Solve $e^{k_1/x}+e^{k_2/x}+\cdots+e^{k_N/x}=1$ for $x$

Question

How can I solve $e^{k_1/x}+e^{k_2/x}+\cdots+e^{k_N/x}=1$ for $x$,

where $N\geq 1, k_1,\ldots,k_N \in \mathbb{R}, k_1,\ldots,k_N < 0, x\in \mathbb{R}$ and $x >0$.

I looked at the basic rules of exponentiation and logarithms and they do not seem to help simplify the equation in this particular case.

As a side comment: the values of $N$ I am working with are $N\approx 10^6$.

Answer 1

Let $f(x) = e^{k_1/x}+e^{k_2/x}+\cdots+e^{k_N/x}-1 =\sum_{i=1}^N e^{k_i/x}-1 $, and let $K = \sum_{i=1}^N k_i $.

The restrictions that $x > 0$ and $k_i < 0$ are important in what follows.

$f'(x) =\sum -\frac{k_i}{x^2}e^{k_i/x} =-\frac1{x^2}\sum k_ie^{k_i/x} =\frac1{x^2}\sum |k_i|e^{k_i/x} $, so $f'(x) > 0$. This means that your function has at most one root. Since $f(x) < 0$ for small $x$ and $f(x) > 0$ for large $x$, $f$ has exactly one positive root.

To get simple bounds on the root of $f(x) = 0$, let $k_{min} = \min(k_i)$ and $k_{max} = \max(k_i)$. Note that, since the $k_i < 0$, $|k_{min}| >|k_{max}| $.

Then $e^{k_{min}/x} \le e^{k_i/x} \le e^{k_{max}/x} $ so $e^{k_{min}/x} \le \dfrac{\sum_{i=1}^N e^{k_i/x}}{N} \le e^{k_{max}/x} $. If $f(x) = 0 $, then $e^{k_{min}/x} \le \dfrac{1}{N} \le e^{k_{max}/x} $ or $k_{min}/x \le -\ln(N) \le k_{max}/x $ or, since $k_i < 0$, $|k_{min}|/x \ge \ln(N) \ge |k_{max}|/x $ or $\frac{|k_{max}|}{\ln(N)} \le x \le \frac{|k_{min}|}{\ln(N)} $.

Another bound can be gotten using $e^z \ge 1+z$. $f(x) \ge \sum (1+k_i/x) - 1 = N+\frac1{x}\sum k_i - 1 = N+\frac{K}{x} - 1 $. If $f(x) = 0$, then $1-N \ge \frac{K}{x} = -\frac{|K|}{x} $ or $N-1 \le \frac{|K|}{x} $ or $x \le \frac{|K|}{N-1} $.

This can be improved by using the inequality between the arithmetic and geometric means.

$\begin{array}\\ \dfrac{\sum_{i=1}^N e^{k_i/x}}{N} &\ge \left(\prod_{i=1}^N e^{k_i/x}\right)^{1/N}\\ &= e^{\sum_{i=1}^N k_i/(Nx)}\\ &= e^{K/(Nx)}\\ \end{array} $

Therefore, if $f(x) = 0$, $\dfrac1{N} \ge e^{K/(Nx)} $ or $-\ln(N) \ge K/(Nx) $ or $\ln(N) \le |K|/(Nx) $ or $x \le \dfrac{|K|}{N\ln(N)} $.

So an initial $x$ could be $x_0 =\frac{|K|}{N\ln(N)} $.

We can then apply Newton's iteration, $x_{n+1} =x_n - \frac{f(x_n)}{f'(x_n)} $ a few times and see what happens. It should be OK.

Answer 2

This is equivalent to finding roots of the polynomial

$$y^{-k_1}+\cdots +y^{-k_N}-1=0$$

(by making $y:= e^{-\frac{1}{x}}$) so your problem is as difficult as finding roots of polynomials. It depends on the degree of that polynomial, in this case, it depends on the max of the $-k_i$.

Answer 3

Write an algorithm to solve the problem...

Let

$$ Q = \left( \sum_{m=1}^{n} \exp(k_m y) - 1 \right)^2. $$

In case we have the solution, we have $$ Q = 0 $$.

For changes in $Q$ we have

$$ \delta Q = 2 Q \left( \sum_{m=1}^{n} k_m \exp(k_m y) \right) \delta y. $$

For each step use

$$ \delta y = - 2 Q \left( \sum_{m=1}^{n} k_m \exp(k_m y) \right), $$

whence

$$ \delta Q \le 0, $$

so we get closer to the solution.

Once you have found $y$, the solution for $x$ is given by

$$ x = \frac{1}{y}. $$

This would do the trick.