Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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}

Thanks!

share|improve this question
add comment

1 Answer

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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