User:MikeB17/sandbox

Proof that the series exhibits log-log growth
Here is another proof that actually gives a lower estimate for the partial sums; in particular, it shows that these sums grow at least as fast as $log log n$. The proof is due to Ivan Niven, adapted from the the product expansion idea of Euler. In the following, a sum or product taken over $p$ always represents a sum or product taken over a specified set of primes.

The proof rests upon the following four inequalities:


 * Every positive integer $p$ can be uniquely expressed as the product of a square-free integer and a square as a consequence of the fundamental theorem of arithmetic. Start with:


 * $$i = q_1^{2{\alpha}_1+{\beta}_1} \cdot q_2^{2{\alpha}_2+{\beta}_2} \cdot\ldots \cdot q_r^{2{\alpha}_r+{\beta}_r},$$

where the &beta;s are 0 (the corresponding power of prime $i$ is even) or 1 (the corresponding power of prime $q$ is odd). Factor out one copy of all the primes whose &beta; is 1, leaving a product of primes to even powers, itself a square. Relabeling:
 * $$i = (p_1 p_2 \ldots p_s) \cdot b^2,$$

where the first factor, a product of primes to the first power, is square free. Inverting all the $q$s gives the inequality
 * $$ \sum_{i=1}^n \frac 1 i \le \left(\prod_{p \le n} \left(1 + \frac 1 p \right)\right) \cdot \left(\sum_{k=1}^n \frac 1 {k^2}\right) = A \cdot B.$$

To see this, note that
 * $$\frac 1 i = \frac 1 {p_1 p_2 \ldots p_s} \cdot \frac 1 {b^2},$$

where
 * $$\begin{align}

\left(1 + \frac 1 p_1\right)\left(1 + \frac 1 p_2\right) \ldots \left(1 + \frac 1 p_s\right) &= \left(\frac 1 p_1\right)\left(\frac 1 p_2\right)\ldots\left(\frac 1 p_s\right) + \ldots\\ &= \frac 1 {p_1 p_2 \ldots p_s} + \ldots.\end{align}$$ That is, $$1/(p_1p_2 \ldots p_s)$$ is one of the summands in the expanded product $i$. And since $$1/b^2$$ is one of the summands of $A$, every $B$ is represented in one of the terms of $i$ when multiplied out. The inequality follows.

$AB$ is transcendental
Lemma 2. $$\left|\tfrac{Q}{k!}\right|<1$$ for sufficiently large $$k$$.

Proof. Note that

$$\begin{align} f_k e^{-x} &= x^{k}[(x-1)(x-2)\cdots(x-n)]^{k+1}e^{-x}\\ &= \left (x(x-1)\cdots(x-n) \right)^k \cdot \left ((x-1)\cdots(x-n)e^{-x}\right)\\ &= u(x)^k \cdot v(x), \end{align}$$

where $$u(x)$$ and $$v(x)$$ are continuous functions of $$x$$ for all $$x$$, so are bounded on the interval $$[0,n]$$. That is, there are constants $$G > 0, \; H > 0$$ such that
 * $$|f_k e^{-x}| \leq |u(x)|^k \cdot |v(x)| < G^k H \quad \text{ for } 0 \leq x \leq n.$$

So each of those integrals composing $$Q$$ is bounded, the worst case being


 * $$\left|\int_{0}^{n}f_{k}e^{-x}\,dx\right| \leq \int_{0}^{n}|f_{k}e^{-x}|\,dx \leq \int_{0}^{n}G^k H\,dx = nG^k H.$$

It is now possible to bound the sum $$Q$$ as well:


 * $$|Q| < G^{k} \cdot nH(|c_1|e+|c_2|e^2+\cdots+|c_n|e^{n}) = G^k \cdot M,$$

where $$M$$ is a constant not depending on $$k$$. It follows that


 * $$\left| \frac{Q}{k!} \right| < M \cdot \frac{G^k}{k!} \rightarrow 0 \text{ as } k \rightarrow \infty,$$

finishing the proof of this lemma.

Choosing a value of $$k$$ satisfying both lemmas leads to a non-zero integer ($$P/k!$$) added to a vanishingly small quantity ($$Q/k!$$) being equal to zero, an impossibility. It follows that that the original assumption, that $$e$$ can satisfy a polynomial equation with integer coefficients, is also impossible; that is, $$e$$ is transcendental.

Tartaglia's Triangle
Tartaglia was proficient with binomial expansions and included many worked examples in Part II of the General Trattato, one a detailed explanation of how to calculate the summands of $$(6 + 4)^7$$, including the appropriate binomial coefficients.

Tartaglia knew of "Pascal's Triangle" one hundred years before Pascal, as shown in this image from the General Trattato. His examples are numeric, but he thinks about it geometrically, the horizontal line $$ab$$ at the top of the triangle being broken into two segments $$ac$$ and $$cb$$, where point $$c$$ is the apex of the triangle. Binomial expansions amount to taking $$(ac+cb)^n$$ for exponents $$n = 2, 3, 4, \cdots$$ as you go down the triangle. The symbols along the outside represent powers at this early stage of algebraic notation: $$ce = 2, cu = 3, ce.ce = 4$$, and so on. He writes explicitly about the additive formation rule, that (for example) the adjacent 15 and 20 in the fifth row add up to 35, which appears beneath them in the sixth row.

Volume of a tetrahedron
Tartaglia was a prodigious calculator and master of solid geometry. In Part IV of the General Trattato he shows by example how to calculate the height of a pyramid on a triangular base, that is, an irregular tetrahedron.

