We give a number of examples of application of the techniques described in the article. We focus on the case *d*_{
n
}(*x*)*e*_{
n
}(*x*) > 0, but examples of application for monotonic systems with *d*_{
n
}(*x*)*e*_{
n
}(*x*) < 0 are also given.

### 4.1 Cases with *d*_{
n
}(*x*)*e*_{
n
}(*x*) > 0

We analyze three families of functions, which have as particular cases some classical orthogonal polynomials outside the interval of orthogonality In all cases Theorem 4 holds, with the exception of Laguerre functions of negative argument. In this case Theorem 4 can not be applied but the characteristic roots are still monotonic and the same analysis is therefore possible. Some monotonicity properties for the determinants of some of the functions analyzed were considered in [15].

#### 4.1.1 Modified bessel functions

These are solutions of *x*^{2}*y*″ + *xy' -* (*x*^{2} + *ν*^{2})*y* = 0. This was the case considered in detail in [1], and most of the results obtained in that paper are direct consequences of the more general results of the present one.

#### 4.1.2 Parabolic cylinder functions

The parabolic cylinder function *U*(*n, x*) is a solution of the differential equation *y*″(*x*) - (*x*^{2}/4 *+ n*)*y*(*x*) = 0, with coefficient *A*(*x*) = - (*x*^{2}/4 *+ n*) depending monotonically on the parameter *n* (Theorem 4 holds).

Considering the DDE satisfied by *U*(*n, x*) [[16], 12.8.2-3] and defining *y*_{
n
}(*x*) = *e*^{iπn}*U*(*n, x*) we have:

\begin{array}{c}\begin{array}{c}{y}_{n}^{\prime}\left(x\right)=\frac{x}{2}{y}_{n}\left(x\right)+{y}_{n-1}\left(x\right),\hfill \end{array}\\ {y}_{n-1}^{\prime}\left(x\right)=-\frac{x}{2}{y}_{n-1}\left(x\right)+\left(n-1/2\right){y}_{n}\left(x\right).\end{array}

(23)

where *n* will be real and positive. For this system

{\eta}_{n}\left(x\right)=-\frac{x}{2\sqrt{n-1/2}},{\stackrel{\u0304}{\eta}}_{n}\left(x\right)={\eta}_{n+1}\left(x\right),{\lambda}_{n}^{\pm}\left(x\right)=\frac{-2}{x\mp \sqrt{4n-2+{x}^{2}}}

(24)

From [[16], 12.9.1] we have *h*_{
n
}(*+∞*) = 0^{-} and {h}_{n}^{\prime}\left(+\infty \right)={0}^{+} and because {\lambda}_{n}^{-}\left(+\infty \right)={0}^{+} then Theorem 3 holds, as well as Theorems 5 and 6. Therefore

**Theorem 9**. *For n* > 1/2 *and x* ≥ 0 *the following holds*

\frac{2}{x+\sqrt{4n+2+{x}^{2}}}<\frac{U\left(n,x\right)}{U\left(n-1,x\right)}<\frac{2}{x+\sqrt{4n-2+{x}^{2}}}

(25)

*The lower bound also holds if n* **∈** (-1/2,1/2) *and it turns to an equality if n =* -1/2.

The lower bound is obtained from the upper bound and the application of the three-term recurrence relation: if *B*_{
m
}(*n, x*) is a positive upper (lower) bound for *U*(*n, x*)/*U*(*n-* 1,*x*),*x* > 0, then

{B}_{m+1}\left(n,x\right)=1/\left(x+\left(n+1/2\right){B}_{m}\left(n+1,x\right)\right)

(26)

is a lower (upper) bound for the same ratio. The process can be continued as *m* → + ∞ and the sequence is convergent (because *U*(*n, x*) is minimal).

