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How do you interpret this equation? I've seen far simpler equations for solving the magnitude, like the square root of all of the elements of a vector squared added to one another. What exactly is going on with the Greek symbol in that equation? Is there something I'm missing?

[vector magnitude calculation][1]

2.6 The lenght (norm) of a vector

The lenght ( or norm ) of a vector a is: $$\left\|{a}\right\|=\sqrt{aa}=\sqrt{\sum_{i=1}^n a_i^2}$$

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  • $\begingroup$ For the general case. $\endgroup$ – Joe Sep 13 '17 at 16:56
  • $\begingroup$ @HansLundmark Thanks :) I'll try and dig a little deeper next time on here. $\endgroup$ – Jason Procka Sep 13 '17 at 17:10
  • $\begingroup$ Don't worry! It's easier to find if one knows the answer already... ;-) $\endgroup$ – Hans Lundmark Sep 13 '17 at 17:15
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Perhaps you're just confused about the notation, $$\sqrt{\sum_{i=1}^n a^2_i} = \sqrt{a^2_1 + a^2_2 \ldots a^2_{n-1} + a^2_n}$$

This is an easy way to generalise the norm for n-dimensional vectors.

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  • $\begingroup$ Awesome, I'm finally getting it :) What does it mean exactly when you wrote n-1 as a subscript? $\endgroup$ – Jason Procka Sep 13 '17 at 17:08
  • $\begingroup$ The (n-1)th component of the vector, ie. $a = (a_1, a_2 \ldots a_{n-1}, a_n$). Let me know if I'm still being unclear. $\endgroup$ – George Coote Sep 13 '17 at 17:16
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The Greek symbol is the capital letter sigma $\Sigma$. In mathematics, it is used to denote a sum. For example, $$\sum_{i=1}^na_i^2$$ means you are summing starting from $i=1$ (the number under the sigma) and adding all the numbers with indices up to some value $n$ (the number on top of the sigma). As an example, let $(a_1, a_2, a_3)=(3,4,5)$. We could write the norm of this vector as $$\sqrt{\sum_{i=1}^3a_i^2}=\sqrt{a_1^2+a_2^2+a_3^2}=\sqrt{3^2+4^2+5^2}=\sqrt{50}$$

As an interesting aside, there is a similar notation used for the product of a set of numbers. Instead of sigma (for a summation), we write pi (for a product). So $$\prod_{i=1}^na_i$$ is the product of the numbers $a_1, a_2, ..., a_n$

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