Comparing rates of change: which function increases faster?I am comparing two functions for x≥1f(x)=ln⁡(⌊x9⌋!)−ln⁡(⌊x10⌋!)−ln⁡(⌊x90⌋!)g(x)=(2.07766)x9+(0.057712778)xComparing their values over a finite range suggests that f(x) grows slightly faster than g(x).Since f(x) is not continuous, I am not able to use its derivative to establish the rate of growth. How can I verify that as x increases, f(x) increases faster than g(x)?It looks like I copied the problem wrong. It should have read (0.057712778)x instead of (0.103883)x. The correct term was (1.03883)⋅x18=(0.057712778)x

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hofirati8y9 · Accepted Answer

The problem is a bit messy since f(x) is a step function. Initially, we can remove the steps with the restriction x=90y, under which conditions f(y)=ln⁡(10y)!−ln⁡(9y)!−ln⁡(y)! We can now use Stirling&#039;s approximation to getf(y)=10yln⁡(10y)−10y−9yln⁡(9y)+9y−yln⁡y+y+O(ln⁡(10y))=10yln⁡10+10yln⁡y−9yln⁡9−9yln⁡y−yln⁡y+O(ln⁡(10y))=y(10ln⁡10−9ln⁡9)+O(ln⁡(10y))≈3.250829733y+O(ln⁡(10y))On the other hand,g(y)=(2.07766)10y+(0.103883)(90y)=9.34947y+O(y)Based on this, for at least those x that are multiples of 90, we can say that g(x) grows faster than f(x) (eventually), although they are both close to linear growth.

Comparing rates of change: which function increases faster? I

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