Racetrack principle

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 ) = h ( x ) − h ( 0 ) x − 0 = 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 ) ≥ g ′ ( x ) for all x > a , and if f ( a ) = g ( a ) , then f ( x ) ≥ 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.