Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Intuitive Explanation of What a p norm is, for any arbitrary p>0

  1. Aug 9, 2011 #1
    Supposing [itex] V [/itex] is a normed vector space, the p-norm of [itex]{\bf x} \in V [/itex] is:
    [itex]\lVert {\bf x} \rVert_p := \left(\sum_{i=1}^n |x|^p \right)^{\frac{1}{p}}[/itex]

    There are 3 special cases:
    [itex] p= 2: [/itex] Euclidean distance - 'as the crow flies'
    [itex] p = 1: [/itex] Taxicab distance - sum the absolute value of components
    [itex] p=\infty: [/itex] Supremum distance - take the maximum component (in abs value)

    I would like to have an intuition for what happens when we change p to values in-between these special cases. In particular, if we interpret x to be a vector of data points, what changes about the way the norm ||.||_p describes the data vector as we change p?

    Also, if we consider p<1, the l-p 'norm' does not satisfy the triangle inequality, but still tells us something about the data. Can anyone give an intuition for what happens with values of 0<p<1?

    The only thing I can think of is that with higher values of p, outliers (components with higher deviation from the mean of all components) contribute more to the value of the norm. Is there anything more to it?


  2. jcsd
  3. Aug 9, 2011 #2
    One good way to think about it is to look at what happens to the unit circle as p varies. When p=∞, the unit circle is just a square. As p decreases, the outer corners begin to move inward until p=1, when they become just straight lines between the cardinal unit vectors, and the unit circle becomes a diamond. For p<1, the diagonal lines start to bulge inward, creating a four-pointed "star". This figure is not convex, which is why the triangle inequality fails for p<1.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook