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Attaching significance to invariants is an effort to recognize what, because of its form or colour or meaning or otherwise, is important or significant in what is only trivial or ephemeral. A simple instance of failing in this is provided by the poll-man at Cambridge, who learned perfectly how to factorize a^2 - b^2 but was floored because the examiner unkindly asked for the factors of p^2 - q^2.

In J. R. Newman (ed.) The World of Mathematics, New York: Simon and Schuster, 1956.

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# Sums of Powers of Positive Integers

## Solutions to Exercises 21-22

Exercise 21.  According to Bernoulli’s formula,

\eqalign{\int n^4 & = {1 \over {4 + 1}}n^{4 + 1} + {1 \over 2}n^4 + {4 \over 2}An^{4 - 1} + {{4 \cdot 3 \cdot 2} \over {2 \cdot 3 \cdot 4}}Bn^{4 - 3} \cr &= {1 \over 5}n^5 + {1 \over 2}n^4 + {4 \over 2} \cdot {1 \over 6}n^3 + \left( { - {1 \over {30}}} \right)n = {1 \over 5}n^5 + {1 \over 2}n^4 + {1 \over 3}n^3 - {1 \over {30}}n}

and

\eqalign{\int n^5 &= {1 \over {5 + 1}}n^{5 + 1} + {1 \over 2}n^5 + {5 \over 2}An^{5 - 1} + {{5 \cdot 4 \cdot 3} \over {2 \cdot 3 \cdot 4}}Bn^{5 - 3} \cr &= {1 \over 6}n^6 + {1 \over 2}n^5 + {5 \over 2} \cdot {1 \over 6}n^4 + {5 \over 2}\left( { - {1 \over {30}}} \right)n^2 = {1 \over 6}n^6 + {1 \over 2}n^5 + {5 \over {12}}n^4 - {1 \over {12}}n^2.}

Students may have noticed already that, since A = 1/6, the coefficient of nc – 1 is c/12.  According to Bernoulli’s formula,

$${\int n^{11}} = {1\over{11+1}}n^{11+1}+ {1\over 2} n^{11}+ {{11}\over 2} A n^{11 - 1}+ {{11.\,10.\,9} \over {2.3.4}}Bn^{11 - 3}+ {{11.\,10.\,9.\,8.\,7} \over {2.3.4.5.6}}Cn^{11 - 5}$$

$$+ {{11.\,10.\,9.\,8.\,7.\,6.\,5} \over {2.3.4.5.6.7.8}}Dn^{11 - 7}+ {{11.\,10.\,9.\,8.\,7.\,6.\,5.4.3} \over {2.3.4.5.6.7.8.9.10}}En^{11 - 9}$$

$$= {1 \over {12}}n^{12} + {1 \over 2}n^{11} + {{11} \over 2} \cdot {1 \over 6}n^{10} + {{11.\,10.\,9} \over {2.3.4}}\left( { - {1 \over {30}}} \right)n^8 + {{11.\,10.\,9.\,8.\,7} \over {2.3.4.5.6}} \cdot {1 \over {42}}n^6$$

$$+ {{11.\,10.\,9}\over{2.3.4}}\left( {-{1\over{30}}}\right) n^4 + {{11}\over 2} \cdot {5\over {66}}n^2$$

$$= {1 \over {12}}n^{12} + {1 \over 2}n^{11} + {{11} \over {12}}n^{10} - {{11} \over 8}n^8 + {{11} \over 6}n^6 - {{11} \over 8}n^4 + {5 \over {12}}n^2.$$

Exercise 22.  According to Bernoulli’s formula,

$${\int n^{12}={1 \over {12 + 1}}n^{12 + 1} + {1 \over 2}n^{12} + {{12} \over 2}An^{12 - 1} + {{12.\,11.\,10} \over {2.3.4}}Bn^{12 - 3} + {{12.\,11.\,10.\,9.\,8} \over {2.3.4.5.6}}Cn^{12 - 5}}$$

$$+ {{12.\,11.\,10.\,9.\,8.\,7.\,6} \over {2.3.4.5.6.7.8}}Dn^{12 - 7}+ {{12.\,11.\,10.\,9.\,8.\,7.\,6.\,5.\,4} \over {2.3.4.5.6.7.8.9.10}}En^{12 - 9}$$

$$+ {{12.\,11.\,10.\,9.\,8.\,7.\,6.\,5.\,4.\,3.\,2} \over {2.3.4.5.6.7.8.9.10.11.12}}Fn^{12 -11}$$

$$= {1 \over {13}}n^{13} + {1 \over 2}n^{12} + {{12} \over 2} \cdot {1 \over 6}n^{11} + {{12.\,11.\,10} \over {2.3.4}}\left( { - {1 \over {30}}} \right)n^9 + {{12.\,11.\,10.\,9.\,8} \over {2.3.4.5.6}} \cdot {1 \over {42}}n^7$$

$$+ {{12.\,11.\,10.\,9} \over {2.3.4.5}}\left( { - {1 \over {30}}} \right)n^5 + {{12.11} \over {2.3}} \cdot {5 \over {66}}n^3 + Fn$$

$$= {1 \over {13}}n^{13} + {1 \over 2}n^{12} + n^{11} - {{11} \over 6}n^9 + {{22} \over 7}n^7 - {{33} \over {10}}n^5 + {5 \over 3}n^3 + Fn,$$

where $${1 \over {13}} + {1 \over 2} + 1 - {{11} \over 6} + {{22} \over 7} - {{33} \over {10}} + {5 \over 3} + F = 1,$$

or

$$F = 1 - {{210 + 1365 + 2730 - 5005 + 8580 - 9009 + 4550} \over {2730}} = {{2730 - 3421} \over {2730}} = - {{691} \over {2730}},$$

whence

$$\int n^{12} = {1 \over {13}}n^{13} + {1 \over 2}n^{12} + n^{11} - {{11} \over 6}n^9 + {{22} \over 7}n^7 - {{33} \over {10}}n^5 + {5 \over 3}n^3 - {{691} \over {2730}}n.$$

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Beery, Janet, "Sums of Powers of Positive Integers," Loci (February 2009), DOI: 10.4169/loci003284

### 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?

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.

### 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 Ã¢â€¹â€¦ (n + 1) / 2) and is therefore symmetric with the center at (Ã¢Ë†â€™Ã‚Â½); and that * sums of even powers can be factored to a function like above multiplied by (2 Ã¢â€¹â€¦ n + 1) and therefore antisymmetric with the center at (Ã¢Ë†â€™Ã‚Â½). 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.