Critical-Point Counts and Sharp Coefficient Bounds for Logarithmic Polynomial Perturbations

1. Introduction

Sharp polynomial bounds for logarithmic functions have a long history. Early examples include Mitrinović [11], Love [6], and Bougoffa and Krasopoulos [4]. Shenton and Kemp [16] obtained rational bounds for ${log}^2(1+t)$, and Sándor ([15], Section 2) gave another proof and refinements. Streeter and Dillon developed sharp Taylor polynomial enclosures ([17], Definition 1 and Proposition 1), while Qi surveyed normalized Taylor remainders [14]. Other related work includes the renormalization results of Abu-Ghuwaleh [1] and the parameter thresholds studied by Suna and Ray [18]. Malešević et al. combine stratified families with Taylor expansions to select best constants [7]. Mićović et al. use stationary points in stratified families to obtain inequalities and minimax approximations [9]. Mićović, Malešević, and Zhu extend constant selection to multi-parameter families [10]. These works concern best-constant selection rather than the critical-point classification below.

Consider the positive-coefficient polynomial family

$$p\left(x\right)=1+x+{\displaystyle \sum _{j=2}^m}{a}_j{x}^j,a_j>0,$$

(1)

and the normalized logarithmic remainder

$$Q_p\left(x\right)={\displaystyle \frac{logp\left(x\right)-x}{x^2}}.$$

(2)

A finite-polynomial reduction of Mao–Tian’s monotonicity-variation rule ([8], Monotonicity Rule 18) gives, for a degree-m polynomial, at most $m-1$ positive critical points. The upper bound therefore follows from established power-series monotonicity theory. For related variation-diminishing and sign-regular results, see Karp, Vishnyakova, and Zhang [5]. The remaining question is sharpness within the restricted family (1). For every $m\ge 3$ there is a positive polynomial with an exact order-$(m-1)$ boundary degeneracy. Its coefficient-to-jet map has full rank, so every prescribed ordered configuration of $m-1$ sufficiently small positive simple critical points can be realized after scaling.

The sharpness construction comes from the probabilists’ Hermite polynomials. Only their classical generating function and zero interlacing enter the argument ([12], Equation 18.12.16 and §18.2(vi)). For $m=3$ the construction gives the explicit positive pair

$${\lambda }_{*}={\displaystyle \frac{1}{\sqrt{6}}},{\eta }_{*}={\displaystyle \frac{1}{\sqrt{6}}}-{\displaystyle \frac{1}{3}},$$

which is the base point for a more detailed two-parameter study of

$$p_{\lambda ,\eta }\left(x\right)=1+x+\lambda x^2+\eta x^3.$$

(3)

For this cubic polynomial, the number and order of positive critical points are determined throughout the positive parameter quadrant. The fold extends to the boundary $\lambda =0$, where a one-variable parametrization gives a unique endpoint ${\eta }_c$. Near $({\lambda }_{*},{\eta }_{*})$, a second analytic curve marks where the positive local minimum equals the value at the origin. Across that curve, the active minimizer for the sharp lower quadratic coefficient changes, and the coefficient loses differentiability in a transverse parameter direction.

The degree-two case $p\left(x\right)=1+x+\lambda x^2$ admits a complementary exact analysis on $[0,1]$. Besides the quadratic profile (2), consider the normalized cubic profile

$$C_{\lambda }\left(x\right)={\displaystyle \frac{log(1+x+\lambda x^2)-x-(\lambda -1/2)x^2}{x^3}}.$$

The two shape classifications give the best quadratic and cubic coefficients in the prescribed polynomial classes, including the parameter ranges where an interior point rather than an endpoint supplies the sharp coefficient. General one-turn criteria include the higher-order fraction rules of Bitsouni, Gialelis, and Marinescu ([3], Theorem 3.2) and the l’Hospital-type results of Pinelis ([13], Propositions 4.1, 4.3, and 4.4). The thresholds below are instead determined by the derivative factorizations and endpoint signs of this family.

For the detailed degree-two sections, write

$$q_{\lambda }\left(x\right)=1+x+\lambda x^2,F_{\lambda }\left(x\right)=log{q}_{\lambda }\left(x\right).$$

As $x\to 0$ with $\lambda$ fixed,

$$F_{\lambda }\left(x\right)=x+\left(\lambda -{\displaystyle \frac{1}{2}}\right)x^2+\left({\displaystyle \frac{1}{3}}-\lambda \right)x^3+\left(-{\displaystyle \frac{1}{2}}{\lambda }^2+\lambda -{\displaystyle \frac{1}{4}}\right)x^4+O\left(x^5\right).$$

(4)

The exact numerical brackets used below follow from

$$logy=2{\displaystyle \sum _{j=0}^{n-1}}{\displaystyle \frac{z^{2j+1}}{2j+1}}+R_n,z={\displaystyle \frac{y-1}{y+1}},0<R_n<{\displaystyle \frac{2z^{2n+1}}{(2n+1)(1-z^2)}},$$

(5)

valid for $y>1$ and $n\ge 1$. The identity and remainder estimate follow by integrating the geometric series for ${(1-z^2)}^{-1}$ and bounding the positive tail.

Sections 2–3 establish the general degree bound and its sharp Hermite realization for (1). Section 4–6 analyze the cubic phase diagram, the unique fold endpoint on $\lambda =0$, and the local switch in the sharp lower coefficient. Section 7–9 give the degree-two sharp-bound analysis.

2. Critical-Point Bound in Terms of Degree

Let

$$p\left(x\right)=1+x+{\displaystyle \sum _{j=2}^m}{a}_j{x}^j,m\ge 2,a_j>0,$$

(6)

and

$$Q_p\left(x\right)={\displaystyle \frac{logp\left(x\right)-x}{x^2}},x>0.$$

(7)

At $x=0$, $Q_p$ and its derivatives mean those of its analytic extension, which exists since $p\left(0\right)=p^{\prime }\left(0\right)=1$. Put

$$h_p\left(x\right)={\displaystyle \frac{{(logp)}^{\prime }\left(x\right)-1}{2x}}.$$

(8)

Then

$$x^2(h_p-Q_p)={\int }_0^x{t}^2{h}_p^{\prime }\left(t\right)dt,Q_p^{\prime }\left(x\right)={\displaystyle \frac{2}{x}}(h_p-Q_p).$$

(9)

Lemma 1. 

Let $A,B$ be real polynomials of degree at most m, and assume every coefficient of B is strictly positive. Then $A/B$ changes monotonicity at most $m-1$ times on $(0,\infty )$.

Proof. 

Fix $R>0$ and $0<\tau <1/R$. Multiply numerator and denominator by ${(1-\tau x)}^{-1}$. The quotient is unchanged, while the new denominator is an infinite power series with strictly positive coefficients. Starting at index m, the new numerator and denominator coefficients are geometric with the same ratio $\tau$, hence their coefficient ratio is constant from that index onward. The resulting coefficient-ratio sequence can therefore change monotonicity at most $m-1$ times. Mao–Tian ([8], Monotonicity Rule 18) gives the same bound for $A/B$ on $(0,R)$. Since R is arbitrary, the claim follows. □

Theorem 1. 

