Ratios of Adjacent Normalized Remainders of tan²x under Finite-Block Coefficient Perturbations

1. Introduction

Throughout, $\mathbb{N}=\{1,2,\dots \}$. For a function analytic near the origin,

$$f\left(x\right)={\displaystyle \sum _{j=0}^{\infty }}{c}_j{x}^j$$

and an integer $m\ge 0$ with $c_{m+1}\ne 0$, the mth normalized Maclaurin remainder is [8]

$$T_m\left[f\left(x\right)\right]=\left\{\begin{array}{cc}{\displaystyle \frac{f\left(x\right)-{\sum }_{j=0}^m{c}_j{x}^j}{c_{m+1}{x}^{m+1}},}\hfill & x\ne 0,\hfill \\ 1,\hfill & x=0.\hfill \end{array}\right.$$

(1)

Normalized remainders have been studied for several elementary and special functions. Qi [8] surveys the subject. Recent work establishes logarithmic absolute monotonicity and series expansions for normalized remainders of $tanx$ [15] and of $arcsinx$ and $arctanx$ [12]. Suna and Ray [14] prove an exact monotonicity threshold for normalized remainders of a degenerate exponential family. For the square of the tangent function, Zhang and Qi [10] proved logarithmic convexity of the normalized tails and expanded their logarithms in power series. Liu and Qi [6] proved that, for every $n\in \mathbb{N}$,

$${\displaystyle \frac{T_{2n+1}\left[{tan}^{2}x\right]}{T_{2n-1}\left[{tan}^{2}x\right]}}$$

is decreasing on $(0,\pi /2)$.

In their proof of Theorem 1.1, Liu and Qi [6] use two ingredients. The first is the classical Biernacki–Krzyż monotonicity principle [4]. A quotient of two power series with positive denominator coefficients inherits the monotonicity of the coefficient-ratio sequence. The second is the strict decrease of the adjacent coefficient ratios in the Maclaurin expansion of ${tan}^{2}x$. Piecewise versions of the power-series quotient principle were developed by Yang, Chu, and Wang [9]. In particular, a coefficient-ratio sequence that rises once and then falls can produce a quotient with one interior turn. For broader unimodality-preservation results, see Karp, Vishnyakova, and Zhang [5].

For the one-coefficient case, write

$${tan}^{2}x={\displaystyle \sum _{j=1}^{\infty }}{a}_j{x}^{2j},\left|x\right|<{\displaystyle \frac{\pi }{2}},$$

(2)

and for $n\in \mathbb{N}$ set

$$f_{n,\lambda }\left(x\right)={tan}^{2}x+\lambda a_n{x}^{2n},\lambda >-1.$$

(3)

Thus $a_n$ is replaced by $(1+\lambda )a_n$, while all other Maclaurin coefficients remain unchanged. The condition $\lambda >-1$ is exactly the range in which the perturbed coefficient remains positive. The perturbation changes the first comparison in the coefficient-ratio sequence for the adjacent normalized tails while preserving the strict decrease of all subsequent ratios. The resulting problem is a parameter version of the monotonicity problem studied by Liu and Qi [6].

The lower perturbed tail in the one-coefficient problem has the form $1+cyH\left(y\right)$ with $c>0$. It is an instance of the preimage family for the normalization operator $\mathcal{S}\left(G\right)\left(y\right)=(G\left(y\right)-1)/\left(y{G}^{\prime }\left(0\right)\right)$ described by Abu-Ghuwaleh [1]. The monotonicity and finite-oscillation ingredients come from the classical Biernacki–Krzyż/Yang line and the general finite-change theorem of Mao and Tian [2].

For the tangent-square coefficients, perturbing r consecutive terms gives at most r interior critical points, and this bound is attained for every r by a full local unfolding of an explicit order-r degeneracy. Two consecutive perturbations admit an exact two-parameter phase diagram. An open region has a local minimum followed by a local maximum, and two switch functions determine when these interior extrema become global. The $r=1$ case is analyzed separately, including its sharp range, maximizer motion, and asymptotics.

Section 2 establishes the coefficient facts and a derivative-quotient lemma. Section 3 treats arbitrary finite consecutive perturbations. Section 4–6 analyze the one-coefficient case. Section 7 gives the two-coefficient critical-point classification and the sharp-extremum phase diagram.

2. Coefficient Ratios and Normalized Tails

Let $B_k$ denote the Bernoulli numbers. The coefficients in (2) are positive and satisfy [6]

$$a_j={\displaystyle \frac{2^{2j+2}\left(2^{2j+2}-1\right)(2j+1)\left|B_{2j+2}\right|}{(2j+2)!}},j\in \mathbb{N}.$$

(4)

In particular,

$$a_1=1,a_2={\displaystyle \frac{2}{3}},a_3={\displaystyle \frac{17}{45}},a_4={\displaystyle \frac{62}{315}},a_5={\displaystyle \frac{1382}{14175}}.$$

(5)

Put

$${\gamma }_j={\displaystyle \frac{a_{j+1}}{a_j}},j\in \mathbb{N}.$$

(6)

Bao, He, and Qi [11] studied monotonicity and bounds for ratios of adjacent nonzero Bernoulli numbers. Further sharp bounds were obtained by Liu, Xu, He, and Qi [13].

The following equivalent form of Liu and Qi’s Lemma 2.1 controls the normalized tail quotients.

Lemma 1 

([6], Lemma 2.1). The sequence ${\left({\gamma }_j\right)}_{j\ge 1}$ is strictly decreasing.

