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A positive midpoint enclosure for Boys functions on bounded arguments

Statement

For $m \in N_{0}$ , $T \geq 0$ , and integer $N \geq 1$ , let $h = 1 / N$ , $t_{j} = (j - \frac{1}{2}) h$ for $1 \leq j \leq N$ , and $Q_{m, N} (T) = h \sum_{j = 1}^{N} t_{j}^{2 m} e^{- T t_{j}^{2}} .$ Then all quadrature weights and summands are positive and $∣ F_{m} (T) - Q_{m, N} (T) ∣ \leq \frac{m + T}{2 N} .$ Consequently, for $0 \leq T \leq T_{\max}$ , the same positive rule has the uniform enclosure error $(m + T_{\max}) / (2 N)$ .

Facts & Assumptions

Given: $F_{m} (T)$ as in Boys functions and lower incomplete-gamma notation, $m \geq 0$ , $T \geq 0$ , and $N \geq 1$ .

[F1]

Boys functions and lower incomplete-gamma notation: $F_{m} (T) = \int_{0}^{1} f (t) d t$ with $f (t) = t^{2 m} e^{- T t^{2}}$ .

[F2]

Mean value theorem on a closed real interval: a differentiable function whose derivative has absolute value at most $L$ varies by at most $L$ times the distance between two points.

[L9]

The exponential function is smooth and $(\exp)^{'} = \exp$ and Sum, product, quotient, and chain rules from difference quotients: the exponential is smooth with derivative itself; finite sums, products and polynomial compositions obey the product, power and chain rules. Applying these rules one coordinate or parameter at a time justifies every displayed finite Gaussian derivative.

Proof

1.1givenF1F2algebraL9

For $m = 0$ , $f^{'} (t) = - 2 T t e^{- T t^{2}}$ , so $∣ f^{'} (t) ∣ \leq 2 T$ . For $m \geq 1$ , differentiation gives $f^{'} (t) = 2 m t^{2 m - 1} e^{- T t^{2}} - 2 T t^{2 m + 1} e^{- T t^{2}}$ ; since $0 \leq t \leq 1$ and $e^{- T t^{2}} \leq 1$ , $∣ f^{'} (t) ∣ \leq 2 m + 2 T$ . Thus the Lipschitz constant $L = 2 m + 2 T$ is valid in both cases by [F2].

2.1givenstep 1.1algebra

On each subinterval of length $h$ , the mean-value bound gives $∣ f (t) - f (t_{j}) ∣ \leq L ∣ t - t_{j} ∣$ . Integrating over that subinterval yields error at most $L h^{2} / 4$ , because the integral of distance from its midpoint is $h^{2} / 4$ . Summing $N$ subintervals gives $L / (4 N) = (m + T) / (2 N)$ . All midpoints and weights are positive and $f (t_{j}) > 0$ .

3.1givenstep 2.1algebra∎

If $T \leq T_{\max}$ , then $(m + T) / (2 N) \leq (m + T_{\max}) / (2 N)$ ; hence the stated uniform bound follows without changing the positive weights or nodes.

Depends on

Used by

Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Trygve Helgaker, Molecular Integral Evaluation, 10th Sostrup Summer School lecture notes (2008) (standard reference, not scraped)
L. E. McMurchie and E. R. Davidson, One and Two Electron Integrals over Cartesian Gaussian Functions, UCRL-13791 (1976) (standard reference, not scraped)