For p in (6), the normalized logarithmic remainder $Q_p$ has at most $m-1$ distinct positive critical points.

Proof. 

The function $h_p$ is a quotient $A/p$, with A of degree at most $m-1$. After padding the numerator by a zero coefficient in degree m, Lemma 1 gives at most $m-1$ monotonicity changes of $h_p$. Let $J\left(x\right)={\int }_0^x{t}^2{h}_p^{\prime }\left(t\right)dt$. The rational function $h_p$ is not constant, since $h_p\left(x\right)\sim -1/\left(2x\right)$ at infinity. On the first monotonicity interval, J starts at zero and then has a fixed nonzero sign. Each later monotonicity interval contributes at most one new zero of J. Hence J has at most $m-1$ positive zeros. Equation (9) gives the result. □

3. Sharpness Via Hermite Polynomials

Let ${\mathrm{He}}_n$ be the probabilists’ Hermite polynomial defined by ([12], Equation 18.12.16)

$$e^{zt-t^2/2}={\displaystyle \sum _{j=0}^{\infty }}{\mathrm{He}}_j\left(z\right){\displaystyle \frac{t^j}{j!}}.$$

(10)

For $m\ge 3$, let $z_m$ be the largest positive zero of ${\mathrm{He}}_{m+1}$ and put ${\sigma }_m=1/z_m$. Define

$$c_j={\displaystyle \frac{{\sigma }_m^j{\mathrm{He}}_j\left(z_m\right)}{j!}},P_m\left(x\right)={\displaystyle \sum _{j=0}^m}{c}_j{x}^j.$$

(11)

By simplicity and interlacing of orthogonal-polynomial zeros ([12], §18.2(vi)), $z_m$ lies to the right of all zeros of ${\mathrm{He}}_j$ for $j\le m$. Since ${\mathrm{He}}_j$ is monic, $c_0=c_1=1$ and $c_j>0$ for $2\le j\le m$.

Theorem 2. 

For every $m\ge 3$, the upper bound in Theorem 1 is sharp. More precisely, $P_m$ has an order-$(m-1)$ critical degeneracy at the origin. For every $0<{\xi }_1<\dots <{\xi }_{m-1}$, there are positive-coefficient polynomials arbitrarily close to $P_m$ whose normalized logarithmic remainders have exactly $m-1$ simple positive critical points $x_i\left(\epsilon \right)$ with

$$x_i\left(\epsilon \right)=\epsilon {\xi }_i+o\left(\epsilon \right).$$

Proof. 

Putting $z=z_m$ and $t={\sigma }_{m}x$ in (10) gives

$$e^{x-{\sigma }_m^2{x}^2/2}={\displaystyle \sum _{j=0}^{\infty }}{c}_j{x}^j.$$

(12)

The Hermite recurrence becomes

$$(j+1)c_{j+1}=c_j-{\sigma }_m^2{c}_{j-1}.$$

(13)

Since $c_{m+1}=0$,

$$c_{m+2}=-{\displaystyle \frac{{\sigma }_m^2{c}_m}{m+2}}<0.$$

(14)

The numerator of $h_{P_m}$ has, after padding by zero in degree m, coefficient-to-denominator ratios

$$-{\displaystyle \frac{{\sigma }_m^2}{2}},\dots ,-{\displaystyle \frac{{\sigma }_m^2}{2}},0.$$

For finite polynomials $A=\sum A_j{x}^j$ and $B=\sum B_j{x}^j$ with $B_j>0$, writing $r_j=A_j/B_j$ gives the exact identity

$$A^{\prime }B-A{B}^{\prime }=\sum _{j>k}(j-k)B_j{B}_k(r_j-r_k)x^{j+k-1}.$$

The displayed ratios are nondecreasing and not all equal, so this identity shows directly that $h_{P_m}^{\prime }\left(x\right)>0$ for every $x>0$. By (9),

$$Q_{P_m}^{\prime }\left(x\right)>0(x>0).$$

(15)

Meanwhile (12)–(14) give

$$Q_{P_m}\left(x\right)=-{\displaystyle \frac{{\sigma }_m^2}{2}}+{\displaystyle \frac{{\sigma }_m^2{c}_m}{m+2}}{x}^m+O\left(x^{m+1}\right),$$

(16)

so

$$Q_{P_m}^{\prime }\left(x\right)={\kappa }_m{x}^{m-1}+O\left(x^m\right),{\kappa }_m={\displaystyle \frac{m{\sigma }_m^2{c}_m}{m+2}}>0.$$

(17)

For $P_{\mathbf{a}}\left(x\right)=1+x+{\sum }_{j=2}^m{a}_j{x}^j$, write $\left[x^k\right]Q_{P_{\mathbf{a}}}^{\prime }=b_k\left(\mathbf{a}\right)$. At the reference vector, $b_0=\dots =b_{m-2}=0$. Let

$$P_m{\left(x\right)}^{-1}={\displaystyle \sum _{j=0}^{\infty }}{d}_j{x}^j.$$

If $L_n=\left[x^n\right]log{P}_{\mathbf{a}}$, then $\partial L_n/\partial a_j=d_{n-j}$. Hence the Jacobian of $(L_3,\dots ,L_{m+1})$ with respect to $(a_2,\dots ,a_m)$ is

$$T_{m-1}=\left(\begin{array}{ccccc}{d}_1& d_0& 0& \dots & 0\\ d_2& d_1& d_0& \ddots & ⋮\\ ⋮& \ddots & \ddots & \ddots & 0\\ d_{m-1}& \dots & d_2& d_1& d_0\end{array}\right).$$

The convolution identity $P_m\sum d_j{x}^j=1$ and induction on the determinant size give

$$det{T}_{m-1}={(-1)}^{m-1}{c}_{m-1}.$$

Since $b_k=(k+1)L_{k+3}$,

$$det{\displaystyle \frac{\partial (b_0,\dots ,b_{m-2})}{\partial (a_2,\dots ,a_m)}}={(-1)}^{m-1}(m-1)!c_{m-1}\ne 0.$$

(18)

The inverse function theorem therefore permits any sufficiently small choice of $(b_0,\dots ,b_{m-2})$.

Fix $0<{\xi }_1<\dots <{\xi }_{m-1}$ and prescribe the coefficients below degree $m-1$ of

$${\kappa }_m{\displaystyle \prod _{i=1}^{m-1}}(x-\epsilon {\xi }_i).$$

Let $\mathbf{a}\left(\epsilon \right)$ denote the resulting coefficient vector. These coefficients converge to the positive reference coefficients. After scaling $x=\epsilon t$, analyticity and (17) give

$${\epsilon }^{-(m-1)}{Q}_{P_{\mathbf{a}\left(\epsilon \right)}}^{\prime }\left(\epsilon t\right)⟶{\kappa }_m{\displaystyle \prod _{i=1}^{m-1}}(t-{\xi }_i)$$

in $C^1$ on compact t-intervals. Simple-root stability gives $m-1$ distinct simple positive critical points. Theorem 1 excludes any others. □

Remark 1. 

