Racetrack principle

In calculus, the racetrack principle describes the movement and growth of two functions in terms of their derivatives.

This principle is derived from the fact that if a horse named Frank Fleetfeet always runs faster than a horse named Greg Gooseleg, then if Frank and Greg start a race from the same place and the same time, then Frank will win. More briefly, the horse that starts fast and stays fast wins.

In symbols:

if $f'(x)>g'(x)$ for all $x>0$, and if $f(0)=g(0)$, then $f(x)>g(x)$ for all $x>0$.

or, substituting ≥ for > produces the theorem

if $f'(x) \ge g'(x)$ for all $x>0$, and if $f(0)=g(0)$, then $f(x) \ge g(x)$ for all $x>0$.

which can be proved in a similar way

Proof

This principle can be proven by considering the function h(x) = f(x) - g(x). If we were to take the derivative we would notice that for x>0

$h'= f'-g'>0.$

Also notice that h(0) = 0. Combining these observations, we can use the mean value theorem on the interval [0, x] and get

$h'(x_0)= \frac{h(x)-h(0)}{x-0}= \frac{f(x)-g(x)}{x}>0.$

Since x > 0 for the mean value theorem to work then we may conclude that f(x) - g(x) > 0. This implies f(x) > g(x).

Generalizations

The statement of the racetrack principle can slightly generalized as follows;

if $f'(x)>g'(x)$ for all $x>a$, and if $f(a)=g(a)$, then $f(x)>g(x)$ for all $x>a$.

as above, substituting ≥ for > produces the theorem

if $f'(x) \ge g'(x)$ for all $x>a$, and if $f(a)=g(a)$, then $f(x) \ge g(x)$ for all $x>a$.

Proof

This generalization can be proved from the racetrack principle as follows:

Given $f'(x)>g'(x)$ for all $x>a$ where a≥0, and $f(a)=g(a)$,

Consider functions $f_2(x)=f(x-a)$ and $g_2(x)=g(x-a)$

$f_2'(x)>g_2'(x)$ for all $x>0$, and $f_2(0)=g_2(0)$, which by the proof of the racetrack principle above means $f_2(x)>g_2(x)$ for all $x>0$ so $f(x)>g(x)$ for all $x>a$.

Application

The racetrack principle can be used to prove a lemma necessary to show that the exponential function grows faster than any power function. The lemma required is that

$e^{x}>x$

for all real x. This is obvious for x<0 but the racetrack principle is required for x>0. To see how it is used we consider the functions

$f(x)=e^{x}$

and

$g(x)=x+1.$

Notice that f(0) = g(0) and that

$e^{x}>1$

because the exponential function is always increasing (monotonic) so $f'(x)>g'(x)$. Thus by the racetrack principle f(x)>g(x). Thus,

$e^{x}>x+1>x$

for all x>0.