Wednesday, January 17, 2018
Heath devotes an entire chapter to the question of the originality of Diophantus. Was algebra his sole invention? Or like Euclid was he the compiler drawing from a number of earlier sources. The method of exhaustion did not originate with Archimedes or Euclid but was developed earlier by Eudoxus who improved on an argument on the squaring of the circle by the orator Antiphon. Eudoxus was a student of Archytas, a Pythagorean, but he also traveled to Heliopolis in Egypt to study there. Archytas is believed to be the founder of mathematical mechanics and is now given credit for the Mechanical Problems in the Corpus of Aristotle rather than Aristotle himself. It is similar in nature to the mechanics of Archimedes.
So ancient learning made its way down from one generation to the next along with new discoveries as they were acquired. The scarcity of papyrus or parchment may have led to the publication of more concise summaries of what was done earlier. In Diophantus one finds there is a concerted effort to reduce formulas. Another means of transmission of ancient knowledge was from teacher to student where the chief writing instrument was the slate. The petroglyph is an even more ancient method for transmission of culture. So could the pyramids also have been a vehicle for the passage of knowledge and skills to future generations either intentionally or unintentionally?
Supplemental (Jan 18): The blocks of the Egyptian pyramids don't appear to be stacked as neatly as they were to show the sum of the power series. This can be seen by looking at the corners. The heights of a layers appear to be uniform but the blocks don't always appear to have the same lengths. This may have been standard practice of construction in ancient Egypt. Also, when Galileo cites Aristole in his dialogues he may be referring to the Mechanical Problems which are now attributed to Archytas. The author of the Mechanical Problems was once referred to as pseudo-Aristotle by the classical scholars. In the Mechanical Problems some of the arguments are found to be lacking in rigor.
Tuesday, January 16, 2018
There's an easier way still to find the coefficients of the formula for Pythagoras' partial sums of integers. One just sets up a set of linear equations by substituting values of n and the partial sum, Σ. The same can be done for series involving higher powers but more equations would be needed to determine the coefficients.
Diophantus certainly could have done this. He was able to solve 6th degree polynomials. His notation for a polynomial was similar to that used for weights and measures. One might contain a multiple of a cube of an unknown plus another multiple of squares of the unknown plus a multiple of the unknown plus a number of units. The Greek alphabet was used to represent numbers so ɑ=1, ιγ=13, ε =5 and β=2 in the example cited. The title of Diophantus' Book was Arithmetica which comes from the Greek work for number, αριθμός, which of course Diophantus used.
Monday, January 15, 2018
Here's an earlier version of the derivation of the coefficients for the partial sums of squares which is easier to follow than the general derivation. The subscript of the Σ indicates that the polynomial used is cubic and the number of layers for the partial sum is k. The polynomial is assumed to work for all values of k so it should work for k-1. The equations for the coefficients are obtained by equating the coefficients of corresponding powers of k. Then one can solve for the unknown coefficients one after another.
It's doubtful that the early Egyptians would have been able to do this. They did have complicated algorithms for solving math problems that are nearly algebraic in nature. The problem with assessing the capabilities of the Egyptians is that their methods may have been restricted to an elite and secluded like the inner sanctums of their temples. Pythagoras himself might have picked up his secretiveness from the Egyptian priests.
Sunday, January 14, 2018
We next consider how much lifting would be required to build the a pyramid a number n layers high. Our unit of elevation is the height of one layer. We number the layers starting from the top and the kth layer has k2 blocks on it and each block there has to be raised n-k levels. The sum total of the changes in elevation is can be designated Σh. Σ(p,n) below is an abbreviation for the partial sum of n level of the series with power p. Using the formulas for the power series from the last post we can simplify the formula for the total amount of hoisting that is required.
In the check sum is 16*0+9*1+4*2+1*3=9+8+3=20. The formula gives 16*15/12=4*5=20.
Supplemental (Jan 14): Stumbled across a humorous satire involving Pythagoras by Lucian while googling him for the last post.
Saturday, January 13, 2018
Ancient Greek philosophers like Thales and Pythagoras in the 7th and 6th centuries B.C. studied under the priests of Egypt in Heliopolis where they learned about geometry and astronomy. Pythagoras appears to have been fascinated by sums of series like the triangular numbers 1+2+3+4=10. Could he have learned this from the Egyptians? If so, how far back can we go?
