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.

Egyptian Mathematics

  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 B_k. 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' Determination of the Volumes of Cones, Cylinders & Spheres

  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.

Galileo and Archimedes on Mechanics

  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.

Hero of Alexandria's Mechanics

    The Mechanics of Hero of Alexandria tells us that the ancient Romans had a fairly good understanding of the use of simple machines. Hero discusses "the wheel and axle, the lever, the pulley, the wedge, and the screw." In his Catoptrica he also shows that light reflecting from a mirrored surface takes a minimum path.

  Hero's Mechanics survived only in an Arabic translation. There doesn't appear to be an English translation available but there are French translations and German translations. In French the simple machines are given as "le treuil, le levier, la  moufle (poulie), le coin et la vis sans fin."

Lagrange in his Analytical Mechanics makes a rather cryptic remark about the Center of Force and for heavy bodies subject to gravity it being the center of the Earth. What is peculiar about it is that he equates the ratio of the potential energy to the sum of the forces acting on a mechanism as the distance to the center of the Earth. This would be true if the distances to the Earth's center are the same or the forces acting on the parts of a body are equivalent to a single force acting on a center of mass. The change in direction of the vertical at sea level would also allow us to compute the distance to the Earth's center. Perhaps this is an allusion to the work in Geodesy that was taking place in the early 19th Century.

Supplemental (Jan 6): The key to Lagrange's system of mechanics are the virtual velocities or equivalently small changes in position which produce no change in their scalar product with the corresponding forces. Perhaps Lagrange was suggesting the problems of the chain and the arch for the "student" of analytical mechanics. In 1829 Gauss proposed "A new general Principle of Mechanics," the principle of least constraint based on least squares.

Minimum for a Horizontal Line

  What happens if the path is limited to a horizontal line instead of an ellipse?

In this case y equals a constant value and there is only one independent variable which we can take to be x. We can easily determine a formula for the potential energy and set its derivative equal to zero
to get an equation that can be solve for x.

On simplifying this equation we get a simple expression for the square of x and we have to be careful about the sign when taking the square root. All the factors are positive except for y so we need a minus sign to make the right side of the equation positive and then we can solve for x. The equation for x allows us to determine that the triangles involving the angles are similar and we see that angles are also equal at equilibrium for this problem. A check using the value of y_min for the ellipse shows that we get the same x_min.

What's remarkable is that for both problems the angles are equal at the equilibrium position. We get the same results whether we use the ellipse or the horizontal line which is tangent at the equilibrium point.

Minimum of the Zip Line Ellipse

  The equation for the ellipse only depends on the relative numerical values of the anchor positions so one can solve for the equilibrium position just by finding the lowest point on elllipse. There are no physical "laws" needed. Let's review the equation for the ellipse and those for formulas for the equilibrium found using Lagrange's undetermined multipliers.

What happens if we try to find the minimum value of y for the ellipse? To do that we need to take the derivative of the formula for it and set that equal to zero and solve for the unknown angle.

The cosine of the minimum angle, θ_min, turns out to be equal to the eccentricity times the cosine of the angle of the line through the focal points of the ellipse, θ₀, and we get the same coordinates as the other formulas gave for the equilibrium position.

So the path that the constraints impose the motion of the weight is all that we really need to solve this problem.

Adding a Counter Weight

 One can modify the zip line problem is the previous post by adding a counterweight m. We learned that the two angles were equal for equilibrium so we can simplify the solution making same assumption here.

 The length of the line is now ℓ=a+b+c so we have eliminated one variable and replaced it with another, c. The potential has an additional term for the height of m as well.

We have to modify the potential again to allow an unconstrained variation of the unknown parameters. The constraints here are constants of the variation and we can write them in a form that doesn't change the potential. We again take the derivative of the modified potential, V', and set it equal to zero to find the minimum of V'. If this is true for arbitrary changes in the unknowns their coefficients have to be equal to zero. The first of the three resulting equations gives us the same value for μ as before which allows us to simplify the second equation. We now have two simultaneous equations which allow us to solve for the undetermined multipliers λ and μ. The value of μ allows us to solve for sinθ.

We now have enough information to solve for the values of a and c.

The values of x, y, and sinθ from the previous problem allows us to find the mass m needed to maintain the equilibrium there. The formulas above allow us to solve for the remaining unknowns.

We can plot the data to help visualize the results.

The counterweight allows us to determine that the tension of the line is T=mg for equilibrium. The action of simple machines was all that Lagrange would have needed to develop general methods for mechanics.

A Zip Line Problem

  The section on finding the minimum of a function in a mechanical problem is a little difficult to follow in Lagrange's Analytical Mechanics so an example to illustrate the procedure may be helpful. Let's consider a zip line with a weight suspended on it and see if we can find the equilibrium position. We can set up a coordinate system with the origin at the lowest anchor for the line and the coordinates for the higher anchor will be (L,h). The length of the cable is ℓ=a+b and a point on the line can be designated by the coordinates (x,y) as shown.

Two unit vectors e₁ and e₂ are needed to simplify the specification of the problem and that the sum of the two vectors a and b will always equal the position of the higher anchor. The function that we want to minimize is the potential V. We can use the length of segment b to show that length of segment a is a linear function of the coordinates.

We can also show that the weight moves along an elliptical path by substituting the formulas for x and y as a function of a and θ into the linear equation for a and then solve the result for a.

The variation of the potential function V is subject to two constraints which are the components of the equation for the sum of the two segments. Multiplying the constraints by two undetermined multipliers assumed to be known constants and adding to the potential V we get a modified potential function V' for which we want the minimum. Taking the derivative of V' an setting it equal to zero we get three equations which we need to solve for the three unknowns a, θ, and φ.

The last equation tells us that the dot product of the second unit vector with a vector formed from the Lagrange multipliers is zero so we know it is perpendicular to this vector. Dividing by the square root of the sum of the squares of the multipliers gives us the second unit vector. The same trick works for the second equation and we get the components of the first unit vector in terms of the multipliers. Substituting these results into the first equation allows us to determine the value of μ which in turn allows us to show that the two angles are equal at the equilibrium position.

This result also allows us to simplify the equations for sum of the segments and solve for the unknowns in terms of the known quantities.

We are now in a position to do a numerical problem and find the values of x and y for the equilibrium position,

and plot the results.