Stable special function recurrences and certified evaluation — Examples

For $a=b=1,c=2$, the Pochhammer products give $(1{)}_k=k!$ and $(2{)}_k=(k+1)!$, so each coefficient is $1\mathrm{/}(k+1)$. For $\mathrm{\mid }z\mathrm{\mid }<1$, the geometric series converges uniformly along the segment $sz$ ($0\le s\le 1$). Parameterizing the path and integrating the real and imaginary component power series termwise by [F4] gives ${\sum }_{k=0}^{\mathrm{\infty }}\frac{z^{k+1}}{k+1}={\int }_0^1\frac{z}{1-sz}ds=-\mathrm{Log}(1-z),$ where [F2], [F3], and [F4] justify the path integral identity since $\mathrm{Re}(1-sz)>0$ along the segment. Dividing by $z$ gives the formula, and the series gives value $1$ at $z=0$.

The principal logarithm is holomorphic when $1-z$ avoids the nonpositive real axis, that is, when $z\mathrm{\notin }[1,\mathrm{\infty })$. The quotient has a removable value at $z=0$ by step 1.1, so it defines the stated continuation domain, which contains $-2$.

Since $\mathrm{\mid }-2\mathrm{\mid }>1$, the defined Gauss-series branch fails its domain check and [F5] returns domain; this does not contradict the continued formula in step 2.1, which is a separately declared branch.

By [F1], the exact output is $\mathrm{erfc}(x)$. Subtracting the perturbed inputs gives $\widehat{1}-\stackrel{}{\mathrm{erf}}=\mathrm{erfc}(x)+{\delta }_1-{\delta }_2\mathrm{.}$ Since [F2] and [F3] give $\mathrm{erfc}(x)=2{\pi }^{-1\mathrm{/}2}{\int }_x^{\mathrm{\infty }}{e}^{-t^2}dt>0$, division by the exact value and the triangle inequality yield the stated relative-error bound.

The same tail integral tends to zero as $x\to \mathrm{\infty }$, because the nonnegative Gaussian is integrable by [F3]. If $\mathrm{\mid }{\delta }_1\mathrm{\mid },\mathrm{\mid }{\delta }_2\mathrm{\mid }\le \eta$ with fixed $\eta >0$, the choice ${\delta }_1=\eta$, ${\delta }_2=-\eta$ attains error $2\eta$, so the relative error is $2\eta \mathrm{/}\mathrm{erfc}(x)$ and is unbounded. By [F1], multiplying containing intervals for $\mathrm{erfcx}(x)$ and $e^{-x^2}$ with outward rounding encloses $\mathrm{erfc}(x)$ and avoids the subtraction. For exact dyadic $x>0$, [F4] and [F5] supply those factors whenever their finite evaluations succeed; this asserts no finite-cap success for arbitrary real $x$.

The positive term ratio is $\frac{a_{k+1}(x)}{a_k(x)}=\frac{k+1\mathrm{/}2}{x^2}\mathrm{.}$ For $0\le k<N$, this ratio is at most $(N-1\mathrm{/}2)\mathrm{/}{x}^2\le 1$, so the terms do not increase up to $a_N$. For $k\ge N$, the ratio is at least $(N+1\mathrm{/}2)\mathrm{/}{x}^2\ge 1$, so terms do not decrease after $a_N$. Hence $a_N$ is a least term.

The upper bound in [F1] equals $(1\mathrm{/}2{)}_N\mathrm{/}(x^{2N+1}\sqrt{\pi })=a_N(x)$, and its sign is fixed by $(-1{)}^N$. Thus the first omitted least-sized term also bounds the signed asymptotic remainder. At $x=3$, $N=9$ satisfies $8.5\le 9\le 9.5$; the same finite certificate applies.

