A closed form solution for this exponential sum inequality $e^{-Ax} + e^{-Bx} \leq C$?

Question

Is there a closed form solution for $e^{-Ax} + e^{-Bx} \leq C$ where $A,B\in\mathbb{R}_{+}$ and $C\in [0,1]$ ?

Answer 1

Without loss of generality, I shall assume $a >b$.

For the zero of function $$f(x)=e^{-a x}+e^{-b x}-c$$ the solution is between $$x_a=\frac{\log \left(\frac{2}{c}\right)}{a} \qquad \text{and} \qquad x_b=\frac{\log \left(\frac{2}{c}\right)}{b}$$

Now, we shall consider the more linear problem of $$g(x)=\log(e^{-a x}+e^{-b x})-\log(c)$$ for which $$g'(x)=-\frac{a e^{-a x}+b e^{-b x}}{e^{-a x}+e^{-b x}}\,\, <0 \qquad \text{and} \qquad g''(x)=\frac{(a-b)^2 e^{ (a+b)x}}{\left(e^{a x}+e^{b x}\right)^2}\,\,>0$$ Now, one iteration of Newton method will give $$x'_a=x_a-\frac{g(x_a)}{g'(x_a)}\,\,> \,\,x_a $$

Since $g(a)>0$, by Darboux theorem, since the second derivative is positive, $x'_a$ is an underestimate of the solution $(x'_a < x_{sol})$. A second iteration $$x''_a=x'_a-\frac{g(x'_a)}{g'(x'_a)}$$ will probably give almost the solution.

Trying for a few values of $a$ and $b$ for $c=\frac 12$, some results $$\left( \begin{array}{ccccccc} a & b & x_a & x_b & x'_a & x''_a & \text{solution} \\ \pi & e & 0.441271200 & 0.509989195 & 0.474860563 & 0.474869172 & 0.474869172 \\ 2 \pi & e & 0.220635600 & 0.509989195 & 0.342888065 & 0.348336941 & 0.348346335 \\ \pi & \frac{e}{2} & 0.441271200 & 1.019978390 & 0.685776130 & 0.696673882 & 0.696692669 \\ 2 \pi & 2 e & 0.220635600 & 0.254994597 & 0.237430282 & 0.237434586 & 0.237434586 \end{array} \right)$$

Edit

There is one case which is easy to check : $b=\frac a2$. For this case, we have $$x'_a=\frac{2 \left(\sqrt{c}+\sqrt{2}\right) \log \left(\sqrt{c}+\sqrt{2}\right)-3 \sqrt{c} \log (c)-2 \sqrt{2} \log (c)-\sqrt{2} \log (2)}{a \left(2 \sqrt{c}+\sqrt{2}\right)}$$ while $$x_{sol}=\frac{1}{a}\log \left(\frac{2 c+1+\sqrt{4 c+1}}{2 c^2}\right)$$ At this point, the ratio $\frac{x'_a}{x_{sol}}$ does not depend on $a$. It starts at $1$ for $c=0$, goes through a minimum of $0.981671$ around $c=0.05$ and grows up to $0.996795$ for $c=1$.

It seems that a better approximation would be given by the first iterate of the original Halley method. This new estimate write $$x_{est}=x_a+\frac{2\, g(x_a)\, g'(x_a)}{g(x_a)\, g''(x_a)-2\, g'(x_a)^2}$$ For the four cases given above, it would lead to $$\{0.474869174,0.348456482,0.696912963,0.237434587\}$$

For the specific case where $b=\frac a2$, the ratio $\frac{x_{est}}{x_{sol}}$ does not depend on $a$. It starts at $1$ for $c=0$, goes through a maximum of $1.00973$ around $c=0.005$ and decreases to $0.999990$ for $c=1$.

