History

Janot de Stainville ascribes this proof to Joseph Fourier.^[1] Start with:

The reason for the inequality is that each factorial is greater than a product of as many twos as there are factors in the factorial, discounting the one; for example, \( 4! = 1 \cdot 2 \cdot 3 \cdot 4 \gt 1 \cdot 2 \cdot 2 \cdot 2 = 2^3. \) The final step results from summing the geometric series. It follows that \( 2 \lt e \lt 3 \) and in particular, that \( e \) is not a whole number.

John Hidinger (Sep 11, 1839 — May 7, 1883). Photo 1864.

Sometimes mom talked about her grandfather's being in the Civil War and how he was captured and spent time in Confederate prisons. It was fascinating, especially considering that she was born and raised in Saskatchewan and was a naturalized American citizen. Her American roots are deep, going back to the founding of New Haven, Connecticut in the 1630s, but all that turned up later. Her grandfather John Hidinger was from Iowa, though born in Saxony and coming to the United States as a boy (the name was probably Heidinger in the old country). Imagine my surprise when rooting through the archives of the Wisconsin Historical Society to find that he served with the 8^th Wisconsin Volunteer Infantry, the Eagle Regiment, so denoted because of their mascot, the eagle Old Abe, a sensation who occupied the Wisconsin state capitol for many years after the war, and whose replica overlooks the Wisconsin Assembly to this day. It tickled me as an anti-war protester on the University of Wisconsin campus circa 1970, because the right-wingers in the legislature would always call us out-of-town agitators. More like coming home I often thought when passing the Civil War cannon at Camp Randall where great-grandfather Hidinger trained, now a little park on the UW campus.

Eratosthenes Measures the Earth

Like Aristotle^[1] before him and in keeping with virtually all educated opinion in ancient Greece^[2], Eratosthenes (c. 276 BC - c. 194 BC) assumed the earth was spherical. He set out to measure its size using a sound method that has stood the test of time. His results were good (about \( 5.4\% \) too high) and no one did better for over a thousand years. His near correct result was ignored well into the modern era, including by Columbus. Snellius and Picard finally nailed down the earth's circumference in the seventeenth century using Eratosthenes' experimental design and taking advantage of advances in mathematics and instrumentation that had accrued over the intervening \( 1900 \) years.

A system of three congruences is shown on the right, but start with the simpler system:
\[ \begin{align*}
x &\equiv 1 \hspace{-.6em} {\pmod{2}}\\
x &\equiv 2 \hspace{-.6em} {\pmod{3}}.
\end{align*} \]
Values congruent \( \text{mod} \; 6 \) are certainly congruent \( \text{mod} \; 2 \) and \( \text{mod} \; 3, \) so in looking for an \( x \) solving both congruences simultaneously, it suffices to consider congruence classes \( \text{mod} \; 6 \) and in particular their smallest positive residues, namely \( 0, 1, 2, 3, 4, 5. \) We're seeking an odd number among those \( 6 \) since \( x \equiv 1{\pmod{2}}, \) one that is also congruent to \( 2 \; \text{mod} \; 3. \) \( x = 1 \) won't do, since \( 1 \equiv 1{\pmod{3}} \) neither will \( x = 3, \) since \( 3 \equiv 0{\pmod{3}}. \) \( x = 5 \) is the solution, since it satisfies both congruences, and it is the only solution \( \text{mod} \; 6. \)

The theorem in question is:

If \( p \) is an odd prime with \( p \equiv 1 \; (\text{mod} \; 4), \) then \( p \) is the sum of two squares.\( (1) \)

Only if is easy, because for all natural numbers \( n, \; n^2 \equiv 0, 1 \; (\text{mod } 4), \) so \( n^2 + m^2 \equiv 0, 1, 2 \; (\text{mod} \; 4) \) and a sum of two squares cannot be congruent to \( 3 \; (\text{mod} \; 4). \) Obviously \( 2 = 1^2 + 1^2 \) as well. The Brahmagupta-Fibonacci identity assures that a product of sums of two squares is itself a sum of two squares:

\[ \begin{equation}{(a^{2}+b^{2})(c^{2}+d^{2})=(ac+bd)^{2}+(ad-bc)^{2}.}\tag{2} \end{equation} \]

Integers under \( 40 \) that are the sum of two squares. \( \color{red}{25} \) is the first that is the sum of two squares in two ways.