For $m=3$, ${\mathrm{He}}_4\left(z\right)=z^4-6z^2+3$ and $z_3=\sqrt{3+\sqrt{6}}$. Formula (11) gives

$$c_2={\displaystyle \frac{1}{\sqrt{6}}},c_3={\displaystyle \frac{1}{\sqrt{6}}}-{\displaystyle \frac{1}{3}}.$$

4. The Cubic Two-Parameter Phase Diagram

Set

$$q_{\lambda ,\eta }\left(x\right)=1+x+\lambda x^2+\eta x^3,Q_{\lambda ,\eta }\left(x\right)={\displaystyle \frac{log{q}_{\lambda ,\eta }\left(x\right)-x}{x^2}}.$$

Put

$${\lambda }_0\left(\eta \right)=\eta +{\displaystyle \frac{1}{3}},{\eta }_{*}={\displaystyle \frac{1}{\sqrt{6}}}-{\displaystyle \frac{1}{3}}.$$

(19)

The auxiliary quotient is

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

(20)

For $\lambda >0$, twice its coefficient-ratio sequence is

$$2\lambda -1,3\eta -\lambda ,-{\displaystyle \frac{\eta }{\lambda }},0.$$

(21)

Also

$${\partial }_{\lambda }{Q}_{\lambda ,\eta }={\displaystyle \frac{1}{q}},{\partial }_{\lambda }{Q}_{\lambda ,\eta }^{\prime }=-{\displaystyle \frac{q^{\prime }}{q^2}}<0,$$

(22)

and

$$Q_{\lambda ,\eta }\left(x\right)=\lambda -{\displaystyle \frac{1}{2}}+(\eta -\lambda +\frac{1}{3})x+\left(-\eta -{\displaystyle \frac{{\lambda }^2}{2}}+\lambda -{\displaystyle \frac{1}{4}}\right)x^2+O\left(x^3\right).$$

(23)

Theorem 3. 

The positive-parameter phase diagram is as follows.

1.

If $0<\eta \le {\eta }_{*}$, then $Q_{\lambda ,\eta }$ is strictly increasing for $0<\lambda \le {\lambda }_0\left(\eta \right)$. For $\lambda >{\lambda }_0\left(\eta \right)$ it has exactly one positive critical point, its unique global minimum on $[0,\infty )$.

2.

If $\eta >{\eta }_{*}$, there is a unique threshold $\Lambda \left(\eta \right)\in [0,{\lambda }_0\left(\eta \right))$ with the following properties.

(a)

If $0<\lambda <\Lambda \left(\eta \right)$, then $Q_{\lambda ,\eta }$ is strictly increasing.

(b)

If $\Lambda \left(\eta \right)>0$, then at $\lambda =\Lambda \left(\eta \right)$ there is a unique positive degenerate critical point.

(c)

If $\Lambda \left(\eta \right)<\lambda <{\lambda }_0\left(\eta \right)$, then there are exactly two positive critical points, a local maximum followed by a local minimum.

(d)

If $\lambda \ge {\lambda }_0\left(\eta \right)$, then there is exactly one simple positive critical point, the unique global minimum.

Proof. 

At $\lambda ={\lambda }_0\left(\eta \right)$, the first two entries of (21) agree. If $0<\eta \le {\eta }_{*}$, then

$$2\eta -{\displaystyle \frac{1}{3}}\le -{\displaystyle \frac{\eta }{\eta +1/3}}<0,$$

so the coefficient ratios in (20) are nondecreasing and not constant. The finite identity used in the proof of Theorem 2 therefore gives $h^{\prime }\left(x\right)>0$ for every $x>0$, and (9) gives $Q^{\prime }>0$. Equation (22) then gives the same conclusion for all smaller positive $\lambda$.

For the boundary and supercritical ranges, direct differentiation gives

$$h^{\prime }\left(x\right)={\displaystyle \frac{H_{\lambda ,\eta }\left(x\right)}{2q_{\lambda ,\eta }{\left(x\right)}^2}},$$

where, in ascending powers of x,

$$\begin{array}{cc}\hfill H_{\lambda ,\eta }\left(x\right)=& (1+3\eta -3\lambda )+(-2\eta -4{\lambda }^2+2\lambda )x\hfill \\ \hfill & +({\lambda }^2+2\eta -9\eta \lambda )x^2+2\eta (\lambda -3\eta )x^3+{\eta }^2{x}^4.\hfill \end{array}$$

If $\lambda \ge \eta +1/3$, these five coefficients have at most one sign change, from negative to positive. If the coefficient of x is positive, then $\eta <\lambda -2{\lambda }^2$, whence $\lambda >1/3$ and

$${\lambda }^2+2\eta -9\eta \lambda >2\lambda {(3\lambda -1)}^2>0.$$

The same inequality gives $\eta <1/6$ and hence $\lambda >3\eta$, so the coefficient of $x^3$ is positive. Conversely, if the coefficient of $x^2$ is positive, then the coefficient of $x^3$ must be positive. Otherwise $\lambda \le 3\eta$, which forces $\eta \ge 1/6$, while the $x^2$ coefficient is decreasing in $\lambda$ on $[\eta +1/3,3\eta ]$ and at the left endpoint equals

$$-{\displaystyle \frac{72{\eta }^2+3\eta -1}{9}}<0,$$

a contradiction. Thus once a positive coefficient occurs, every later nonzero coefficient is positive.

If $\lambda >{\lambda }_0\left(\eta \right)$, the constant coefficient of H is negative and the leading coefficient is positive. Descartes’ rule ([2], Proposition 1) therefore gives exactly one positive zero of H, so h decreases and then increases, with a unique strict minimum. At $\lambda ={\lambda }_0\left(\eta \right)$ and $\eta >{\eta }_{*}$, the constant coefficient vanishes and the coefficient of x is

$$-{\displaystyle \frac{2(18{\eta }^2+12\eta -1)}{9}}<0.$$

After removing the factor x, the same sign argument gives exactly one positive zero. Hence h again decreases and then increases, with a unique strict minimum.

If $\lambda >{\lambda }_0\left(\eta \right)$, (23) gives $Q^{\prime }(0+)<0$, while

$$Q^{\prime }\left(x\right)={\displaystyle \frac{1}{x^2}}+O\left({\displaystyle \frac{logx}{x^3}}\right)>0(x\to \infty ).$$

(24)

The integral in (9) first decreases and then increases, so it crosses zero exactly once, and the crossing is simple. At $\lambda ={\lambda }_0\left(\eta \right)$ with $\eta >{\eta }_{*}$, the quadratic coefficient in (23) is $-(18{\eta }^2+12\eta -1)/36<0$, so $Q^{\prime }<0$ for small positive x. The preceding decrease-increase argument gives one simple positive zero here as well.

