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Given that $f(n,k) = \sum \limits_{i=0}^{n} i^k $, can we calculate the value of $f(n,k+1)$ using $f(n,k)$ or any recursive formula ? ,

I searched the net with no success, also there is a "closed formulas, but they use Bernoulli's numbers" so..., does this makes it hard to come up with recursive equation, also if proven that there are no recursive equation please cite the paper, or a brief proof would be good.

Thanks

2 Answers2

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Let $ P_k(n)=\sum_{i=1}^ni^k. $ We outline a method to compute $P_{k}(n)$ recursively from $P_j(n)$ for $j=1,\dotsb, k-1$.

Let $X$ be a uniformly distributed random variable on $\{1,2,\dotsc. n\}$. Then $n+1-X$ is equal in distribution to $X$. In particular $$ E(n+1-X)^k=EX^k=P_{k}(n)/n $$ But $$ E(n+1-X)^k=\sum_{m=0}^k\binom{k}{m}(n+1)^{k-m}(-1)^{m}\frac{P_{m}(n)}{n} $$ by the binomial theorem so we have that $$ \sum_{m=0}^k\binom{k}{m}(n+1)^{k-m}(-1)^{m}\frac{P_{m}(n)}{n}=\frac{P_{k}(n)}{n}\tag{0}. $$ Equation (0) together with the initial condition $P_{0}(n)=n$ specifies the desired recurrence.

Sri-Amirthan Theivendran
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    In (0), $(-1)^kP_k(n)/n$ is on the LHS, and $P_k(n)/n$ is on the RHS. When $k$ is even, they cancel, so you don't get a formula for $P_k(n)$. – Mike Earnest Jun 14 '18 at 00:53
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Some possibilities, possibly useful.

Let $f(n,k) = \sum \limits_{i=0}^{n} i^k $.

Then

$\begin{array}\\ f(n+1,k) &= \sum \limits_{i=0}^{n+1} i^k\\ &=0^k+ \sum \limits_{i=1}^{n+1} i^k\\ &=0^k+ \sum \limits_{i=0}^{n} (i+1)^k \qquad (0^k=1 \text{ if }k=0, 0 \text{ otherwise})\\ &=0^k+ \sum \limits_{i=0}^{n}\sum_{j=0}^k \binom{k}{j} i^j\\ &=0^k+ \sum_{j=0}^k \binom{k}{j}\sum \limits_{i=0}^{n} i^j\\ &=0^k+ \sum_{j=0}^k \binom{k}{j}f(n, j)\\ \end{array} $

Another.

$\begin{array}\\ 2f(n, k) &=\sum \limits_{i=0}^{n} i^k+\sum \limits_{i=0}^{n} i^k\\ &=\sum \limits_{i=0}^{n} i^k+\sum \limits_{i=0}^{n} (n-i)^k\\ &=\sum \limits_{i=0}^{n} (i^k+(n-i)^k)\\ \text{so}\\ 2f(n, 2k+1) &=\sum \limits_{i=0}^{n} (i^{2k+1}+(n-i)^{2k+1})\\ &=\sum \limits_{i=0}^{n} (i^{2k+1}+\sum_{j=0}^{2k+1}\binom{2k+1}{j}n^j(-1)^{2k+1-j}i^{2k+1-j})\\ &=\sum \limits_{i=0}^{n} (i^{2k+1}+(-1)^{2k+1}i^{2k+1}+\sum_{j=1}^{2k+1}\binom{2k+1}{j}n^j(-1)^{2k+1-j}i^{2k+1-j})\\ &=\sum \limits_{i=0}^{n} \sum_{j=1}^{2k+1}\binom{2k+1}{j}n^j(-1)^{2k+1-j}i^{2k+1-j}\\ &=\sum \limits_{i=0}^{n} \sum_{j=0}^{2k}\binom{2k+1}{2k+1-j}n^{2k+1-j}(-1)^{j}i^{j}\\ &= \sum_{j=0}^{2k}\binom{2k+1}{2k+1-j}n^{2k+1-j}(-1)^{j}\sum \limits_{i=0}^{n}i^{j}\\ &= \sum_{j=0}^{2k}\binom{2k+1}{2k+1-j}n^{2k+1-j}(-1)^{j}f(n, j)\\ \end{array} $

What happens if you use $2k$ instead of $2k+1$?

marty cohen
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