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Mark Spahn science forum addict
Joined: 07 Jul 2005
Posts: 62

Posted: Mon Jun 26, 2006 6:21 pm Post subject:
Re: 1^e+2^e+...+n^e = ?



Correction: "edegree polynomial" should be "(e+1)degree polynomial".
"Mark Spahn" <mspahn@localnet.com> wrote in message news:...
It is easy enough to prove by mathematical induction that
1 + 2 + ... + n = (1/2)n(n+1)
1^2 + 2^2 + ... + n^2 = (1/6)n(n+1)(2n+1)
It is also easy to prove that
1^3 + 2^3 + ... + n^3 = (1+2+...+n)^2
from which it follow that
1^3 + 2^3 + ... + n^3 = [(1/2)n(n+1)]^2 = (1/4)n^2(n+1)^2.
But is there a general formula for the value of
S(e,n) = 1^e + 2^e + ... + n^e,
where e is an arbitrary positive integer?
Do we know that whatever S(e,n) is, it must be
an edegree polynomial in n (rather than some
nonpolynomial expression involving n and e)?
What is the name of the field of mathematics that
deals with such questions, that is, the sum of the
eth powers of the first n positive integers?
I suspect that if there is a simple formula for S(e,n),
Euler or Gauss must have found it long ago.
 Mark Spahn 

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Stephen J. Herschkorn science forum Guru
Joined: 24 Mar 2005
Posts: 641


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Virgil science forum Guru
Joined: 24 Mar 2005
Posts: 5536

Posted: Mon Jun 26, 2006 9:24 pm Post subject:
Re: 1^e+2^e+...+n^e = ?



In article <12a097cq4mkvk67@corp.supernews.com>,
"Mark Spahn" <mspahn@localnet.com> wrote:
Quote:  It is easy enough to prove by mathematical induction that
1 + 2 + ... + n = (1/2)n(n+1)
1^2 + 2^2 + ... + n^2 = (1/6)n(n+1)(2n+1)
It is also easy to prove that
1^3 + 2^3 + ... + n^3 = (1+2+...+n)^2
from which it follow that
1^3 + 2^3 + ... + n^3 = [(1/2)n(n+1)]^2 = (1/4)n^2(n+1)^2.
But is there a general formula for the value of
S(e,n) = 1^e + 2^e + ... + n^e,
where e is an arbitrary positive integer?

There is a separate formula for each positive integer, e, and some
general techniques for generating those formulae.
Quote:  Do we know that whatever S(e,n) is, it must be
an edegree polynomial in n (rather than some
nonpolynomial expression involving n and e)?

For each positive natural number e, S(e,n) is a polynomial in n of
degree e+1.
Quote:  What is the name of the field of mathematics that
deals with such questions, that is, the sum of the
eth powers of the first n positive integers?
I suspect that if there is a simple formula for S(e,n),
Euler or Gauss must have found it long ago.

Google for "sum of powers" or see
http://mathworld.wolfram.com/PowerSum.html 

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Mark Spahn science forum addict
Joined: 07 Jul 2005
Posts: 62

Posted: Mon Jun 26, 2006 10:55 pm Post subject:
Re: 1^e+2^e+...+n^e = ?



Thanks for this pointer. It is just what I was looking for.
To find 1^4 + 2^4 + ... + n^4, I had the idea to set this
equal to (an+b)(cn+d)(en+f)(gn+h)(in+j), plug in the
computed values for many different n's, then
solve for a,b,...,j (if possible). But I had no assurance that the
answer would be factorizable like this, and in fact it is not:
1^4 + 2^4 + ... + n^4 = (1/30)n(n+1)(2n+1)(3n^2+3n1).
 Mark Spahn
"Stephen J. Herschkorn" <sjherschko@netscape.net> wrote in message
news:44A027E0.9020305@netscape.net...
http://mathworld.wolfram.com/PowerSum.html

Stephen J. Herschkorn sjherschko@netscape.net
Math Tutor on the Internet and in Central New Jersey and Manhattan 

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Christopher Night science forum beginner
Joined: 30 May 2005
Posts: 29

Posted: Tue Jun 27, 2006 9:07 pm Post subject:
Re: 1^e+2^e+...+n^e = ?



On Mon, 26 Jun 2006, Mark Spahn wrote:
Quote:  Thanks for this pointer. It is just what I was looking for.
To find 1^4 + 2^4 + ... + n^4, I had the idea to set this
equal to (an+b)(cn+d)(en+f)(gn+h)(in+j), plug in the
computed values for many different n's, then
solve for a,b,...,j (if possible). But I had no assurance that the
answer would be factorizable like this, and in fact it is not:
1^4 + 2^4 + ... + n^4 = (1/30)n(n+1)(2n+1)(3n^2+3n1).

Your technique should work with a slight modification. Just use an
unfactored polynomial instead of a factored one:
Sum(x^4,x=0...n) = a + bn + cn^2 + dn^3 + en^4 + fn^5
Then simply take the six linear equations you get for n = 0 to 5 to solve
for {a,b,c,d,e,f}. This is not that much more difficult than the technique
they give on the MathWorld page.
Christopher 

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