# How to establish regions of convexity/concavity of a ratio of exponential polynomials?

Problem:

Let $$f\colon \mathopen[0,1\mathclose] \to \mathbb{R}$$ be defined as $$f(x) = \frac{e^{\rho x}-1}{e^{\rho x}-1+e^{\rho (1-\gamma x)}-e^{\rho (1-\gamma) x}}$$ where $$\rho$$ and $$\gamma$$ are strictly positive real parameters. I am interested in showing that, for any fixed $$\rho$$, there exists a value $$\gamma_{\rho}$$ such that for any $$\gamma\geq \gamma_{\rho}$$, the function $$f$$ is first convex on an interval $$[0,\bar{x}[$$ and then concave on $$[\bar{x},1]$$, where $$\bar{x}$$ is the inflection point of $$f$$. This essentially boils down to showing that $$f''$$ is positive before $$\bar{x}$$ and negative afterwards.

What I have tried:

I observed that for all $$x \in \mathopen]0,1\mathclose[$$, the function $$f$$ satisfies the differential equation $$f'(x) = \alpha(x) f(x) (1-f(x))$$ where $$\alpha(x) = \rho \left( \frac{1}{e^{\rho x}-1} + \frac{1}{e^{\rho (1-x)}-1} + (1+\gamma) \right).$$ From this, I deduced that $$f(x) = \frac{f(x_{0})}{ f(x_{0}) + \big( 1-f(x_{0}) \big) \, e^{ -\int_{x_{0}}^{x} \alpha(s) \mathrm{d}s } }$$ for some $$x_{0} \in \mathopen]0,1\mathclose[$$. Consequently, $$f''(x)$$ has the same sign as $$\frac{\sigma''(A(x))}{\sigma'(A(x))} + \frac{\alpha'(x)}{\alpha(x)^2}$$ where $$\sigma(x) = \frac{f(x_{0})}{ f(x_{0}) + ( 1-f(x_{0}) ) e^{ -x} }$$ is the sigmoid function (which is a convex-concave function) and $$A(x) = \int_{x_{0}}^{x} \alpha(s) \mathrm{d}s.$$ The challenge is that the signs of $$\frac{\sigma''(A(x))}{\sigma'(A(x))}$$ and $$\frac{\alpha'(x)}{\alpha(x)^2}$$ go in opposite directions, making it difficult to determine the sign of $$f''$$. A similar problem arises when you directly take the second derivative of $$f$$.

Question:

How can we determine the sign of $$f''(x)$$ and thus establish the regions of convexity and concavity of $$f$$? Are there any other techniques or insights that can help navigate this problem?

• When you say "for sufficiently large $\gamma$", do you mean "regardless of $\rho$" or "for fixed $\rho$ and sufficiently large in terms of $\rho$"? Aug 12, 2023 at 14:10
• Thank you for pointing out this imprecision and I apologize for my late reply. What I meant was that for any $\rho$, there exists a value $\gamma_{\rho}$ such that for any $\gamma\geq \gamma_{\rho}$ the function $f''$ is strictly positive and then strictly negative on the interval $\mathopen[0,1\mathclose]$. What my simulations suggest is that, for a fixed $\rho$, if $\gamma$ is less than a certain $\gamma_{\rho}$, then $f''$ is positive everywhere on $\mathopen[0,1\mathclose]$ so that $f$ is convex. I've corrected the original post.
– vico
Aug 21, 2023 at 15:33
• Do you have a response to the answer below? Aug 22, 2023 at 15:24

This conjecture is true.

Indeed, letting $$r:=\rho$$, $$a:=\gamma$$, $$t:=rx$$, and $$u:=e^r-1$$, we see
$$\begin{equation*} f(x)=R(t):=\frac{e^t-1}{(1+u)e^{-a t}-e^{t-a t}+e^t-1}. \tag{10}\label{10} \end{equation*}$$ So, the problem can be restated as follows:

Show that for each real $$u>0$$ there is some real $$a_u>0$$ such that for each $$a\ge a_u$$ there is some $$t_{u,a}\in(0,\ln(1+u))$$ such that $$R''\ge0$$ on $$[0,t_{u,a}]$$ and $$R''\le0$$ on $$[t_{u,a},\ln(1+u)]$$.

In what follows, it is assumed that $$u>0$$, $$a$$ is a large enough (depending on $$u$$) positive real number, and $$t\in(0,\ln(1+u)]$$ -- unless otherwise specified. Note that $$(1+u)e^{-a t}-e^{t-a t}+e^t-1>e^{-a t}-e^{t-a t}+e^t-1=(e^t-1)(1-e^{-a t})>0$$, so that \eqref{10} makes sense.

We have \begin{equation*} \begin{aligned} g(t)&:=R''(t)e^{-a t} \left(-e^{a t}+e^{a t+t}-e^t+u+1\right)^3 \\ &=-a^2 (u+1) e^{a t}+e^{a t+2 t} \left(-\left(a^2 (u+3)\right)-2 a u-u\right) \\ &+e^t (u+1) \left(a^2 (u+3)+2 a u+u\right)+e^{2 t} \left(-\left(a^2 (2 u+3)\right)-2 a u+u\right) \\ &+e^{a t+t} \left(a^2 (2 u+3)+2 a u-u\right)+a^2 e^{3 t}+a^2 e^{a t+3 t}-a^2 (u+1)^2. \end{aligned} \end{equation*} Note that $$-e^{a t}+e^{a t+t}-e^t+u+1>-e^{a t}+e^{a t+t}-e^t+1=(e^t-1)(e^{a t}-1)>0$$, so that $$g(t)$$ equals $$R''(t)$$ in sign. So, the problem can be further restated as follows:

Show that for each real $$u>0$$ there is some real $$a_u>0$$ such that for each $$a\ge a_u$$ there is some $$t_{u,a}\in(0,\ln(1+u))$$ such that $$g\ge0$$ on $$[0,t_{u,a}]$$ and $$g\le0$$ on $$[t_{u,a},\ln(1+u)]$$.

