Need bound for absolute value of complex-valued special functions (Taylor coefficients of Faddeeva's w(z))

Question

To guarantee accuracy for code [1] that computes Faddeeva's w(z) [2] using Taylor expansions around different centers, I would need upper bounds for the absolute values $|w_n(z)|$ of the coefficients

$w_n(z) := \frac{w^{(n)}(z)}{n!}$

in the first quadrant for $|z|<7$. A bound for $|w_n/w_{n-1}|$ for $n$ larger than some $N$ would be particular helpful.

The following is known about the $w_n$:

(1) With $w_0(z) = w(z)$ and $w_{-1}(z) = -i/\sqrt{\pi}$, the recursion

$w_n = -\frac{2}{n}\left( z w_{n-1} + w_{n-2} \right)$

holds for $n=1,2,3,\ldots$

(2) For $\text{Im} z>0$,

$w_n(z) = (-1)^n \frac{i}{\pi} \int_{-\infty}^{+\infty}dt\frac{e^{-t^2}}{(z-t)^n}$

(3) With the parabolic cylinder function $U$ [3],

$w_n(z) = \frac{i^n\sqrt{2}^{n+1}}{\sqrt{\pi}}\,e^{-z^2/2}\,U\left(n+\frac{1}{2}, -i\sqrt{2}z\right).$

(4) From numerical tests (always for $\text{Re} z \ge0$, $\text{Im} z \ge0$, $|z|<7$) it seems pretty certain, but is not proven:

• $\text{max}_z |w_0(z)| = 1$, reached at $z=0$.
• $|w_1(z)|$ may be larger than $|w_0(z)|$, but for $n>1$ the $|w_n(z)|$ are strictly decreasing functions of $n$.

Short of a full answer, I would also be grateful for hints which road is most promising.

References:

• [1] https://jugit.fz-juelich.de/mlz/libcerf
• [2] https://dlmf.nist.gov/7.2.E3
• [3] https://dlmf.nist.gov/12

Answer 1

For $n=1,2,\dots$, using the inequalities $$|w_n|\le\frac{2}{n}( |z| |w_{n-1}| + |w_{n-2}| )$$ and $$r_n:=\frac{\Gamma(n/2+1)}{n \Gamma(n/2+1/2)}\le r_1=\frac{\sqrt\pi}{2},$$ by induction on $n$ we see that $$|w_n|\le W_n:=c\frac{a^n}{\Gamma(n/2+1)},$$ where $$a:=\frac{|z|\sqrt\pi+\sqrt{|z|^2\pi+4}}{2},\quad c:=\max\Big(|w_0|,\frac{\sqrt\pi\,|w_1|}{2a}\Big).$$

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Indeed, the inequality $r_n\le r_1$ holds for $n=1,2,\dots$ because $r_2<r_1$ and $r_{n+2}=\frac n{n+1}\,r_n<r_n$ for $n=1,2,\dots$.

Next, it is easy to see that $|w_n|\le W_n$ for $n=0,1$. Suppose now that $|w_k|\le W_k$ for $n=1,2,\dots$ and $k=0,\dots,n-1$. Then for $n=1,2,\dots$ we have $$\begin{aligned} |w_n|&\le\frac{2}{n}( |z| |w_{n-1}| + |w_{n-2}| ) \\ &\le\frac{2}{n}( |z| W_{n-1} + W_{n-2} ) \\ &=\Big(2\frac{|z|}a\,r_n+\frac1{a^2}\Big)W_n \\ &\le\Big(2\frac{|z|}a\,r_1+\frac1{a^2}\Big)W_n =W_n, \end{aligned}$$ so that $|w_n|\le W_n$. $\quad\Box$

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The upper bound $W_n$ on $|w_n|$ converges fast down to $0$ for large $n$.