The sequence in the cited lemma, evaluated at $j+1$, is ${\gamma }_j/2$, as follows from (4). The classical Bernoulli–zeta identity [7] is

$$|B_{2m}|={\displaystyle \frac{2\left(2m\right)!}{{\left(2\pi \right)}^{2m}}}\zeta \left(2m\right),m\ge 1,$$

where $\zeta$ denotes the Riemann zeta function. Substitution in (4) gives

$$a_j={\displaystyle \frac{2(2^{2j+2}-1)(2j+1)\zeta (2j+2)}{{\pi }^{2j+2}}}.$$

(7)

Hence

$${\gamma }_j={\displaystyle \frac{1}{{\pi }^2}}{\displaystyle \frac{(2^{2j+4}-1)(2j+3)\zeta (2j+4)}{(2^{2j+2}-1)(2j+1)\zeta (2j+2)}}.$$

(8)

Since $\zeta \left(s\right)\to 1$ as $s\to \infty$,

$$\underset{j\to \infty }{lim}{\gamma }_j={\displaystyle \frac{4}{{\pi }^2}}.$$

(9)

Lemma 1 therefore gives

$${\gamma }_j>{\displaystyle \frac{4}{{\pi }^2}}(j\in \mathbb{N}).$$

(10)

For $j\in \mathbb{N}$ define a normalized tail in the variable $y=x^2$ by

$$H_j\left(y\right)={\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{a_{j+k}}{a_j}}{y}^k,0\le y<\rho ,\rho ={\displaystyle \frac{{\pi }^2}{4}}.$$

(11)

Thus

$$H_j\left(x^2\right)=T_{2j-1}\left[{tan}^{2}x\right].$$

(12)

The adjacent tails satisfy the elementary recursion

$$H_j\left(y\right)=1+{\gamma }_{j}y{H}_{j+1}\left(y\right).$$

(13)

The Biernacki–Krzyż rule [4] is used in the following form. Let $A\left(y\right)={\sum }_{k\ge 0}{A}_k{y}^k$ and $B\left(y\right)={\sum }_{k\ge 0}{B}_k{y}^k$ converge for $\left|y\right|<r$, with $B_k>0$. If $q_k=A_k/B_k$ is nonincreasing and not constant, then ${(A/B)}^{\prime }\left(y\right)<0$ for $0<y<r$. The strict derivative inequality follows directly from

$$A^{\prime }B-A{B}^{\prime }=\sum _{j>k\ge 0}(j-k)B_j{B}_k(q_j-q_k)y^{j+k-1}<0.$$

The double series converges absolutely on this interval.

For the perturbed tails, the Biernacki–Krzyż rule gives the following derivative consequence.

Lemma 2. 

Let $H\left(y\right)={\sum }_{k\ge 0}{h}_k{y}^k$ have positive coefficients, $h_0=1$, and radius of convergence $r>0$. If ${(h_{k+1}/h_k)}_{k\ge 0}$ is strictly decreasing, then

$$D\left(y\right):={\displaystyle \frac{H^{\prime }\left(y\right)}{H{\left(y\right)}^2}}\mathrm{satisfies}{D}^{\prime }\left(y\right)<0(0<y<r).$$

(14)

Proof. 

Apply the Biernacki–Krzyż rule [4] to

$$W\left(y\right)={\displaystyle \frac{H^{\prime }\left(y\right)}{H\left(y\right)+y{H}^{\prime }\left(y\right)}}={\displaystyle \frac{{\sum }_{k\ge 0}(k+1)h_{k+1}{y}^k}{{\sum }_{k\ge 0}(k+1)h_k{y}^k}}.$$

The coefficient ratios are $h_{k+1}/h_k$, so $W^{\prime }\left(y\right)<0$. Differentiation gives

$$W^{\prime }\left(y\right)={\displaystyle \frac{H\left(y\right)H^{\prime \prime }\left(y\right)-2H^{\prime }{\left(y\right)}^2}{{(H\left(y\right)+y{H}^{\prime }\left(y\right))}^2}},D^{\prime }\left(y\right)={\displaystyle \frac{H\left(y\right)H^{\prime \prime }\left(y\right)-2H^{\prime }{\left(y\right)}^2}{H{\left(y\right)}^3}}.$$

Both denominators are positive, proving the assertion. □

3. Finite Consecutive Coefficient Perturbations and a Sharp Critical-Point Bound

Fix $n,r\in \mathbb{N}$. Let $b_0,\dots ,b_{r-1}>0$, put $b_j=1$ for $j\ge r$, and define

$$f_{n,\mathbf{b}}\left(x\right)={tan}^{2}x+{\displaystyle \sum _{j=0}^{r-1}}(b_j-1)a_{n+j}{x}^{2(n+j)}.$$

(15)

Let

$$R_{n,\mathbf{b}}\left(x\right)={\displaystyle \frac{T_{2n+1}\left[f_{n,\mathbf{b}}\left(x\right)\right]}{T_{2n-1}\left[f_{n,\mathbf{b}}\left(x\right)\right]}}.$$

With $y=x^2$, write

$$G_{\mathbf{b}}\left(y\right)=T_{2n+1}\left[f_{n,\mathbf{b}}\right]\left(\sqrt{y}\right)=\sum _{k\ge 0}{g}_k{y}^k,g_0=1,$$

and put

$$c={\gamma }_n{\displaystyle \frac{b_1}{b_0}},q_k={\displaystyle \frac{g_{k+1}}{g_k}}={\gamma }_{n+1+k}{\displaystyle \frac{b_{k+2}}{b_{k+1}}},$$

(16)

using $b_j=1$ for $j\ge r$. The tail recursion gives

$$R_{n,\mathbf{b}}\left(\sqrt{y}\right)={\displaystyle \frac{G_{\mathbf{b}}\left(y\right)}{1+cy{G}_{\mathbf{b}}\left(y\right)}}.$$

(17)

Set

$$D_{\mathbf{b}}\left(y\right)={\displaystyle \frac{G_{\mathbf{b}}^{\prime }\left(y\right)}{G_{\mathbf{b}}{\left(y\right)}^2}},W_{\mathbf{b}}\left(y\right)={\displaystyle \frac{G_{\mathbf{b}}^{\prime }\left(y\right)}{G_{\mathbf{b}}\left(y\right)+y{G}_{\mathbf{b}}^{\prime }\left(y\right)}}.$$

Then

$${\displaystyle \frac{d}{dy}}{R}_{n,\mathbf{b}}\left(\sqrt{y}\right)={\displaystyle \frac{G_{\mathbf{b}}{\left(y\right)}^2(D_{\mathbf{b}}\left(y\right)-c)}{{(1+cy{G}_{\mathbf{b}}\left(y\right))}^2}},$$

(18)

and

$$W_{\mathbf{b}}^{\prime }={\displaystyle \frac{G{G}^{\prime \prime }-2{\left(G^{\prime }\right)}^2}{{(G+y{G}^{\prime })}^2}},D_{\mathbf{b}}^{\prime }={\displaystyle \frac{G{G}^{\prime \prime }-2{\left(G^{\prime }\right)}^2}{G^3}}.$$

(19)

The coefficient-ratio sequence of $W_{\mathbf{b}}$ is exactly $\left(q_k\right)$. Since $q_k={\gamma }_{n+1+k}$ for $k\ge r-1$, this sequence has at most $r-1$ changes of monotonicity. Mao and Tian’s general theorem [2] therefore gives at most $r-1$ monotonicity changes of $W_{\mathbf{b}}$ and hence of $D_{\mathbf{b}}$.

