Saturday, January 13, 2018
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 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.
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