Now, consider *y*_{
n
}(*x*) = *U*(*n, -x*), which is also solution of (23). Using the values of *U*(*n*, 0) and *U'*(*n*, 0) [[16], 12.2.6-7] it is easy to prove that *h*_{
n
}(0^{+}) > 0, {h}_{n}^{\prime}\left({0}^{+}\right)>0, *n* > 1/2, *x* ≥ 0 and then {h}_{n}^{\prime}\left({0}^{+}\right)d{\lambda}_{n}^{+}\left({0}^{+}\right)/dx>0 and Theorem 1 holds. The corresponding Perron-Kreuser bounds (Theorems 5 and 7), give:

**Theorem 10**. *For n* > 3/2 *and x* ≥ 0 *the following holds*

\frac{x+\sqrt{4n-6+{x}^{2}}}{2n-1}<\frac{U\left(n,-x\right)}{U\left(n-1,-x\right)}<\frac{x+\sqrt{4n-2+{x}^{2}}}{2n-1}

(27)

*The upper bound is also valid if n* ∈ (1/2, 3/2).

The upper bound in (25) has the same expression as (27) but with *x* replaced by -*x*.

Therefore:

**Remark 4**. *Theorems* 9 *and* 10 *hold for all real x, but for x* < 0 *the lower bound of Theorem* 9 *only holds for all x* < 0 *if n* > 1/2. *The lower bounds are sharper when x* > 0.

The following Turán-type inequalities are obtained from Theorems 9 and 10:

**Theorem 11**. *Let F*(*x*) = *U*(*n, x*)^{2}/(*U*(*n -* 1,*x*)*U*(*n* + 1,*x*)). *Then, for all real x:*

\sqrt{\frac{n-3/2}{n+1/2}}<\frac{n-1/2}{n+1/2}F\left(x\right)<1<F\left(x\right)<\sqrt{\frac{n+3/2}{n-1/2}}

(28)

*The first inequality holds for n* > 3/2 *and the rest for n* > 1/2. *For x* < 0 *the third inequality also holds if n* ∈ (-1/2, 1/2).

Finally, considering Theorem 8 and writing together the results for *U*(*n, x*) and *U*(*n, -x*) we have the next result.

**Theorem 12**. *For all real x and n* ≥ 1/2 *the following holds:*

-\sqrt{{x}^{2}/4+n+1/2}<\frac{{U}^{\prime}\left(n,x\right)}{U\left(n,x\right)}<-\sqrt{{x}^{2}/4+n-1/2}

(29)

*The left inequality also holds for n* > -1/2.

These type of bounds are useful for studying the attainable accuracy of methods for computing the functions. In [17], the following estimation for large *x* and/or *n* was

considered for the condition number with respect to *x:*

{C}_{x}\left(U\left(a,x\right)\right)=\left|x{U}^{\prime}\left(a,x\right)/U\left(a,x\right)\right|~x\sqrt{{x}^{2}/4+a},

(30)

and similarly for *V*(*a, x*). The bounds (29) prove that this a good estimation because it lies between the upper and lower bounds. From the previous discussion on the *V*(*a, x*) function, one can prove that similar bounds are valid for moderate *x* (*x* > 1 is enough); we consider later this function.

Integrating (29) we have

\begin{array}{c}{F}_{n+1/2}\left(x\right)/{F}_{n+1/2}\left(y\right)<\frac{U\left(n,y\right)}{U\left(n,x\right)}<{F}_{n-1/2}\left(x\right)/{F}_{n-1/2}\left(y\right),\\ {F}_{\alpha}\left(x\right)=\text{exp}\left(\frac{x}{2}\sqrt{{x}^{2}/4+\alpha}\right){\left(x+2\sqrt{{x}^{2}/4+\alpha}\right)}^{\alpha}\end{array}

(31)

and, in particular,

{F}_{n+1/2}\left(x\right)<\frac{U\left(a,x\right)}{U\left(a,0\right)}<{F}_{n-1/2}\left(x\right)

(32)

where

{F}_{\alpha}\left(x\right)=\text{exp}\left(-\frac{x}{2}\sqrt{\frac{{x}^{2}}{4}+\alpha}\right){\left(\frac{x}{2\sqrt{\alpha}}+\sqrt{\frac{{x}^{2}}{4\alpha}+1}\right)}^{-\alpha}

(33)

The bounds (32) are useful for obtaining the range of parameters for which function values are computable within the arithmetic capabilities of a computer (overflow and underflow limits). These results confirms the estimations based on the Liouville-Green approximation used in [18].

° **Iterated coerror functions and Mill's ratio:** In particular, considering Theorem 9 and the relation of parabolic cylinder functions *U*(*n* + 1/2, *x*) with the iterated coerror functions *i*^{n}erfc(*x*) [[16], 12.7.7], *n* ∈ ℕ, the following follows:

{M}_{n+1}\left(x\right)<\frac{{i}^{n}\mathsf{\text{erfc(}}x\mathsf{\text{)}}}{{i}^{n-1}\mathsf{\text{erfc}}\left(x\right)}<{M}_{n}\left(x\right),\phantom{\rule{1em}{0ex}}n=1,2,\dots ;{M}_{n}\left(x\right)={\left(x+\sqrt{2n+{x}^{2}}\right)}^{-1}.

(34)

These inequalities appear in [19].

Theorem 9 also gives bounds on Mill's ratio (*n =* 1/2). From the lower bound in Theorem (9) and the upper bound obtained by iterating with (26) we have

**Theorem 13**. *Let* r\left(x\right)={e}^{{x}^{2}/2}{\int}_{x}^{+\infty}{e}^{-{t}^{2}/2}dt, *then*

\frac{2}{x+\sqrt{{x}^{2}+4}}<r\left(x\right)<\frac{4}{3x+\sqrt{{x}^{2}+8}}

(35)

The lower bound was obtained in [20] and the upper bound in [21]. In our case, these results follow from a more general result. See also [22] for an alternative proof.

Further iterations (see (26)) give additional sharper bounds:

**Theorem 14**.

{R}_{2k+1}<r\left(x\right)<{R}_{2k}\left(x\right)

(36)

{R}_{n}\left(x\right)=\frac{1}{x+}\frac{1}{x+}\frac{2}{x+}\dots \frac{n}{{T}_{n}\left(x\right)},{T}_{n}=\left(x+\sqrt{4n+{x}^{2}}\right)/2

(37)

where, as usual we denote \frac{1}{a+}\frac{1}{b+}\cdots =1/\left(a+1/\left(b+\cdots \phantom{\rule{0.3em}{0ex}}\right)\right)

° **Hermite polynomials of imaginary variable:** A similar analysis to that for *U*(*n, -x*) can be carried for the PCF *V*(*n, x*). Indeed, *y*_{
n
}(*x*) = *V*(*n, x*)/Γ(*n* + 1/2) is a solution of (23) and *h*_{
n
}(*x*) = *y*_{
n
}(*x*)/*y*_{
n
}_{-1}(*x*) is such that *h*_{
n
}(0^{+}) > 0. Two situations take place depending on the values of *n*. First, if *n* ∈ (2*k -* 1, 2*k*), *k* ∈ ℕ, then {h}_{n}^{\prime}\left({0}^{+}\right)>0 the upper bound of Theorem 10 holds for all *x* > 0 while the lower bound will hold for *n* ∈ (2*k*, 2*k* + 1). Contrarily, if *n* ∈ (2*k*, 2*k* + 1) then {h}_{n}^{\prime}\left({0}^{+}\right)<0, while *h*_{
n
}(*+*∞) > 0, and the upper bound only holds for large enough *x;* a similar situation occurs with the lower bound when *n* ∈ (2*k -* 1, 2*k*).

