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So far, I'm stuck on this problem of converting a series to a polynomial and showing that it exhibits certain properties.

PROBLEM

Show that the polynomial formula for $P_k(n) = \sum_{j=1}^n j^k$ is characterized by the following two properties:

  1. $P_k(0) = 0$ for all $k$
  2. $P_k(x)-P_k(x-1) = x^k$ for all $k$ and all $x$

Then show that $P_{k+1}(x) = (k+1)\int_0^x P_k(t) \, dt + C_{k+1} x$ for some constant $C_{k+1}$.

By "polynomial formula", we mean that $P_1(n) = 1 + 2 + ... + n = \frac 12 n^2 + \frac 12 n$ for all positive integer $n$, so we extend $P_1(x) = \frac 12 x^2 + \frac 12 x$ for all real $x$.

MY APPROACH

I showed that any polynomial that satisfies those two properties is unique, so that, if that integral expression (call it $R_{k+1}(x)$) satisfies them with $R_{k+1}(x)-R_{k+1}(x-1)=x^{k+1}$, then we must have $P_{k+1} = R_{k+1}$.

I'm just not sure how to show that $R_{k+1}$ exhibits the second property. Here's what I have:

$R_{k+1}(x) - R_{k+1}(x-1) = (k+1)\int_{x-1}^x P_k(t) \, dt + C_{k+1}$

Expanding $P_k(t) = p_0 + p_1 t + p_2 t^2 + ...$ gives

$R_{k+1}(x) - R_{k+1}(x-1) = (k+1)\int_{x-1}^x \sum_{i=0}^\infty p_i t^i \, dt + C_{k+1}$

$ = (k+1)\sum_{i=0}^\infty \frac{p_i}{i+1}(x^{i+1}-(x-1)^{i+1}) + C_{k+1}$

Any advice on where to go from here? Thanks in advance.

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  • $\begingroup$ Use Euler McLaurin formula. It gives quickly the final form for $P_{k + 1} (x)$. $\endgroup$ – user98186 Dec 13 '15 at 20:18
  • $\begingroup$ How do you mean? We never learned about Bernoulli numbers or polynomials. $\endgroup$ – Drew Foster Dec 13 '15 at 21:02
  • $\begingroup$ Don't worry about the remainder terms. Denote the sum of all such by $C_{k + 1}$ $\endgroup$ – user98186 Dec 13 '15 at 21:06
  • $\begingroup$ With this formula? $\sum_{j=m+1}^n f(j) = \int_m^n f(t) \, dt + \frac 12 (f(n)-f(m)) + Cn$ with $f(j) = i^{k+1}$, $m=0$ and $n=x$? Then $P_{k+1}(x) = \sum_{j=1}^x j^{k+1} = \int_0^x t^{k+1} \, dt + C_{k+1}x$ $= \frac{1}{k+2} x^{k+1} + Cx$? $\endgroup$ – Drew Foster Dec 13 '15 at 22:46
  • $\begingroup$ And that is what you wanted if you iterate one step back. $\endgroup$ – user98186 Dec 14 '15 at 14:44
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I got it! (I think.)

SOLUTION

Let $R_{k+1}(x) = (k+1)\int_0^x P_k(t) \,dt + C_{k+1}x$. Then note that $R_{k+1}(0) = (k+1)\int_0^0 P_k(t) \,dt + C_{k+1} \cdot 0 = 0$.

For the second property:

By hypothesis, we have $P_k(t) - P_k(t-1) = t^k$

Integrating from $0$ to $x$: $\int_0^x P_k(t) \, dt - \int_0^x P_k(t-1) \, dt = \int_0^x t^k \, dt = \frac{1}{k+1}x^{k+1}$

Multiplying by $k+1$ and letting $u=t-1$ in the second integral:

$(k+1)\int_0^x P_k(t) \, dt - (k+1)\int_{-1}^{x-1} P_k(u) \, du = x^{k+1}$

Splitting the second integral at 0:

$(k+1)\int_0^x P_k(t) \, dt + \underset{C_{k+1}}{\underbrace{^- (k+1)\int_{-1}^{0} P_k(u) \, du}} - (k+1)\int_0^{x-1} P_k(u) \, du = x^{k+1}$

But $C_{k+1} = C_{k+1} \times \left[x - (x-1) \right]$, so

$(k+1)\int_0^x P_k(t) \, dt + C_{k+1}x - \left[(k+1)\int_0^{x-1} P_k(u) \, du + C_{k+1}(x-1)\right] = x^{k+1}$

But this is precisely

$R_{k+1}(x) - R_{k+1}(x-1) = x^{k+1}$

Hence, by the uniqueness of $P_{k+1}$, we must have $P_{k+1}(x) = (k+1)\int_0^x P_k(t) \,dt + C_{k+1}x$.

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