Theorem 1. 

For every $n,r\in \mathbb{N}$ and every positive multiplier vector $\mathbf{b}$, the function $R_{n,\mathbf{b}}$ has at most r distinct critical points in $(0,\pi /2)$. This bound is sharp for every n and r. More precisely, for any $0<{\xi }_1<\dots <{\xi }_r$ there are positive vectors $\mathbf{b}\left(\epsilon \right)$, for all sufficiently small $\epsilon >0$, such that $R_{n,\mathbf{b}\left(\epsilon \right)}$ has exactly r simple critical points satisfying

$$x_i{\left(\epsilon \right)}^2=\epsilon {\xi }_i+o\left(\epsilon \right)(1\le i\le r).$$

Proof. 

The upper bound follows from (19). The analytic function $D_{\mathbf{b}}$ has at most r monotonicity intervals. It cannot be constant on any nonempty interval. Analyticity would make $D_{\mathbf{b}}\equiv d$ on $(0,\rho )$, and the differential equation $G^{\prime }=d{G}^2$ with $G\left(0\right)=1$ would force $G\left(y\right)={(1-dy)}^{-1}$, so every coefficient ratio would equal d, contrary to the eventually strictly decreasing tangent-square ratios. Hence each monotonicity interval contributes at most one solution of $D_{\mathbf{b}}=c$, and (18) gives at most r critical points.

For sharpness put

$$s={\gamma }_{n+r},t={\gamma }_{n+r+1}<s.$$

Starting with $b_r^{*}=1$, define

$$b_j^{*}={\displaystyle \frac{{\gamma }_{n+j}}{s}}{b}_{j+1}^{*}(1\le j\le r-1),b_0^{*}={\displaystyle \frac{{\gamma }_n}{s}}{b}_1^{*}.$$

(20)

Then $c=s$ and $q_0=\dots =q_{r-1}=s$, while $q_r=t$. Hence

$$G_{{\mathbf{b}}^{*}}\left(y\right)={\displaystyle \frac{1}{1-sy}}-s^r(s-t)y^{r+1}+O\left(y^{r+2}\right),$$

and direct substitution in $D=G^{\prime }/G^2$ gives

$$D_{{\mathbf{b}}^{*}}\left(y\right)-s=-(r+1)s^r(s-t)y^r+O\left(y^{r+1}\right).$$

(21)

Thus the boundary zero at $y=0$ has exact order r.

Use $q_0,\dots ,q_{r-2}$ as local coordinates for $b_1,\dots ,b_{r-1}$ and write $D\left(y\right)=d_0+d_{1}y+\dots$. At the point $q_0=\dots =q_{r-2}=s$, varying $q_j$ changes all later coefficients multiplicatively. Since the unperturbed coefficients through degree r are $1,s,\dots ,s^r$, its first variation satisfies

$$\delta G=\delta q_j{\displaystyle \frac{s^j{y}^{j+1}}{1-sy}}+O\left(y^{r+1}\right).$$

Linearizing $D=G^{\prime }/G^2$ around ${(1-sy)}^{-1}$ therefore gives, through terms of degree $r-1$,

$$\delta D=\delta q_j{s}^j\left((j+1)y^j-(j+2)s{y}^{j+1}\right)+O\left(y^r\right).$$

Hence the Jacobian

$${\displaystyle \frac{\partial (d_1,\dots ,d_{r-1})}{\partial (q_0,\dots ,q_{r-2})}}$$

is upper bidiagonal with determinant

$${(-1)}^{r-1}r!s^{r(r-1)/2}\ne 0.$$

(22)

For $r=1$, the displayed determinant is the empty determinant 1. Let $L=-(r+1)s^r(s-t)$. By the inverse function theorem, for small $\epsilon$ one can choose $q_0,\dots ,q_{r-2}$ so that $d_1,\dots ,d_{r-1}$ equal the corresponding coefficients of

$$L{\displaystyle \prod _{i=1}^r}(y-\epsilon {\xi }_i),$$

and then choose the independent coordinate c so that the constant coefficient agrees as well. The triangular relations in (16) determine a positive vector $\mathbf{b}\left(\epsilon \right)\to {\mathbf{b}}^{*}$. If $F_{\epsilon }=D_{\mathbf{b}\left(\epsilon \right)}-c$, analyticity in both the coefficient parameters and y gives, on every bounded real z-interval,

$${\epsilon }^{-r}{F}_{\epsilon }\left(\epsilon z\right)⟶L{\displaystyle \prod _{i=1}^r}(z-{\xi }_i)\mathrm{in}{C}^1.$$

Each ${\xi }_i$ is a simple zero of the limiting polynomial. Simple-root stability (equivalently, the implicit function theorem on disjoint small neighborhoods of the ${\xi }_i$) therefore gives a unique simple zero $z_i\left(\epsilon \right)\to {\xi }_i$ in each such neighborhood. Hence $y_i\left(\epsilon \right)=\epsilon z_i\left(\epsilon \right)$ gives at least r simple positive zeros with the stated asymptotics, and the upper bound proved above excludes any others. □

Remark 1. 

The general Mao–Tian monotonicity-change theorem gives the bound “at most r”. The explicit order-r tangent-square degeneracy, the full-rank jet calculation (22), and the realization within the restricted coefficient-perturbation family establish sharpness in Theorem 1.