We only consider the first case. Using the relation of *V*(*n* + 1/2,*x*), *n* ∈ ℕ, with Hermite polynomials [[16], 12.7.3] we get:

**Theorem 15**.

\frac{V\left(n,x\right)}{V\left(n-1,x\right)}<\frac{x+\sqrt{4n-2+{x}^{2}}}{2},\phantom{\rule{1em}{0ex}}x>0,\phantom{\rule{1em}{0ex}}n\in \left(2k-1,2k\right),\phantom{\rule{1em}{0ex}}k\in \mathbb{N}

(38)

\frac{x+\sqrt{4n-6+{x}^{2}}}{2}<\frac{V\left(n,x\right)}{V\left(n-1,x\right)},\phantom{\rule{1em}{0ex}}x>0,\phantom{\rule{1em}{0ex}}n\in \left(2k,2k+1\right),\phantom{\rule{1em}{0ex}}k\in \mathbb{N}

(39)

-i\frac{{H}_{2k+1}\left(ix\right)}{{H}_{2k}\left(ix\right)}<x+\sqrt{4k+2+{x}^{2}}\phantom{\rule{1em}{0ex}}x>0,\phantom{\rule{1em}{0ex}}k=0,1,2\dots

(40)

i\frac{{H}_{2k-1}\left(ix\right)}{{H}_{2k}\left(ix\right)}<{\left(x+\sqrt{4k-2+{x}_{2}}\right)}^{-1},\phantom{\rule{2.77695pt}{0ex}}\phantom{\rule{2.77695pt}{0ex}}x>0,\phantom{\rule{1em}{0ex}}k\in \mathbb{N}

(41)

\frac{{H}_{2k}{\left(ix\right)}^{2}}{{H}_{2k-1}\left(ix\right){H}_{2k+1}\left(ix\right)}>\sqrt{\frac{k-1/2}{k+1/2}},\phantom{\rule{1em}{0ex}}k\in \mathbb{N},\phantom{\rule{1em}{0ex}}x\in \mathbb{R}.

(42)

Hermite polynomials of imaginary argument were also considered in [15]. The well-known Turàn-type inequality for Hermite polynomials [23] *H*_{
n
}(*x*)^{2} - *H*_{
n-
}_{1}(*x*)*H*_{
n+
}_{1}(*x*) > 0, *x* ∈ ℝ, does not hold on the imaginary axis, but a similar property {H}_{n}{\left(ix\right)}^{2}-\sqrt{\left(n-1\right)/\left(n+1\right)}{H}_{n-1}\left(ix\right){H}_{n+1}\left(ix\right)>0 holds true for all *x* > 0 if *n* is even.

#### 4.1.3 Oblate legendre functions

These are Legendre functions of imaginary argument, which are functions appearing in the solution of Dirichlet problems in oblate spheroidal coordinates [6]. Denoting

{p}_{n}\left(x\right)={e}^{-in\pi /2}{P}_{n}^{m}\left(ix\right)

(43)

and using the differential relations [[16], 14.10.4-5] we have

\begin{array}{c}{p}_{n}^{\prime}\left(x\right)=\frac{1}{1+{x}^{2}}\left\{nx{p}_{\nu}\left(x\right)+\left(n+m\right){p}_{\nu -1}\left(x\right)\right\}\\ {p}_{n-1}^{\prime}\left(x\right)=\frac{1}{1+{x}^{2}}\left\{-nx{p}_{\nu -1}\left(x\right)+\left(n-m\right){p}_{\nu}\left(x\right)\right\}\end{array}

(44)

and {q}_{n}\left(x\right)={Q}_{n}^{m}\left(ix\right), {Q}_{n}^{m} being the second kind Legendre function, satisfies the same system. We consider *n > m and x* > 0. This is again an example for which Theorem 4 holds. The roles played in this case by the functions {Q}_{n}^{m}\left(ix\right) and {P}_{n}^{m}\left(ix\right) are very similar to the roles of *U*(*n, x*) and *V*(*n, x*) in the previous section. We omit details and only summarize the main results.

