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

Posted: Mon Jun 26, 2006 5:47 pm Post subject:
1^e+2^e+...+n^e = ?



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 

Back to top 


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 

Back to top 


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


Back to top 


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 

Back to top 


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 

Back to top 


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 

Back to top 


Google


Back to top 



The time now is Tue Aug 30, 2016 6:49 pm  All times are GMT

