Sampling 1

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Just a template for now — will update soon enough.

\(\gcd(a^n + b, b^n + a)\) is eventually constant. So will $(1,1)$ work?

Imagine you want to compute the area under a tricky curve, like $f(x) = \sin(x^2)$, from $x = 0$ to $x = 1$. Analytically, it’s a mess. Monte Carlo makes it simple:

  1. Generate random points $(x, y)$ in a rectangle containing the curve (e.g., $x \in [0, 1]$, $y \in [0, 1]$).
  2. Check if each point lies below the curve ($y \leq f(x)$).
  3. The fraction of points below the curve, multiplied by the rectangle’s area, approximates the integral.

Mathematically, if you sample $N$ points and $M$ fall below the curve, the area is roughly:

\[\text{Area} \approx \frac{M}{N} \cdot \text{Rectangle Area}\]

As $N$ grows, the approximation gets better, thanks to the law of large numbers.

Why Are They Cool?

Monte Carlo methods shine in high-dimensional problems, like those in machine learning or physics. For example:

  • Bayesian Inference: Estimating posterior distributions in probabilistic models.
  • Optimization: Approximating gradients in complex loss landscapes.
  • Physics Simulations: Modeling particle interactions in statistical mechanics.

The catch? They can be slow to converge (think $O(1/\sqrt{N})$ error), but their simplicity and flexibility make them a go-to tool.

A Fun Example

Suppose you want to estimate $\pi$ using Monte Carlo. Draw a unit square ($1 \times 1$) and a quarter-circle inside it (radius 1, centered at $(0, 0)$). Generate random points in the square and count how many fall inside the quarter-circle. Since the quarter-circle’s area is $\pi/4$, the ratio of “hits” to total points, multiplied by 4, approximates $\pi$:

\[\pi \approx 4 \cdot \frac{\text{Points inside quarter-circle}}{\text{Total points}}\]