The base of the pyramid is a $$13-14-15$$ triangle $$bcd$$, with edges of length $$20, 18$$, and $$16$$ rising up to the apex $$a$$ from points $$b$$, $$c$$, and $$d$$ respectively. Base triangle $$bcd$$ partitions into $$5-12-13$$ and $$9-12-15$$ triangles by dropping the perpendicular from point $$d$$ to side $$bc$$. He proceeds to erect a triangle in the plane perpendicular to line $$bc$$ through the pyramid's apex, point $$a$$, calculating all three sides of this triangle and noting that its height is the height of the pyramid. At the last step, he applies what amounts to this formula for the height $$h$$ of a triangle in terms of its sides $$p, q, r$$ (the height from side $$p$$ to its opposite vertex):


 * $$h^2 = r^2 - \left({{{p^2 + r^2 - q^2}} \over {2p}}\right)^2,$$

a formula deriving from the Law of Cosines (not that he cites any justification in this section of the General Trattato).

Tartaglia drops a digit early in the calculation, taking $$305 \frac{31}{49}$$ as $$305 \frac{3}{49}$$, but his method is sound. The final (correct) answer is:


 * $$\text{ height of pyramid } = \sqrt{240 \frac{615}{3136}}.$$

The volume of the pyramid is easily gotten after that (not that Tartaglia gives it):


 * $$\begin{align}

V &= 1/3 \times \text{ base } \times \text{ height }\\ &= 1/3 \times \text{ Area } (\triangle bcd) \times \text{ height }\\ &= 1/3 \times 84 \times \sqrt{240 \frac{615}{3136}}\\ &= 433.9513222 \end{align}$$

Simon Stevin invented decimal fractions later in the sixteenth century, so the last figure would have been foreign to Tartaglia, who always used fractions. All the same, his approach is in some ways a modern one, suggesting by example an algorithm for calculating the height of most or all irregular tetrahedra, but (as usual for him) he gives no explicit formula.

Ballistics
Nova scientia (1537) was Tartaglia's first published work, described by Matteo Valleriani as:

"... one of the most fundamental works on mechanics of the Renaissance, indeed, the first to transform aspects of practical knowledge accumulated by the early modern artillerists into a theoretical and mathematical framework."

Then dominant Aristotelian physics preferred categories like "heavy" and "natural" and "violent" to describe motion, generally eschewing mathematical explanations. Tartaglia brought mathematical models to the fore, "eviscerat[ing] Aristotelian terms of projectile movement" in the words Mary J. Henninger-Voss. One of his findings was that the maximum range of a projectile was achieved by directing the cannon at a 45° angle to the horizon.

Tartaglia's model for a cannonball's flight was that it proceeded from the cannon in a straight line, then after a while started to arc towards the earth along a circular path, then finally dropped in another straight line directly towards the earth — examine the image above to get a sense of this for different elevations of the gun. At the end of Book 2 of Nova scientia, Tartaglia proposes to find the length of that initial rectilinear path for a projectile fired at an elevation of 45°, engaging in a Euclidean-style argument, but one with numbers attached to line segments and areas, and eventually proceeds algebraically to find the desired quantity (procederemo per algebra in his words).

Mary J. Henninger-Voss notes that "Tartaglia's work on military science had an enormous circulation throughout Europe", being a reference for common gunners into the eighteenth century, sometimes through unattributed translations. He influenced Galileo as well, who owned "richly annotated" copies of his works on ballistics as he set about solving the projectile problem once and for all.

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Tartaglia's biographer Arnoldo Masotti writes that:

"At the age of about fourteen, he [Tartaglia] went to a Master Francesco to learn to write the alphabet; but by the time he reached “k,” he was no longer able to pay the teacher. “From that day,” he later wrote in a moving autobiographical sketch, “I never returned to a tutor, but continued to labor by myself over the works of dead men, accompanied only by the daughter of poverty that is called industry” (Quesiti, bk. VI, question 8)."

Tartaglia moved to Verona around 1517, then to Venice in 1534, a major European commercial hub and one of the great centers of the Italian renaissance at this time. Also relevant is Venice's place at the forefront of European printing culture in the sixteenth century, making early printed texts available even to poor scholars if sufficiently motivated or well-connected — Tartaglia knew of Archimedes' work on the quadrature of the parabola, for example, from Guarico's Latin edition of 1503, which he had found "in the hands of a sausage-seller in Verona in 1531" (in mano di un salzizaro in Verona, l'anno 1531 in his words).

Tartaglia eked out a living teaching practical mathematics in abacus schools and earned a penny where he could:

"This remarkable man [Tartaglia] was a self educated mathematics teacher who sold mathematical advice to gunners and architects, ten pennies one question, and had to litigate with his customers when they gave him a worn out cloak for his lectures on Euclid instead of the payment agreed on."

Translations
Archimedes' works began to be studied outside the universities in Tartaglia's day as exemplary of the notion that mathematics is the key to understanding physics, Federigo Commandino reflecting this notion when saying in 1558 that "with respect to geometry no one of sound mind could deny that Archimedes was some god". Tartaglia published a 71-page Latin edition of Archimedes in 1543, Opera Archimedis Syracusani philosophi et mathematici ingeniosissimi, containing Archimedes' works on the parabola, the circle, centers of gravity, and floating bodies. Guarico had published Latin editions of the first two in 1503, but the works on centers of gravity and floating bodies had not been published before. Tartaglia published Italian versions of some Archimedean texts later in life, his executor continuing to publish his translations after his death. Galileo probably learned of Archimedes' work through these widely disseminated editions.

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