What n is required to make n > a^n

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In summary, the conversation discusses finding a bound for n in terms of a to satisfy the inequality n! >= a^n. One suggestion is to use Sterling's formula to determine that n = \lceil e\, a \rceil will work. Another suggestion is to choose n = a^2, which can be proven by comparing the terms in the expansion of the left and right sides of the inequality. This provides a relatively easy proof and allows for choosing a larger bound, such as n = a^2 + a.
  • #1
uart
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I'm looking for a bound on n ( that is, "n" as a function of "a") to satisfy,

[tex]n! \geq a^n[/tex]

It doesn't have to be the lowest possible bound, preferably one that's relatively easy to prove.

Thanks.
 
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  • #2
OK I've just found Sterling's formula, looking at that I can deduce that [tex]n = \lceil e\, a \rceil[/tex] will definitely work.

Does anyone have a bound that is very easy proof. Maybe something like n = 3a or even larger.
 
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  • #3
Well, up until n=a, you clearly can't get anywhere (and have a-1 more a's on the RHS than the LHS)

Then you add up an a (actually, a number greater than a against a)for every one added to n until you hit a2

Then you have a-2 a's on the RHS compared to the LHS. From then on, every time you add one to n, you multiply the LHS by something greater than a2, and the RHS by a. So if n=a2+a (so you pick up a extra a's from the a2 and larger terms to mitigate the a-1 you felll behind by at the start), then n! >= an

That should work. If it needs clarification, I'll try to write it out a bit neater
 
  • #4
Ok good, going as high as sqaures makes it pretty easy.

Take n = a^2 (for a>=2). You can just compare the terms (in the expansion of LHS and RHS) in pairs and show that every chosen pair-product on the LHS is greater than or equal to that of each RHS pair (which is of course always a*a on the RHS).

eg

1 * a^2 = a*a
2 * (a^2-1) > a*a
3 * (a^2-2) > a*a
...
k ^ (a^2+1-k) > a*a : { down to k = floor(a^2 / 2) }

That's the type of thing I was looking for, something that's like a "handwaving + by inspection" type of "proof". ;)
 
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1. What is the concept of "n > a^n"?

The concept of "n > a^n" refers to an inequality in which the value of n is greater than the value of a raised to the power of n. In other words, n is being compared to a^n, where a is a constant number.

2. What is the significance of this inequality in scientific research?

This inequality can be used to understand the growth rate of a function. It helps in determining the minimum value of n required to make n > a^n true, which can provide insight into the efficiency and potential of a certain process or phenomenon.

3. How is the value of n calculated in this inequality?

The value of n can be calculated through various mathematical methods such as logarithms, differentiation, or trial and error. It depends on the specific equation or scenario in which the inequality is being applied.

4. What is the relationship between n and a in this inequality?

The relationship between n and a in this inequality is that n must be greater than a^n for the inequality to be true. This means that as n increases, a^n must increase at a faster rate in order for n to be greater.

5. Can this inequality be applied to all scientific fields?

Yes, this inequality can be applied to various scientific fields such as biology, physics, chemistry, and economics. It can be used to analyze the growth rate of different phenomena and processes in these fields.

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