Now assume $\eta >{\eta }_{*}$. Since a negative value of $Q^{\prime }$ exists at ${\lambda }_0\left(\eta \right)$, define

$$\Lambda \left(\eta \right)=inf\{\lambda \in [0,{\lambda }_0\left(\eta \right)):Q_{\lambda ,\eta }^{\prime }\left(x\right)<0\mathrm{for}\mathrm{some}x>0\}.$$

By (22), this negativity set is upward closed. Hence below the threshold $Q^{\prime }$ is strictly positive. If it vanished, a sufficiently small increase of $\lambda$ would make it negative while remaining below the threshold.

If $\Lambda \left(\eta \right)>0$, choose parameters decreasing to the threshold and corresponding points where the derivative is negative. Uniform positivity near the origin follows from (23), and uniform positivity at infinity follows from (24). Hence, after passing to a subsequence, these points have a positive finite limit at which $Q^{\prime }=0$. At the threshold $Q^{\prime }\ge 0$. Two separated zeros would, under a small increase of $\lambda$, create at least four positive zeros, contrary to Theorem 1 for $m=3$. Thus the threshold zero is unique. Above the threshold but below ${\lambda }_0$, positivity at both ends and a negative interior value give exactly two sign-changing zeros, hence a local maximum followed by a local minimum. This proves the classification. □

5. The Unique Fold Endpoint on $\mathbf{\lambda }=\mathbf{0}$

Allow $\lambda =0$. Let $t_0\in (2,3)$ be the unique solution of

$$2tlogt=3(t-1).$$

(25)

For $t>t_0$, set

$$X\left(t\right)={\displaystyle \frac{2tlogt-3(t-1)}{t-2}},E\left(t\right)={\displaystyle \frac{t-1-X\left(t\right)}{X{\left(t\right)}^3}}.$$

(26)

Theorem 4. 

The function E has exactly one critical point $t_c$ on $(t_0,\infty )$, and it is a strict global minimum. Put

$${\eta }_c=E\left(t_c\right),x_c=X\left(t_c\right).$$

Then

$$\Lambda \left(\eta \right)>0\iff {\eta }_{*}<\eta <{\eta }_c,\Lambda \left(\eta \right)=0\iff \eta \ge {\eta }_c.$$

At $(\lambda ,\eta )=(0,{\eta }_c)$, $Q^{\prime }$ has one positive double zero $x_c$. For $\eta >{\eta }_c$, $Q_{0,\eta }^{\prime }$ has exactly two simple positive zeros. The fold extends analytically through $(0,{\eta }_c)$ and crosses $\lambda =0$ transversely with $d\lambda /d\eta <0$.

Proof. 

For $\lambda =0$ put $t=1+x+\eta x^3$. The equation $Q^{\prime }=0$ becomes

$$x(t-2)+3(t-1)-2tlogt=0.$$

The function $2tlogt-3(t-1)$ decreases and then increases on $(1,\infty )$, and it has exactly one zero beyond 2. Its values at 2 and 3 have opposite signs. This proves the asserted uniqueness of $t_0\in (2,3)$. For $t>t_0$, $X\left(t\right)>0$, while

$$t-1-X\left(t\right)={\displaystyle \frac{t^2-1-2tlogt}{t-2}}>0,$$

because $t^2-1-2tlogt$ is strictly increasing for $t>1$ and vanishes at 1. Conversely, $1<t<2$ would give $E\left(t\right)<0$, and $2<t<t_0$ would give $X\left(t\right)<0$. Thus all positive critical points with $\eta >0$, and only those, have the parametrization $x=X\left(t\right)$, $\eta =E\left(t\right)$ with $t>t_0$. Moreover

$$X^{\prime }\left(t\right)={\displaystyle \frac{2t-4logt-1}{{(t-2)}^2}}>0.$$

(27)

A calculation gives

$$E^{\prime }\left(t\right)={\displaystyle \frac{(t-2)\mathcal{R}\left(t\right)}{{[2tlogt-3t+3]}^4}},$$

(28)

where

$$\begin{array}{cc}\hfill \mathcal{R}\left(t\right)=& 2t^{3}logt-9t^3+12t^{2}logt+24t^2\hfill \\ \hfill & -16t{(logt)}^2-8tlogt-27t+12.\hfill \end{array}$$

(29)

At $a=t_0$, using (25),

$$\begin{array}{cc}\hfill \mathcal{R}\left(a\right)& =-{\displaystyle \frac{3{(a-2)}^2(a-1)(2a-3)}{a}}<0,\hfill \\ \hfill {\mathcal{R}}^{\prime }\left(a\right)& =-{\displaystyle \frac{(a-2)(a-1)(16a^2-39a+18)}{a^2}}<0,\hfill \\ \hfill {\mathcal{R}}^{\prime \prime }\left(a\right)& =-{\displaystyle \frac{2(a-2)(a-1)(13a-12)}{a^2}}<0,\hfill \\ \hfill {\mathcal{R}}^{\prime \prime \prime }\left(a\right)& =-{\displaystyle \frac{2(a-1)(7a^2+4a-24)}{a^3}}<0,\hfill \\ \hfill {\mathcal{R}}^{\prime \prime \prime \prime }\left(a\right)& ={\displaystyle \frac{4(3a^3-6a^2-20a+24)}{a^4}}<0.\hfill \end{array}$$

The numerator polynomial in the last line is negative on $2<a<3$. Furthermore

$${\mathcal{R}}^{\prime \prime \prime \prime }\left(t\right)={\displaystyle \frac{4[3t^2-6t-16logt+4]}{t^3}}.$$

(30)

The bracket is convex. It is negative at $t_0$ and at 3, while for $t\ge 3$ its derivative is positive and increasing. Hence ${\mathcal{R}}^{\prime \prime \prime \prime }$ changes sign exactly once. The leading asymptotics are

$${\mathcal{R}}^{\prime \prime \prime }\left(t\right)=12logt+O\left(1\right),{\mathcal{R}}^{\prime \prime }\left(t\right)=12tlogt+O\left(t\right),$$

$${\mathcal{R}}^{\prime }\left(t\right)=6t^{2}logt+O\left(t^2\right),\mathcal{R}\left(t\right)=2t^{3}logt+O\left(t^3\right),$$

so all four functions are eventually positive. Starting from the five negative values above, successive integration shows that ${\mathcal{R}}^{\prime \prime \prime }$, ${\mathcal{R}}^{\prime \prime }$, ${\mathcal{R}}^{\prime }$, and $\mathcal{R}$ each change sign exactly once. In particular $\mathcal{R}$ has a unique zero $t_c$. Equation (28) and the limits $E\left(t\right)\to \infty$ at both ends show that $t_c$ is the unique global minimum.

