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  • Simple derivation of the Euler-Boole type summation formula and examples of its use
    Lampret, Vito
    Using a simple technique, from the classical Euler-Maclaurin formula is derived the equality ▫$$ \begin{aligned} \sum_{k=1}^n(-1)^{k+1} f(k) =&\sum_{k=1}^{2m-1}(-1)^{k+1} f(k) + ... \frac{1}{2}\Bigl[f(2\lfloor \frac{n+1}{2}\rfloor) - f(2m)\Bigr]\\ &-\sum_{i=1}^{\lfloor p/2\rfloor} \Bigl(4^i-1\Bigr) \frac{B_{2i}}{(2i)!}\Bigl[f^{(2i-1)}(2\lfloor \frac{n+1}{2}\rfloor) - f^{(2i-1)}(2m)\Bigr]\\ &+r_p(m, n), \end{aligned} $$▫ true for integers ▫$m, n, p\ge 1$▫, such that ▫$n\ge 2m+1$▫, and for a function ▫$f\in C^p[2m,n+1]$▫, where the remainder ▫$r_p(m, n)$▫ is estimated as ▫$$ |r_p(m, n)| \le \frac{1+2^{-p}}{3\pi^{p-2}}\int_{2m}^{2\left\lfloor \frac{n+1}{2}\right\rfloor} \left|f^{(p)}(x)\right| \,\mathrm{d}x. $$▫ This formula implies also an infinite alternating summation rule having an integral in the error term only.
    Vir: Mediterranean journal of mathematics. - ISSN 1660-5446 (Vol. 19, iss. 2, Apr. 2022, art. 77 (20 str.))
    Vrsta gradiva - članek, sestavni del ; neleposlovje za odrasle
    Leto - 2022
    Jezik - angleški
    COBISS.SI-ID - 105758211

    Povezava(-e):

    https://link.springer.com/article/10.1007/s00009-022-02000-x
    DOI

vir: Mediterranean journal of mathematics. - ISSN 1660-5446 (Vol. 19, iss. 2, Apr. 2022, art. 77 (20 str.))

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