Inverse Function Rule

One of the most well known theorems in mathematics is the Prime Number Theorem (PNT). The Prime Number Theorem has the same form as the trivial statement that was proved about $E(n)$ = the number of evens, in the previous subsection: $$ \lim_{n\rightarrow \infty} \frac{E(n)}{ n/2 } = 1 $$ Let $P(n)$ be the number of primes less than $n$, where $n$ is a natural number. Then this is the Prime Number Theorem: $$ \lim_{n\rightarrow \infty} \frac{P(n)}{ n / \ln(n) } = 1 $$ Quite another theory says that the cardinality of the prime numbers is the same as the cardinality of the natural numbers, which seems contradictory to the asymptotic (natural) density found: $P(n) \approx n / \ln(n)$ .
Likewise, the numerosity (cardinality) of the squares equals the numerosity of the naturals. There are "as many" squares as there are naturals. This is known as Galileo's Paradox. Because there exists a bijection between the naturals and the squares:

1 2 3  4  5  6  7  8  9  10  11  12  13  ..
| | |  |  |  |  |  |  |   |   |   |   |
1 4 9 16 25 36 49 64 81 100 121 144 169  ..

Despite our respect for Galileo's work, let's have another way to look at it:

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 .. 25 .. n
 |     |         |                    |     |
1^2   2^2       3^2                  4^2   5^2

Let $Q(n)$ be the number of squares less than or equal to $n$, where $n$ is a natural. Then we have a PNT look-alike (though much less difficult to prove) theorem: $$ \lim_{n\rightarrow \infty} \frac{Q(n)}{ \sqrt{n} } = 1 $$ Proof. It's easy to see that $\sqrt{n} - 1 < Q(n) \le \sqrt{n}$ . Now divide by $\sqrt{n}$ to get $1 - 1/\sqrt{n} < Q(n) / \sqrt{n} \le 1$ . For $n\rightarrow \infty$ the result follows.
The numerosity (cardinality) of the powers of $7$ equals the numerosity of the naturals. There are as many powers of $7$ as there are naturals. Proof: there exists a bijection between the naturals and these powers. As depicted here:

1  2   3    4     5      6      7       8        9        10  ..
|  |   |    |     |      |      |       |        |         | 
7 49 343 2401 16807 117649 823543 5764801 40353607 282475249  ..

But let's have another way to look at it:

 1   2 3 4 5 6   7   8 9 10 .. 48  49   50 .. 343 .. 2401 ..
 |               |                  |           |       |
7^0             7^1                7^2         7^3     7^4

Let $Z(n)$ be the number of $7$-powers less than or equal to $n$, where $n$ is a natural number. Then we have the theorem: $$ \lim_{n\rightarrow \infty} \frac{Z(n)}{ \ln(n) / \ln(7) } = 1 $$ Proof. From a picture [ graph of $\ln(x) / \ln(7)$ ] it's easy to see that: $$ Z(n) \le \frac{\ln(n)}{\ln(7)} < Z(n)+1 \quad \Longrightarrow \quad \frac{\ln(n)}{\ln(7)}-1 < Z(n) \le \frac{\ln(n)}{\ln(7)} $$ Divide by $\ln(n)/\ln(7)$ to get: $$1 - \frac{1}{\ln(n)/\ln(7)} < \frac{Z(n)}{\ln(n)/\ln(7)} \le 1 $$ For $n\rightarrow \infty$ the result follows easily.
Written as an asymptotic equality: $Z(n) \approx \ln(n)/\ln(7)$ . Now suppose that $\aleph_0$ is an incredibly large number - did I suggest anything? - then what if we write: $$Z(\aleph_0) = \frac{\ln(\aleph_0)}{\ln(7)}$$

We seek to generalize the above results.

   1     2     3     4     5     6    7     8     9   .. 
   |     |     |     |     |     |    |     |     |
 A(1)  A(2)  A(3)  A(4)  A(5)  A(6) A(7)  A(8)  A(9)  ..

