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A time will however come (as I believe) when physiology will invade and destroy mathematical physics, as the latter has destroyed geometry.
Daedalus, or Science and the Future, London: Kegan Paul, 1923.
Loci: Convergence
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Sums of Powers of Positive Integers
Solutions to Exercises 4-6
Exercise 4. For n = 3, three shells constructed of 6 • 12 = 6 cubes, 6 • 22 = 24 cubes, and 6 • 32 = 54 cubes fit together to form a 3 x 4 x 7 rectangular solid, as shown in Figure 16A. This construction illustrates that $$6\left(1^2 + 2^2 + 3^2\right) = {3\cdot 4\cdot 7},$$ or $$1^2 + 2^2 + 3^2 = {{3\cdot 4\cdot 7} \over 6},$$ or, more generally, $$1^2 + 2^2 + 3^2 + \cdots + n^2 = {{n(n + 1)(2n + 1)} \over 6}$$ for any positive integer n.
Nilakantha reasoned that the outside shell contained 6 • 32 = 54 cubes as follows (see Figure 16B):
Exercise 5. For n = 3, three shells constructed of 6 • (1 • 2)/2 = 6 cubes, 6 • (2 • 3)/2 = 18 cubes, and 6 • (3 • 4)/2 = 36 cubes fit together to form a 3 x 4 x 5 rectangular solid, as shown in Figure 17A. This construction illustrates that $$6\left(1 + 3 + 6\right) = {3\cdot 4\cdot 5},$$ or $$1 + 3 + 6 = \frac {3\cdot 4\cdot 5}{6},$$ or, more generally, $$1 + 3 + 6 + \cdots + {{n(n + 1)} \over 2} = {{n(n + 1)(n + 2)} \over 6}$$ for any integer n.
Nilakantha may have reasoned that the outside shell contained 6 • (3 • 4)/2 = 36 cubes as follows (see Figure 17B):
Exercise 6. They may have made the following computations.
1 + 3 = 4 = 22
3 + 6 = 9 = 32
6 + 10 = 16 = 42
10 + 15 = 25 = 52
15 + 21 = 36 = 62
The general relationship can be expressed as
Tn + Tn+1 = (n+1)2,
where Tn is the nth triangular number, or as
$${{n(n + 1)} \over 2} + {{(n + 1)(n + 2)} \over 2} = (n + 1)^2.$$
The latter equation can be checked easily using algebra.
Next page >> Solutions to Exercises 7-8
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Beery, Janet, "Sums of Powers of Positive Integers," Loci (February 2009), DOI: 10.4169/loci003284
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thread #1:
Karaji's solution
I just wonder if the Karaji Generalized his method of solutions? Is there any information about that. For instance did he found sum of fifth powers?
Reply:
Al-Karaji and higher powers by Janet Beery (posted: 01/20/2010 )
My understanding is that there is no evidence that al-Karaji derived a "formula" for the sum of the fourth, fifth, or higher powers. His justification for his formula for the sum of the cubes was ingenious but al-Haytham's idea seems more readily generalizable.
thread #2:
On the symmetry of the resulting polynomials
== Conjecture == It would be interesting to know that * sums of odd powers can be factored as a polynomial on (n ⋅ (n + 1) / 2) and is therefore symmetric with the center at (−½); and that * sums of even powers can be factored to a function like above multiplied by (2 ⋅ n + 1) and therefore antisymmetric with the center at (−½). It would be interesting to know whether any rule concerning the polynomials thus obtained.
Replies:
Re: On the symmetry of the resulting polynomials by Rick Mabry (posted: 12/30/2011 )
Christopher, see the following article, especially section 3 (Faulhaber Polynomials). Although my browsers don't let me view the code you pasted, I think the article confirms some of what you're getting at (some of which might be gleaned from page 8 of the article we're viewing).
A. F. Beardon, Sums of Powers of Integers,
American Mathematical Monthly,
Vol. 103, No. 3 (Mar., 1996), pp. 201-213.
More about symmetry: by Janet Beery (posted: 01/03/2013 )
The most succinct, beautiful, and perhaps even historically plausible explanation I've seen for the symmetry of the Faulhaber polynomials about -1/2 was in a recent article by Reuben Hersh in the College Math Journal 43:4 (Sept. 2012), pp. 322-4. His approach? Extend the polynomials to negative values of n.