The Minimum Value of Pi

This problem is much easier easier to understand with a little help from desmos . To start, we will derive the general curve length formula for distance of order $p$ for function $f$ . This formula is,

$L_{p}=\lim_{n \to \infty} \sum_{i=1}^{n} d(p_{i},p_{i+1})=\lim_{n \to \infty} \sum_{i=1}^{n} (|x_{i} - x_{i+1}|^p+|y_{i}-y_{i+1}|^p)^{\frac{1}{p}}$

Now, from the Mean Value Theorem, we know that on the interval $[x_{i},x_{i+1}]$ there exists a point $x_{i}^{*}$ such that, $f'(x_{i}^{*}) = \frac{f(x_{i+1}) - f(x_{i})}{x_{i+1}-x_{i}}=\frac{\Delta y}{\Delta x}$

Substituting in $x_{i} - x_{i+1} = \Delta x$ and $y_{i}-y_{i+1} = \Delta y = \Delta x f'(x_{i}^{*})$ ,

$L_{p}=\lim_{n \to \infty} \sum_{i=1}^{n} (|\Delta x|^p+|\Delta x f'(x_{i}^{*})|^p)^{\frac{1}{p}}=\lim_{n \to \infty} \sum_{i=1}^{n} (1 + (f'(x^{*})^{p})^{\frac{1}{p}} \Delta x$

This can be simplified by rewriting the equation as a definite integral,

$L_{p} = \int_{a}^{b} (1 + (f'(x^{*})^{p})^{\frac{1}{p}} dx = \int_{a}^{b} (1 + (\frac{d}{dx})^{p})^{\frac{1}{p}} dx$

Using this general arc length equation, we can evaluate $\pi_{p}$ as the ratio of a circle of order $p$ and radius 1's circumference, for we can use the equation $y = (1-|x|^p)^\frac{1}{p}$ from $[0,1]$ and multiply the arc length by 4, and the diameter 2. This gives us the general formula for $\pi_{p}$ , $\pi_{p}=2 \int_{0}^{1} (1 + (\frac{d}{dx} [(1-|x|^p)^\frac{1}{p}])^{p})^{\frac{1}{p}} dx$

By analyzing this graph on desmos , we find that the minimum value of pi is at $p=2$ and $\pi = \boxed{3.14159}$ , which is our "normal" value of $\pi$ using euclidean distance.

This problem doesn't exactly have a solution that can fit here. However, if anyone is interested, I recommend checking out π is the Minimum Value for Pi by C. L. Adler and James Tanton.