Along the critical curve,

$${\partial }_{\eta }{Q}_{0,E\left(t\right)}^{\prime }\left(X\left(t\right)\right)=-{\displaystyle \frac{2t-4logt-1}{t(t-2)}}<0.$$

(31)

Together with (27), this proves simplicity away from $t_c$. At $t_c$, $E^{\prime \prime }\left(t_c\right)>0$ gives ${\partial }_x^2{Q}^{\prime }\left(x_c\right)>0$, so the zero is exactly double. The assertions about $\Lambda$ now follow from Theorem 3.

Finally let $\Phi (x,\lambda ,\eta )=Q_{\lambda ,\eta }^{\prime }\left(x\right)$. At the endpoint, $\Phi ={\Phi }_x=0$, ${\Phi }_{xx}>0$, ${\Phi }_{\lambda }<0$, and ${\Phi }_{\eta }<0$. The Jacobian of $(\Phi ,{\Phi }_x)$ with respect to $(\lambda ,x)$ has determinant ${\Phi }_{\lambda }{\Phi }_{xx}\ne 0$. The analytic implicit function theorem gives the continuation, and $d\lambda /d\eta =-{\Phi }_{\eta }/{\Phi }_{\lambda }<0$. □

Numerically,

$$t_c\approx 44.0907343560,x_c\approx 4.86104259185,{\eta }_c\approx 0.332822376598.$$

6. Local Switch in the Sharp Lower Quadratic Coefficient

The $m=3$ Hermite point is

$${\lambda }_{*}={\displaystyle \frac{1}{\sqrt{6}}},{\eta }_{*}={\displaystyle \frac{1}{\sqrt{6}}}-{\displaystyle \frac{1}{3}}.$$

Put

$$u=\eta -\lambda +{\displaystyle \frac{1}{3}},v=-\eta -{\displaystyle \frac{{\lambda }^2}{2}}+\lambda -{\displaystyle \frac{1}{4}}.$$

(32)

The Jacobian of $(u,v)$ with respect to $(\lambda ,\eta )$ equals $1/\sqrt{6}$ at the base point, and

$$Q\left(x\right)=Q\left(0\right)+ux+v{x}^2+d_0{x}^3+O\left(x^4\right),d_0={\displaystyle \frac{5\sqrt{6}-12}{90}}>0.$$

(33)

Theorem 5. 

For all sufficiently small negative v, there are two distinct analytic curves $u=H\left(v\right)$ and $u=M\left(v\right)$ with

$$H\left(v\right)={\displaystyle \frac{v^2}{3d_0}}+O\left(v^3\right),M\left(v\right)={\displaystyle \frac{v^2}{4d_0}}+O\left(v^3\right),0<M\left(v\right)<H\left(v\right).$$

The first is the fold where two positive critical points are created. On the second, the positive local minimum has value $Q\left(0\right)$. In a sufficiently small parameter neighborhood, the global minimum of Q on $[0,1]$ is $Q\left(0\right)$ for $M\left(v\right)<u$, is attained both at the origin and at a positive point when $u=M\left(v\right)$, and is attained uniquely at the positive local minimum for $0<u<M\left(v\right)$. The optimal lower quadratic coefficient is continuous but not differentiable across the switch curve in a transverse parameter direction.

Proof. 

The equations $Q^{\prime }=Q^{\prime \prime }=0$ have an invertible Jacobian with respect to $(u,x)$ at the origin, giving

$$x_{\mathrm{f}}\left(v\right)=-{\displaystyle \frac{v}{3d_0}}+O\left(v^2\right),H\left(v\right)={\displaystyle \frac{v^2}{3d_0}}+O\left(v^3\right).$$

Solving $\left[Q\right(x)-Q(0\left)\right]/x=0$ together with $Q^{\prime }=0$ gives

$$x_{\mathrm{s}}\left(v\right)=-{\displaystyle \frac{v}{2d_0}}+O\left(v^2\right),M\left(v\right)={\displaystyle \frac{v^2}{4d_0}}+O\left(v^3\right).$$

At the base point, set $N=x(F^{\prime }-1)-2(F-x)$. Then

$$N^{\prime \prime }\left(x\right)={\displaystyle \frac{2x}{q{\left(x\right)}^3}}{\displaystyle \sum _{j=0}^6}{c}_j{x}^j,$$

where $c_0=c_1=0$ and $c_j>0$ for $2\le j\le 6$. Hence $Q^{\prime }>0$ on the positive half-line at the base point. Compactness on bounded intervals and the positive tail $Q^{\prime }\left(x\right)\sim x^{-2}$ confine all nearby critical points to a small neighborhood of zero. Fix $\delta >0$ sufficiently small. At the base point, Q is strictly increasing, so ${min}_{[\delta ,1]}[Q\left(x\right)-Q\left(0\right)]>0$. After shrinking the parameter neighborhood, this uniform gap persists while all critical points remain in $(0,\delta )$. Thus the local comparison near the origin determines the global minimum on $[0,1]$. The stated alternatives follow. Finally,

$${\partial }_{\eta }Q\left(0\right)=0,{\partial }_{\eta }Q\left(x\right)={\displaystyle \frac{x}{q\left(x\right)}}>0(x>0),$$

so the two active branches have different transverse derivatives when they meet. □

7. The Quadratic Coefficient Profile

For $x>0$, define

$$Q_{\lambda }\left(x\right)={\displaystyle \frac{F_{\lambda }\left(x\right)-x}{x^2}}.$$

By (4), set

$$Q_{\lambda }\left(0\right)=\lambda -{\displaystyle \frac{1}{2}}.$$

Thus $Q_{\lambda }$ is continuous on $[0,1]$, and

$$Q_{\lambda }\left(0\right)=\lambda -{\displaystyle \frac{1}{2}},Q_{\lambda }\left(1\right)=log(\lambda +2)-1.$$

(34)

The sign of its derivative at 1 determines the transition to a decreasing profile.

Lemma 2. 

The equation

$$2(\lambda +2)log(\lambda +2)=3(\lambda +1)$$

(35)

has a unique solution $\Lambda >0$. Moreover,

$$0.39<\Lambda <0.4,\Lambda \approx 0.3969988263.$$

Proof. 

Set

$$\nu \left(\lambda \right)={\displaystyle \frac{3(\lambda +1)-2(\lambda +2)log(\lambda +2)}{\lambda +2}}.$$

Then

$${\nu }^{\prime }\left(\lambda \right)=-{\displaystyle \frac{2\lambda +1}{{(\lambda +2)}^2}}<0.$$

Using (5) with $n=2$ at $\lambda =0.39$ and $n=4$ at $\lambda =0.4$ gives $\nu \left(0.39\right)>0$ and $\nu \left(0.4\right)<0$. Hence the intermediate value theorem and strict monotonicity give the assertion. The displayed decimal is a numerical approximation to this root. □

Theorem 6. 

Let $\lambda \ge 0$.

1.

If $0\le \lambda \le 1/3$, then $Q_{\lambda }$ is strictly increasing on $[0,1]$.

