The Euler–Maclaurin formula states an interdependency between

\begin{align} I\quad:=&\quad\int_m^nf(x) \, dx;\ m,n\in\mathbb{Z}\\[6pt] S\quad:=&\quad\sum_{k=m}^n f(k) \\[6pt] D\quad:=&\quad\left\lbrace \frac{d^j}{dx^j} f(m) \right\rbrace \cup \left\lbrace \frac{d^j}{dx^j} f(n) \right\rbrace \end{align}

The Euler–Maclaurin formula is "traditionally" used to estimate the value of $S$ from $I$ and $D$ and also to estimate $I$ from $S$ and $D$


are there examples where the Euler–Maclaurin formula has been beneficially used to fix the the values of the derivatives of a function in the endpoints of its integration interval in order to fix the values of the elements of $D$ via $\lbrace I,\left(x_i,y_i\right)_{i=m}^n \rbrace$?

To be specific: have there been attempts to determine the values of the elements of $D$ via the Euler–Maclaurin formula to find "ideal" derivatives for use in clamped spline-interpolation, i.e. where the disambiguation of the interpolation is done via fixing derivatives at the ends of the interpolation interval by plugging the (estimated) value of $I$ and/or the coordinate values of the $\left(x_i,f(x_i)\right)$ into the Euler–Maclaurin formula?

  • Why do you format this as $$\begin{align} I\quad:=&&\int_m^nf(x) \, dx;\ m,n\in\mathbb{Z}\\ \\ S\quad:=&&\sum_{k=m}^n f(k) \\ \\ D\quad:=&&\left\lbrace \frac{d^j}{dx^j} f(m) \right\rbrace \cup \left\lbrace \frac{d^j}{dx^j} f(n) \right\rbrace \end{align}$$ rather than as seen below? $$\begin{align} I\quad:=& \quad \int_m^nf(x) \, dx;\ m,n\in\mathbb{Z}\\ \\ S\quad:=& \quad \sum_{k=m}^n f(k) \\ \\ D\quad:=&\quad \left\lbrace \frac{d^j}{dx^j} f(m) \right\rbrace \cup \left\lbrace \frac{d^j}{dx^j} f(n) \right\rbrace \end{align}$$ – Michael Hardy Oct 11 at 18:32
  • @MichaelHardy unfortunately I don't have the chance to do frequent mathematical type setting and thus lack some expertise in that respect. I will adopt your suggested formatting, which looks much nicer. – Manfred Weis Oct 12 at 5:07

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