Servois' 1817 "Memoir on Quadratures" – Challenge Problems for Numerical Analysis Students

Servois' “Quadratures” paper also contains several gems that could be incorporated as historically-based projects within a numerical analysis course.

The complete English translation of Servois' paper itself can be found here (pdf).

We recall that the equation numbers denoted by Roman numerals below refer back to equations that appeared in earlier sections of this article, in order to distinguish them from the original equation numbers used by Servois in his memoir. There is no correlation between these Roman numerals and his Hindu Arabic numerals.

1. Servois derived the Composite Trapezoidal Rule early in his memoir, in his equation (9). It appears again in a more recognizable form, a few lines later in the displayed equation shortly after equation (10): $$T = \frac{1}{2}\omega\left(y + E^{-1}y\right) + \frac{1}{2}\omega\left(E^{-1}y + E^{-2}y\right) + \ldots+ \frac{1}{2}\omega\left(Ev + v\right).$$ Bearing in mind Servois' notation ($x_k = E^k(a)$, $v=f(a)$, $y=f(x)$), show that this translates into the Composite Trapezoidal Rule as expressed in our equation (I).
2. Similarly, in his equation (12), Servois set $$R = \omega E^{-\frac{1}{2}}y + \omega E^{-\frac{3}{2}}y + \ldots + \omega E^{\frac{1}{2}}v.$$ Show that this is an expression of the Composite Midpoint Rule.
3. In his equation (43), Servois gave the Lagrange Interpolating Polynomial, or at least a special case of it. Show that if $x=u$ and $x_k=k$ for $k=0, 1, \ldots , n$, then the Lagrange Interpolating Polynomial (II) reduces to Servois' formula (43).
4. Servois used the notation $B_n=|b_{2n}|$ for the absolute value of the $2n$th Bernoulli number. Show that making this substitution in equation (V) gives the expression (VI). Show further that the substitution $x=\frac{\omega}{2}$ in equation (VI) gives equation (7) in Servois' "Memoir on Quadratures": $$1 - \frac{1}{2}\omega \cot\left(\frac{1}{2} \omega\right) = \frac{\omega^2 B_1}{2!} + \frac{\omega^4 B_2}{4!} + \frac{\omega^6 B_3}{6!} + \cdots .$$
5. Perform the calculations that give the values of $A, B, C, \ldots$ in equation (VIII).
6. Equation (10) in Servois' "Memoir on Quadratures" can be written in modern notation as $$\begin{align*} \int_{x_0}^{x_n} F(x) dx = & \omega\left(\sum_{k=0}^{n-1} F(x_k) + \frac{1}{2} F(x_n) -\frac{1}{2} F(x_0) \right) \\ & - \frac{\omega^2 B_1}{2!}\left(F'(x_n) - F'(x_0)\right) + \frac{\omega^4 B_2}{4!}\left(F^{(3)}(x_n) - F^{(3)}(x_0)\right)\\ & - \frac{\omega^6 B_3}{6!}\left(F^{(5)}(x_n) - F^{(5)}(x_0)\right) + \cdots. \end{align*}$$ To see that this is equivalent to the Euler-MacLaurin Formula (X), we first note that Servois uses $B_n = |b_{2n}|$, which accounts for the alternating signs of the terms involving the Bernoulli numbers. <br>More substantively, we let $g(t) = F(a + \omega t)$ and $x=x_0 + \omega t$. Using this notation, show that $$\int_0^n g(t)\, dt = \frac{1}{\omega} \int_{x_0}^{x_n} F(x)\, dx \quad \mbox{and} \quad  g^{(k)}(t) = \omega^k F^{(k)}(x).$$