2.

If $1/3<\lambda <\Lambda$, then there is a unique $x_{\lambda }^{\left(2\right)}\in (0,1)$ such that $Q_{\lambda }$ is strictly decreasing on $[0,x_{\lambda }^{\left(2\right)}]$ and strictly increasing on $[x_{\lambda }^{\left(2\right)},1]$.

3.

If $\lambda \ge \Lambda$, then $Q_{\lambda }$ is strictly decreasing on $[0,1]$.

In the middle regime, $x_{\lambda }^{\left(2\right)}$ is the unique solution in $(0,1)$ of

$$2\left(log{q}_{\lambda }\left(x\right)-x\right)={\displaystyle \frac{x^2(2\lambda -1-\lambda x)}{q_{\lambda }\left(x\right)}}.$$

(36)

Proof. 

For $x>0$, write

$$Q_{\lambda }^{\prime }\left(x\right)={\displaystyle \frac{N_{\lambda }\left(x\right)}{x^3}},$$

where

$$N_{\lambda }\left(x\right)=x\left(F_{\lambda }^{\prime }\left(x\right)-1\right)-2\left(F_{\lambda }\left(x\right)-x\right).$$

(37)

Since

$$F_{\lambda }^{\prime }\left(x\right)={\displaystyle \frac{1+2\lambda x}{q_{\lambda }\left(x\right)}},$$

equation $N_{\lambda }\left(x\right)=0$ is exactly (36). Also $N_{\lambda }\left(0\right)=N_{\lambda }^{\prime }\left(0\right)=0$, and a direct differentiation gives

$$N_{\lambda }^{\prime \prime }\left(x\right)={\displaystyle \frac{2x(1+2\lambda x)\left({\lambda }^2{x}^2+\lambda x+1-3\lambda \right)}{q_{\lambda }{\left(x\right)}^3}}.$$

(38)

The quadratic factor

$$p_{\lambda }\left(x\right)={\lambda }^2{x}^2+\lambda x+1-3\lambda$$

is increasing in x for $\lambda >0$.

If $0\le \lambda \le 1/3$, then $p_{\lambda }\left(x\right)\ge 0$ on $[0,1]$, and it is strictly positive for $x>0$. Hence $N_{\lambda }^{\prime \prime }\left(x\right)>0$ for $x>0$, so $N_{\lambda }^{\prime }\left(x\right)>0$ and $N_{\lambda }\left(x\right)>0$. Thus $Q_{\lambda }^{\prime }\left(x\right)>0$.

Now suppose $\lambda >1/3$. Since

$$p_{\lambda }\left(0\right)=1-3\lambda <0,p_{\lambda }\left(1\right)={(\lambda -1)}^2\ge 0,$$

the function $p_{\lambda }$ has exactly one zero in $(0,1]$. Consequently $N_{\lambda }^{\prime \prime }$ is negative before that zero and nonnegative after it, with the second interval possibly degenerate. Thus $N_{\lambda }^{\prime }$ starts from 0, first decreases, and then increases if the second interval is nondegenerate. It can therefore change sign at most once, from negative to positive. The same is true of $N_{\lambda }$, which also starts from 0 and is initially negative. Finally,

$$N_{\lambda }\left(1\right)={\displaystyle \frac{3(\lambda +1)-2(\lambda +2)log(\lambda +2)}{\lambda +2}}=\nu \left(\lambda \right).$$

(39)

By Lemma 2, this endpoint value is positive for $1/3<\lambda <\Lambda$, zero for $\lambda =\Lambda$, and negative for $\lambda >\Lambda$. This gives, respectively, one interior zero of $N_{\lambda }$, no interior zero with $N_{\lambda }<0$ up to the endpoint, and strict negativity throughout $(0,1]$. The three asserted shapes follow. □

Remark 2. 

Theorems 6 and 7 give explicit instances of the one-turn alternatives in Pinelis ([13], Propositions 4.1, 4.3, and 4.4). The endpoint values in (39) and (48) determine whether a turn occurs once the polynomial sign changes have been established.

The shape theorem determines the sharp quadratic inequality.

Corollary 1. 

For each $\lambda \ge 0$, the inequality

$$x+A{x}^2\le F_{\lambda }\left(x\right)\le x+B{x}^2,0\le x\le 1,$$

holds if and only if

$$A\le A_2\left(\lambda \right),B\ge B_2\left(\lambda \right),$$

where

$$A_2\left(\lambda \right)=\left\{\begin{array}{cc}\lambda -{\displaystyle \frac{1}{2}},\hfill & 0\le \lambda \le {\displaystyle \frac{1}{3}},\hfill \\ Q_{\lambda }\left(x_{\lambda }^{\left(2\right)}\right),\hfill & {\displaystyle \frac{1}{3}}<\lambda <\Lambda ,\hfill \\ log(\lambda +2)-1,\hfill & \lambda \ge \Lambda ,\hfill \end{array}\right.$$

(40)

and

$$B_2\left(\lambda \right)=max\left\{\lambda -{\displaystyle \frac{1}{2}},log(\lambda +2)-1\right\}.$$

(41)

Proof. 

For $x>0$, subtract x and divide by $x^2$. The largest admissible lower coefficient is ${min}_{[0,1]}{Q}_{\lambda }$, and the smallest admissible upper coefficient is ${max}_{[0,1]}{Q}_{\lambda }$. Theorem 6 gives the minimum in (40). The same theorem shows that $Q_{\lambda }$ has no interior maximum in any parameter regime, so its maximum is the larger of the two endpoint values in (34). □

Remark 3. 

The two entries in (41) are equal at the unique solution ${\mu }_2$ of

$$log(\lambda +2)=\lambda +{\displaystyle \frac{1}{2}}.$$

Set $h\left(\lambda \right)=\lambda +1/2-log(\lambda +2)$. Then $h^{\prime }\left(\lambda \right)=1-1/(\lambda +2)>0$ for $\lambda \ge 0$, while $h\left(0\right)=1/2-log2<0$ and $h\left(\lambda \right)\to \infty$ as $\lambda \to \infty$. Hence this solution exists and is unique. Numerically,

$${\mu }_2\approx 0.3576766739.$$

Moreover, (5), with $n=3$ at $\lambda =0.35$ and $n=2$ at $\lambda =0.36$, gives $h\left(0.35\right)<0<h\left(0.36\right)$. Hence $0.35<{\mu }_2<0.36<\Lambda$. Thus ${\mu }_2$ lies in the interior-minimum window $(1/3,\Lambda )$. Within that window, the global maximum of $Q_{\lambda }$ switches from $x=1$ to $x=0$ while the global minimum remains interior.

