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Intuitively clearer proofs of the sum of squares formula. Jonathan A. Cox SUNY Fredonia Sigma Xi December 7, 2007. Riemann sums. Area under curve ≈ sum of areas of rectangles Δx=width of each rectangle Area ≈ Σ f ( x i )Δx. Handy formulas for computing Riemann sums.
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Intuitively clearer proofs of the sum of squares formula Jonathan A. Cox SUNY Fredonia Sigma Xi December 7, 2007
Riemann sums Area under curve ≈ sum of areas of rectangles Δx=width of each rectangle Area ≈ Σf(xi)Δx
Handy formulas for computing Riemann sums • Sum of integers
Handy formulas for computing Riemann sums • Sum of integers • Sum of squares
Handy formulas for computing Riemann sums • Sum of integers • Sum of squares • Sum of cubes
Handy formulas for computing Riemann sums • Sum of integers • Sum of squares • Sum of cubes • Even fourth powers!* *D. Varberg and E. Purcell. Calculus with Analytic Geometry. Sixth Ed.
Why is ? Gauss Legend 1 + 2 + 3 + ∙ ∙ ∙ + 49 + 50 + 51 + 52 + ∙ ∙ ∙ + 98 + 99 + 100
Why is ? Gauss Legend 1 + 2 + 3 + ∙ ∙ ∙ + 49 + 50 + 51 + 52 + ∙ ∙ ∙ + 98 + 99 + 100
Why is ? Gauss Legend 1 + 2 + 3 + ∙ ∙ ∙ + 49 + 50 + 51 + 52 + ∙ ∙ ∙ + 98 + 99 + 100 50 pairs, each with sum 101
Standard proofs of sum of squares • Induction
Induction Proof of . • Base case: Let n=1. Then • Induction step: Assume the formula is holds for n and show that it works for n+1.
Standard proofs of sum of squares • Induction • Telescoping sum of cubes
Standard proofs of sum of squares • Induction • Telescoping sum of cubes These proofs are not intuitively clear!
Regrouping the sum by odds • Every perfect square is a sum of consecutive odd numbers. • 36=1+3+5+7+9+11 • Write each square in the sum as a sum of odds. • Regroup all like odds together and add these first. • These two different ways of summing the squares give an equality which can be solved for the desired sum.
Regrouping the sum by odds • Better, but not intuitively clear
Regrouping the sum by odds • Better, but not intuitively clear • Involves algebraic acrobatics
Regrouping the sum by odds • Better, but not intuitively clear • Involves algebraic acrobatics • Had this explanation been discovered previously?
Regrouping the sum by odds • Better, but not intuitively clear • Involves algebraic acrobatics • Had this explanation been discovered previously? • Martin Gardner’s skyscraper construction (Knotted Doughnuts and other Mathematical Entertainments)
The quest for an intuitively clear proof • Benjamin, Quinn and Wurtz give a proof by counting squares on an n x n chessboard in 2 different ways. (College Math. J., 2006) • Benjamin and Quinn give another “purely combinatorial” proof in Proofs that Really Count. • There are more than 10 different proofs! • And…………………. • SOME OF THEM ARE INTUITIVELY CLEAR! • We’ll look at up to five of the remaining proofs (as time permits).
Solving a linear system • By Don Cohen, from www.mathman.biz
Solving a linear system • By Don Cohen, from www.mathman.biz • Assume the formula is a polynomial in n
Solving a linear system • By Don Cohen, from www.mathman.biz • Assume the formula is a polynomial in n • First, what’s the degree of the polynomial?
Solving a linear system • By Don Cohen, from www.mathman.biz • Assume the formula is a polynomial in n • First, what’s the degree of the polynomial? • The third differences are constant, so the formula will be cubic.
