Math 2.0

Numerical analysis Definition

Euler–Boole summation

Builds on Smooth function

Definition

The Euler polynomials Ek(x)E_k(x), for k≥0k \ge 0, are defined by the generating function 2extet+1=∑k≥0Ek(x)tkk!.\frac{2 e^{xt}}{e^t + 1} = \sum_{k \ge 0} E_k(x) \frac{t^k}{k!}. The periodic Euler functions E~k\widetilde{E}_k are the functions on ℝ\R with E~k(x)=Ek(x)\widetilde{E}_k(x) = E_k(x) for 0≤x<10 \le x < 1 and E~k(x+1)=−E~k(x)\widetilde{E}_k(x + 1) = -\widetilde{E}_k(x) for all xx.

Let m≥1m \ge 1 and n≥1n \ge 1 be integers, let 0≤h≤10 \le h \le 1, and let ff be a smooth function on an open interval that contains [0,n][0, n]. Euler–Boole summation is the identity ∑j=0n−1(−1)jf(j+h)=12∑k=0m−1Ek(h)k!((−1)n−1f(k)(n)+f(k)(0))+12(m−1)!∫0nf(m)(x) E~m−1(h−x) dx,\sum_{j=0}^{n-1} (-1)^j f(j + h) = \frac12 \sum_{k=0}^{m-1} \frac{E_k(h)}{k!} \left( (-1)^{n-1} f^{(k)}(n) + f^{(k)}(0) \right) + \frac{1}{2 (m-1)!} \int_0^n f^{(m)}(x) \, \widetilde{E}_{m-1}(h - x) \, dx, and the finite sum over kk is the Euler–Boole expansion of the alternating sum to order mm.

Remarks

The first Euler polynomials are E0(x)=1E_0(x) = 1 and E1(x)=x−12E_1(x) = x - \frac12. They satisfy Ek(x+1)+Ek(x)=2xkE_k(x + 1) + E_k(x) = 2x^k, which is the source of the formula.

Example: take f(x)=xf(x) = x, h=0h = 0, n=2n = 2 and m=2m = 2. The left side is 0−1=−10 - 1 = -1. The term k=0k = 0 gives 12(−f(2)+f(0))=−1\frac12 (-f(2) + f(0)) = -1, the term k=1k = 1 gives 12⋅(−12)⋅(−f′(2)+f′(0))=0\frac12 \cdot (-\frac12) \cdot (-f'(2) + f'(0)) = 0, and the integral vanishes because f′′=0f'' = 0. Both sides equal −1-1.

Non-example: the identity is for alternating sums. For the plain sum ∑j=0n−1f(j)\sum_{j=0}^{n-1} f(j) the right side is wrong in general, and the Euler–Maclaurin formula applies instead.

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  11. Euler–Boole summation definition

Further reading

  1. Web https://en.wikipedia.org/wiki/Euler–Boole_summation
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MSC65B15, 11B68, 40A25
DefinesEuler–Boole summation
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Changed2 November 2026

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