Let there be defined a function $A : N \rightarrow N$ on the naturals $\mathbb{N}$ . Examples are the squares [ $A(n) = n^2$ ] and the powers of seven [ $A(n) = 7^n$ ] . Assume, in general, that $A(n)$ is a sequence monotonically increasing with $n$ .
The numerosity / cardinality of the function $A(n)$ equals the numerosity of the naturals; there are as many values of $A$ as there are naturals.
Proof: there exists a bijection between the naturals and these values, due to the fact that $A(n)$ is monotonically increasing with $n$, thus it has no upper bound, as is the case with the naturals too.
But let's have another way to look at it [ e.g with $A(n) = 7^n$ ]:

  1   2 3 4 5 6   7  8 9 10 .. 48  49  50 .. 343 .. 2401 ..
  0   0 0 0 0 0   1  1 1  1     1   2   2      3       4   : D(n)
A(0)            A(1)              A(2)       A(3)    A(4)

Let $D(n)$ be the number of $A(m)$ values (count) less than or equal to $n$, where $(m,n)$ are natural numbers. Then we have the following Theorem. $$ \lim_{n\rightarrow \infty} \frac{D(n)}{ A^{-1}(n) } = 1 $$ Proof. $A(n)$ is monotonically increasing with $n$ , therefore the inverse sequence $A^{-1}(n)$ exists in the first place. And it is monotonically increasing as well. Furthermore, we see that $D(n)$ is $m$ for $A(m) \le n < A(m+1)$ .
Consequently: $A(D(n)) \le n < A(D(n)+1)$ hence $D(n) \le A^{-1}(n) < D(n) + 1$ hence $A^{-1}(n) - 1 < D(n) \le A^{-1}(n)$ .
Divide by $A^{-1}(n)$ to get: $1 - 1/A^{-1}(n) < D(n) / A^{-1}(n) \le 1$ . For $n\rightarrow \infty$ now the theorem follows, because $A^{-1}(n) \rightarrow \infty$ .

My latest information is that this Inverse Function Rule - term coined up by Tony Orlow at the internet - goes all the way back to Carl Friedrich Gauss.
Written as an asymptotic equality: $D(n) \approx A^{-1}(n) $ . If $\aleph_0$ is a very large number, then we shall write symbolically (and ironically): $Z(\aleph_0) = A^{-1}(\aleph_0)$.
Examples.
$A(n) = n^2 \quad \Longrightarrow \quad \lim_{n\rightarrow\infty} D(n)/\sqrt{n} = 1$
$A(n) = 7^n \quad \Longrightarrow \quad \lim_{n\rightarrow\infty} D(n)/[\,\ln(n)/\ln(7)\,] = 1$
$A(n) = 2.n \quad \Longrightarrow \quad \lim_{n\rightarrow\infty} D(n)/[\,n/2\,] = 1$
Still another example at the Wikipedia page about Fibonacci numbers. For large $n$, the Fibonacci numbers are approximately $F_n \approx \phi^n/\sqrt{5}$ . Giving a limit similar to the one for the powers of seven: $$ A(n) = F_n \quad \Longrightarrow \quad \lim_{n\rightarrow\infty} \frac{D(n)}{\ln(n.\sqrt{5}) / \ln( \phi )} = 1 \qquad \mbox{where} \quad \phi = \frac{1}{2}(1+\sqrt{5}) $$ Notes.
If the intended use of the Inverse Function Rule would be to find for example a formula for all of the prime numbers, then you would be disappointed. Let it be suggested that $A^{-1}(n) = n / \ln(n)$ - but the inverse of this function is useless, because it is not a bijection $\mathbb{N} \rightarrow \mathbb{N}$.
Another example. Let $B(n)$ be the number of positive fractions (reducible or not) with denominators and numerators less than or equal to a natural number $n$ . Then the following is rather trivial - we would rather like to have a formula for irreducible fractions instead of this: $$ \lim_{n\rightarrow \infty} \frac{B(n)}{ n^2 } = 1 $$ With $A^{-1}(n) = n^2$, therefore $A(n) = \sqrt{n}$, which is also useless because it is not a bijection $\mathbb{N} \rightarrow \mathbb{N}$.