Binomial

\((1 + x)^n\)

\[ (1 + x)^0 = 1 \]

\[ (1 + x)^1 = 1 + x \]

\[ (1 + x)^2 = 1 + 2x + x^2 \]

\[ (1 + x)^3 = 1 + 3x + 3x^2 + x^3 \]

\[ (1 + x)^n = \sum\limits_{k = 0}^{n} f(n, k) x^k = \sum\limits_{k = -\infty}^{\infty} f(n, k) x^k \]

\[ f(n, 0) = 1 \]

\[ f(n, k) = 0, \]

for \(k\) bigger than \(n\) or smaller than \(0\)

\[ (1 + x)^n = (1 + x) (1 + x)^{n - 1} = (1 + x) \sum\limits_{k = -\infty}^{\infty} f(n - 1, k) x^k = \sum\limits_{k = -\infty}^{\infty} f(n - 1, k) x^k + \sum\limits_{k = -\infty}^{\infty} f(n - 1, k) x^{k + 1} = \sum\limits_{k = -\infty}^{\infty} f(n - 1, k) x^k +\sum\limits_{k = -\infty}^{\infty} f(n - 1, k - 1) x^k \]

\[ \sum\limits_{k = -\infty}^{\infty} f(n, k) x^k = \sum\limits_{k = -\infty}^{\infty} (f(n - 1, k) + f(n - 1, k - 1)) x^k \]

\[ f(n, k) = f(n - 1, k) + f(n - 1, k - 1) \]

discrete calculus

\[ f(n, k) - f(n - 1, k) = f(n - 1, k - 1) \]

\[ \Delta f(x) = : f(x + 1) - f(x) \]

\[ x^{\frac{n}{}} = : \prod\limits_{k = x - n + 1}^{x} k = x \cdot (x - 1) \cdot (x - 2) ... \text{n times} = \frac{x!}{(x - n)!} \]

\[ \Delta x^{\frac{n}{}} = \frac{(x + 1)!}{(x - n + 1)!} - \frac{x!}{(x - n)!} = (x + 1) \frac{x!}{(x - n + 1)!} - \frac{x!}{\frac{(x - n + 1)!}{x - n + 1}} = (x + 1) x^{\frac{n - 1}{}} - (x - n + 1) x^{\frac{n - 1}{}} = x^{\frac{n - 1}{}} ((x + 1) - (x - n + 1)) \]

\[ \Delta x^{\frac{n}{}} = n x^{\frac{n - 1}{}} \]

\[ \Delta f(x, k) = f(x + 1, k) - f(x, k) = f((x + 1), k) - f((x + 1) - 1, k) = f((x + 1) - 1, k - 1) \]

\[ \Delta f(x, k) = f(x, k - 1) \]

\[ \Delta \frac{x^{\frac{n}{}}}{n!} = \frac{x^{\frac{n - 1}{}}}{(n - 1)!} \]

\[ f(x, k) = \frac{x^{\frac{k}{}}}{k!} = \frac{x!}{k!(x - k)!} = \begin{pmatrix} x \\ k \\ \end{pmatrix} \]

\[ (1 + x)^n = \sum\limits_{k = 0}^{\infty} \begin{pmatrix} n \\ k \\ \end{pmatrix} x^k = \sum\limits_{k = 0}^{\infty} \frac{n! x^k}{k!(n - k)!} \]

if \(x\) is closer to \(0\), than this will converge faster!

\(\pi\)

\[ \pi = 2 \int_{-1}^{1} \sqrt{1 - x^2} dx = 4 \int_{0}^{1} \sqrt{1 - x^2} dx \]

but at \(0\), this is \(1\), and at \(\frac{1}{2}\), this will converge really fast!

But

\[ \int_{0}^{\frac{1}{2}} \sqrt{1 - x^2} dx = \frac{\sqrt{3}}{8} + \frac{\pi}{12} \]

\(\sqrt{3}\)

\[ \sqrt{3} = \sqrt{4 - 1} = \sqrt{4(1 - \frac{1}{4})} = 2 \sqrt{1 - \frac{1}{4}} = 2 (1 - \frac{1}{4})^{\frac{1}{2}} \]

which converges really fast!

thus...

\[ \pi = 12 ((\sum\limits_{k = 0}^{\infty} \frac{\begin{pmatrix} \frac{1}{2} \\ k \\ \end{pmatrix} (-\frac{1}{4})^{k + 1}}{k + 1}) - \frac{\sum\limits_{k = 0}^{\infty} \begin{pmatrix} \frac{1}{2} \\ k \\ \end{pmatrix} (-\frac{1}{4})^k}{4}) \]

where

\[ \begin{pmatrix} \frac{1}{2} \\ k \\ \end{pmatrix} = \frac{\prod\limits_{m = \frac{1}{2} - k + 1}^{\frac{1}{2}} m}{k!} = \frac{\prod\limits_{m = 1 - k}^{0} m + \frac{1}{2}}{k!} \]