4. The Monotonicity Threshold

Fix $n\in \mathbb{N}$ and $\lambda >-1$. Consider

$$R_{n,\lambda }\left(x\right)={\displaystyle \frac{T_{2n+1}\left[f_{n,\lambda }\left(x\right)\right]}{T_{2n-1}\left[f_{n,\lambda }\left(x\right)\right]}},0\le x<{\displaystyle \frac{\pi }{2}}.$$

(23)

The perturbation in (3) occurs exactly at the leading term of $T_{2n-1}$ and disappears from the numerator $T_{2n+1}$. Using (13) gives

$$\begin{array}{cc}\hfill T_{2n-1}\left[f_{n,\lambda }\left(x\right)\right]& =1+{\displaystyle \frac{{\gamma }_n}{1+\lambda }}{x}^2{H}_{n+1}\left(x^2\right),\hfill \end{array}$$

(24)

$$\begin{array}{cc}\hfill T_{2n+1}\left[f_{n,\lambda }\left(x\right)\right]& =H_{n+1}\left(x^2\right).\hfill \end{array}$$

(25)

Therefore, with $y=x^2$,

$${\mathcal{R}}_{n,\lambda }\left(y\right):=R_{n,\lambda }\left(\sqrt{y}\right)={\displaystyle \frac{H_{n+1}\left(y\right)}{1+{\displaystyle \frac{{\gamma }_n}{1+\lambda }}y{H}_{n+1}\left(y\right)}}.$$

(26)

In particular, ${\mathcal{R}}_{n,\lambda }\left(0\right)=1$.

Define the first threshold

$${\alpha }_n={\displaystyle \frac{{\gamma }_n}{{\gamma }_{n+1}}}-1.$$

(27)

Lemma 1 gives ${\alpha }_n>0$.

Theorem 2. 

Let $n\in \mathbb{N}$ and $\lambda >-1$.

1.

If $-1<\lambda \le {\alpha }_n$, then $R_{n,\lambda }$ is strictly decreasing on $(0,\pi /2)$.

2.

If $\lambda >{\alpha }_n$, then there is a unique $x_{n,\lambda }\in (0,\pi /2)$ such that $R_{n,\lambda }$ is strictly increasing on $(0,x_{n,\lambda })$ and strictly decreasing on $(x_{n,\lambda },\pi /2)$. Hence $x_{n,\lambda }$ is the unique interior critical point and the unique global maximizer.

Proof. 

Put $H=H_{n+1}$, $c_{\lambda }={\gamma }_n/(1+\lambda )$, and $D=H^{\prime }/H^2$. The coefficient ratios of H are ${\gamma }_{n+1+k}$. Lemmas 1 and 2 give $D^{\prime }<0$ and $D\left(0\right)={\gamma }_{n+1}$.

At the other endpoint, the exact identity

$$H\left(y\right)={\displaystyle \frac{{tan}^2\sqrt{y}-{\sum }_{j=1}^n{a}_j{y}^j}{a_{n+1}{y}^{n+1}}}$$

has a pole of order two. The Laurent expansion of $tan\sqrt{y}$ at $\rho ={\pi }^2/4$ and differentiation of that expansion give

$$H\left(y\right)\sim C_n{(\rho -y)}^{-2},H^{\prime }\left(y\right)\sim 2C_n{(\rho -y)}^{-3},C_n={\displaystyle \frac{{\pi }^2}{a_{n+1}{\rho }^{n+1}}}.$$

(28)

Consequently $D\left(y\right)\to 0$ as $y\uparrow \rho$. Differentiating (26) now gives

$${\mathcal{R}}_{n,\lambda }^{\prime }\left(y\right)={\displaystyle \frac{H{\left(y\right)}^2\left(D\left(y\right)-c_{\lambda }\right)}{{(1+c_{\lambda }yH\left(y\right))}^2}}.$$

(29)

If $\lambda \le {\alpha }_n$, then $c_{\lambda }\ge D\left(0\right)>D\left(y\right)$ for $y>0$, so the derivative is strictly negative. If $\lambda >{\alpha }_n$, then $0<c_{\lambda }<D\left(0\right)$, and the strictly decreasing function D crosses $c_{\lambda }$ exactly once. The derivative is positive before that point and negative after it. The substitution $x=\sqrt{y}$ preserves these signs. □

The local expansion

$$R_{n,\lambda }\left(x\right)=1+\left({\gamma }_{n+1}-{\displaystyle \frac{{\gamma }_n}{1+\lambda }}\right)x^2+O\left(x^4\right)$$

explains ${\alpha }_n$. Lemma 2 supplies the global exclusion of further critical points. In the interior-maximum regime, write $y_{n,\lambda }=x_{n,\lambda }^2$. The critical-point equation is

$$D\left(y_{n,\lambda }\right)={\displaystyle \frac{{\gamma }_n}{1+\lambda }},D={\displaystyle \frac{H_{n+1}^{\prime }}{H_{n+1}^2}}.$$

(30)

5. Sharp Range and Motion of the Maximizer

The endpoint at $\pi /2$ produces a second parameter threshold. Since $H_{n+1}\left(y\right)\to \infty$ as $y\uparrow \rho$, equation (26) gives

$$\underset{x\to {(\pi /2)}^{-}}{lim}{R}_{n,\lambda }\left(x\right)={\displaystyle \frac{1+\lambda }{\rho {\gamma }_n}}={\displaystyle \frac{4(1+\lambda )}{{\pi }^2{\gamma }_n}}.$$

(31)

Extend $R_{n,\lambda }$ continuously to $x=\pi /2$ by this value and define

$${\beta }_n=\rho {\gamma }_n-1={\displaystyle \frac{{\pi }^2}{4}}{\gamma }_n-1.$$

(32)

The two thresholds satisfy $0<{\alpha }_n<{\beta }_n$. The inequality ${\alpha }_n>0$ follows from Lemma 1, and

$${\beta }_n-{\alpha }_n={\gamma }_n\left(\rho -{\displaystyle \frac{1}{{\gamma }_{n+1}}}\right)>0$$

by (10).

Corollary 1. 

