- #1
ktoz
- 171
- 12
While responding to another thread about summing nth powers of integers, I came up with what might be a new method. There is an old one (Faulhabers formula) http://mathworld.wolfram.com/FaulhabersFormula.html" , but mine seems to be considerably simpler.
Most likely, someone has already discovered this, but I wouldn't know, no math training beyond high school...
[tex]
\sum_{j=0}^m j^p \quad = \quad p! \sum_{j=0}^m Q_{pj} R_{pmj}
[/tex]
[tex]
Q_{pk} \quad = \quad \left\{
\begin{array}{ll}
\sum_{j=0}^k -1^j \frac{ (k - j + 1)^p}{j!(p - j)!} \quad & k < p \\
1 & k \geq p
\end{array}
[/tex]
[tex]
R_{pmk} \quad = \quad \prod_{l=1}^p 1 + \frac{m - l}{k}
[/tex]
Ken
Added later: With a slight variation, equation also yeilds the following interesting relation
[tex]
x^n \quad = \quad p! \sum_{j=0}^x Q_{nj} R_{n-1xj}
[/tex]
Most likely, someone has already discovered this, but I wouldn't know, no math training beyond high school...
[tex]
\sum_{j=0}^m j^p \quad = \quad p! \sum_{j=0}^m Q_{pj} R_{pmj}
[/tex]
[tex]
Q_{pk} \quad = \quad \left\{
\begin{array}{ll}
\sum_{j=0}^k -1^j \frac{ (k - j + 1)^p}{j!(p - j)!} \quad & k < p \\
1 & k \geq p
\end{array}
[/tex]
[tex]
R_{pmk} \quad = \quad \prod_{l=1}^p 1 + \frac{m - l}{k}
[/tex]
Ken
Added later: With a slight variation, equation also yeilds the following interesting relation
[tex]
x^n \quad = \quad p! \sum_{j=0}^x Q_{nj} R_{n-1xj}
[/tex]
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