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I'm reading this book in Classical Mechanics and they derive an equation for the time a projectile takes to reach the ground once is fired (accounting for air resistance):

$$T=\frac{kV+g}{gk}(1-e^{-kT})$$

I do not have any questions on how they derive the equation or about the equation itself.My question is on what they say after formulating such equation:

"This is a transcendental equation, and therefore we cannot obtain an analytic expression for T."

I know transcendental equations are logarithm functions, trig functions or exponential functions, but I still do not understand when they say "you cannot obtain an analytical expression".

What do they mean ?

Hermes Chirino
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  • For example the equation $e^x=5-x$ doesn't have a closed form solution in terms of elementary functions. – user5402 Jun 30 '15 at 21:09
  • http://math.stackexchange.com/questions/568976/how-to-solve-the-transcendental-equation-ah-bhc-with-a-parameter#568994 – user5402 Jun 30 '15 at 21:18
  • A closed-formed expression for the time of flight of a projectile in a linear resisting medium in terms of the Lambert W function can be found here – omegadot Jul 03 '15 at 01:07

1 Answers1

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Typically, equations which involve mixtures of polynomial, trigonometric, exponential terms do not show explicit solutions in terms of elementary functions (the solution of $x=\cos(x)$ is the simplest example I have in mind).

However, you will be pleased (I hope !) to know that any equation which can write or rewrite $$A+Bx+C\log(D+Ex)=0$$ has solutions which express in terms of Lambert function $W(x)$ which is defined by $x=W(x)e^{W(x)}$. As metacompactness commented, the solution of the equation you posted is given by $$T=a+\frac{W\left(-a\, k\, e^{-a k}\right)}{k}$$ where $a=\frac{kV+g}{gk}$.

You will learn about Lambert function which is a beautiful one (Lambert and Euler worked together). In the real domain, $W(x)$ requires $x \geq -\frac 1e$. So, for your case with $a\, k\, e^{-a k} \leq \frac 1e$, no problem since this always holds.

Claude Leibovici
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