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I found the following formula in my previous question. This differs from my previous question in that I want an alternative proof of the below recursive formula for calculating $\displaystyle\sum_{k=1}^nk^p$.


Suppose I had a function recursively defined as

$$f(x,p)=a_px+p\int_0^xf(t,p-1)dt$$

$$a_p=1-p\int_0^1f(t,p-1)dt$$

For $p\in\mathbb N$. For $p=0$, we trivially get $f(x,0)=x$, which shall be our initial condition.

It can then be noticed that

$$a_1=1-\int_0^1tdt=\frac12$$

$$f(x,1)=\frac12x+\int_0^xtdt=\frac12x+\frac12x^2$$

$$a_2=1-2\int_0^1\frac12t+\frac12t^2dt=\frac16$$

$$f(x,2)=\frac16x+\int_0^x\frac12t+\frac12t^2dt=\frac16x+\frac14x^2+\frac16x^3$$

And the general pattern is $f(x,p)=\sum_{k=1}^xk^p$ whenever $x\in\mathbb N\quad(?)$


How do I prove that whenever $x\in\mathbb N$

$$f(x,p)=\sum_{k=1}^xk^p$$

without applying the methods mentioned in the link above?

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    $\begingroup$ To those users who will jump to close this question as a duplicate: the OP is looking for an alternative proof to the one he already knows. $\endgroup$ – Alex M. Nov 28 '16 at 14:55
  • $\begingroup$ Differentiating $f(x,p)$, rearrange the expression and then substitute the integral terms. Repeat the process and then rearrange and substitute the remaining integral seemed to allow the recurrence relation to be rearranged into the following recurrence differential equation: $$f(x,p)-x\partial_xf(x,p)+x^2\partial^2_xf(x,p)=x^2p(\partial_xf(x,p-1)+f(x,p-1))$$ however it is unsure if there are any symmetry can be exploited here $\endgroup$ – Secret Nov 30 '16 at 11:50
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Define $$B_p(x):=p\left(f(x,p-1)-x^{p-1}\right)+a_p$$ then $$B'_p(x)=p\left(f'(x,p-1)-(p-1)x^{p-2}\right)=p\left [\left(a_{p-1}x+(p-1)\int_0^xf(t,p-2)\textrm{d}t\right)^{'}-(p-1)x^{p-2} \right]=p\left(a_{p-1}+(p-1)f(x,p-2)-(p-1)x^{p-2}\right)=pB_{p-1}(x)$$ and $$\int_0^1B_p(t)\textrm{d}t=\int_0^1\left[p\left(f(t,p-1)-t^{p-1}\right)+a_p\right]\textrm{d}t=p\int_0^1f(t,p-1)\textrm{d}t-p\int_0^1t^{p-1}\textrm{d}t+a_p=1-a_p-p\frac{1}{p}+a_p=0$$ With the starting value $B_0(w)=1$ this uniquely determines the sequence of Bernoulli polynomials see Koenigsberger, K. (2003) Analysis 1: Springer.

It follows that $$B_p(0)=p\left(f(0,p-1)-0^{p-1}\right)+a_p=a_p=B_p$$ and $$\frac{B_{p+1}(x+1)-B_{p+1}}{p+1}=\frac{(p+1)(f(x+1,p)-(x+1)^p)+a_{p+1}-a_{p+1}}{p+1}=f(x+1,p)-(x+1)^p=\sum_{k=1}^xk^p.$$

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