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Loci: Convergence
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Sums of Powers of Positive Integers
Solutions to Exercises 7-8
Exercise 7.
Figure 18. (1 + 2 + 3 + 4)2 = 13 + 23 + 33 + 43
$$(1 + 2 + 3 + 4)^2 = (1 + 2 + 3)^2 + \left( {2 \cdot 4(1 + 2 + 3) + 4^2 } \right) \quad\quad {\rm (Step\ 1)}$$
$$= (1 + 2 + 3)^2 + \left( {2 \cdot 4{{3 \cdot 4} \over 2} + 4^2 } \right)$$
$$= (1 + 2 + 3)^2 + \left( {4^2 (3 + 1)} \right)$$
$$= (1 + 2 + 3)^2 + 4^3$$
$$= (1 + 2)^2 + \left( {2 \cdot 3(1 + 2) + 3^2 } \right) + 4^3 \quad\quad {\rm (Step\ 2)}$$
$$= (1 + 2)^2 + \left( {2 \cdot 3{{2 \cdot 3} \over 2} + 3^2 } \right) + 4^3$$
$$= (1 + 2)^2 + \left( {3^2 (2 + 1)} \right) + 4^3$$
$$= (1 + 2)^2 + 3^3 + 4^3$$
$$= 1 + (2 \cdot 2 \cdot 1 + 2^2 ) + 3^3 + 4^3 \quad\quad\quad {\rm (Step\ 3)}$$
$$= 1 + (2 \cdot 2^2 ) + 3^3 + 4^3$$
$$(1 + 2 + 3 + 4)^2= 1^3 + 2^3 + 3^3 + 4^3$$
Each cubic term represents a gnomon. In particular, 13, 23, 33, and 43 are, respectively, the areas of the yellow, green, blue, and red gnomons in Figure 18.
Exercise 8. $$1^4 + 2^4 + 3^4 + \cdots + n^4 = {1 \over 5}n^5 + {1 \over 2}n^4 + c_3 n^3 + c_2 n^2 + c_1 n + c_0,$$
where $$c_3 + c_2 + c_1 + c_0 = 1 - \left( {{1 \over 5} + {1 \over 2}} \right) = {3 \over {10}}.$$
Three possible formulas are
$${1 \over 5}n^5 + {1 \over 2}n^4 + {1 \over {10}}n^3 + {1 \over {10}}n^2 + {1 \over {10}}n,$$
$${1 \over 5}n^5 + {1 \over 2}n^4 + {1 \over 5}n^3 + {1 \over {10}}n, \quad {\rm and}$$
$${1 \over 5}n^5 + {1 \over 2}n^4 + {1 \over 5}n^3 + {1 \over {15}}n^2 + {1 \over {30}}n.$$
The correct formula is $$1^4 + 2^4 + 3^4 + \cdots + n^4 = {1 \over 5}n^5 + {1 \over 2}n^4 + {1 \over 3}n^3 - {1 \over {30}}n.$$
Surprised?
Next page >> Solutions to Exercises 9-12
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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.