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This question already has an answer here:

If we take the definition of π in the form:

π is the ratio of a circle's circumference to its diameter.

There implicitly assumed that the norm is Euclidian:

\begin{equation} \|\boldsymbol{x}\|_{2} := \sqrt{x_1^2 + \cdots + x_n^2} \end{equation}

And if we take the Chebyshev norm:

\begin{equation} \|x\|_\infty=\max\{ |x_1|, \dots, |x_n| \} \end{equation}

The circle would transform into this:

Circle in the Chebyshev norm

And the π would obviously change it value into $4$.

Does this lead to any changes? Maybe on other definitions of π or anything?

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marked as duplicate by Dennis Gulko, Ayman Hourieh, William, Dominic Michaelis, Micah Mar 17 '13 at 15:48

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

It wouldn't be $\pi$. The symbol $\pi$ defines the smallest positive real number such that $e^{\pi i} \in \mathbb{R}$. – dtldarek Mar 16 '13 at 21:34
@dtldarek Alternatively, we can define it as $$2\int_{-1}^1 \sqrt{1-x^2}dx$$ or as the smallest positive real such that $$\sin x=0$$ (which is pretty much what you say) – Pedro Tamaroff Mar 16 '13 at 22:26
I think you already answered your question. Changing something changes things: if you change the norm, the circumference/diameter changes (though what do you even mean by "diameter" here?). The usual $\pi$ is wonderful because it equals many different things simultaneously; if you changed things, some of these different things might no longer be equal. – Greg Martin Mar 16 '13 at 22:30
up vote 1 down vote accepted

Under Euclidean metric there are number of constants that their values coincide and are collectively denoted by the symbol $\pi$.

How ever some of the coinciding values are independent from the metric and some are coupled with the metric and the geometry under consideration.

In your example, would the calculation of areas remain the same? How does the value of calculation under new metric need to be adjusted for unit square? Would the $\pi$ of calculation of area be the same one as the one for calculation of perimeter?

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