Let $f(x)=\frac{x\ln x}{1+x}$,evaluate $\lim _{m\to f\left(x_0\right)+}\frac{x_1+x_2-2x_0}{m+x_0}$.

Question

Let $f(x)=\frac{x\ln x}{1+x}$,suppose $x_0$is the extreme point of $f$.For $m\in \mathbb{R} $,it's easy to see that if $m>f(x_0)=-x_0$,then the equation $f(x)=m$ has exactly two real roots $x_1,x_2$,$x_1<x_0<x_2$.The question is how to evaluate the limit \begin{align*} \lim _{m\to f\left(x_0\right)+}\frac{x_1+x_2-2x_0}{m+x_0}？ \end{align*} To this problem,we can see that if $m\to f(x_0)+$,then $x_1\to x_0^-$,and $x_2\to x_0^+$.By Taylor's formula we have $$f(x_1)-f(x_0)=f'(x_0)(x_1-x_0)+\frac12f''(\xi)(x_1-x_0)^2=\frac12f''(\xi)(x_1-x_0)^2,$$ where $\xi \in (x_1,x_0)$.So $$m+x_0\sim \frac{f''(x_0)}{2}(x_1-x_0)^2.$$ But I can't handle $x_1+x_2-2x_0$.If we can expand $x_1+x_2-2x_0$ as terms of power of $(x_1-x_0)$,then it must be $$x_1+x_2-2x_0\sim k(x_1-x_0)^2.$$ But how can I get it?Maybe there are other ways.

Answer 1

Considering $$f(x)=\frac{x \log (x)}{x+1}$$ (which trivially cancels at $x=0$ and $x=1$), we have $$f'(x)=\frac{x+\log (x)+1}{(x+1)^2}\qquad \text{and} \qquad f''(x)=-\frac{x^2+2 x \log (x)-1}{x (x+1)^3}$$ The first derivative cancels at $$x_*=W\left(\frac{1}{e}\right)$$ where $W(.)$ is Lambert function.

We have $$f(x_*)=-W\left(\frac{1}{e}\right)\qquad \text{and} \qquad f''(x_*)=\frac{1}{W\left(\frac{1}{e}\right) \left(W\left(\frac{1}{e}\right)+1\right)}$$ So, $x_*$ corresponds to a minimum of the function.

Now expand $f(x)$ as a series around $x_*$ and get $$f(x)=f(x_*)+\frac 12 f''(x_*) (x-x_*)^2 + O\left((x-x_*)^3\right)$$ and then the approximation of the roots $(x_1,x_2)$ when $f(x)=m$ $$x_1=x_*-\sqrt{2 x_* \left(x_*+1\right)} \sqrt{m+x_*}$$ $$x_2=x_*+\sqrt{2 x_* \left(x_*+1\right)} \sqrt{m+x_*}$$

I think that now, you have a sufficient number of elements to finish properly your work.

Edit

The above approximation is not sufficient at all; it is only acceptable very close to $x_*$.

A much better approximation is given by the $[2,2]$ Padé approximant $$g(x)=\frac{18 x_*^3 (x_*+1)+6 x_*^2 (4 x_*+1) (x-x_*)+5 (x_*-2) x_* (x-x_*)^2 } {-18 \left(x_*^2 (x_*+1)\right)-6 (x_* (4 x_*+1)) (x-x_*)+(1-5 x_*) (x-x_*)^2 }$$

$$\Phi=\int_0^1 \big[f(x)-g(x)\big]^2\, dx= 2.56 \times 10^{-5}$$

and we are left with a quadratic equation in $(x-x_*)$. This gives $$\Delta=x_1+x_2-2x_*=\frac{6 x_* (4 x_*+1) (x_*+m) } {9 x_*+(1-5 x_*) (x_*+m)}$$ which would give $$\lim _{m\to f\left(x_0\right)+}\frac{x_1+x_2-2x_0}{m+x_0}=\frac{2}{3} (4 x_*+1)$$

For illustration purposes, I used $m=-0.15$. The approximation gives $$x_1=0.05166\qquad \qquad x_2=0.69003\qquad \qquad \Delta=0.18475$$ while the exact solutions are $$x_1=0.05428\qquad \qquad x_2=0.69324\qquad \qquad \Delta=0.19059$$

Numerical checks

Using the above was computed the value of $$R_k=\frac{x_1+x_2-2x_0}{m+x_0}\qquad\text{with}\qquad m=(1-10^{-k}) f(x_0)$$

$$\left( \begin{array}{cc} k & R_k \\ 1 & 1.424602322711298857 \\ 2 & 1.410758744957374844 \\ 3 & 1.409390617211381501 \\ 4 & 1.409253962752398454 \\ 5 & 1.409240298885867560 \\ 6 & 1.409238932515004374 \\ 7 & 1.409238795878075951 \end{array} \right)$$ while $$\frac{2}{3} (4 x_*+1)=1.409238780696196787$$

Answer 2

Some thoughts

Remark: Actually, $x_1 = \mathrm{e}^{W_{-1}(m\mathrm{e}^{-m}) + m}$ and $x_2 = \mathrm{e}^{W_0(m\mathrm{e}^{-m}) + m}$. But we avoid using it.