Let $n\in \mathbb{N}$ and $\lambda >-1$. Put

$$E_{n,\lambda }={\displaystyle \frac{4(1+\lambda )}{{\pi }^2{\gamma }_n}}.$$

Then the minimum of $R_{n,\lambda }$ on $[0,\pi /2]$ is

$$m_{n,\lambda }=\left\{\begin{array}{cc}{E}_{n,\lambda },\hfill & -1<\lambda \le {\beta }_n,\hfill \\ 1,\hfill & \lambda \ge {\beta }_n,\hfill \end{array}\right.$$

(33)

where at $\lambda ={\beta }_n$ both endpoints are minimizers. Its maximum is

$$M_{n,\lambda }=\left\{\begin{array}{cc}1,\hfill & -1<\lambda \le {\alpha }_n,\hfill \\ R_{n,\lambda }\left(x_{n,\lambda }\right),\hfill & \lambda >{\alpha }_n.\hfill \end{array}\right.$$

(34)

Consequently,

$$m_{n,\lambda }\le R_{n,\lambda }\left(x\right)\le M_{n,\lambda }\left(0\le x\le {\displaystyle \frac{\pi }{2}}\right),$$

and both constants are best possible.

Proof. 

The endpoint values are 1 and $E_{n,\lambda }$. Theorem 2 gives the maximum and excludes an interior minimum. Comparing the endpoints gives $E_{n,\lambda }\le 1$ exactly when $\lambda \le {\beta }_n$. The constants are sharp because they are the actual extrema of the continuous extension. □

The two thresholds therefore have different geometric roles. Table 1 summarizes the resulting phase diagram.

The maximizer varies with the parameter as follows.

Corollary 2. 

For fixed $n\in \mathbb{N}$, the maximizer $x_{n,\lambda }$ is analytic and strictly increasing for $\lambda >{\alpha }_n$, with

$$\underset{\lambda \downarrow {\alpha }_n}{lim}{x}_{n,\lambda }=0,\underset{\lambda \to \infty }{lim}{x}_{n,\lambda }={\displaystyle \frac{\pi }{2}}.$$

(35)

The maximum $M_{n,\lambda }$ is strictly increasing on this interval, and its corresponding limits are 1 and ∞.

Proof. 

Equation (30) is the inverse of the analytic map

$$\lambda \left(y\right)={\displaystyle \frac{{\gamma }_n}{D\left(y\right)}}-1,{\lambda }^{\prime }\left(y\right)=-{\displaystyle \frac{{\gamma }_n{D}^{\prime }\left(y\right)}{D{\left(y\right)}^2}}>0.$$

This map sends $(0,\rho )$ onto $({\alpha }_n,\infty )$, proving all assertions about the maximizer by the inverse function theorem. For fixed $y>0$, equation (26) is strictly increasing in $\lambda$. Evaluating at the maximizing point for the smaller parameter proves the strict increase of $M_{n,\lambda }$. Its limit at ${\alpha }_n$ is 1 by continuity and (35). Finally, ${\mathcal{R}}_{n,\lambda }\left(y\right)\to H_{n+1}\left(y\right)$ for each $y<\rho$, and $H_{n+1}$ is unbounded. The maximum therefore tends to infinity. □

The maximizer has the following asymptotic formulas near $\lambda ={\alpha }_n$ and as $\lambda \to \infty$.

Corollary 3. 

Fix $n\in \mathbb{N}$, and put

$$A_n={\displaystyle \frac{{\gamma }_{n+1}}{2{\gamma }_n({\gamma }_{n+1}-{\gamma }_{n+2})}}.$$

As $\lambda \downarrow {\alpha }_n$,

$$y_{n,\lambda }=A_n(\lambda -{\alpha }_n)+O\left({(\lambda -{\alpha }_n)}^2\right),$$

(36)

and hence

$$x_{n,\lambda }\sim \sqrt{A_n}{(\lambda -{\alpha }_n)}^{1/2}.$$

(37)

As $\lambda \to \infty$,

$$\begin{array}{cc}\hfill \rho -y_{n,\lambda }& \sim {\displaystyle \frac{{\pi }^2}{2a_n{\rho }^{n+1}(1+\lambda )}},\hfill \end{array}$$

(38)

$$\begin{array}{cc}\hfill {\displaystyle \frac{\pi }{2}}-x_{n,\lambda }& \sim {\displaystyle \frac{\pi }{2a_n{\rho }^{n+1}(1+\lambda )}},\hfill \end{array}$$

(39)

$$\begin{array}{cc}\hfill M_{n,\lambda }& \sim {\displaystyle \frac{1+\lambda }{\rho {\gamma }_n}}.\hfill \end{array}$$

(40)

Proof. 

Let

$$F(y,\lambda )=H^{\prime }\left(y\right)-c_{\lambda }H{\left(y\right)}^2,c_{\lambda }={\displaystyle \frac{{\gamma }_n}{1+\lambda }},H=H_{n+1}.$$

The critical-point equation (30) is $F(y,\lambda )=0$. At $(0,{\alpha }_n)$,

$$H\left(0\right)=1,H^{\prime }\left(0\right)={\gamma }_{n+1},H^{\prime \prime }\left(0\right)=2{\gamma }_{n+1}{\gamma }_{n+2},c_{{\alpha }_n}={\gamma }_{n+1}.$$

Therefore

$$F_y(0,{\alpha }_n)=-2{\gamma }_{n+1}({\gamma }_{n+1}-{\gamma }_{n+2})<0$$

and

$$F_{\lambda }(0,{\alpha }_n)={\displaystyle \frac{{\gamma }_{n+1}^2}{{\gamma }_n}}>0.$$

Since H is analytic near 0, the implicit function theorem gives a smooth local root through $(0,{\alpha }_n)$, and Corollary 2 shows that, on the right of ${\alpha }_n$, this root is the unique critical point $y_{n,\lambda }$. Thus

$${\left.{\displaystyle \frac{d{y}_{n,\lambda }}{d\lambda }}\right|}_{\lambda ={\alpha }_n^{+}}=-{\displaystyle \frac{F_{\lambda }(0,{\alpha }_n)}{F_y(0,{\alpha }_n)}}=A_n,$$

which proves (36) and (37).