A still better approximation would be given by the first iterate of the original Householder method. This new estimate write $$x_{est}=x_a+\frac{3 \,g(x_a) \left(g(x_a) \,g''(x_a)-2\, g'(x_a)^2\right)}{g(x_a)^2 \,g'''(x_a)+6\, g'(x_a)^3-6 \,g(x_a) \, g'(x_a)\, g''(x_a)}$$

For the four cases given above, it would lead to $$\{0.474869172,0.348390812,0.696781624,0.237434586\}$$

For the specific case where $b=\frac a2$, the ratio $\frac{x_{est}}{x_{sol}}$ starts at $1$ for $c=0$, goes through a maximum of $1.00014$ around $c=0.155$ and decreases to $0.999990$ for $c=1$

Answer 2

WLOG, assume $A > B > 0$ and $0 < C \le 1$. Let $p = \frac{B}{A} \in (0, 1)$. Let $a = C^{p-1}$. Let $u = \frac{1}{C}\mathrm{e}^{-Ax}$. We need to solve the equation $u + a u^p = 1$ which admits an infinite series solution (see [1]) $$u = \sum_{k=0}^\infty \frac{\Gamma(pk+1)a^k (-1)^k}{\Gamma((p-1)k+2) k!}.$$ Thus, the solution of $\mathrm{e}^{-Ax} + \mathrm{e}^{-Bx} = C$ is given by $$x = - \frac{\ln C}{A} -\frac{1}{A}\ln \left(\sum_{k=0}^\infty \frac{\Gamma(pk+1)a^k(-1)^k}{\Gamma((p-1)k+2) k!}\right). \tag{1}$$

For example, $A = \sqrt{5}, B = \sqrt{2}$, $C = \frac{2}{3}$, (1) gives $x \approx 0.619497866$.

Reference

[1] Nikos Bagis, Solution of Polynomial Equations with Nested Radicals, https://arxiv.org/pdf/1406.1948.pdf

Answer 3

You have better to recast the equation into a symmetric form, by putting $$ \left\{ \matrix{ s = \left( {A + B} \right)/2 \hfill \cr d = \left( {A - B} \right)/2 \hfill \cr} \right.\quad \Leftrightarrow \quad \left\{ \matrix{ A = s + d \hfill \cr B = s - d \hfill \cr} \right. $$ so as to get $$ e^{\, - Ax} + e^{\, - Bx} = e^{\, - sx} \left( {e^{\, - dx} + e^{\,dx} } \right) = 2e^{\, - sx} \cosh (dx) $$ and so $$ \cosh (dx) \le {C \over 2}e^{\,sx} $$

Then you can perform on this the various approximation processes already indicated.

Answer 4

I prefer to add another answer instead of adding to the previous one which is already too long.

Instead of the previous starting point, let us use $$x_0=\frac{2\log \left(\frac{2}{c}\right)}{a+b} $$ which is obtained by the first iteration of Newton method starting at $x=0$. By Darboux theorem, this is an underestimate of the solution; its advantage is that it takes into account both $a$ and $b$.

The results for the previous four cases ($x_1$ being the first iterate of Newton method starting at $x_0$). $$\left( \begin{array}{cccccc} a & b & x_0 & x_1 & \text{solution} \\ \pi & e & 0.473148142 & 0.474869150 & 0.474869172 \\ 2 \pi & e & 0.308015202 & 0.347822293 & 0.348346335 \\ \pi & \frac{e}{2} & 0.616030405 & 0.695644586 & 0.696692669 \\ 2 \pi & 2 e & 0.236574071 & 0.237434575 & 0.237434586 \end{array} \right)$$ The results are much better.

For the case where $b=\frac a2$, the ratio $\frac {x_1}{x_{sol}}$ starts at $1$ for $c=0$, goes through a minimum of $0.996777$ around $c=0.04$ and grows up to $0.999935$ for $c=1$. Much better again.

For the same case, using one iteration of Halley method, the ratio $\frac {x_1}{x_{sol}}$ starts at $1$ for $c=0$, goes through a maximum of $1.00091$ around $c=0.01$ and grows up to $1$ for $c=1$. Much better again.

For the same case, using one iteration of Householer method, the ratio $\frac {x_1}{x_{sol}}$ starts at $1$ for $c=0$, goes through a maximum of $1.000001$ around $c=0.21$ and grows up to $1$ for $c=1$. Much better again.