Is there any way to solve this equation for $x$?: $\sum_{k\in K} e^k \left(\frac{a}{(1+be^{c(x+k)})}-d\right)=0$

Question

I'm trying to obtain the $x$ that satisfies the following equality where all pronumerals are constants except $k$ and $x$ but can't see a way to do it

$$\sum_{k\in K} e^k \left(\frac{a}{(1+be^{c(x+k)})}-d\right)=0$$

$K$ is an arbitrary finite subset of $\mathbb N$

Answer 1

You want to solve (numerically) the equation $$\sum_{k=i}^{k=j} e^k \left(\frac{a}{1+b\,e^{c(x+k)}}-d\right)=0$$ that is to say $$\sum_{k=i}^{k=j}\frac{e^k}{1+b\,e^{ck}\,e^{c x}}=A \qquad \text{with} \qquad A=\frac d a\, \frac{e^{j+1}-e^i}{e-1}$$ $$y=e^{cx}\qquad \implies \qquad \sum_{k=i}^{k=j}\frac{e^k}{1+b\,e^{ck}\,y}=A$$

Now, the problem is totally different depending on the sign of $b$.

If $b>0$, there is no vertical asymptote and the lhs is a decreasing function of $y$ which will have a single solution if $d<a$ (otherwise, no solution).

If $b<0$ you will have something like $(j-i)$ vertical asymptotes and one root between each pair and one root after the largest asymptote.

If $b>0$, since $$\sum_{k=i}^{k=j}\frac{e^k}{1+b\,e^{ck}\,y} < \sum_{k=i}^{k=j}\frac{e^k}{b\,e^{ck}\,y}=\frac{e^{-c i+c+i}-e^{-c j+j+1}}{b y( e^c -e)}$$ we have an upper bound for $y$ to be used as starting guess for Newton-like methods.