**Theorem 16**. *The following holds for x* > 0 *and real n* > *m* > 0

0<i\frac{{Q}_{n}^{m}\left(ix\right)}{{Q}_{n-1}^{m}\left(ix\right)}<\frac{n+m}{n}{\left[x+\sqrt{1+{x}^{2}-\frac{{m}^{2}}{{n}^{2}}}\right]}^{-1}<\sqrt{\frac{n+m}{n-m}}

(45)

i\frac{{Q}_{n}^{m}\left(ix\right)}{{Q}_{n-1}^{m}\left(ix\right)}>\frac{n+m}{nx+\left(n+1\right)\sqrt{1+{x}^{2}-\frac{{m}^{2}}{{\left(n+1\right)}^{2}}}}

(46)

1<\frac{n+m+1}{n+m}\frac{{Q}_{n}^{m}\left(ix\right)}{{Q}_{n-1}^{m}\left(ix\right){Q}_{n+1}^{m}\left(ix\right)}<\sqrt{\frac{{\left(n+2\right)}^{2}-{m}^{2}}{{n}^{2}-{m}^{2}}}

(47)

**Theorem 17**. *The following holds for x* > 0 *and integer n, m, n > m:*

0<-i\frac{{P}_{n}^{m}\left(ix\right)}{{P}_{n-1}^{m}\left(ix\right)}<\frac{n}{n-m}\left[x+\sqrt{1+{x}^{2}-\frac{{m}^{2}}{{n}^{2}}}\right],\phantom{\rule{1em}{0ex}}n-m\phantom{\rule{2.77695pt}{0ex}}odd

(48)

\frac{1}{n-m}\left[nx+\left(n-1\right)\sqrt{1+{x}^{2}-\frac{{m}^{2}}{{\left(n-1\right)}^{2}}}\right]<-i\frac{{P}_{n}^{m}\left(ix\right)}{{P}_{n-1}^{m}\left(ix\right)},n-m\phantom{\rule{2.77695pt}{0ex}}even

(49)

\frac{{P}_{n}^{m}{\left(ix\right)}^{2}}{{P}_{n-1}^{m}\left(ix\right){P}_{n+1}^{m}\left(ix\right)}<1+\frac{1}{n-m},\phantom{\rule{1em}{0ex}}n-m\phantom{\rule{2.77695pt}{0ex}}odd

(50)

For *m* = 0 we have Legendre polynomials. If *n* is odd, we have *P*_{
n
}(*ix*)^{2} < 0 and therefore *P*_{
n
}(*ix*)^{2} - (1 + 1/*n*)*P*_{
n
}_{-1}(*ix*)*P*_{
n
}_{+1}(*ix*) > 0. It appears, as numerical experiments show, that in this case the same Turán inequality that holds in the real interval (-1,1) [24] also holds in the imaginary axis if *n* is odd: *P*_{
n
}(*ix*)^{2} - *P*_{
n-
}_{1}(*ix*)*P*_{
n+
}_{1}(*ix*) > 0; the same is not true if *m ≠* 0.

#### 4.1.4 Laguerre functions of negative argument

Next we consider an example for which Theorem 4 can not be applied but the analysis is possible because the characteristic roots are monotonic.

Consider the Laguerre functions {y}_{\nu ,\alpha}\left(x\right)={L}_{\nu}^{\alpha}\left(-x\right),x>0. Using well known recurrences and differentiation formulas, we have

\begin{array}{c}{y}_{\nu +1,\alpha -1}^{\prime}\left(x\right)={y}_{\nu ,\alpha}\left(x\right)\\ x{y}_{\nu ,\alpha}^{\prime}\left(x\right)=-\left(\alpha +x\right){y}_{\nu ,\alpha}\left(x\right)+\left(v+1\right){y}_{\nu +1,\alpha -1}\end{array}

(51)

and

\left(\nu +1\right){y}_{\nu +1,\alpha -1}\left(x\right)=\left(\alpha +x\right){y}_{\nu ,\alpha}\left(x\right)+x{y}_{\nu -1,\alpha +1}

(52)

Considering [[25], Theorem 2] it follows that *y*_{
v,α
} is a dominant solution of the recurrence (52) in the direction of increasing *ν* (and decreasing *α*).