For the large-parameter limit, write ${\Delta }_{\lambda }=\rho -y_{n,\lambda }$. The proof of Theorem 2 gives

$$H\left(y\right)\sim C_n{(\rho -y)}^{-2},H^{\prime }\left(y\right)\sim 2C_n{(\rho -y)}^{-3},C_n={\displaystyle \frac{{\pi }^2}{a_{n+1}{\rho }^{n+1}}}.$$

Since $y_{n,\lambda }\uparrow \rho$, the critical equation implies

$${\displaystyle \frac{{\gamma }_n}{1+\lambda }}={\displaystyle \frac{H^{\prime }\left(y_{n,\lambda }\right)}{H{\left(y_{n,\lambda }\right)}^2}}\sim {\displaystyle \frac{2{\Delta }_{\lambda }}{C_n}}.$$

Hence

$${\Delta }_{\lambda }\sim {\displaystyle \frac{C_n{\gamma }_n}{2(1+\lambda )}}={\displaystyle \frac{{\pi }^2}{2a_n{\rho }^{n+1}(1+\lambda )}},$$

which is (38). Since

$$\rho -y_{n,\lambda }=\left({\displaystyle \frac{\pi }{2}}-x_{n,\lambda }\right)\left({\displaystyle \frac{\pi }{2}}+x_{n,\lambda }\right)$$

and $x_{n,\lambda }\to \pi /2$, equation (39) follows.

Finally, at the critical point,

$${\displaystyle \frac{M_{n,\lambda }}{(1+\lambda )/\left(\rho {\gamma }_n\right)}}={\displaystyle \frac{c_{\lambda }\rho H\left(y_{n,\lambda }\right)}{1+c_{\lambda }{y}_{n,\lambda }H\left(y_{n,\lambda }\right)}}.$$

The preceding asymptotics give $c_{\lambda }{y}_{n,\lambda }H\left(y_{n,\lambda }\right)\to \infty$, while $y_{n,\lambda }\to \rho$. The last quotient therefore tends to 1, proving (40). □

6. Threshold Asymptotics and the Case $N=1$

The two thresholds tend to zero at different rates.

Corollary 4. 

As $n\to \infty$,

$$\begin{array}{cc}\hfill {\alpha }_n& ={\displaystyle \frac{4}{(2n+1)(2n+5)}}+O\left(4^{-n}\right),\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill {\beta }_n& ={\displaystyle \frac{2}{2n+1}}+O\left(4^{-n}\right).\hfill \end{array}$$

(42)

In particular,

$${\alpha }_n\sim n^{-2},{\beta }_n\sim {\displaystyle \frac{1}{n}},{\displaystyle \frac{{\alpha }_n}{{\beta }_n}}⟶0.$$

Proof. 

From (8),

$${\gamma }_n={\displaystyle \frac{4}{{\pi }^2}}{\displaystyle \frac{2n+3}{2n+1}}{\epsilon }_n,$$

where

$${\epsilon }_n={\displaystyle \frac{1-2^{-2n-4}}{1-2^{-2n-2}}}{\displaystyle \frac{\zeta (2n+4)}{\zeta (2n+2)}}=1+O\left(4^{-n}\right).$$

The estimate $0<\zeta \left(s\right)-1\le 2^{-s}+2^{1-s}/(s-1)$ for $s>1$ gives the stated error term. Therefore

$${\beta }_n={\displaystyle \frac{2n+3}{2n+1}}{\epsilon }_n-1={\displaystyle \frac{2}{2n+1}}+O\left(4^{-n}\right),$$

which proves (42). Also

$${\displaystyle \frac{{\gamma }_n}{{\gamma }_{n+1}}}={\displaystyle \frac{{(2n+3)}^2}{(2n+1)(2n+5)}}{\displaystyle \frac{{\epsilon }_n}{{\epsilon }_{n+1}}}.$$

Since

$${\displaystyle \frac{{(2n+3)}^2}{(2n+1)(2n+5)}}=1+{\displaystyle \frac{4}{(2n+1)(2n+5)}}$$

and ${\epsilon }_n/{\epsilon }_{n+1}=1+O\left(4^{-n}\right)$, equation (41) follows. □

The first few threshold values are listed in the next table.

Table 2. The first shape and endpoint thresholds (decimal values rounded to ten places).

n	${\mathit{\gamma }}_{\mathit{n}}$	${\mathit{\alpha }}_{\mathit{n}}$	${\mathit{\beta }}_{\mathit{n}}$
1	$2/3$	$0.1764705882$	$0.6449340668$
2	$17/30$	$0.0876344086$	$0.3981939568$
3	$62/119$	$0.0518187014$	$0.2855367077$
4	$691/1395$	$0.0341740947$	$0.2222036991$

Table 2. The first shape and endpoint thresholds (decimal values rounded to ten places).

n	${\mathit{\gamma }}_{\mathit{n}}$	${\mathit{\alpha }}_{\mathit{n}}$	${\mathit{\beta }}_{\mathit{n}}$
1	$2/3$	$0.1764705882$	$0.6449340668$
2	$17/30$	$0.0876344086$	$0.3981939568$
3	$62/119$	$0.0518187014$	$0.2855367077$
4	$691/1395$	$0.0341740947$	$0.2222036991$

For $n=1$, the ratio has an explicit expression. Here $a_1=1$, ${\gamma }_1=2/3$, and ${\gamma }_2=17/30$, so

$${\alpha }_1={\displaystyle \frac{3}{17}},{\beta }_1={\displaystyle \frac{{\pi }^2}{6}}-1.$$

(43)

The perturbation is

$$f_{1,\lambda }\left(x\right)={tan}^{2}x+\lambda x^2,$$

and substitution into (1) gives

$$R_{1,\lambda }\left(x\right)={\displaystyle \frac{3(1+\lambda )({tan}^{2}x-x^2)}{2x^2({tan}^{2}x+\lambda x^2)}}.$$

(44)

Thus $R_{1,\lambda }$ decreases for $-1<\lambda \le 3/17$ and has one interior maximum for $\lambda >3/17$. Its minimum is $min\{1,6(1+\lambda )/{\pi }^2\}$. At $\lambda =0$ this recovers the unperturbed ratio in [6].

