How do I know if this equation can be solved symbolically?

Question

Can these equations be solved symbolically for $x$?

$$ \begin{align} x &= \frac{p - p_m(x)}{p_m(x) - p_m(x)^2} \\ \\ p_m(x) &= \frac{e^x}{e^x + e^y} \\ \end{align} $$

If not (which I assume), how do I know?

Answer 1

If it could be done, it would be most likely done with the Lambert W function. However, with the amount of addition in there, it is unlikely you could solve it like so.

So putting this all in, we have

$$x=\frac{p-p_m(x)}{p_m(x)[1-p_m(x)]}$$

$$=\frac{(e^x+e^y)\left(p-\frac{e^x}{e^x+e^y}\right)}{e^x\left(1-\frac{e^x}{e^x+e^y}\right)}$$

$$=\frac{(e^x+e^y)[p(e^x+e^y)-e^x]}{e^{x+y}}$$

$$=e^{-x-y}[p(e^{2x}+2e^{x+y}+e^{2y})-e^{2x}-e^{x+y}]$$

$$=e^{-x-y}[(p-1)e^{2x}+(2p-1)e^{x+y}+2e^{2y}]$$

$$x=(p-1)e^{x-y}+(2p-1)+2e^{y-x}\tag1$$

At this point, I am fairly confident there is no way to solve for $x$. We can, however, approach the problem numerically to get close values for $x$.

For Newton's method, we have

$$a_{n+1}=a_n-\frac{(p-1)e^{a_n-y}+(2p-1)+2e^{y-a_n}-a_n}{(p-1)e^{a_n-y}-2e^{y-a_n}+1}$$

And $x=\lim_{n\to\infty}a_n$.

Perhaps if you make $p$ or $y$ known, the problem could be handled better, but I am unsure if there is much you can do here.

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Furthermore, if we assume the answer is some very large $x$, then we have

$$\lim_{x\to\infty}p_m(x)=1$$

We would also have

$$\lim_{x\to\infty}x=\lim_{x\to\infty}\frac{p-p_m(x)}{p_m(x)[1-p_m(x)]}\approx\lim_{p_m(x)\to1}\frac1{1-p_m(x)}$$

Upon further inspection, we find that $p_m$ will approach $1$ from the negative side, and thus it can be seen that for $x$ to be large, we must have $p<p_m=1$ for both sides to be positive and large.

Assume negative and large $x$, and we will find

$$\lim_{x\to-\infty}p_m(x)=0$$

$$\lim_{x\to-\infty}x=\lim_{p_m\to0^+}\frac{p-p_m(x)}{p_m(x)[1-p_m(x)]}$$

For that to be true, we must have $p<0$.