With *h*(*x*) = *y*_{
ν
}_{,α}(*x*)/*y*_{
v+
}_{1,α-1}(*x*), the positive characteristic root λ^{+}(*x*) of the associated Riccati equation turns out to be increasing if *ν* > -1 and *α* > 0. On the other hand, it is easy to check that for these values *h*(0^{+}) > 0 and *h'*(0^{+}) > 0. Theorem 1 holds and λ^{+}(*x*) is a bound:

**Theorem 18**. *For any α* > 0, *ν* > -1 *and x* > 0 *the following holds*

0<\frac{{L}_{\nu +1}^{\alpha -1}\left(x\right)}{{L}_{\nu}^{\alpha}\left(-x\right)}<\frac{\alpha +x+\sqrt{{\left(\alpha +x\right)}^{2}+4\left(\nu +1\right)x}}{2\left(\nu +1\right)}

(53)

On the other hand, from the recurrence (52) we have

\frac{{L}_{\nu}^{\alpha}\left(-x\right)}{{L}_{\nu +1}^{\alpha -1}\left(-x\right)}={\left(\frac{\alpha +x}{\nu +1}+\frac{x}{\nu +1}\frac{{L}_{\nu -1}^{\alpha +1}\left(-x\right)}{{L}_{\nu}^{\alpha}\left(-x\right)}\right)}^{-1}

(54)

and from this we obtain the second Perron-Kreuser bound:

**Theorem 19**. *For any α* > -1, *ν* > 0 *and x* > 0 *the following holds*

\frac{{L}_{\nu +1}^{\alpha -1}\left(-x\right)}{{L}_{\nu}^{\alpha}\left(-x\right)}>\frac{\alpha +x-1+\sqrt{{\left(\alpha +x+1\right)}^{2}+4\nu x}}{2\left(\nu +1\right)}

(55)

And from these bounds we get the following Turán-type inequalities:

**Theorem 20**. *For any ν* ≥ 0 *and α* ≥ 0, *x* > 0 the *following holds:*

\frac{\nu}{\nu +1}\frac{\alpha}{\alpha +1}<\frac{{L}_{\nu +1}^{\alpha -1}\left(-x\right)}{{L}_{\nu}^{\alpha}\left(-x\right)}\frac{{L}_{\nu -1}^{\alpha +1}\left(-x\right)}{{L}_{\nu}^{\alpha}\left(-x\right)}<\frac{\nu}{\nu +1}

(56)

A second independent solution of (51) which is a minimal solution of (52) as *ν* → +∞ follows from [[25], Theorem 2]. Bounds can be also obtained for this solution. We omit the details.

Other bounds and inequalities can be obtained using other recursions or using relations between contiguous functions. For example, using [[16], 18.9.13], we have:

\frac{{L}_{\nu +1}^{\alpha -1}\left(x\right)}{{L}_{\nu}^{\alpha}\left(x\right)}=1+\frac{{L}_{\nu +1}^{\alpha}\left(x\right)}{{L}_{\nu}^{\alpha}\left(x\right)}

(57)

and upper and lower bounds for {L}_{n}^{\alpha}\left(-x\right)/{L}_{n-1}^{\alpha}\left(-x\right) follow from the previous results. As a consequence of this new bounds, one can prove the following

**Theorem 21**.

\frac{\nu}{\nu +1}<\frac{{L}_{\nu -1}^{\alpha}\left(-x\right)}{{L}_{\nu}^{\alpha}\left(-x\right)}\frac{{L}_{\nu +1}^{\alpha}\left(-x\right)}{{L}_{\nu}^{\alpha}\left(-x\right)}<\frac{\nu}{\nu +1}\frac{\nu +\alpha +1}{\nu +\alpha -1}

(58)

*where the first inequality holds for ν* > 0, *α* > -1 *and the second for ν* > 0, *ν* + *α* > 1.

For positive *x*, it is known that {L}_{n-1}^{\alpha}\left(x\right){L}_{n+1}^{\alpha}\left(x\right)/{L}_{n}^{\alpha}{\left(x\right)}^{2}<1 [23]. For negative argument we have an upper bound greater that 1, which suggests that the Turán-type inequality for positive *x* does not hold for negative *x*, as numerical experiments show.