8. The Cubic Coefficient Profile

For $x>0$, define

$$C_{\lambda }\left(x\right)={\displaystyle \frac{F_{\lambda }\left(x\right)-x-(\lambda -\frac{1}{2})x^2}{x^3}}.$$

By (4), extend it continuously by

$$C_{\lambda }\left(0\right)={\displaystyle \frac{1}{3}}-\lambda .$$

The endpoint values are

$$C_{\lambda }\left(0\right)={\displaystyle \frac{1}{3}}-\lambda ,C_{\lambda }\left(1\right)=log(\lambda +2)-\lambda -{\displaystyle \frac{1}{2}}.$$

(42)

Two algebraic thresholds arise from the sign at the origin.

$$\alpha =1-{\displaystyle \frac{1}{\sqrt{2}}}\approx 0.2928932188,\omega =1+{\displaystyle \frac{1}{\sqrt{2}}}\approx 1.7071067812.$$

(43)

The remaining shape threshold is defined by a transcendental equation.

Lemma 3. 

The equation

$$6(\lambda +2)log(\lambda +2)=2{\lambda }^2+11\lambda +8$$

(44)

has a unique solution $\delta >0$. Moreover,

$$0.31<\delta <0.312,\delta \approx 0.3116840733.$$

In particular, $\alpha <\delta <1/3<\omega$.

Proof. 

Set

$$m\left(\lambda \right)={\displaystyle \frac{2{\lambda }^2+11\lambda +8-6(\lambda +2)log(\lambda +2)}{2(\lambda +2)}}.$$

Then

$$m^{\prime }\left(\lambda \right)={\displaystyle \frac{{\lambda }^2+\lambda +1}{{(\lambda +2)}^2}}>0.$$

Using (5) with $n=4$ at $\lambda =0.31$ and $n=3$ at $\lambda =0.312$ gives $m\left(0.31\right)<0<m\left(0.312\right)$, proving existence and uniqueness of a root in $(0.31,0.312)$. Since $\alpha =1-1/\sqrt{2}<0.3<0.31$ and $0.312<1/3$, the stated order follows. □

Theorem 7. 

Let $\lambda \ge 0$.

1.

If $0\le \lambda \le \alpha$, then $C_{\lambda }$ is strictly decreasing on $[0,1]$.

2.

If $\alpha <\lambda <\delta$, then there is a unique $x_{\lambda }^{(3,+)}\in (0,1)$ such that $C_{\lambda }$ is strictly increasing on $[0,x_{\lambda }^{(3,+)}]$ and strictly decreasing on $[x_{\lambda }^{(3,+)},1]$. Thus $C_{\lambda }$ attains its global maximum only at $x_{\lambda }^{(3,+)}$.

3.

If $\delta \le \lambda \le \omega$, then $C_{\lambda }$ is strictly increasing on $[0,1]$.

4.

If $\lambda >\omega$, then there is a unique $x_{\lambda }^{(3,-)}\in (0,1)$ such that $C_{\lambda }$ is strictly decreasing on $[0,x_{\lambda }^{(3,-)}]$ and strictly increasing on $[x_{\lambda }^{(3,-)},1]$. Thus $C_{\lambda }$ attains its global minimum only at $x_{\lambda }^{(3,-)}$.

The interior critical points are precisely the zeros in $(0,1)$ of

$$M_{\lambda }\left(x\right)=x\left[F_{\lambda }^{\prime }\left(x\right)-1-2\left(\lambda -{\displaystyle \frac{1}{2}}\right)x\right]-3\left[F_{\lambda }\left(x\right)-x-\left(\lambda -{\displaystyle \frac{1}{2}}\right)x^2\right].$$

(45)

Proof. 

For $x>0$,

$$C_{\lambda }^{\prime }\left(x\right)={\displaystyle \frac{M_{\lambda }\left(x\right)}{x^4}}.$$

The function $M_{\lambda }$ extends continuously to 0 with $M_{\lambda }\left(0\right)=0$. Differentiation and simplification give

$$M_{\lambda }^{\prime }\left(x\right)={\displaystyle \frac{x^3{K}_{\lambda }\left(x\right)}{q_{\lambda }{\left(x\right)}^2}},$$

(46)

where

$$K_{\lambda }\left(x\right)={\lambda }^2(2\lambda -1)x^2+2\lambda (3\lambda -1)x-2{\lambda }^2+4\lambda -1.$$

(47)

The constant term factors as

$$K_{\lambda }\left(0\right)=-2{\lambda }^2+4\lambda -1=-2(\lambda -\alpha )(\lambda -\omega ).$$

Also

$$M_{\lambda }\left(1\right)={\displaystyle \frac{2{\lambda }^2+11\lambda +8-6(\lambda +2)log(\lambda +2)}{2(\lambda +2)}}=m\left(\lambda \right).$$

(48)

If $0\le \lambda \le \alpha$, then all three coefficients in (47) are nonpositive, and for $x>0$ the polynomial $K_{\lambda }\left(x\right)$ is strictly negative. Hence $M_{\lambda }^{\prime }\left(x\right)<0$, so $M_{\lambda }\left(x\right)<0$ and $C_{\lambda }^{\prime }\left(x\right)<0$.

Suppose $\alpha <\lambda <\delta$. Since $\delta <1/3$, the quadratic and linear coefficients of $K_{\lambda }$ are negative while $K_{\lambda }\left(0\right)>0$. Thus $K_{\lambda }$ is strictly decreasing on $[0,1]$. The endpoint relation $m\left(\lambda \right)<0$ implies $M_{\lambda }\left(1\right)<0$, whereas $M_{\lambda }$ is positive for small positive x because $M_{\lambda }^{\prime }\left(x\right)>0$ there. Therefore $K_{\lambda }$ changes sign once, $M_{\lambda }$ first increases and then decreases, and $M_{\lambda }$ crosses zero exactly once. This gives the unique interior maximum of $C_{\lambda }$.

Now let $\delta \le \lambda <1/3$. Again $K_{\lambda }$ is strictly decreasing from a positive initial value. If it stays nonnegative, then $M_{\lambda }$ is increasing and positive. If it crosses zero, then $M_{\lambda }$ first increases and then decreases. But (48) and $m\left(\lambda \right)\ge 0$ show that it cannot reach zero before $x=1$. Hence $M_{\lambda }\left(x\right)>0$ for $0<x<1$.

For $1/3\le \lambda \le 1/2$, the polynomial $K_{\lambda }$ is concave in x. Both endpoint values are positive. One has $K_{\lambda }\left(0\right)>0$ because $\lambda <\omega$, and

$$K_{\lambda }\left(1\right)=2{\lambda }^3+3{\lambda }^2+2\lambda -1>0.$$

The polynomial on the right is strictly increasing for $\lambda \ge 0$ and, at $\lambda =1/3$, has value $2/27$. A concave function lies above its chord, so $K_{\lambda }\left(x\right)>0$ on $[0,1]$. For $1/2\le \lambda \le \omega$, all coefficients in (47) are nonnegative, and $K_{\lambda }\left(x\right)>0$ for $x>0$. Consequently $M_{\lambda }\left(x\right)>0$ and $C_{\lambda }^{\prime }\left(x\right)>0$ for $0<x<1$ throughout the regime $\delta \le \lambda \le \omega$.

