Racetrack principle

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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[edit]

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[edit]

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[edit]

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[edit]

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.

External links[edit]