\( 5 = 1^2 + 2^2 \) is the sum of two squares, \( 3 \) is not. Dealing with whole numbers only, including \( 0, \) it's a bit of a riddle coming up with the criterion distinguishing the two situations. Based on empirical investigations, mathematicians in the \( 17^\text{th} \) century found the key. According to Leonard Dickson^[1]:

A. Girard (Dec 9, 1632) had already made a determination of the numbers expressible as a sum of two integral squares: every square, every prime \( 4n + 1, \) a product formed of such numbers, and the double of one of the foregoing.

The part about primes \( p \equiv 1 \; (\text{mod} \; 4) \) is central, because a product of two numbers each of which is the sum of two squares is itself the sum of two squares. Since \( 5 = 1^2 + 2^2 \) and \( 13 = 2^2 + 3^2, \) for example, \( 65 = 5 \cdot 13 \) is also the sum of two squares: \( 65 = 4^2 + 7^2. \) In fact there is a second representation: \( 65 = 1^2 + 8^2, \) and the number of representations is of interest too (this exact example is from Diophantus).

Progressive differences of the first few cubes.

Write down the first few cubes, then put their differences \( \Delta \) in the second column, the differences of those differences \( \Delta^2 \) in the third column, and so on. Remarkably, \( \Delta^3 = 6 \), and that is true for any contiguous sequence of cubes (obviously \( \Delta^4 = 0 \)). Do that with the fourth powers and you find that \( \Delta^4 = 24, \) and in general for contiguous \( n^{th} \) powers, \( \Delta^n = n!. \) The key to unlocking this mystery is the Calculus of Finite Differences, out of vogue now apparently, but with a hallowed history going back to Newton and before and studied in depth by George Boole in 1860.^[1] His book can still be read with profit, as can C. H Richardson's little text from 1954. ^[2]

Euler used \( \Delta^n x^n = n! \) in 1755 to prove the two squares theorem. Boole and those following him employed the term "calculus" advisedly, many theorems in the finite case matching similar ones in the familiar infinitesimal calculus. Which stands to reason, considering all there is in common, it's just that now \( \Delta x = 1. \)

Click image for original.

When a problem contains several unknowns whose relationships are so complicated that one is obliged to form several equations; then, to discover the values of the unknowns, one makes all of them vanish, except one, which combined only with known quantities, gives, if the problem is determined, a final Equation, whose resolution reveals this first unknown, and then by this means all the others.

Robert E. Lee Moore (1882-1974)

It's a statement when someone names their child after Robert E. Lee, a man who did his best to destroy the United States in order to preserve slavery. Robert E. Lee was lionized more in death than in life, a paragon of the Lost Cause, the glorious if doomed rebellion of a brave people who wanted nothing but to be left alone, crushed by the soulless and brutal industrial juggernaut (Sherman's march to the Sea!). It's the big lie, forwarded for 150 years to defend the indefensible. What a wretched history of oppression, assiduously rebuilt over the generations by people like Moore, Sr. and his illustrious and vicious son Robert E. Lee Moore. The Compromise of 1877, peonage, disenfranchisement, lynching, Plessy v. Ferguson, Jim Crow, the Dunning school false flag on reconstruction. Read the old classics by W. E. B. Du Bois, Eric Foner, and C. Vann Woodward (himself a son of the south), among others, if you still doubt the long-standing construction and reconstruction of anti-black racism in this country down through the generations since 1865.

Frigyes Riesz.

The Riesz Representation Theorem is a foundation stone of 20^th century functional analysis. Generalized almost beyond recognition, Frigyes (Frédéric) Riesz originally proved the theorem in 1909 for \( C[0,1] \), the continuous real-valued functions on \( [0,1] \):

If \( \mathcal{A} \) is a bounded linear functional on \( C[0,1] \), then there is a function \( \alpha \) of bounded variation on \( [0,1] \) such that for all \( f \in C[0,1] \): \[ \mathcal{A}[f(x)] = {\int_0^1 f(x)d\alpha(x).} \hskip{60pt} (1) \]

In this article, I propose to retrace Riesz's original proof in Sur les opérations fonctionnelles linéaires^[1] in 1909, augmenting with his discussion in Sur certains systèmes singuliers d'équations intégrales^[2] in 1911 where appropriate.