Finally, let $\lambda >\omega$. Then $K_{\lambda }\left(0\right)<0$, while both the linear and quadratic coefficients in (47) are positive. Hence $K_{\lambda }$ is strictly increasing and has at most one zero. Since m is increasing and $\lambda >\omega >\delta$, one has $M_{\lambda }\left(1\right)>0$, whereas $M_{\lambda }$ is negative for small positive x. It follows that $K_{\lambda }$ changes sign once and that $M_{\lambda }$ subsequently crosses zero exactly once. Therefore $C_{\lambda }$ strictly decreases and then strictly increases, as claimed. □

Corollary 2. 

For each $\lambda \ge 0$, the inequality

$$x+\left(\lambda -{\displaystyle \frac{1}{2}}\right)x^2+C{x}^3\le F_{\lambda }\left(x\right)\le x+\left(\lambda -{\displaystyle \frac{1}{2}}\right)x^2+D{x}^3,0\le x\le 1,$$

holds if and only if

$$C\le A_3\left(\lambda \right),D\ge B_3\left(\lambda \right),$$

where

$$A_3\left(\lambda \right)=\left\{\begin{array}{cc}min\left\{{\displaystyle \frac{1}{3}}-\lambda ,log(\lambda +2)-\lambda -{\displaystyle \frac{1}{2}}\right\},\hfill & 0\le \lambda \le \omega ,\hfill \\ C_{\lambda }\left(x_{\lambda }^{(3,-)}\right),\hfill & \lambda >\omega ,\hfill \end{array}\right.$$

(49)

and

$$B_3\left(\lambda \right)=\left\{\begin{array}{cc}{\displaystyle \frac{1}{3}}-\lambda ,\hfill & 0\le \lambda \le \alpha ,\hfill \\ C_{\lambda }\left(x_{\lambda }^{(3,+)}\right),\hfill & \alpha <\lambda <\delta ,\hfill \\ log(\lambda +2)-\lambda -{\displaystyle \frac{1}{2}},\hfill & \lambda \ge \delta .\hfill \end{array}\right.$$

(50)

Proof. 

The optimal lower and upper cubic coefficients are respectively the minimum and maximum of $C_{\lambda }$ on $[0,1]$. Theorem 7 gives all interior extrema. When $\lambda \le \omega$, no interior minimum occurs, so the minimum is the smaller endpoint value in (42). When $\lambda >\omega$, the unique interior critical point is the global minimum. For the maximum, the decreasing regime gives the value at 0, the interior-maximum regime gives $C_{\lambda }\left(x_{\lambda }^{(3,+)}\right)$, and the increasing regime gives the value at 1. If $\lambda >\omega$, the maximum is again an endpoint value. Here $\lambda >\omega >1$, so $\lambda +2>3>e$ and $log(\lambda +2)>1>5/6$, which implies $C_{\lambda }\left(1\right)>C_{\lambda }\left(0\right)$. □

Remark 4. 

The endpoint values in (42) are equal exactly when

$$log(\lambda +2)={\displaystyle \frac{5}{6}}.$$

Thus the switch occurs at

$${\mu }_3=e^{5/6}-2\approx 0.3009758909.$$

For the exponential series through degree 6, the partial sum exceeds $2.3009$, while the remaining positive tail is at most

$${\displaystyle \frac{{(5/6)}^7}{7!}}{\displaystyle \frac{1}{1-5/48}}.$$

The resulting upper bound is smaller than $2.3010$. Hence $2.3009<e^{5/6}<2.3010$, and therefore $0.3009<{\mu }_3<0.3010$. Together with $\alpha <0.3<0.301<0.31<\delta$, this proves $\alpha <{\mu }_3<\delta$. Hence, during the interior-maximum regime, the global minimum changes from $x=1$ to $x=0$ even though the global maximum stays at the unique interior point.

9. Shape Transitions and the Benchmark Case $\mathbf{\lambda }=\mathbf{1}$

Table 1 shows the shape transitions. The endpoint-switch values ${\mu }_2$ and ${\mu }_3$ affect which endpoint supplies the non-interior extremum, but they do not change the number of interior critical points.

The principal critical values are

$$\begin{array}{c}\alpha =1-{\displaystyle \frac{1}{\sqrt{2}}}\approx 0.292893,\delta \approx 0.311684,{\displaystyle \frac{1}{3}},\\ \Lambda \approx 0.396999,\omega =1+{\displaystyle \frac{1}{\sqrt{2}}}\approx 1.707107.\end{array}$$

The secondary endpoint switches are

$${\mu }_3=e^{5/6}-2\approx 0.300976,{\mu }_2\approx 0.357677.$$

Corollary 3. 

For $0\le x\le 1$,

$$x+(log3-1)x^2\le log(1+x+x^2)\le x+{\displaystyle \frac{1}{2}}{x}^2,$$

and

$$x+{\displaystyle \frac{1}{2}}{x}^2-{\displaystyle \frac{2}{3}}{x}^3\le log(1+x+x^2)\le x+{\displaystyle \frac{1}{2}}{x}^2+\left(log3-{\displaystyle \frac{3}{2}}\right)x^3.$$

All four displayed coefficients are best possible within the indicated polynomial classes.

Proof. 

Since $1>\Lambda$, Theorem 6 shows that $Q_1$ is strictly decreasing. Hence its minimum is $Q_1\left(1\right)=log3-1$ and its maximum is $Q_1\left(0\right)=1/2$. Since $\delta <1<\omega$, Theorem 7 shows that $C_1$ is strictly increasing. Its endpoint values are $C_1\left(0\right)=-2/3$ and $C_1\left(1\right)=log3-3/2$. □

Remark 5. 

At $\lambda =1$, Taylor and Hermite interpolation remainders also give the displayed bounds. The parameter classification shows where this endpoint behavior changes to an interior extremum.

10. Conclusions

For positive-coefficient polynomials of degree m, the normalized logarithmic remainder has at most $m-1$ positive critical points. This variation bound is the finite-polynomial consequence of Mao and Tian ([8], Monotonicity Rule 18). Within the present logarithmic family the bound is sharp. The Hermite construction gives an explicit positive order-$(m-1)$ degeneracy, and the nonzero jet determinant gives arbitrary ordered configurations of $m-1$ small positive simple critical points after unfolding.

For the cubic polynomial $1+x+\lambda x^2+\eta x^3$, the positive critical points are classified throughout the positive parameter quadrant. The fold has a unique endpoint on $\lambda =0$, and a separate local curve marks the switch in the sharp lower quadratic coefficient. The degree-two case gives exact sharp quadratic and cubic coefficient bounds on $[0,1]$, with every endpoint/interior transition determined explicitly or by a unique one-variable equation. The arguments use classical monotonicity, Descartes’ rule, Hermite theory, and the implicit function theorem. The sharp Hermite realization and the exact parameter phase structure are specific to the present logarithmic family.