### 4.2 Two examples with *d*_{
n
}(*x*)*e*_{
n
}(*x*) < 0

The DDEs corresponding to a pair {*p*_{
n
}(*x*),*p*_{
n
}_{-1}(*x*)} of classical orthogonal polynomials satisfy *d*_{
n
}(*x*)*e*_{
n
}(*x*) < 0 in their interval of orthogonality because this is a necessary condition for oscillation [[8], Lemma 2.4]. However, for values of the variable for which the polynomials are free of zeros, one can expect that *η*_{
n
}(*x*)^{2} > 1 and that the DDE becomes monotonic (*η*_{
n
}(*x*)^{2} < 1 is also a necessary condition for oscillation [[9], Theorem. 2.1]). This is the case of Laguerre and Hermite polynomials for large enough *x* > 0. We consider these two examples.

#### 4.2.1 Hermite polynomials

Hermite polynomials satisfy

\begin{array}{c}{H}_{n}^{\prime}\left(x\right)=2n{H}_{n-1}\left(x\right),\\ {H}_{n-1}^{\prime}\left(x\right)=2x{H}_{n-1}-{H}_{n}\left(x\right)\end{array}

(59)

We have {\eta}_{n}\left(x\right)=x/\sqrt{2n} and *η*_{
n
}(*x*) > 1 if x>\sqrt{2n} (monotonic case). The characteristic roots are both of them positive

{\lambda}_{n}^{\pm}\left(x\right)=x\pm \sqrt{{x}^{2}-2n.}

(60)

Defining *h*_{
n
}(*x*) = *H*_{
n
}(*x*)/*H*_{
n
}_{-1}(*x*) we have that *h*_{
n
}(*+*∞) = +∞ and {h}_{n}^{\prime}\left(+\infty \right)>0 because the coefficient of degree *n* of *H*_{
n
}(*x*) is positive. Then {h}_{n}\left(x\right)>{\lambda}_{n}^{+}\left(x\right) for enough *x* > 0 because {h}_{n}^{\prime}\left(x\right)>0 only if {h}_{n}\left(x\right)<{\lambda}_{n}^{-}\left(x\right) or {h}_{n}\left(x\right)>{\lambda}_{n}^{+}\left(x\right), but {h}_{n}\left(+\infty \right)>{\lambda}_{n}^{-}\left(+\infty \right)={0}^{+}, and therefore {h}_{n}\left(x\right)>{\lambda}_{n}^{+}\left(x\right) for large *x*. And because {\lambda}_{n}^{+\prime}\left(x\right)>0 if x>\sqrt{2n}, then, necessarily:

{h}_{n}\left(x\right)=\frac{{H}_{n}\left(x\right)}{{H}_{n-1}\left(x\right)}>x+\sqrt{{x}^{2}-2n},x\ge \sqrt{2n.}

(61)

We can iterate the recurrence relation. Contrary to the case *e*_{
n
}(*x*)*d*_{
n
}(*x*) > 0, we will not obtain sequences of lower and upper bounds, but only lower bounds. Writing

{h}_{n+1}\left(x\right)=2x-2n/{h}_{n}\left(x\right)

(62)

and using (61) we get a lower bound for *h*_{
n+
}_{1}(*x*). We shift the parameter *n* and get

{h}_{n}\left(x\right)>x+\sqrt{{x}^{2}-2\left(n-1\right)},x\ge \sqrt{2\left(n-1\right).}

(63)

This improves Eq. (61) and enlarges the range of validity of the bound with respect to *x*, but reduces the range of validity with respect to *n* (*n* ≥ 2).

The next iteration gives a bound for *n* ≥ 3:

\begin{array}{ll}\hfill {h}_{n}\left(x\right)& >F\left(n,x\right),x\ge \phantom{\rule{2.77695pt}{0ex}}\sqrt{2\left(n-2\right)}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}F\left(n,x\right)={\left(n-2\right)}^{-1}\left[\left(n-3\right)x+\left(n-1\right)\sqrt{{x}^{2}-2\left(n-2\right)}\right]\phantom{\rule{2em}{0ex}}\end{array}

(64)

Because the largest zero of *H*_{
n
}(*x*) is larger than that of *H*_{
n-
}_{1}(*x*), Eq. (64) implies that the largest zero of *H*_{
n
}(*x*) is smaller than \sqrt{2\left(n-2\right)}, *n* ≥ 3.

