3
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ You can approximate solutions, but I’m not sure about exact solutions. If nobody posts anything, I’ll try to write up an approximation. $\endgroup$ – Clayton Sep 30 at 13:48
  • 2
    $\begingroup$ In general no. If $\frac BA$ or $\frac AB$ is an integer $\leq 4$, yes. $\endgroup$ – Claude Leibovici Sep 30 at 13:49
  • $\begingroup$ @Clayton Thanks, I would be happy with an approximation better than solving the system $e^{-Ax}\leq C/2$ and $e^{-Bx}\leq C/2$. $\endgroup$ – Nocturne Sep 30 at 13:51
  • 1
    $\begingroup$ Still better ! We can consider that we have an explicit non-iterative solution. $\endgroup$ – Claude Leibovici Oct 2 at 3:00
  • 1
    $\begingroup$ Have a look at my second answer. Much better. Cheers and thanks for the problem. $\endgroup$ – Claude Leibovici Oct 3 at 3:26
3
$\begingroup$

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 $c=1$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.