To prove this, kill the exponentials of the form $$e^{ct}$$ for constant $$c$$'s one by one, by letting
$$\begin{equation*} D(t):=\frac{g'(t)}{e^t},\quad D_2(t):=\frac{D'(t)}{e^t},\quad D_3(t):=\frac{D_2'(t)}{e^t}, \end{equation*}$$ $$\begin{equation*} D_4(t):=\frac{D_3'(t)}{e^{(a-3)t}},\quad D_5(t):=\frac{D_4'(t)}{e^t},\quad D_6(t):=\frac{D_5'(t)}{e^t}. \end{equation*}$$ Then $$\begin{equation*} D_6(t)=6 a^3 (a+1) (a+2) (a+3) e^t \\ -2 a (a+1) \left(a^2+a-2\right) \left(a^2 (u+3)+2 a u+u\right), \end{equation*}$$ which is clearly increasing in $$t$$, from $$D_6(0)=-2 a (a+1) (a+2) \left(a^3 u+a^2 (u-12)-a u-u\right)<0$$ eventually (that is, for all large enough $$a$$, depending on $$u>0$$) and $$D_6(\ln(1+u))=2 a (a+1) (a+2) \left(2 a^3 u+4 a^2 (2 u+3)+a u+u\right)>0$$ eventually (and, in fact, for all real $$a>0$$ and $$u>0$$).

So, eventually $$D_6$$ is $$-+\,$$: that is, for some $$t_{6;a,u}\in(0,\ln(1+u))$$ we have $$D_6<0$$ on $$[0,t_{6;u,a})$$ and $$D_6>0$$ on $$(t_{6;u,a},\ln(1+u)]$$.

So, eventually $$D_5$$ is down-up -- that is, for some $$t_{6;a,u}\in(0,\ln(1+u))$$ we have $$D_5$$ is decreasing on $$[0,t_{6;u,a}]$$ and increasing on $$[t_{6;u,a},\ln(1+u)]$$. Also, eventually $$D_5(0)=-a (a+1) \left(10 a^3 u+a^2 (5 u-36)-13 a u-2 u\right)<0$$ and $$D_5(\ln(1+u))=a (a+1) \left(a^4 u^2+a^3 u (9 u+14)+a^2 \left(16 u^2+43 u+36\right)+a u (6 u+13)+2 u (2 u+1)\right)>0$$. So, eventually $$D_5$$ is $$-+$$.

So, eventually $$D_4$$ is down-up. Also, eventually $$D_4(0)=2 a^2 \left(-11 a^2 u+6 a (u+2)+5 u\right)<0$$ and $$D_4(\ln(1+u))=a (u+1) \left(2 a^4 u^2+7 a^3 u (u+2)+8 a^2 \left(u^2+3 u+3\right)+5 a u (u+2)+2 u^2\right)>0$$. So, eventually $$D_4$$ is $$-+$$.

So, eventually $$D_3$$ is down-up. Also, eventually $$D_3(0)=-a \left(14 a^2 u+3 a (3 u-4)+u\right)<0$$ and $$D_3(\ln(1+u))\frac{(1+u)^2}{a}=(u+1)^a\left(a^3 u^2+2 a^2 u (u+2)+a \left(-u^2+3 u+6\right)-u (2 u+1)\right) +6 a (u+1)^2>0$$. So, eventually $$D_3$$ is $$-+$$.

So, eventually $$D_2$$ is down-up. Also, eventually $$D_2(0)=-12 a (a+1) u<0$$ and $$D_2(\ln(1+u))(1+u)^2 =(u+1)^a\left(-a^2 u (u+1) (3 u+2)-a u (u+1) (7 u+8)-2 u (u+1)^2\right) +2 a^2 u (u+1)^2-4 a u (u+1)^2+2 u (u+1)^2<0$$. So, eventually $$D_2<0$$.

So, eventually $$D$$ is decreasing. Also, eventually $$D(0)=u \left(a^2 u+2 a (u-3)+u\right)>0$$ and $$D(\ln(1+u))(1+u)^2 =(u+1)^a\left(-a^2 u^2 (u+1)^2-a u (5 u+4) (u+1)^2-u (2 u+3) (u+1)^2\right) -2 a u (u+1)^3+3 u (u+1)^3<0$$. So, eventually $$D$$ is $$+-\,$$: that is, for some $$t_{1;a,u}\in(0,\ln(1+u))$$ we have $$D>0$$ on $$[0,t_{1;u,a})$$ and $$D<0$$ on $$(t_{1;u,a},\ln(1+u)]$$.

So, eventually $$g$$ is up-down -- that is, for some $$t_{1;a,u}\in(0,\ln(1+u))$$ the function $$g$$ is increasing on $$[0,t_{1;u,a})$$ and decreasing on $$(t_{1;u,a},\ln(1+u)]$$. Also, eventually $$g(0)=(2 a+1) u^2>0$$ and $$g(\ln(1+u))(1+u)^2 =(u+1)^a\left(-2 a u^2 (u+1)^3-u (u+2) (u+1)^3\right) +2 u (u+1)^4<0$$.

So, eventually $$g$$ is $$+-$$. $$\quad\Box$$

• Thanks a lot for the answer!
– vico
Aug 22, 2023 at 18:55