (-p/2) is less than qx-py which is less than 0 since:

(a) From Gauss's Lemma (see here), we have:

where μ is the number if elements in qP ∩ N where N = of integers -P, that is, {-1, -2, ... -(p-1)/2}

(b) qP ∩ N → qx ≡ n (mod p) where n ∈ N, x ∈ P. [See Gauss's lemma for details if needed, see here for review of modular arithmetic]

(c) There exists y such that (-p/2) is less than qx - py which is less than 0. [By Lemma 1 above]

(d) We know that y is greater than 0 [See Lemma 2 above]

(e) Also, y is less than (q+1)/2. [See Lemma 3 above]

(f) So, y ≤ (q-1)/2 [ since there is no integer greater than (q-1)/2 and less than (q+1)/2 ]

(g) So, y ∈ Q since (d) gives us y ≥ 1 and (f) gives us y ≤ (q-1)/2.

(h) Since x ∈ P from assumption, we have (x,y) ∈ PxQ [see here if you need a review of PxQ notation]

(2) We can use the same reasoning as (#1) to get a value ν such that:

where ν is the number of pairs (y,x) ∈ QxP such that (-q/2) is less than py-qx which is less than 0 .

(3) Combining (1) and (2) gives us:


where if x,y are taken from (#1), we have:

(x,y) ∈ PxQ such that (-p/2) is less than qx-py which is less than 0

If x,y are taken from (#2), then we have:

(y,x) ∈ QxP such that (-q/2) is less than py-qx which is less than 0

(4) Combining both possibilities, we have:

(x,y) ∈ PxQ, (-p/2) is less than qx-py which is less than (q/2).

QED

Euler's Criterion

In today's blog I continue the proof for the Principle of Quadratic Reciprocity. If you are not familiar with the concepts of quadratic reciprocity, quadratic residues, or the Legendre Symbol start here.

If you would like to start this proof at the beginning, start here.

If you are not familiar with the Legendre Symbol, you should review here.

Today's proof is taken from Gareth A. Jones and J. Mary Jones' Elementary Number Theory.

Definition 1: Q_n: the set of quadratic residues

That is, these are the integers a mod n where a ≡ s² (mod n) and s ∈ U_n

NOTE: See here for a review of U_n

Examples:

Q₇ = { 1, 2, 4 }

1² ≡ 1 (mod 7) [ 1 ∈ U₇ since 1 ≡ 1 (mod 7) ]
2² ≡ 4 (mod 7) [ 4 ∈ U₇ since (4)(2) ≡ 1 (mod 7)]
4² ≡ 2 (mod 7) [ 2 ∈ U₇ since (2)(4) ≡ 1 (mod 7)]

Q₈ = { 1 }

1² ≡ 1 (mod 8)

Lemma 1:


Proof:

This is true by the the definition of the Legendre Symbol (see here)

QED

Lemma 2: if p is an odd prime and g is a primitive root for U_p, then gⁱ ∈ Q_p if and only if i is even.

Proof:

(1) if i is even, then g^i/2 ∈ U_p [See here for definition of primitive root]

(2) But g^i/2 ∈ U_p, then gⁱ ≡ (g^i/2)² (mod p) and therefore gⁱ ∈ Q_p [See above for definition of Q_p]

(3) if gⁱ ∈ Q_p, then there exists j such that gⁱ ≡ (g^j)² (mod p) [Based on definition of Q_p]

(4) so, i ≡ 2j (mod φ(p)) since:

(a) φ(p) is the order of g. [See detail here]

(b) There exists q,r such that [from the Division Algorithm, see here]:
i = q(φ(p)) + r
0 ≤ r which is less than φ(p)

(c) And, i ≡ r (mod φ(p))

(d) So, gⁱ ≡ g^q(φ(p))+r ≡ (g^φ(p))^q*g^r ≡ (1)^q*g^r ≡ g^r (mod p)

(e) There exists, q', r' such that:
2j = q'(φ(p)) + r'
0 ≤ r' which is less than φ(p)

2j ≡ r' (mod φ(p))

(f) Now it follows that r = r' since:
g^2j ≡ (1)^q'*g^r' ≡ g^r' ≡ g^r (mod p)

(g) Putting this all together, gives us:
2j ≡ i ≡ r (mod p)

(5) Now φ(p) is even. [since φ(p) = p-1 and p is an odd prime -- see here for details.]

(6) So i is even since:

(a) There exists a such that φ(p) = 2a [Since φ(p) is even]

(b) There exists b such that i - 2j = (2a)b [By definition of modular arithmetic, see here]

(c) So, 2 must divide i.

QED

Lemma 3: If p is an odd prime and g is a primitive root mod p, then:


Proof:

(1) Both values are either equal to 1 or -1. [See here for definition of Legendre Symbol]

(2) So, I will prove that:


In other words, it is 1 only if i is even.

(3) In order for it to be 1, gⁱ must be an element of Q_p. [See Lemma 1 above]

(4) From Lemma 2, we know that gⁱ ∈ Q_p if and only if i is even.

QED

Theorem 1: Euler's Criterion: if p is an odd prime and a is any integer then:


Proof:

(1) We can assume that p does not divide a since if it did:


and



which proves the conclusion.

(2) Now, if p doesn't divide a, then gcd(p,a)=1 since p is a prime.

(3) Also, there exists a' such that a ≡ a' (mod p) and a' is a unit mod p since:

(a) gcd(p,a)=1 → there exists u,v such that au + pv = 1 (By Bezout's Identity, see here for proof)

(b) So, au - 1 = pv → au ≡ 1 (mod p)

(c) But then a is a unit mod p (see here for definition of a unit mod p)

(4) The set of units mod p is cyclic (see here for proof)

(5) Then, there exists an element g such that g is the primitive root for the set of units mod p. (see here for definition of cyclic)

(6) So, there exists i such that a ≡ gⁱ (mod p) (see here for definition of primitive root)

(7) Let h ≡ g^(p-1)/2 (mod p)

(8) h² ≡ g^p-1 (mod p) [See here for a review of exponents if needed]

(9) φ(p) = p-1 [See here for details.]

(10) g^p-1 ≡ g^φ(p) ≡ 1 (mod p) [See Euler's Theorem]

(11) So p divides g^p-1-1 → p divides h² - 1 → p divides (h-1)(h+1)

(12) So, h ≡ 1 or h ≡ -1 (mod p)

(13) But h cannot be ≡ 1 (mod p) since (p-1)/2 is less than p-1 and p-1 is the order [See here for the definition of order]

(14) So h ≡ -1 (mod p)

(15) From (#6), we have: a^(p-1)/2 ≡ (gⁱ)^(p-1)/2 ≡ (g^(p-1)/2)ⁱ ≡ hⁱ ≡ (-1)ⁱ (mod p)

(16) Now from Lemma 3 above:


(17) Since a ≡ gⁱ (mod p), we also have:


(18) Finally, since a^(p-1)/2 ≡ (-1)ⁱ (mod p) from#15, we have:



QED