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

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

• You can approximate solutions, but I’m not sure about exact solutions. If nobody posts anything, I’ll try to write up an approximation. – Clayton Sep 30 at 13:48
• In general no. If $\frac BA$ or $\frac AB$ is an integer $\leq 4$, yes. – Claude Leibovici Sep 30 at 13:49
• @Clayton Thanks, I would be happy with an approximation better than solving the system $e^{-Ax}\leq C/2$ and $e^{-Bx}\leq C/2$. – Nocturne Sep 30 at 13:51
• Still better ! We can consider that we have an explicit non-iterative solution. – Claude Leibovici Oct 2 at 3:00
• Have a look at my second answer. Much better. Cheers and thanks for the problem. – Claude Leibovici Oct 3 at 3:26

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 $$c=1$$

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

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.

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.