Let $a_k=(1\mathrm{/}2{)}^{2k+n}\mathrm{/}(k!(k+n)!)$. By [F2], the positive terms $a_k$ decrease strictly to zero. Grouping the alternating series in consecutive pairs gives $J_n(1)=(a_0-a_1)+(a_2-a_3)+\cdots >0$. Also $(k+n)!\ge n!$, so $0<J_n(1)\le {\sum }_{k\ge 0}{a}_k\le \frac{2^{-n}}{n!}{\sum }_{k\ge 0}\frac{(1\mathrm{/}4{)}^k}{k!}=\frac{2^{-n}{e}^{1\mathrm{/}4}}{n!}⟶0.$ In particular, $J_0(1)\ge 1-1\mathrm{/}4=3\mathrm{/}4$ and $J_1(1)\ge 1\mathrm{/}2-1\mathrm{/}16=7\mathrm{/}16$.

By [F4], $D_0(1)=-J_1(1)<0$ and $D_1(1)=J_0(1)>0$; hence $D_2=2D_1-D_0>D_1>0$. If $n\ge 2$ and $D_n\ge D_{n-1}>0$, then $D_{n+1}=2n{D}_n-D_{n-1}\ge (2n-1)D_n\ge 3D_n$. Induction gives $D_n\ge 3^{n-2}{D}_2$ for $n\ge 2$, so $D_n\to \mathrm{\infty }$.

The exact sequence $J_n-\epsilon D_n$ obeys the same linear recurrence by [F3] and [F4], so it is the forward-recurrence solution from the displayed perturbed seeds. Since $J_n\to 0$ and $D_n\to \mathrm{\infty }$ by steps 1.1 and 2.1, some finite $n$ satisfies $\epsilon D_n>J_n$. At that index ${\tilde{J}}_n<0<J_n$. The seed differences have magnitudes $\epsilon \mathrm{\mid }{D}_0\mathrm{\mid }$ and $\epsilon \mathrm{\mid }{D}_1\mathrm{\mid }$, which tend to zero with $\epsilon$.

At $x=\xi$, every enclosure containing the exact value also contains $0$. Therefore the zero-exclusion condition in [F3] fails, while [F4] still permits an absolute-radius success enclosure.

For any $\eta >0$, continuity in [F2] at the zero [F1] gives a neighborhood of $\xi$ on which $\mathrm{\mid }\mathrm{Ai}(x)\mathrm{\mid }<\eta$. Whenever a relative certificate succeeds for an enclosure $B$ of such a value, [F3] implies $\mathrm{rad}(B)\le {\epsilon }_{r}d(B,0)\le {\epsilon }_r\mathrm{\mid }\mathrm{Ai}(x)\mathrm{\mid }<{\epsilon }_r\eta$. Thus the required absolute radius tends to zero as the input approaches $\xi$.

Let $h(t)=t-\sin t$. By [F2], $h(0)=0$ and $h^{\mathrm{\prime }}(t)=1-\cos t\ge 0$ for $t\ge 0$ by [F3], so $\sin t\le t$. Again by [F2] and [F4], $1-\cos \theta ={\int }_0^{\theta }\sin tdt\le {\int }_0^{\theta }tdt={\theta }^2\mathrm{/}2(0\le \theta \le 1)\mathrm{.}$

For $x\ge 1$ and $0\le \theta \le 1\mathrm{/}\sqrt{x}$, step 1.1 gives $x\cos \theta \ge x-1\mathrm{/}2$. Restricting the integral in [F1] to this interval therefore yields $I_0(x)\ge \frac{e^{x-1\mathrm{/}2}}{\pi \sqrt{x}}\mathrm{.}$ The exponential series gives $e^x\ge x^2\mathrm{/}2$ for $x\ge 0$, so this lower bound is at least $e^{-1\mathrm{/}2}{x}^{3\mathrm{/}2}\mathrm{/}(2\pi )$ and tends to infinity.

The scaled range in [F1] holds for every $x\ge 0$, while step 2.1 shows the unscaled value exceeds any fixed finite range for sufficiently large $x$. The scaled integral method of Composite midpoint enclosure for the scaled modified Bessel integral evaluates the bounded scaled quantity directly.