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For $p\geq 1,$ the $p$-norm of a vector $(x,y)\in\Bbb R^2$ is the number $\|(x,y)\|_p=(|x|^p+|y|^p)^{1/p}.$

I learned this definition some time ago, but I never really understood it. Is there a useful and not too difficult to understand geometric intuition that explains it (at least for natural $p$)?

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    Draw a picture of $||x||_p = 1$.2012-07-15
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    @Thomas I know what the unit circles look like, more or less. I've used Wolphram Alpha to draw the pictures. But I'm not sure how it explains what the norms actually mean.2012-07-15
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    I know that for $p=1$ we can think of it as a distance we have to walk to get somewere in a city with orthogonal streets. And of course I have a pretty good intuition of the $p=2$ case.2012-07-15
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    @Bartek, right, that's why the $1$-norm is sometimes called the "Manhattan norm", since the city of Manhattan is divided into blocks... for the $\infty$-norm, see [this](http://en.wikipedia.org/wiki/Chebyshev_distance).2012-07-15
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    @J.M. Oh, cool! Does the knight also have its normed space?2012-07-15
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    Bartek: No since the points (0,0) and (2,0) are at knight-distance 2 but the points (0,0) and (1,0) are at distance 3 (instead of 1). Hence the knight-distance is not a norm distance.2012-07-15
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    I recently found a nice article, about [$π_p$, the value of $π$ in $ℓ_p$](http://math.stanford.edu/~vakil/files/07-0333.pdf)2012-07-16

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