What's The Discrete Math Derivative Equivalent?

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$$ƒ = b^n$$
$$ b,n,I ∈ ℤ $$

Condition: Upon choosing a base value [itex]b[/itex]..

$$ n | b^n ≤ I $$
(n is determined based off the value of [itex]b[/itex] to yield the highest ƒ without going over [itex]I[/itex])
$$1<b<L , L<<I$$
where [itex]I[/itex] is some large number, and [itex]L[/itex] is also sufficiently large such that we want to avoid going through each base integer via trial and error....

How might I determine the base value that yields a value [itex]ƒ[/itex] that is closest to [itex]I[/itex]?
 

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  • #2
andrewkirk
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I'm not sure exactly what you are asking.
Are you asking how to, given ##L,I\in\mathbb{Z}## with ##0<<L<<I##, find the combination ##b,n\in\mathbb{Z}## that maximizes ##b^n## subject to the constraints
1. ##1<b<L##; and
2. ##b^n<I##

If so then you could set ##Hi=\lfloor\log_2 I\rfloor## and ##Lo=\lceil \log_L I\rceil##, then find
$$b^n=max_{Lo\leq m\leq Hi}\bigg( \lfloor I^\tfrac{1}{m}\rfloor\bigg)^m$$

where ##n## is the value of ##m## that delivers that maximum.
That will give a much smaller search space than ##1,...,L##.
 
Last edited:

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