Could Pythagoras Have Derived the Formulas for a Sum of Two Rotations?

  In the last post we derived formulas for rotations from trigonometric formulas but we may have gotten a little ahead of ourselves. So we need to ask if Pythagoras could have derived the same formulas using geometry. Let's suppose we have two right triangles with rational coordinates (x,y) and (u,v) respectively. To add the rotations we can construct the second triangle, ΔOBE, on the hypotenuse of the first triangle, ΔOAP.

So we need to determine the coordinates of the point B on the unit circle. The major difficulty is to find the sides of triangle ΔEFB. To do so we construct a copy of ΔEFB at the origin, namely, triangle ΔOGD. Line BE is perpendicular to line OA so to find the direction of BE from the origin we need to construct a copy of triangle ΔOAP based on the vertical axis, namely, ΔOCQ. We can then mark off point D a distance v away from the origin. Using the proportions for the similar triangles in triangles ΔOAP and ΔOCQ we can deduce the formulas for OH, EH, GO, and OR. The new coordinates are just sums and differences of these lengths.

So one would not need trigonometry to deduce the formulas for the addition of the two triangles and Pythagoras should have been able to to this. The remaining question is whether he was motivated to do so or not.

Merging the Two Sets of Integer Sided Triangles

  One can reconcile the two sets of triangles by allowing λ to have rational values which are half integers in this case. We can still index the points with integers by using μ=2λ.

Both sets of points will then map onto the same curve.

We can also map all these points onto a unit circle by letting x=a/c and y=b/c which are rational numbers p/q with p and q integers.

The series of points with rational coordinates on the unit circle are not unique. Starting with point on the circle with rational coordinates near the horizontal axis we can use the trigonometric relations for the sum of two angles to step along the circle and get another point with rational coordinates since the rational numbers are closed under the arithmetic operations of addition, subtraction, multiplication and division (excluding zero).

For Pythagoras the proportions of the sides of the triangles and the calculation of lengths and areas would have been his main concern. In ancient Egypt angles were determined by right triangles and was expressed as a seked. The best the Pythagoreans appear to have done in terms of angles were the standard angles of an arc, 30°, 45°, 60° and 90°, and their coordinates.

Triangles With Integer Sides

  In his book on Diophantus' work Arithmetic or Numbers Heath states that Pythagoras gives a family of solutions which provide integer values for the sides of a right triangle. How could Pythagoras have accomplished this? The Pythagorean Theorem states that the sum of the squares of the sides are equal to the square of the hypotenuse or a²+b²=c². We can subtract the square of one side from the square of the hypotenuse and then solve for this side using the other side and an auxiliary variable x.

We get a rational expression for side a and can factor the numerator. Side a is presumed to be an integer so lets assume that b-x is divisible by 2x with the quotient being equal to another auxiliary variable λ. We can then express all the sides of the triangle in terms of  the variables x and λ. Noting that all three sides have a common factor, x, we can reduce the expressions for the sides by ignoring the larger similar triangles.

This is the family of solutions attributed to Pythagoras. Evaluating the formulas for integer values of λ does indeed give integer solutions. Note that b is always an odd number. We can find another set of solutions for which b is an even number by manipulating the formulas a little.

We can compare the two sets of numbers by plotting them.

Multiplying all three sides of these triangles by a common integer factor gives more integer solutions. We can also multiply by a rational number to rescale and find smaller similar triangles.

 That Pythagoras was able to arrive at the first set of solutions indicates that he had a fairly good grasp of algebra.

Questions of Originality in Greek Mathematics

  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.

Diophantus of Alexandria

  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.

Easier to Follow Derivation of the Coefficients for p=2

  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.

The Amount of Elevation Required

  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.