Euler’s identity is $e^{i\pi} + 1 = 0$, which packs five fundamental constants into a single equation.
A classic — the Gaussian integral:
\[\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}.\]Claim. For all $n \in \mathbb{N}$, $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$.
Proof. By induction on $n$.
Base case. For $n = 1$, the left side is $1$ and the right side is $\frac{1 \cdot 2}{2} = 1$. ✓
Inductive step. Assume the identity holds for some $n \geq 1$. Then
\[\sum_{k=1}^{n+1} k = \left(\sum_{k=1}^{n} k\right) + (n+1) = \frac{n(n+1)}{2} + (n+1) = \frac{(n+1)(n+2)}{2},\]which is the formula with $n+1$ in place of $n$. $\blacksquare$
Posts here can mix prose, math, and code blocks freely. A short code sample:
def gauss_sum(n: int) -> int:
return n * (n + 1) // 2
That’s all for the first post.
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