7. The Two-Parameter Phase Diagram for Perturbations of Two Consecutive Coefficients

For two consecutive coefficients, Theorem 1 supplies the critical-point bound. The phase boundaries can then be determined exactly. Put

$$f_{n,\lambda ,\mu }\left(x\right)={tan}^{2}x+\lambda a_n{x}^{2n}+\mu a_{n+1}{x}^{2n+2},\lambda ,\mu >-1.$$

Let

$$R_{n,\lambda ,\mu }\left(x\right)={\displaystyle \frac{T_{2n+1}\left[f_{n,\lambda ,\mu }\left(x\right)\right]}{T_{2n-1}\left[f_{n,\lambda ,\mu }\left(x\right)\right]}}.$$

With $y=x^2$, $H=H_{n+2}$, define

$$c_{\mu }={\displaystyle \frac{{\gamma }_{n+1}}{1+\mu }},C_{\mu }={\gamma }_n(1+\mu ),A_{\lambda ,\mu }={\displaystyle \frac{C_{\mu }}{1+\lambda }},$$

$$U_{\mu }\left(y\right)=1+c_{\mu }yH\left(y\right),D_{\mu }\left(y\right)={\displaystyle \frac{U_{\mu }^{\prime }\left(y\right)}{U_{\mu }{\left(y\right)}^2}}.$$

Then

$$R_{n,\lambda ,\mu }\left(\sqrt{y}\right)={\displaystyle \frac{U_{\mu }\left(y\right)}{1+A_{\lambda ,\mu }y{U}_{\mu }\left(y\right)}},$$

(45)

and its derivative has the sign of $D_{\mu }\left(y\right)-A_{\lambda ,\mu }$.

The adjacent coefficient ratios of $U_{\mu }$ are

$$c_{\mu },{\gamma }_{n+2},{\gamma }_{n+3},\dots .$$

The Biernacki–Krzyż rule and the one-turn theorem [3] applied to $U_{\mu }^{\prime }/(U_{\mu }+y{U}_{\mu }^{\prime })$ show that $D_{\mu }$ is strictly decreasing when $\mu \le {\alpha }_{n+1}$, whereas for $\mu >{\alpha }_{n+1}$ it has at most one turn. In the latter case

$$D_{\mu }^{\prime }\left(0\right)=2c_{\mu }({\gamma }_{n+2}-c_{\mu })>0,D_{\mu }\left(y\right)⟶0(y\uparrow \rho ),$$

so it increases and then decreases, with a unique strict maximum. In this case let

$$s_{n,\mu }=\underset{0<y<\rho }{arg\; max}{D}_{\mu }\left(y\right),M_n\left(\mu \right)=D_{\mu }\left(s_{n,\mu }\right).$$

Define

$$L_n\left(\mu \right)={\displaystyle \frac{{\gamma }_n}{{\gamma }_{n+1}}}{(1+\mu )}^2-1,$$

(46)

and, for $\mu >{\alpha }_{n+1}$,

$${\Lambda }_n\left(\mu \right)={\displaystyle \frac{C_{\mu }}{M_n\left(\mu \right)}}-1.$$

(47)

Then $-1<{\Lambda }_n\left(\mu \right)<L_n\left(\mu \right)$.

Theorem 3. 

If $-1<\mu \le {\alpha }_{n+1}$, then $R_{n,\lambda ,\mu }$ is strictly decreasing for $\lambda \le L_n\left(\mu \right)$ and has one interior critical point, a strict global maximum, for $\lambda >L_n\left(\mu \right)$.

If $\mu >{\alpha }_{n+1}$, then it is strictly decreasing for $\lambda <{\Lambda }_n\left(\mu \right)$. At $\lambda ={\Lambda }_n\left(\mu \right)$ it is still strictly decreasing but has one stationary point where the derivative does not change sign. For

$${\Lambda }_n\left(\mu \right)<\lambda <L_n\left(\mu \right)$$

it has exactly two interior critical points and is strictly decreasing, then increasing, then decreasing. For $\lambda \ge L_n\left(\mu \right)$ it has one interior critical point, a strict global maximum.

Proof. 

The critical points are exactly the intersections of $D_{\mu }$ with the horizontal line $A_{\lambda ,\mu }$. The two parameter boundaries are $A=M_n\left(\mu \right)$ and $A=c_{\mu }$, which are respectively $\lambda ={\Lambda }_n\left(\mu \right)$ and $\lambda =L_n\left(\mu \right)$. □

Theorem 3 determines the local shape but not whether the local extrema are global after comparison with the endpoints. Put

$$B_n\left(\mu \right)=\rho {\gamma }_n(1+\mu )-1.$$

(48)

At $x=\pi /2$, extend $R_{n,\lambda ,\mu }$ continuously by its left-hand limit. The endpoint values are

$$R_{n,\lambda ,\mu }\left(0\right)=1,R_{n,\lambda ,\mu }\left({\displaystyle \frac{\pi }{2}}\right)={\displaystyle \frac{1+\lambda }{\rho {\gamma }_n(1+\mu )}}.$$

(49)

For $\mu >{\alpha }_{n+1}$ and $c_{\mu }<A<M_n\left(\mu \right)$, let $y_{-}\left(A\right)<s_{n,\mu }<y_{+}\left(A\right)$ be the two solutions of $D_{\mu }\left(y\right)=A$. These branches extend continuously to the relevant endpoints. Set

$$y_{-}\left(c_{\mu }\right)=0,y_{-}\left(M_n\left(\mu \right)\right)=y_{+}\left(M_n\left(\mu \right)\right)=s_{n,\mu },$$

and let $y_{+}\left(c_{\mu }\right)\in (s_{n,\mu },\rho )$ denote the unique positive solution of $D_{\mu }\left(y\right)=c_{\mu }$. Set

$$V_{\mu }={\displaystyle \frac{1}{U_{\mu }}},F_{\mu ,A}\left(y\right)=V_{\mu }\left(y\right)+Ay.$$

Then $1/R=F$ and ${\partial }_{y}F=A-D_{\mu }$.