Let's look at what can be learned by counting the number of uniform blocks in a pyramid. Take a number of rectangular blocks with sides a, b and c.
We can stack them to form a small pyramid.
The blue blocks along the left edge are in a line and we can determine sums and subtotals just by counting. The blocks along a side form the series 1, 2, 3, 4 and their partial sums are 1, 3, 6, 10. Next going down vertically counting the number of blocks in each lever we get a series of squares 1, 4, 9, 16 whose partial sums are 1, 5, 14, 30. The first series is the sum of the heights of the individual blocks. The second series is the sum of the areas of the sides of the blocks. And the third is a sum of volumes of the levels.
Note that in the Moscow papyrus there is a formula for the frustum of a square pyramid. How could we find such a formula? We can start by searching for formulas that give the partial sums of the series above. Let's assume that the formulas for the partial sums are polynomials of the number, n, of layers included in the partial sum with the coefficients Bk. Above we noted that a partial sum of a series is equal to the previous partial sum for the plus the number of blocks in each layer for the particular sum. Noting that for zero layers we have zero for a partial sum we can use this as a starting point and setting n=0 in the polynomial we deduce that B0=0 so we can eliminate the constant term from the polynomial. Below a formula is derived for the remaining unknown coefficients.
This formula isn't as complicated as it would seem at first. Note that we get a set of linear equations and the last equation just has one unknown which is easily determined. The penultimate equation involves two of the coefficients with only one unknown and so on. So we can easily determine the coefficients one at a time. We can also use matrix methods to solve for the coefficients with Excel.
The tables above give the coefficients for the partial sums of the first few power series and the formulas Σk=n2/2+n/2, Σk2=n3/3+n2/2+n/6 and Σk3=n4/4+n3/2+n2/4. With some factoring these are the product formulas for the sum of the respective power series. The formula for Σk2 can be used to find the volume of the pyramid. If the number of layers in increased while keeping the total height, total width and total length constant we get the volume of a pyramid with plane sides.
Sliding the blocks around on each layer doesn't change the number of blocks or the height, width or length of a pyramid so the volume remains unchanged so the formula is quite general.
So it would seem there is a "spirit" of sorts stored in the Egyptian pyramids.
Archimedes used the method of exhaustion to determine the volumes of a number of geometrical solids. It's considered a precursor to calculus which is used to sum infinitesimally small elements of volume, area and length. The method of exhaustion uses a series of inscribed and circumscribed geometrical solids to fill and surround a volume and focuses on the difference between the two. As the difference becomes smaller and smaller the remainder is reduced to zero. Archimedes determined that the volume of a sphere is four time that of a cone whose base is equal to the area cut by a plane through the center of the sphere and whose height is the radius of the sphere. The volume of a cone was known prior to this and can be found in Euclid's Elements to be 1/3 of the cylinder that just surrounds it. Archimedes then uses this to determine that the volume of a cylinder that just circumscribes a sphere is 3/2 that of the sphere. The ratio of the volume of a triangular prism to that of a prism that just contains it was found in Euclid to 1/3.
Some of these formulas were known even earlier as can be seen from the Moscow papyrus and the Rhind papyrus with various approximations used for the value of π.
Friday, January 12, 2018
Galileo's last work, Mathematical Discourses Concerning Two New Sciences, was published in 1632 a few years before his death and contains four dialogues concerned primarily with a discussion of the resistance of matter to breakage, the mechanics of levers, uniform and accelerated motion and the motion of projectiles. He speaks highly of Archimedes' Mechanics which is mathematical in nature and prefers this to Aristotle who mentions the lever in the last book of his Physics (parts 4 and 6). One of the characters in the dialogue, Simplicius, speaks for Aristotle. Another is Salviati, a deceased friend of Galileo, expounds the author's views and the third, Sagredo, another deceased friend of Galileo's, represents an intermediate position.
Archimedes' work On Floating Bodies contains a discussion which is similar to the cryptic passage in Lagrange's Analytical Mechanics in which the lines of the weights meet at the Earth's center of gravity.
Supplemental (Jan 12): Galileo describes the chain as being very nearly a parabola and uses an inverted chain (arch?) in an argument.