Solving $-1=e^a-2e^{av}$ as part of an equation system

Question

Problem

Given $f_2(x)=e^{ax-b}+c$ with $x \in \left(0,1\right)$, I am trying to calculate the parameters $a,b,c$ in respect to the following constraints:

$$ \begin{align} f_2(0) &= 0 \\ f_2(1) &= 1 \\ f_2(v) &= \frac{1}{2} \end{align} $$ Where $v \in \left(0.5,1\right)$ is a fixed parameter.

Current Situation

I have tried solving this analytically and have reached several other representations, including $-1=e^a-2e^{av}$, which should lead to a solution for $a$ but have not found one. Wolfram-alpha can calculate one and given $v=0.6$ as an example, a solution is $a \approx 0.822163$, $b \approx 0.24373$, $c \approx -0.784057$ (see graph below).

Substitution of $e^a=u$ yields $-1 = u - 2u^v$ which seems doable ($v$ is constant), but I can't seem to find a solution myself.

Question

I would like to hear pointers as to how (and if) this can be solved analytically.

Context

I am trying to find a monotonic and continuous function $f$ that satisfies the above constraints in order to stretch the interval $\left(0,1\right)$ to have it's new center at $v$. Such a function exists for the case $v \in \left(0,0.5\right)$ using the logarithm:

$$f_1(x) = a \cdot \ln(x+b)+c$$

The parameters can be determined analytically and given that $\exp$ is the inverse of $\ln$, I expect a similiar solution for $v > 0.5$ using $\exp$.

Answer 1

In other words, you are looking for the inverse function of $$v=\frac 1 a \log \left(\frac{e^a+1}{2}\right)\tag 1$$ with $a \in (0,\infty)$.

As you supposed, it does not exist. From a numerical point of view, there is no difficulties.

Using series, we have $$\color{red}{w}=2v-1=\sum_{n=0}^\infty \alpha_n \, a^{2n+1}$$ which can be transformed into a $[2p+1,2p]$ Padé approximant $P_p$. For example, the simplest $$P_2=\frac{3 a \left(a^2+40\right)}{32 \left(a^2+15\right)}$$ whose error is $\frac{31 a^7}{4838400}$.

It is quite decent for $0 \leq a < \frac 52 $ that is to say $\frac 12 \leq v \leq \frac{3}{4}$.

Solving for $a$, we need the solution of the cubic equation $$3a^3-32w\,a^2+120\,a-480w=0$$ which has only one real root for $v \in [0.5,1.0]$.

$$a=\frac{4}{9} \left(8w+\sqrt{270-256 w^2} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{2 w \left(2048 w^2+405\right)}{\left(270-256 w^2\right)^{3/2}}\right)\right)\right)$$

Plugging this expression in $(1)$, we have, using Taylor expansion, $$v=v-\frac{31744 }{4725}\left(v-\frac{1}{2}\right)^7+O\left(\left(v-\frac{1}{2}\right)^{9}\right)$$ For $v=\frac 56$, the error term is $-0.00307$.

Some values

$$\left( \begin{array}{ccc} v & \text{estimate}& \text{solution} \\ 0.500 & 0.00000 & 0.00000 \\ 0.525 & 0.20033 & 0.20033 \\ 0.550 & 0.40269 & 0.40269 \\ 0.575 & 0.60919 & 0.60919 \\ 0.600 & 0.82216 & 0.82216 \\ 0.625 & 1.04423 & 1.04426 \\ 0.650 & 1.27852 & 1.27865 \\ 0.675 & 1.52881 & 1.52925 \\ 0.700 & 1.79977 & 1.80107 \\ 0.725 & 2.09719 & 2.10077 \\ 0.750 & 2.42824 & 2.43751 \end{array} \right)$$

Now, for large values of $a$, we can write $$v=1-\frac{\log (2)}{a}+\frac{e^{-a}}{a}+\cdots$$ which leads to $$a=\frac{\log (2)}{1-v}+W\left(-\frac{2^{-\frac{1}{1-v}}}{1-v}\right)$$ where $W(.)$ is Lambert function.

$$\left( \begin{array}{ccc} v & \text{estimate}& \text{solution} \\ 0.750 & 2.41519 & 2.43751 \\ 0.775 & 2.81420 & 2.82444 \\ 0.800 & 3.27704 & 3.28128 \\ 0.825 & 3.83774 & 3.83923 \\ 0.850 & 4.55057 & 4.55097 \\ 0.875 & 5.51290 & 5.51297 \\ 0.900 & 6.92161 & 6.92161 \\ 0.925 & 9.24067 & 9.24067 \\ 0.950 & 13.8629 & 13.8629 \\ 0.975 & 27.7259 & 27.7259 \end{array} \right)$$

Trying to improve for values of $v$ around $\frac 34$, we can write $$\color{red}{x}=v-\frac{2}{5} \log \left(\frac{1}{2} \left(1+e^{5/2}\right)\right)=\sum_{n=1}^\infty \beta_n \, \left(a-\frac{5}{2}\right)^n$$ and build the simplest $[1,1]$ corresponding Padé approximant. The $\beta_n$ being quite messy, they have been rationalized to give for the rhs

$$x=\frac {\left(a-\frac{5}{2}\right) }{\frac{8912 }{3127}\left(a-\frac{5}{2}\right)+\frac{12114}{823}} \implies a=\frac{5}{2}-\frac{37880478 x}{823 (8912 x-3127)}$$

Using the exact expansion, much better results for the "critical" zone $$\left( \begin{array}{ccc} v & \text{estimate}& \text{solution} \\ 0.70 & 1.80789 & 1.80107 \\ 0.71 & 1.92107 & 1.91716 \\ 0.72 & 2.04012 & 2.03821 \\ 0.73 & 2.16553 & 2.16480 \\ 0.74 & 2.29780 & 2.29764 \\ 0.75 & 2.43752 & 2.43751 \\ 0.76 & 2.58533 & 2.58534 \\ 0.77 & 2.74197 & 2.74222 \\ 0.78 & 2.90824 & 2.90943 \\ 0.79 & 3.08506 & 3.0885 \\ 0.80 & 3.27346 & 3.28128 \end{array} \right)$$

Answer 2

1)

$$f_2(x)=e^{ax-b}+c$$

Your constraints yield the equation system

$$\left\{e^{-b}+c=0,e^{a-b}+c=1,e^{av-b}+c=\frac{1}{2}\right\}.$$

This equation system has solutions if

$$b=\ln(e^{a}-1)+2k_1\pi i\ \land\ c=-(e^a-1)^{-1}\ \land\ v=\frac{1}{a}\left(\ln(\frac{1}{2}e^a+\frac{1}{2})+2k_2\pi i\right)$$

for $k_1,k_2\in\mathbb{Z}$.

2)

$$-1=e^a-2e^{av}$$

$$-1=u-2u^v$$

For rational $v$, this equation is related to an algebraic equation and we can use the known solution formulas and methods for algebraic equations.
For rational $v\neq 0,1$, the equation is related to a trinomial equation.
For real or complex $v\neq 0,1$, the equation is in a form similar to a trinomial equation. A closed-form solution can be obtained using confluent Fox-Wright Function $\ _1\Psi_1$ therefore.
$\ $

Szabó, P. G.: On the roots of the trinomial equation. Centr. Eur. J. Operat. Res. 18 (2010) (1) 97-104

Belkić, D.: All the trinomial roots, their powers and logarithms from the Lambert series, Bell polynomials and Fox–Wright function: illustration for genome multiplicity in survival of irradiated cells. J. Math. Chem. 57 (2019) 59-106