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I'm going through the proof from the book 'Bernoulli Numbers and Zeta Functions' and there's a step that I don't understand. My best guess at the minute is Taylor series. (It's the line highlighted in yellow in the image) Thanks

Similarly, taking the derivatives successively, we obtain $$ S_{k}^{(j)}(0)=k(k-1) \cdots(k-j+2) b_{k-j+1} \quad(2 \leq j \leq k+1) $$ Finally, we have $$ \begin{aligned} \color{red}{S_{k}(x)} &=\sum_{j=0}^{k+1} \frac{S_{k}^{(j)}(0)}{j !} \color{red}{x}^{j} \\ &=\sum_{j=1}^{k+1} \frac{1}{k+1}\left(\begin{array}{c} k+1 \\ j \end{array}\right) b_{k-j+1} x^{j} \quad\left(S_{k}^{(0)}(0)=0\right) \\ &=\frac{1}{k+1} \sum_{j=0}^{k}\left(\begin{array}{c} k+1 \\ j \end{array}\right) b_{j} x^{k+1-j} \end{aligned} $$

Wolgwang
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Jacamo
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  • It appears a Taylor expansion was used of sorts, but you haven't shown us what $S_k(x)$ is. Another approach would be to apply the Euler-Maclaurin summation formula, which also works since the $n$th derivative eventually becomes identically zero. – Simply Beautiful Art Mar 10 '17 at 01:35
  • S(k,x) = sum from i=1 to x of (i^k) – Jacamo Mar 10 '17 at 01:44
  • Well, you have a problem then. You can't differentiate with respect to $x$. – Simply Beautiful Art Mar 10 '17 at 01:46
  • Sure you can. $S_1(x) = x(x+1)/2$, $S_2(x) = x(x+1)(2x+1)/6$, $S_3(x) = x^2(x+1)^2/4$, etc. For each natural value of $k$, the sum can be simplified to a closed form which is a polynomial in $x$, and you can differentiate the polynomial with respect to $x$. The formula in question is just the usual Taylor series expansion of the polynomial centered at $0$ – Barry Smith Mar 10 '17 at 02:09
  • @BarrySmith But its not directly differentiable from the definition of $S$. Note that$$S_1(x)=\frac{(1+\sin(2\pi x))x(x+1)}2$$is also valid, according to the original definition. – Simply Beautiful Art Mar 10 '17 at 02:12
  • I'm using the number-theoretic context here. Presumably, there's a comment in the omitted part of the text that these are to be interpreted as polynomials in $x$. – Barry Smith Mar 10 '17 at 02:16
  • Yes, earlier on in the book he proves that S is a polynomial in x with leading term of degree (k+1) (my apologies) – Jacamo Mar 10 '17 at 18:12

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I suppose $S_k(n)=\sum_{i=1}^n i^k$, which is a polynomial in $n$ of degree $k+1$. This is just Taylor's formula applied to this polynomial (remeber Taylor's formula at order $k+1$ is a exact formula for polynomials of degree $\le k+1$.

Bernard
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  • Yes, but how does one effectively define this for non-integer $n$ for differentiation purposes? – Simply Beautiful Art Mar 10 '17 at 01:53
  • The differentiation of the S function is with respect to i rather than x so S'k(x)= sum of (ki^(k-1)) for instance. – Jacamo Mar 10 '17 at 01:56
  • @Jacamo That can't possibly make sense. $i$ is fixed, and set to be $i={0,1,2,\dots,n}$ – Simply Beautiful Art Mar 10 '17 at 01:57
  • @Simply Beautiful Art: You replace $n$ with an indeterminate and apply formal differentiation of polynomials. – Bernard Mar 10 '17 at 01:58
  • @Bernard Could you expand on that? I'd really like to see how that'd work out fully. – Simply Beautiful Art Mar 10 '17 at 02:00
  • It's quite simple. For instance, since $S_3(n)=\frac14n2(n+1)^2$, you set $S_3(X)=\frac14X^2(X+1)^2$, and have $S'_3(X)=\frac12X(X+1)^2+\frac12X^2(X+1)$, $S''_3(X)=\frac12((X+1)^2+4X(X+1)+X^2$, $S'''(X)=3(2X+1)$, $S^{(4)}(X)=6$, so $S(0)=S'(0)=0$, $S''(0)=\frac12$, $S'''(0)=3$. – Bernard Mar 10 '17 at 02:10