The desired limit is $\frac83 W_0(\mathrm{e}^{-1}) + \frac23$ where $W_0(\cdot)$ is the principal branch of the Lambert W function, as @Claude Leibovici pointed out.

Fact 1: $f(x) = \frac{x\ln x}{x + 1}$ is strictly decreasing on $(0, x_0)$ and strictly increasing on $(x_0, \infty)$, where $x_0 = W_0(\mathrm{e}^{-1})$ is the unique minimizer of $f(x)$ on $(0, \infty)$.
(Note: From $f'(x) = 0$, we have $\ln x + x + 1 = 0$ that is $x\mathrm{e}^x = \mathrm{e}^{-1}$ which results in $x = W_0(\mathrm{e}^{-1})$.)

Let $m \in (-x_0, 0)$. The equation $f(x) = m$ has exactly two real solution $0 < x_1 < x_0 < x_2 < 1$.

WLOG, assume $m \in (-x_0, -x_0 + 1/10)$.

Let \begin{align*} y_1 &= x_0 - \sqrt{2x_0(1 + x_0)(m + x_0)} + \left(\frac43 x_0 + \frac13\right)(m + x_0) - (m + x_0)^{3/2}, \\ y_2 &= x_0 - \sqrt{2x_0(1 + x_0)(m + x_0)} + \left(\frac43 x_0 + \frac13\right)(m + x_0), \\ z_1 &= x_0 + \sqrt{2x_0(1 + x_0)(m + x_0)} + \left(\frac43 x_0 + \frac13\right)(m + x_0), \\ z_2 &= x_0 + \sqrt{2x_0(1 + x_0)(m + x_0)} + \left(\frac43 x_0 + \frac13\right)(m + x_0) + (m + x_0)^{3/2}. \end{align*} It is easy to prove that $0 < y_1 < y_2 < x_0 < z_1 < z_2 < 1$.

Fact 2: $f(y_1) > m$, $f(y_2) < m$, $f(z_1) < m$, and $f(z_2) > m$ for all $m \in (-x_0, -x_0 + 1/10)$.

By Facts 1-2, we have $y_1 < x_1 < y_2$ and $z_1 < x_2 < z_2$. Thus, $y_1 + z_1 < x_1 + x_2 < y_2 + z_2$. Then, we easily obtain $\lim_{m \to (-x_0)^{+}} \frac{x_1 + x_2 - 2x_0}{m + x_0} = \frac83 W_0(\mathrm{e}^{-1}) + \frac23$.

We are done.

$\phantom{2}$

Proof of Fact 2:

We prove $f(y_1) > m$ only. The rest is similar.

It suffices to prove that $\ln y_1 > m + \frac{m}{y_1}$. With the substitution $u^2 = m + x_0$ ($0 < u < 1/\sqrt{10}$), it suffices to prove that \begin{align*} &\ln \left(x_0 - u\sqrt{2x_0(1 + x_0)} + \left(\frac43 x_0 + \frac13\right)u^2 - u^3\right)\\ &> u^2 - x_0 + \frac{u^2 - x_0}{x_0 - u\sqrt{2x_0(1 + x_0)} + (\frac43 x_0 + \frac13)u^2 - u^3}. \end{align*} Denote $\mathrm{LHS} - \mathrm{RHS}$ by $F(u)$. We have, for all $u \in (0, 1/\sqrt{10})$, \begin{align*} F'(u) &= \frac{u^3(q_4u^4 + q_3u^3 + q_2u^2 + q_1u + q_0)}{9[x_0 - u\sqrt{2x_0(1 + x_0)} +(\frac43 x_0 + \frac13)u^2 - u^3]^2}\\ &> 0 \end{align*} where \begin{align*} q_4 &= -18, \\ q_3 &= 48\,x_0+12, \\ q_2 &= -36\sqrt{2x_0 \left( 1+x_0 \right) }-32\,x_0^{2}-16\,x_0+25, \\ q_1 &= 48x_0\sqrt{2x_0 \left( 1+x_0 \right) }+12\sqrt {2x_0 \left( 1+x_0 \right) }-24\,x_0-24, \\ q_0 &= 36\sqrt{2x_0 \left( 1+x_0 \right) }-52\,x_0^{2}-32\,x_0+2. \end{align*} Also, $F(0) = 0$. Thus, $F(u) > 0$ for all $u \in (0, 1/\sqrt{10})$.

We are done.