There is a unique $A_{+}\left(\mu \right)\in (c_{\mu },M_n\left(\mu \right))$ satisfying

$$F_{\mu ,A_{+}}\left(y_{+}\left(A_{+}\right)\right)=1.$$

(50)

For $K_{+}\left(A\right)=F_{\mu ,A}\left(y_{+}\left(A\right)\right)-1$, the envelope identity gives $K_{+}^{\prime }\left(A\right)=y_{+}\left(A\right)>0$. At the fold $A=M_n\left(\mu \right)$ the ratio R is strictly decreasing apart from its stationary point, so $F\left(s_{n,\mu }\right)>F\left(0\right)=1$ and $K_{+}\left(M_n\left(\mu \right)\right)>0$. At $A=c_{\mu }$ the ratio increases immediately to its positive interior maximum, hence $F\left(y_{+}\left(c_{\mu }\right)\right)<F\left(0\right)=1$ and $K_{+}\left(c_{\mu }\right)<0$. Define

$${\Theta }_n\left(\mu \right)={\displaystyle \frac{C_{\mu }}{A_{+}\left(\mu \right)}}-1,$$

(51)

so

$${\Lambda }_n\left(\mu \right)<{\Theta }_n\left(\mu \right)<L_n\left(\mu \right).$$

A second global switch exists exactly when $\mu >{\beta }_{n+1}$. In that case there is a unique $A_{-}\left(\mu \right)\in (c_{\mu },M_n\left(\mu \right))$ such that

$$F_{\mu ,A_{-}}\left(y_{-}\left(A_{-}\right)\right)=A_{-}\rho .$$

(52)

For $K_{-}\left(A\right)=F_{\mu ,A}\left(y_{-}\left(A\right)\right)-A\rho$ one has $K_{-}^{\prime }\left(A\right)=y_{-}\left(A\right)-\rho <0$. At the fold the ratio is strictly decreasing, so its stationary value is larger than the right endpoint value. Equivalently $K_{-}\left(M_n\left(\mu \right)\right)<0$. Moreover,

$$\underset{A\downarrow c_{\mu }}{lim}{K}_{-}\left(A\right)=1-c_{\mu }\rho .$$

This limit is positive exactly when $\mu >{\beta }_{n+1}$. In this case $c_{\mu }<1/\rho$. Since $V_{\mu }=1/U_{\mu }$ satisfies $V_{\mu }^{\prime }=-D_{\mu }$, $V_{\mu }\left(0\right)=1$, and $V_{\mu }\left(y\right)\to 0$ as $y\uparrow \rho$, one has

$${\int }_0^{\rho }{D}_{\mu }\left(y\right)dy=1.$$

Thus the strict maximum $M_n\left(\mu \right)$ is larger than the average value $1/\rho$, so $A=1/\rho$ lies in the two-critical interval $(c_{\mu },M_n\left(\mu \right))$. Define

$${\Gamma }_n\left(\mu \right)={\displaystyle \frac{C_{\mu }}{A_{-}\left(\mu \right)}}-1.$$

(53)

Then

$${\Lambda }_n\left(\mu \right)<{\Gamma }_n\left(\mu \right)<B_n\left(\mu \right)<L_n\left(\mu \right).$$

(54)

At $A=1/\rho$ both endpoint values of F are 1, while $F_y\left(0\right)=1/\rho -c_{\mu }>0$ and the first critical point is a strict local maximum of F. Hence $K_{-}(1/\rho )>0$. Since $K_{-}$ is strictly decreasing and vanishes at $A_{-}\left(\mu \right)$, it follows that $A_{-}\left(\mu \right)>1/\rho$, which is exactly ${\Gamma }_n\left(\mu \right)<B_n\left(\mu \right)$. The remaining inequalities follow from $A_{-}\left(\mu \right)\in (c_{\mu },M_n\left(\mu \right))$ and $\mu >{\beta }_{n+1}$.

The two switch functions control different competitions. The classification does not require a global ordering between ${\Theta }_n\left(\mu \right)$ and ${\Gamma }_n\left(\mu \right)$. When both are defined, the former governs the global maximum and the latter the global minimum.

Theorem 4. 

The global maximum of $R_{n,\lambda ,\mu }$ is as follows. If $\mu \le {\alpha }_{n+1}$, it is 1 at $x=0$ for $\lambda \le L_n\left(\mu \right)$ and is attained uniquely at the interior maximum for $\lambda >L_n\left(\mu \right)$. If $\mu >{\alpha }_{n+1}$, it is 1 at $x=0$ for $\lambda <{\Theta }_n\left(\mu \right)$. At $\lambda ={\Theta }_n\left(\mu \right)$ the left endpoint and interior local maximum are tied. For $\lambda >{\Theta }_n\left(\mu \right)$ the interior maximum is the unique global maximum.

For the global minimum, if $\mu \le {\beta }_{n+1}$ it is attained at $x=\pi /2$ for $\lambda <B_n\left(\mu \right)$, at both endpoints for $\lambda =B_n\left(\mu \right)$, and at $x=0$ for $\lambda >B_n\left(\mu \right)$. If $\mu >{\beta }_{n+1}$, it is attained at $x=\pi /2$ for $\lambda <{\Gamma }_n\left(\mu \right)$. At $\lambda ={\Gamma }_n\left(\mu \right)$ the right endpoint and interior local minimum are tied. For ${\Gamma }_n\left(\mu \right)<\lambda <L_n\left(\mu \right)$ the interior local minimum is the unique global minimum. For $\lambda \ge L_n\left(\mu \right)$ the unique global minimum is $x=0$.

Proof. 

In the two-critical regime the local minimum of R is the local maximum of F, and the local maximum of R is the local minimum of F. The local maximum always exceeds the right endpoint, so its only global competition is with $R\left(0\right)=1$, giving the threshold ${\Theta }_n\left(\mu \right)$. The local minimum of R is always below the left endpoint, while its competition with the right endpoint is exactly the equation defining ${\Gamma }_n\left(\mu \right)$. Outside the two-critical region, endpoint comparison and Theorem 3 finish the classification. □

Remark 2. 

The one-turn quotient argument used in this section is classical. Theorems 3 and 4 specialize it to the tangent-square coefficient sequence, proving that the two-critical region is nonempty and that the global-extremum switch functions are unique.

8. Conclusions

Finite consecutive coefficient perturbations of the normalized remainder ratio satisfy a sharp critical-point bound in the tangent-square family. After r consecutive coefficients are changed, the ratio has at most r interior critical points. The upper bound comes from Mao and Tian [2], and the tangent-square family attains it for every r through an explicit order-r degeneracy with a full-rank local unfolding.

The one-coefficient case has two exact thresholds at every remainder order, together with closed sharp ranges, monotone motion of the maximizer, and parameter and order asymptotics. With two consecutive perturbations, the ratio has a local minimum followed by a local maximum throughout an open parameter region. Unique switch functions determine when these extrema overtake the relevant endpoints. Together, the finite-block bound, the one-coefficient sharp analysis, and the two-parameter phase diagram describe the critical-point geometry at three levels of the same perturbation problem.