# Does π depends on the norm? [duplicate]

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:

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

And if we take the Chebyshev norm:

$$\|x\|_\infty=\max\{ |x_1|, \dots, |x_n| \}$$

The circle would transform into this:

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, MicahMar 17 '13 at 15:48

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

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