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Let $n \geq 0$ and consider $\sum\limits_{k=0}^x k^n$. It's a theorem that there is a polynomial $P_n(x)$ of degree $n+1$ such that $P_n(x) = \sum\limits_{k=0}^x k^n$. One way to determine these polynomials for each $n$ is to consider their derivatives. In particular, we have that

Let $n > 0$. There is a constant $b_n$ such that $P_n'(x) = nP_{n-1}(x) + b_n$

(These $b_n$ will be the Bernoulli numbers). I'm trying to follow Smoryński's proof of this theorem in his Logical Number Theory, but I still have a few doubts about the proof, and I'd be very grateful if someone could help me to sort them out. The proof is by complete induction on n. I'll detail the proof below; but first, let me state an equation that will be used at some steps in the proof:

(A) $P_{n+1}(x) = \frac{1}{n+2}[(x+1)^{n+2} - 1 - \sum\limits_{j=0}^n \binom{n+2}{j} P_j(x)]$

So here's Smoryński's proof:

(1) $P_{n+1}'(x) = \frac{1}{n+2}[(n+2)(x+1)^{n+1} - 1 - \sum\limits_{j=0}^n \binom{n+2}{j} P_j'(x)]$

By using the powers rule, the constant rule and the sum rule for derivatives.

(2) $\frac{1}{n+2}[(n+2)(x+1)^{n+1} - 1 - \sum\limits_{j=0}^n \binom{n+2}{j} P_j'(x)] = (x+1)^{n+1} - \frac{1}{n+2}P_0'(x) - \sum\limits_{j=1}^n \binom{n+2}{j} P_j'(x)$

By multiplying by $\frac{1}{n+2}$ and changing the index in the summation (that is, by taking out the case $j=0$ from the summation).

(3) $(x+1)^{n+1} - \frac{1}{n+2}P_0'(x) - \sum\limits_{j=1}^n \binom{n+2}{j} P_j'(x) = (x+1)^{n+1} - \frac{1}{n+2}P_0'(x) - \sum\limits_{j=1}^n \binom{n+2}{j} j P_{j-1}(x) + b_j$

By the complete induction hypothesis.

(4) $(x+1)^{n+1} - \frac{1}{n+2}P_0'(x) - \sum\limits_{j=1}^n \binom{n+2}{j} (j P_{j-1}(x) + b_j = (x+1)^{n+1} - \frac{1}{n+2}\sum\limits_{j=1}^n \binom{n+2}{j} j P_{j-1}(x) + b_j$

Here's one step that I didn't understand. The derivative $P_0'(x) = 1$, so how did he cancel $\frac{1}{n+2}P_0'(x)$ from the equation?

(5) $(x+1)^{n+1} - \sum\limits_{j=1}^n \binom{n+2}{j} j P_{j-1}(x) + b_j = (x+1)^{n+1} - \sum\limits_{j=1}^{n} \binom{n+1}{j-1}P_{j-1} + b_j$

Using the identity $\binom{n+2}{j}j = (n+2)\binom{n+1}{j-1}$.

(6) $(x+1)^{n+1} - \sum\limits_{j=1}^{n} \binom{n+1}{j-1}P_{j-1} + b_j = (x+1)^{n+1} - \sum\limits_{j=0}^{n-1} \binom{n+1}{j}P_{j} + b_j$

Changing the index in the summation.

(7) $(x+1)^{n+1} - \sum\limits_{j=0}^{n-1} \binom{n+1}{j}P_{j} + b_j = (x+1)^{n+1} - 1 - \sum\limits_{j=0}^{n-1} \binom{n+1}{j}P_{j} + b_j $

I also didn't get this step. Where did that $(-1)$ came from?

(8) $(x+1)^{n+1} - 1 - \sum\limits_{j=0}^{n-1} \binom{n+1}{j}P_{j} + b_j = (n+1) P_n(x) + b_n$

Using (A) above.

So, as I explained, my questions are basically pertaining to steps (4) and (7) above. Any help?

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  • $\begingroup$ Bernoulli numbers are usually denoted by $B_{n}$/ $\endgroup$ – Aleksandar Jul 13 '15 at 22:52
  • $\begingroup$ @Aleksandar - Thanks for the notation tip. I'm using the notation from Smoryński's book. $\endgroup$ – Nagase Jul 13 '15 at 22:54
  • $\begingroup$ Okay. Been there my book was translated in English from German and rather than $\epsilon$ and $\delta$ they used $v$ and $\mu$. $\endgroup$ – Aleksandar Jul 13 '15 at 22:56
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You have a trivial identity

$$P_n(x)-P_n(x-1)=x^n,$$ which together with $P_n(0)$ uniquely determines $P_n$. Now, deriving the above identity yields

$$P'_n(x)-P'_n(x-1)=n x^{n-1},$$ which determines $P'_n(x)-P'_n(0)$ uniquely, i.e. shows by the above that

$$P'_n(x)-P'_n(0)=nP_{n-1}(x).$$ I would say the obtention of the Bernoulli numbers through this identity follows from the definition of the Bernoulli polynomials from a power series in two variables. Have you looked into this? I'd come back when I have more time, but I'd say that it's not that difficult, given the above.

I.e. take the above and divide by, say, $z^n/n!$, take the sum over $n\geq 0$ and multiply or divide by by $e^{xz}$ (sthg like that). My $n$ is your $n+1$, but the argument works just the same :)

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  • $\begingroup$ That's interesting. Isn't it missing the constant, though? And I'd still like to know the answers to my specific questions regarding steps (4) and (7) of Smoryński's proof. $\endgroup$ – Nagase Jul 13 '15 at 23:01
  • $\begingroup$ It is nice to upvote other people's contributions, and it encourages further assistance. $\endgroup$ – Theon Alexander Jul 18 '15 at 21:33

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