Solving a linear system • By Don Cohen, from www.mathman.biz • Assume the formula is a polynomial in n • First, what’s the degree of the polynomial? • The third differences are constant, so the formula will be cubic. • Want to find the 4 coefficients (variables)
Solving a linear system • By Don Cohen, from www.mathman.biz • Assume the formula is a polynomial in n • First, what’s the degree of the polynomial? • The third differences are constant, so the formula will be cubic. • Want to find the 4 coefficients (variables) • First 4 sums of squares give 4 equations
Solving a linear system • By Don Cohen, from www.mathman.biz • Assume the formula is a polynomial in n • First, what’s the degree of the polynomial? • The third differences are constant, so the formula will be cubic. • Want to find the 4 coefficients (variables) • First 4 sums of squares give 4 equations • Solve system of 4 linear equations in 4 variables
Looking to geometry A sum of squares≈volume of a pyramid with square base Source: David Bressoud, Calculus Before Newton and Leibniz: Part II,http://www.macalester.edu/~bressoud/pub/CBN2.pdf
“Fiddling with the bits that stick out" • Sum of squares = volume of the pyramid • Why not just use volume formula for a pyramid? • It’s not a true pyramid, more like a staircase…. "A very pleasant extension to stacking oranges is to consider the relationship between the volume of the indicative pyramid and the sum of squares, taking cubic oranges of one unit of volume. This, eventually, after some fiddling to account for bits that stick out and bits that stick in, generates the formula for summing squares."(A.W., UK) http://nrich.maths.org/public/viewer.php?obj_id=2497
“Fiddling with the bits that stick out" • Underlying pyramid has volume n3/3 • Add in the half-cubes (triangular prisms) above the slice • Subtract off volumes of n little pyramids added twice
Archimedes’ proof with pyramids(D. Bressoud -- http://www.macalester.edu/~bressoud/pub/CBN2.pdf)
Archimedes’ proof with pyramids(D. Bressoud -- http://www.macalester.edu/~bressoud/pub/CBN2.pdf)
Winner 1: The Greek rectangle method Doug Williams, http://www.mav.vic.edu.au/PSTC/cc/pyramids.htm This is the same construction that Martin Gardner described using skyscrapers!
Winner 1: The Greek rectangle method Doug Williams, http://www.mav.vic.edu.au/PSTC/cc/pyramids.htm
Winner 1: The Greek rectangle methodHow Martin Gardner described it as a skyscraper
Winner 1: The Greek rectangle method Doug Williams, http://www.mav.vic.edu.au/PSTC/cc/pyramids.htm • This is the same construction that Martin Gardner described using skyscrapers! • Flesh out the skyscaper with a sequence of squares on each side to make a rectangle. • Each sequence of squares has area ! • The rectangle has dimensions n(n+1)/2 and 2n+1 !
Winner 1: The Greek rectangle method Doug Williams, http://www.mav.vic.edu.au/PSTC/cc/pyramids.htm
By the way… • The sum of integers formula can be proved with a similar geometric construction. (D. Bressoud -- http://www.macalester.edu/~bressoud/pub/CBN2.pdf)
Winner 2: The six-pyramid constructionDoug Williams, http://www.mav.vic.edu.au/PSTC/cc/pyramids.htm • 6 identical “sum of first n squares” pyramids • Fit them together to form a rectangular prism • It has dimensions n, n+1, and 2n+1 • Thus • It’s fairly easy to see that this works for n+1 if it works for n.
Winner 2: The six-pyramid constructionDoug Williams, http://www.mav.vic.edu.au/PSTC/cc/pyramids.htm 6 identical “sum of first n squares” pyramids Fit them together to form a rectangular prism It has dimensions n, n+1, and 2n+1 Thus It’s fairly easy to see that this works for n+1 if it works for n. (Induction?!)
What makes a good proof? • It should convince the intended audience that the statement is true.
Appendix: Ways of proving sum of squares formula • Induction • Telescoping cubic sum • Regrouping as odds • Pyramid of cubes--fiddling to account for bits • Three lines on chessboard (Benjamin-Quinn-Wurtz) • Combinatorial (Benjamin-Quinn) • Archimedes: Fitting together three pyramids • Solving a system of 4 linear equations in 4 variables • Fitting together six pyramids • Greek rectangles (Martin Gardner's skyscraper) • Integration
Sum of cubes How many rectangles are there on a chessboard? 9 horizontal lines 9 vertical lines Choose two of each rectangles on an 8 x 8 board In general, there are rectangles on an n x n board.
Sum of cubes But the number of rectangles is also ! There are k3rectangles with maximum coordinate k. Source: A. Benjamin, J. Quinn and C. Wurtz. Summing cubes by counting rectangles, CMJ, 2006.