# Norm to evaluate precision

I have two vectors x and y with their respective n coordinates/components and this norm between them:

$$norm = \sqrt{\frac{\displaystyle\sum_{i=1}^{n} (x_{i}-y_{i})^2}{\displaystyle\sum_{i=1}^{n} (x_{i})^2}}$$

This should evaluate if two vectors are equal(more or less). What kind of norm is this?

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I wouldn't call it a norm, at all. I might call it a relative difference. – Gerry Myerson Feb 12 '13 at 12:03
You may be correct, although I got this from a friend and he insisted on calling it "a norm". Unfortunately I can't understand why this should be a good way to compare if two vectors are "nearly" equal – John Smith Feb 12 '13 at 12:24
So, is your question, "What kind of norm is this?" or is your question, "Why is this a good way to see whether two vectors are nearly equal?" – Gerry Myerson Feb 12 '13 at 23:28
The second would be better, thanks!! – John Smith Feb 13 '13 at 0:49

The $\sum_1^n(x_i-y_i)^2$ will be small if and only if the vectors $x=(x_1,\dots,x_n)$ and $y=(y_1,\dots,y_n)$ are close --- indeed, it measures the square of the Euclidean distance between the two vectors. One could use $\sum_1^n|x_i-y_i|$ instead, but the formula with the squares in it has the advantage of being differentiable, which you lose when you go to absolute values.
Now, $\sum_1^n(x_i-y_i)^2$ gives the same value to $x=(2,0,0),y=(1,0,0)$ and to $x=(2,1,0),y=(1,1,0)$ as it gives to $x=(101,100,100),y=(100,100,100)$. This may be OK for some purposes, but for other purposes, it's absurd. The first pair of vectors, one is twice the other, so it seems strange to say they are nearly equal. The second pair, the vectors don't point at all in the same direction (they differ by about $18$ degrees). The third pair are nearly the same length, and point in nearly the same direction, so they better fit the notion of "nearly equal". And the way to take that into account is to divide by the square of the length of the vector $x$, and that's what the $\sum_1^nx_i^2$ in the denominator does.