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In N. Rose Mathematical Maxims and Minims, Raleigh NC: Rome Press Inc., 1988.
Loci: Convergence
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Sums of Powers of Positive Integers
Solutions to Exercises 13-14
Exercise 13. The constant difference table for sums of fifth powers is in Figure 20. Blue entries are obtained from knowing that the next highest entry in the fifth powers column is 0; that is, from knowing that 05 = 0. Red entries are obtained from knowing the constant difference is 120. Blue entries may be obtained in this way, too.
Figure 20. Constant difference table for sums of fifth powers
Substituting the values from the first row of the table in Figure 20 into Harriot’s formula for v6, we obtain the following:
$$\eqalign{720 v^6 &= 120nnnnnn + 1080nnnnn + 3000nnnn + 1800nnn - 3120nn - 2880n \cr & \quad \quad \quad \quad \quad \quad \;\;\; - 720nnnnn - 3600nnnn - 3600nnn + 3600nn + 4320n \cr & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;\quad \;\;\; + 900nnnn + 1800nnn - 900nn - 1800n \cr & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;\;\; + 0nnn\;\;\; - 0nn\;\;\; - 0n \cr & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;\quad \quad \quad \quad \quad \quad \; + 360nn - 360n \cr & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;\quad \;\quad \quad \quad \quad \quad \quad \quad \quad \quad + 720n - 720 \cr & \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \;\;\; + 720\cr & = 120nnnnnn + 360nnnnn + 300nnnn + 0nnn - 60nn + 0n + 0\cr &= 120nnnnnn + 360nnnnn + 300nnnn - 60nn,}$$ or $$v^6 = {1 \over {720}}\left( {120nnnnnn + 360nnnnn + 300nnnn - 60nn} \right)$$ $$= {1 \over 6}nnnnnn + {1 \over 2}nnnnn + {5 \over {12}}nnnn - {1 \over {12}}nn,$$
or, in modern notation,
$$\sum_{k = 1}^n {k^5 } = {1 \over 6}n^6 + {1 \over 2}n^5 + {5 \over {12}}n^4 - {1 \over {12}}n^2.$$
Exercise 14.
We have
$$\eqalign{v^5 = & {{(n + 3)(n + 2)(n + 1)n(n - 1)} \over {5 \cdot 4 \cdot 3 \cdot 2}}e + {{(n + 2)(n + 1)n(n - 1)} \over {4 \cdot 3 \cdot 2}}p\cr & + {{(n + 1)n(n - 1)} \over {3 \cdot 2}} p^2 + {{n(n - 1)} \over {2}} p^3 + (n - 1) p^4 + p^5,}$$
$$\eqalign{v^6 & = {{(n + 4)(n + 3)(n + 2)(n + 1)n(n - 1)} \over {6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}}e + {{(n + 3)(n + 2)(n + 1)n(n - 1)} \over {5 \cdot 4 \cdot 3 \cdot 2}}p\cr &+ {{(n + 2)(n + 1)n(n - 1)} \over {4 \cdot 3 \cdot 2}} p^2 + {{(n + 1)n(n - 1)} \over {3 \cdot 2}} p^3 + {{n(n - 1)} \over 2} p^4 + (n - 1) p^5 + {p^6},}$$ and
$$\eqalign{v^7 & = {{(n + 5)(n + 4)(n + 3)(n + 2)(n + 1)n(n - 1)} \over {7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}}e\cr & + {{(n + 4)(n + 3)(n + 2)(n + 1)n(n - 1)} \over {6 \cdot 5 \cdot 4 \cdot 3 \cdot 2}}p + {{(n + 3)(n + 2)(n + 1)n(n - 1)} \over {5 \cdot 4 \cdot 3 \cdot 2}} p^2\cr & + {{(n + 2)(n + 1)n(n - 1)} \over {4 \cdot 3 \cdot 2}} p^3 + {{(n + 1)n(n - 1)} \over {3 \cdot 2}} p^4 + {{n(n - 1)} \over 2} p^5 + (n - 1) p^6 + p^7}$$ or $$v^7 = {n+5\choose 7}e + {n+4\choose 6}p + {n+3\choose 5}p^2 + {n+2\choose 4}p^3 + {n+1\choose 3}p^4$$
\(\displaystyle{+ {n \choose 2}p^5 + (n - 1) p^6 + p^7.}\)
Next page >> Solutions to Exercises 15-17
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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.