Answer 3

The essence of this question is to expand the roots (technically functional inverses for the multivalued function presented above) in a series around the minimum, where they coincide. As demonstrated by @ClaudeLeibovici above, a second-order approximation is required to evaluate the limit, and we will find that for a general function $f(z)$. Suppose that at $z=z_0$ the function attains a critical point with $f(z_0)=\zeta,f'(z_0)=0, f''(z_0)\neq 0$. Denote the two local branches of the inverse with $f^{-1}_{\pm}(z),f^{-1}_{\pm}(\zeta)=z_0$, where the plus/minus branch indicates the root that in some disk around the critical point obeys $(f^{-1}_+(z)>z_0)/(f^{-1}_-(z)<z_0)$.

How do we find successive monomial approximations to these? It is not a priori clear what the form is close to the critical point, so we try to evaluate the first-order limit

$$L_{1,\pm}=\lim_{z\to \zeta}\frac{f_{\pm}^{-1}(z)-z_0}{(z-\zeta)^a}$$

where $a\in \mathbb{R}$ is to be defined such that the limit above exists and is finite. Perform the change of variables $z=f(t)$:

$$L_{1,\pm}=\lim_{t\to z_0^{\pm}}\frac{t-z_0}{(f(t)-\zeta)^a}$$ Given the Taylor series expansion $f(t)-\zeta=\frac{f''(z_0)}{2}(t-z_0)^2+\mathcal{O}((t-z_0)^3)$ around $z_0$ we readily deduce that for this limit to be finite the exponent needs to take on the value $a=1/2$ and the limit evaluates to

$$L_{1,\pm}=\pm\sqrt{\frac{2}{f''(z_0)}}$$

Following the same thought process we can continue to compute succesive approximations by subtracting the last approximation from the series and dividing by an appropriate power of $z-\zeta$. For the second-order approximation it looks like this:

$$L_{2,\pm}=\lim_{z\to \zeta}\frac{f^{-1}_{\pm}(z)-z_0-(\pm)\sqrt{\frac{2}{f''(z_0)}}\sqrt{z-\zeta}}{(z-\zeta)^a}=\lim_{z\to z_0^{\pm}}\frac{t-z_0-(\pm)\sqrt{\frac{2}{f''(z_0)}}\sqrt{f(t)-\zeta}}{(f(t)-\zeta)^a}$$

We expand the numerator to second order to find that

$$t-z_0-\pm\sqrt{\frac{2}{f''(z_0)}(f(t)-\zeta)}=\frac{f'''(z_0)}{6f''(z_0)}(t-z_0)^2+\mathcal{O}((t-z_0)^{3})$$

Thus $a=1$ and the limit evaluates to

$$L_{2,\pm}=-\frac{f'''(z_0)}{3(f''(z_0))^2}$$

We have computed enough terms to calculate the desired limit. In this notation it can be written

$$\mathcal{L}=\lim_{m\to f(x_0)}\frac{f^{-1}_{+}(m)+f^{-1}_{-}(m)-2x_0}{m-f(x_0)}=2L_{2,\pm}=-\frac{2}{3}\frac{f'''(x_0)}{(f''(x_0))^2}\approx 1.40924 $$

This calculation hints at the fact that there is a generalization of Lagrange's inversion theorem around a critical point. I have not been able to find a general formula for the coefficients of the branches of the inverse, but I think it is possible to obtain such a formula.

It also turns out that there is an exact solution to this equation! Set in the original equation $z=\ln x-m$ which transforms it to the form $ze^z=m e^{-m}$ which admits two solutions

$$x_1(m)=\frac{m}{W_0(me^{-m})}~,~ x_2(m)=\frac{m}{W_{-1}(me^{-m})}$$

However this form is not particularly useful for obtaining expansions.

Answer 4

I do not expect any bounty for this, just wanted to share my thoughts.

Note that in this problem we have implicitly defined functions $x_1(m)$ and $x_2(m)$.

Here is a graph displaying $x_1(m)$ and $x_2(m)$. (Their graphs lie in the second quadrant and are traced out in black as we use the slider to move $m$ closer to $-x_0$.)

The limit we seek, $\displaystyle L:= \lim_{m \rightarrow f(x_0)^+} \frac{x_1(m)+x_2(m)-2x_0}{m+x_0}$, is of the form $\frac{0}{0}$ so we can try to apply L'Hopital's Rule. Unfortunately, doing so yields $\displaystyle L= \lim_{m \rightarrow f(x_0)^+} \frac{x_1'(m) + x_2'(m)}{1}$ which evaluates to $-\infty + \infty$.

Note that we can also split the limit into two parts to obtain $$\displaystyle L= \lim_{m \rightarrow f(x_0)^+} \frac{x_2(m)-x_0}{m - (-x_0)} +\lim_{m \rightarrow f(x_0)^+} \frac{x_1(m)-x_0}{m - (-x_0)}.$$

The difference quotients that appear in the above expression are the slopes of secant lines from $(-x_0,x_0)$ to $(m, x_2(m))$ and from $(-x_0,x_0)$ to $(m, x_1(m))$, respectively. The first represents a positive slope while the second a negative one. Perhaps these slopes cancel in a controlled way as $m \rightarrow -x_0$ so the limit of their sum exists (as a real number), but I expect they do not.