Why is this equation only solvable by approximation?

Question

In the following equation I am trying to solve for $\alpha$, where $F$ and $P$ are constants.

$$P = e^{-\alpha/2}-\frac{1}{1+F}e^{-\alpha(1+F)/2}$$

Normally when I encounter an equation like this that I can't solve by simple algebra, I just stick it in Mathematica/Wolfram Alpha and get out an equation. However, when I did this all of the solutions where determined using approximation, so I couldn't just get a symbolic formula for $\alpha$.

I have two questions:

1. Is there a way to symbolically solve this equation?
2. If not, what makes this equation only solvable by approximation? How can I recognize this property in an equation?

Answer 1

In fact, most equations you write down randomly will not be solvable symbolically. A symbolic solution is an exceptional state, not the normal one.

Furthermore, whether a equation is solvable "symbolically" depends on what "symbols" you're allowed to use. For example, if you didn't have a symbol for square root, you won't be able to solve $x^2=2$ symbolically. And if a equation has a unique solution and you can't solve it symbolically with existing symbols, you can always introduce a new symbol for its solution. That's how many mathematical symbols get introduced.

Answer 2

In the most general case, equation$$P = e^{-\alpha/2}-\frac{e^{-\alpha(1+F)/2}}{1+F}$$ will not show analytical solutions.

However, for some very specific values of $F$, this equation would result in a polynomial in $x=e^{-\alpha/2}$. $$x^{F+1}-(F+1) x+(F+1) P=0$$ This woud be the case for $F=\pm \frac 12,1,\pm 2,\pm 3$.

Answer 3

On a related note, if you want to solve this equation graphically, you can plot it as a (degenerate) nomogram with oriented transparency.

In that form, the equation looks like:

$$0 = (e^{-\alpha/2}-P) - \frac{1}{F+1}e^{-\alpha/2 \cdot (F+1)}$$

or putting $Q \equiv \exp{(-\alpha/2)}$,

$$0 = (Q-P) - \frac{1}{F+1}Q^{F+1}$$

To make the plot,

1. Create a coordinate chart with an x and y axis.

2. The values on the $x$ axis will correspond to values of $Q$.

3. At select values of $x=Q$ along the $x$ axis, solve for the corresponding value of $\alpha$. Namely, $\alpha = 2\log{Q}$. Include that $\alpha$ value at that point, next to the label for the $Q$ value. These values will let you translate between $Q$ values and $\alpha$ values.

   For example, at the point $x=1$, write $\langle Q=1, \alpha=0\rangle$. At $x=5$, write $\langle Q=5, \alpha \approx 1.40\rangle$.

4. Choose various values of $P$ of interest. For each value of $P$, draw the curve $y(x) = x-P$. Label the curve with that value of $P$. (These curves are straight lines.)

5. Choose various values of $F$ of interest. For each value of $F$, draw the curve $y(x) = \frac{1}{1+F} x^{1+F}$. Label the curve with the value of $F$. (These curves are polynomials/exponentials.)

6. Each $P$ curve meets each $F$ curve in exactly one place, and you can read off the $\alpha$ value by looking at the x-coordinate of that intersection.

7. Because of these strong geometric constraints, you can actually solve for any one of $\langle P, F, \alpha\rangle$ given the other two. (For example, to solve for $P$ given $F$ and $\alpha$, locate the $F$ curve and see which $P$ curve it intersects at the $x$ coordinate corresponding to $\alpha$.)