We consider just one more iteration and get

{h}_{n}\left(x\right)\ge 2x-2\left(n-1\right)/F\left(n-1,x\right)=G\left(n,x\right),x>\sqrt{2\left(n-3\right)}

(65)

and if G(n,\sqrt{2(n-3))}>0 then *G*(*n, x*) > 0 if x>\sqrt{2\left(n-3\right)}, and the largest zero will be smaller than \sqrt{2\left(n-3\right)}; this condition is met if *n* ≥ 7. A sharper bound has recently appeared in the literature [26] valid for all *n*. However, our result is sharper than previous results, like for instance those in [27], which is interesting given the simplicity of the analysis. This reflects the fact that the bounds on function ratios (our main topic) are sharp.

#### 4.2.2 Laguerre polynomials

We give some results for Laguerre polynomials omitting details. Defining {h}_{n}^{\alpha}\left(x\right)=-{L}_{n}^{\alpha}\left(x\right)/{L}_{n-1}^{\alpha}\left(x\right), we have {h}_{n}^{\alpha}\left(+\infty \right)=+\infty and {h}_{n}^{\alpha \prime}\left(+\infty \right)=+\infty and, proceeding similarly as before:

\begin{array}{ll}\hfill 2n{h}_{n}^{\alpha}\left(x\right)>& x-\left(2n+\alpha \right)+\sqrt{{\left(x-2n-\alpha \right)}^{2}-4n\left(n+\alpha \right)},\phantom{\rule{2em}{0ex}}\\ x\ge 2n+\alpha +2\sqrt{n\left(n+\alpha \right)}\phantom{\rule{2em}{0ex}}\end{array}

(66)

and after the first iteration of the recurrence we have:

\begin{array}{c}2n{h}_{n}^{\alpha}\left(x\right)>f\left(x\right),x\ge 2{n}^{*}+\alpha +2\sqrt{{n}^{*}\left({n}^{*}+\alpha \right)},{n}^{*}=n-1,\\ \phantom{\rule{2.77695pt}{0ex}}f\left(x\right)=x-\left(2n+\alpha \right)+\sqrt{{\left(x-2{n}^{*}-\alpha \right)}^{2}-4{n}^{*}\left({n}^{*}+\alpha \right).}\end{array}

(67)

This proves that the largest zero of {L}_{n}^{\alpha}\left(x\right) is smaller than {x}^{*}=2n+\alpha -2+\sqrt{\left(n-1\right)\left(n-1+\alpha \right)}, provided that *f*(*x**) > 0, which is true if *α* > (*n -* 1)^{-1} - (*n -* 1), *n* ≥ 2; notice that values *α* < -1 are allowed for large enough *n*. The bound in [28] is slightly sharper, and it is improved in [26].

Further iterations are possible, but not so easy to analyze. The next iteration will give a bound

2n{h}_{n}^{\alpha}\left(x\right)>g\left(x\right),x\ge 2\left(n-2\right)+\alpha +2\sqrt{\left(n-2\right)\left(n-2+\alpha \right)}={x}^{*}.

(68)

*x** is an upper bound for the largest zero provided that *g*(*x**) > 0. This condition is met for a larger range of values of *α* as *n* becomes larger. For *n* ≥ 10, this holds for any *α* > -1. The bound (68) is of more limited validity in terms of *n* but numerical experiments show that it is sharper than the bound in [26] for *α ≤* 12

We expect that lower bounds for the smallest zero can be also obtained with a similar analysis.

The main message, as before, is that the bounds on function ratios are sharp for large *x* because they give the correct asymptotic behavior as *x* → +∞, but also for moderate *x* given the sharpness of the bounds on the largest zero.