I have the following equation and I was wondering if I can solve for x given that it appears both as an exponent and a base:

$[\frac{1}{\sqrt {2\pi}.S}.e^{-\frac{(x-M)^2}{2S^2}}-0.5\frac{1}{\sqrt {2\pi}.S}.e^{-\frac{(x-M)^2}{2S^2}}.\frac{4.(\frac{x-N}{\sqrt {2}.D})}{\sqrt {\pi}.e^{-\frac{(x-N)^2}{2D^2}}+\sqrt {\pi.e^{-2\frac{(x-N)^2}{2D^2}}+16.\frac{(x-N)^2}{2D^2}}} ]+[ \frac{1}{\sqrt {2\pi}.D}.e^{-\frac{(x-N)^2}{2D^2}}-0.5\frac{1}{\sqrt {2\pi}.D}.e^{-\frac{(x-N)^2}{2D^2}}.\frac{4.(\frac{x-M}{\sqrt {2}.S})}{\sqrt {\pi}.e^{-\frac{(x-M)^2}{2S^2}}+\sqrt {\pi.e^{-2\frac{(x-M)^2}{2S^2}}+16.\frac{(x-M)^2}{2S^2}}}] = \frac{1}{v}$

In other words, can I find an expression for x as a function of all the other variables?

  • $\begingroup$ Most likely not. $\endgroup$ – 1-___- Aug 19 '15 at 19:03
  • 2
    $\begingroup$ Wow! Your equation is very complicated! How did you get it, and why do you want to solve it? $\endgroup$ – Crostul Aug 19 '15 at 19:04
  • $\begingroup$ It's an approximation of the derivative of a joint probability function that I would like to equate to the value $\frac{1}{v}$ through x $\endgroup$ – tvbc Aug 19 '15 at 19:07

Let's simplify the equation a bit so we can better see its form:
$$C_{1}e^{X_{1}} + C_{2}e^{X_{1}}\cdot \frac{X_{2}}{C_{3}e^{X_{3}} \sqrt{C_{4}e^{X_{3}} + C_{5}X_{3}}} + C_{6}e^{X_{4}} + C_{7}e^{X_{4}}\cdot \frac{X_{5}}{C_{8}e^{X_{6}} \sqrt{C_{9}e^{X_{6}} + C_{10}X_{6}}} = C_{11}$$ With constants as $C_{n}$ above and expressions involving x as $X_{n}$ above.
Even if all of the constants were 1, we would get:
$$e^{X_1} + e^{X_4} + X_{2}\frac{e^{X_1}e^{-X_{3}}}{\sqrt{e^{X_{3}} + X_{3}}} + X_{5}\frac{e^{X_4}e^{-X_{6}}}{\sqrt{e^{X_{6}} + X_{6}}} = 1$$ Which could, with some further magical simplifications (letting all of the expressions in x be the same), be written as
$$e^{2X} + \frac{2X}{\sqrt{e^{X} + X}} = 1$$
Heck, even if the denominator was 1, we would get (letting $Y = 2X$):
$$e^{Y} + Y = 1$$
Which in and of itself is not solvable for Y.

All of this was just to show that even with certain simplifications your equation reduces to a form which is very much unsolvable, so it is extremely unlikely that your original equation is solvable for x.

  • $\begingroup$ Good enough for me, thanks. $\endgroup$ – tvbc Aug 20 '15 at 20:45

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