The Euler-Maclaurin summation formula is \begin{eqnarray} \sum_{k = a}^{b} f(k) = \int_{a}^{b} f(t) \, dt + B_1 (f(a) + f(b)) + \sum_{n = 1}^{N} \frac{B_{2n}}{(2n)!} ( f^{(2n-1)}(b) - f^{(2n-1)}(a) ) + R_{N}, \end{eqnarray} where $B_{n}$ is the $n^{\text{th}}$-Bernoulli number taking $B_{1} = \tfrac{1}{2}$, and the remainder term is bounded by the following \begin{align} |R_{N}| \leq \frac{|B_{2N} |}{(2n)!} \int_{a}^{b} | f^{(2N)}(t) | \, dt. \end{align} for any arbitrary positive integer $N$. Is there a similar formula for nested sums of the form, \begin{eqnarray} \sum_{k_1 = a_1}^{b_1} \cdots \sum_{k_n = a_n}^{b_n} f(k_1, \dots, k_n). \end{eqnarray}



Yes! There's a whole chapter about it in this book.

  • 1
    $\begingroup$ The formula there (say in Theorem 2 there) is stated only for the case when $f$ is an exponential function. In fact, the Todd operator (in terms of which the formula there is stated), seems to be only defined on functions $f$ of the form $f(z)=\sum_{n=0}^\infty c_n^n z^n/n!$ with bounded $c_n$'s -- so that such functions are somewhat like the exponential functions.Strangely enough, the domain of the operator is not specified there. $\endgroup$ – Iosif Pinelis May 15 '17 at 4:27
  • $\begingroup$ I meant Theorem 10.2 there in the book. $\endgroup$ – Iosif Pinelis May 15 '17 at 14:52

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.