lpetrich
Contributor
Let's consider sums of powers of positive integers.
\( \sum_{k=1}^n 1 = n \\ \sum_{k=1}^n k = \frac12 n (n+1) \\ \sum_{k=1}^n k^2 = \frac16 n (n+1) (2n+1) \\ \sum_{k=1}^n k^3 = \frac14 n^2 (n+1)^2 \\ \sum_{k=1}^n k^4 = \frac{1}{30} n (n+1) (2n+1) (3n^2 + 3n - 1) \)
Not very much of a pattern. Let's look at something similar.
\( \sum_{k=1}^n k (k+1) \cdots (k+m) = \frac{1}{m+2} n (n+1) \cdots (n+m+1) \)
Easy to prove. But Abramowitz and Stegun have some general formulas for positive-integer powers, though they use something called Bernoulli polynomials, Bn(x). These are defined by
\( \frac{t e^{x t}}{e^t - 1} = \sum_{n=0}^{\infty} B_n(x) \frac{t^n} {n!} \)
Bernoulli numbers Bn are Bn(0). Some related polynomials are the Euler polynomials:
\( \frac{2e^{x t}}{e^t + 1} = \sum_{n=0}^{\infty} E_n(x) \frac{t^n} {n!} \)
However, the corresponding Euler numbers are En = 2n En(1/2).
The sum-of-powers formula:
\( \sum_{k=0}^n k^m = \frac{ B_{m+1}(n+1) - B_{m+1} }{m+1} \)
\( \sum_{k=1}^n 1 = n \\ \sum_{k=1}^n k = \frac12 n (n+1) \\ \sum_{k=1}^n k^2 = \frac16 n (n+1) (2n+1) \\ \sum_{k=1}^n k^3 = \frac14 n^2 (n+1)^2 \\ \sum_{k=1}^n k^4 = \frac{1}{30} n (n+1) (2n+1) (3n^2 + 3n - 1) \)
Not very much of a pattern. Let's look at something similar.
\( \sum_{k=1}^n k (k+1) \cdots (k+m) = \frac{1}{m+2} n (n+1) \cdots (n+m+1) \)
Easy to prove. But Abramowitz and Stegun have some general formulas for positive-integer powers, though they use something called Bernoulli polynomials, Bn(x). These are defined by
\( \frac{t e^{x t}}{e^t - 1} = \sum_{n=0}^{\infty} B_n(x) \frac{t^n} {n!} \)
Bernoulli numbers Bn are Bn(0). Some related polynomials are the Euler polynomials:
\( \frac{2e^{x t}}{e^t + 1} = \sum_{n=0}^{\infty} E_n(x) \frac{t^n} {n!} \)
However, the corresponding Euler numbers are En = 2n En(1/2).
The sum-of-powers formula:
\( \sum_{k=0}^n k^m = \frac{ B_{m+1}(n+1) - B_{m+1} }{m+1} \)