Gaussian basis integral recurrences and boys error — Examples

There are $2^4=16$ primitive entries, each perturbed by $\epsilon$, and each product of four coefficients is $K^4$. Thus $\mathrm{\Delta }{G}_{1111}^{\mathrm{\prime }}=16K^4\epsilon$ and $\mathrm{\Delta }E=\frac{1}{2}\mathrm{\Delta }{G}_{1111}^{\mathrm{\prime }}=8K^4\epsilon$. The perturbed energy is $\frac{1}{2}\epsilon (C_1+C_2{)}^4$, so each gradient component is $2\epsilon (C_1+C_2{)}^3=16K^3\epsilon$; its two-entry Frobenius norm is $16\sqrt{2}{K}^3\epsilon$.

Here $B=\mathrm{\parallel }C{\mathrm{\parallel }}_F=\sqrt{2}K$ and $M=\mathrm{\parallel }D{\mathrm{\parallel }}_F=1$. Substitution into [F2] gives energy bound $\frac{1}{2}(1{)}^2(1{)}^2(2{)}^2(\sqrt{2}K{)}^4\epsilon =8K^4\epsilon$ and gradient bound $2(2{)}^2(1{)}^2(\sqrt{2}K{)}^3(1{)}^2\epsilon =16\sqrt{2}{K}^3\epsilon$, equal to the errors in step 1.1.

By [F1]–[F2], the unnormalized overlap is $(\pi \mathrm{/}(a+b){)}^{3\mathrm{/}2}$ and each norm squared is $(\pi \mathrm{/}(2a){)}^{3\mathrm{/}2}$ or $(\pi \mathrm{/}(2b){)}^{3\mathrm{/}2}$. Multiplying by the normalization constants gives $S(h_a,h_b)=(2\sqrt{ab}\mathrm{/}(a+b){)}^{3\mathrm{/}2}$.

The Gram matrix is $\left(\begin{array}{cc}1& S\\ S& 1\end{array}\right)$, whose characteristic polynomial is $(1-\lambda {)}^2-S^2$, so its eigenvalues are $1\pm S$. At $a=b$, $S=1$ and the lower eigenvalue is zero. For $a\mathrm{\ne }b$, $2\sqrt{ab}<a+b$, hence $0<S<1$; continuity of the displayed formula gives $S\to 1$ as $b\to a$, and therefore $(1+S)\mathrm{/}(1-S)\to \mathrm{\infty }$.

For $A=B=0$, [F5] gives $g_a{g}_b=e^{-p\mathrm{\parallel }x{\mathrm{\parallel }}^2}$, so [F1] evaluates the overlap as $(\pi \mathrm{/}p{)}^{3\mathrm{/}2}$. For kinetic energy, substitute [F2] and use both moments in [F1]: $3b(\pi \mathrm{/}p{)}^{3\mathrm{/}2}-2b^2\frac{3}{2p}(\pi \mathrm{/}p{)}^{3\mathrm{/}2}=\frac{3ab{\pi }^{3\mathrm{/}2}}{p^{5\mathrm{/}2}}\mathrm{.}$

In [F4], set $C=P=0$; then the Boys argument is zero and [F3] gives $V_0=2\pi \mathrm{/}p$. For the two-electron integral, both pair centers are zero and their only Hermite coefficient is $1$, so [F4] with $F_0(0)=1$ gives $2{\pi }^{5\mathrm{/}2}\mathrm{/}(pq\sqrt{p+q})$.

The exact identity in [F1] divides a difference of two terms. By [F1] and [F3], $F_0(T)=1-T\mathrm{/}3+R_F$ with $\mathrm{\mid }{R}_F\mathrm{\mid }\le T^2\mathrm{/}10$. By [F2], $e^{-T}=1-T+R_e$ with $\mathrm{\mid }{R}_e\mathrm{\mid }\le T^2$ for $T\le 1$: for $k\ge 2$, $k!\ge 2^{k-1}$, so ${\sum }_{k\ge 2}1\mathrm{/}k!\le {\sum }_{k\ge 2}{2}^{1-k}=1$. Therefore $\mid (F_0(T)-e^{-T})-\frac{2T}{3}\mid \le \frac{11}{10}{T}^2,$ so for $T\le 1\mathrm{/}4$ the positive difference is at least $T\mathrm{/}4$. The two terms being subtracted each tend to one as $T\downarrow 0$.

Applying [F4] with $m=0$ gives a factor $1\mathrm{/}(2T)$ on both seed radii, which diverges as $T\downarrow 0$ for fixed nonzero radii. The stated stored-radius formula follows only for a full affine image with one finite conversion per endpoint; a primitive subtraction error must also be divided by $2T$ before adding the final division error. In contrast, [F3] gives a direct Taylor enclosure with error $T^2\mathrm{/}10$. The algorithm [F5] can therefore select the direct small-argument branch and return only a certified interval; the example makes no claim that every requested tolerance succeeds under every finite cap.

The first bound is $0.4\tau \le \tau$, and the cumulative sum after two is $0.8\tau \le \tau$, so [F3] permits both skips. The selected policy therefore skips both and, with the assumed exact accumulation, records $0.8\tau$.

Adding the third gives $0.8\tau +0.4\tau =\frac{6}{5}\tau >\tau$. By [F3], the screening guard therefore rejects skipping that third block and requires its evaluation or retention. By [F1]–[F2], the reported $0.8\tau$ bounds only the first two omitted contributions and does not assert exact sparsity.