We all know the Monte Carlo Method to estimate $\pi$ (https://en.wikipedia.org/wiki/Monte_Carlo_method). Explanation: The method works because
$4 \cdot \frac{AreaOf:(x,y\in(\mathbb R+)|x^2+y^2<1)}{Total Area =1} = \pi$
This is correct because the area of the positive real numbers that satisfy $x^2+y^2<1$ (also satisfy $y<\sqrt{1-x^2}$) is the integral:
$\int_0^1 {(y=\sqrt{1-x^2})dx} = \left[\dfrac{\arcsin\left(x\right)+x\sqrt{1-x^2}}{2}\right]_0^1 = \frac{\frac{\pi}{2}-0}{2} - 0 = \frac{\pi}{4}$
Then, when we pick random numbers where x and y are between 0 and 1 we can multiply 4 by the probability and get: $4 \cdot \frac{\frac{\pi}{4}}{1} = \pi$
I have also thought of a way to do the same method for ln(2):
the numbers that satisfy $y\cdot (x+1)<1$ also satisfy: $y<\frac{1}{x+1}$
$\int_0^1 {(y=\frac{1}{x+1})dx} = \left[\ln (x+1) \right]_0^1 =\ln(2)-\ln(1) = \ln(2)$
Therefore, when we pick random points where x,y are between 1 and 0, the probability of a point satisfying the inequality is: $ \frac{AreaOf:(x,y\in(\mathbb R+)|y\cdot(x+1)<1)}{Total Area =1} = \ln(2)$
I'm wondering about a lot of things, I will formulate them as formally as possible:
a. Is there a binomial inequality with rational coefficients, where if we randomly choose $a$ random points where $x$ and $y$ are between 0 and 1, and $b$ of those points satisfy the inequality, is the next statement true:
$\exists P(x) \in \mathbb Q: \lim_{a \to \infty }P(\frac{b}{a}) = e$ (or any other transcendental number that I haven't shown)
By $\exists P(x) \in \mathbb Q$ I mean a polynomial with coefficients from $\mathbb Q $. Notice that $\lim_{a \to \infty } \frac{b}{a} $ is just the probability of a number to be in the area (like with the $\ln(2)$ or $\pi$. Their polynomials were $4x$ and $x$)
b. We can add/substract more dimensions to the question, for example we can similarly estimate $\sqrt {2}$ with a polynomial inequality: $x^2<1$ (and put random points on the axis...)
Note: Is there info out there that's already known about this?
Update: we could have $\ln(a)$ where a is any positive rational and $a>1$, with this nice binomial inequality: $y \cdot ((a-1) \cdot x + 1)<1$
and the nice rational polynomial $(a-1)\cdot x$
Proof:
$y \cdot ((a-1) \cdot x + 1)<1 \rightarrow y<\frac{1}{(a-1)\cdot x +1}$
$\int_0^1 {(y=\frac{1}{(a-1)\cdot x+1})dx} = \left[\frac{1}{(a-1)}\cdot \ln [(a-1)\cdot x+1] \right]_0^1 =\frac{1}{(a-1)}\cdot [\ln(a-1 +1)-\ln(1)] = \frac{1}{(a-1)}\cdot \ln(a)$
Then, after the polynomial $P(x) = (a-1) \cdot x$ It's: $P(\frac{a}{b}) = P(\frac{1}{(a-1)}\cdot \ln(a)) = (a-1) \cdot \frac{1}{(a-1)}\cdot \ln(a) = \ln(a)$
Update 2:
We can prove you cant estimate every transcendental number this way:
Proof:
Let's assume by contradiction we can estimate every transcendental number this way, so there is a surjective function $f$ between the set of all binomial inequalities with rational coefficients (A) and the set of all transcendental numbers (B):
$ f: A\rightarrow B$
we know that $|B| = \aleph$ and that $|A| = \aleph_0$ because those binomial inequalities have rational coefficients, so because $f$ is a surjective function:
$|B| \le |A| \rightarrow \aleph \le \aleph_0$
Contradiction, so we cant estimate every transcendental number this way.
Is this proof even correct?