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Well I've got a bit problem with ESN so let me introduce my problem :

I have in my lesson the demonstration of this propriety without ESN :

If $(x_i)_{1 \le i \le n}$ is an orthogonal familly, then $$ \left|\left| \sum_{i=1}^{n} x_i \right|\right|^2 = \sum_{i=1}^{n} ||x_i||^2 $$

Now my problem is to note this two part of equality with ESN :/

$$ ||x^i||^2= ??? $$

And I also want to know how to note

$$ \sum_{i=1}^n\left(x_i \mid \sum_{j=1}^n x_j\right) $$

where $(.|.)$ is dot product :)

Shadock

ParaH2
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    Why would you want to use ENS in the first place? Everything looks ok with regular sums. – TZakrevskiy Feb 04 '15 at 22:27
  • Why not, I like this notation, so I would like to know if it's possible :) – ParaH2 Feb 04 '15 at 22:28
  • You can't really write this in Einstein notation. If you really wanted to you could write $g_{i j} \left( \sum_{i=1}^n x_i^j \right) \left( \sum_{k=1}^n x_k^l \right) = \sum_{m=1}^n g_{j l} x_m^k x_m^l$. – Zhen Lin Feb 04 '15 at 22:45
  • It's so difficult to understand now :P – ParaH2 Feb 04 '15 at 22:50

2 Answers2

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Update:

Sorry if there was any confusion from my initial post. I've edited this to more fully explain my answer. If a vector $x$ can be written as $x = (x_1, \ldots, x_k)$ then in Einstein summation notation $x$ can be denoted as $x_i$ if it's covariant or $x^i$ if it's contravariant (I'm not assuming the Lorentz metric for this kind of vector). The norm of $x$ can then be written as $||x||^2 = x^ix_i$ where the summation is implied by the double index. To write it out fully, we have

$$ \underbrace{x^ix_i}_{Not \; components} \;\; =\;\; \sum_{j=1}^n \underbrace{x_j^2}_{components} \;\; =\;\; ||x||^2. $$

Unfortunately if you have multiple vectors $\textbf{x}^{(1)}, \ldots, \textbf{x}^{(m)}$ where I use the bold-face for emphasis, then I don't think there is a way to express the quantity

$$ \sum_{j=1}^m ||\textbf{x}^{(j)}||^2 $$

In terms of the Einstein summation convention. As Zhen Lin pointed out in the comments below, the notation convention is used to manipulate the components of a vector, not multiple vectors at one time.

@Zhen - thanks for your critique. I would've missed this otherwise.

Mnifldz
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    That would be fine if each $x_i$ were the component of a vector. But if you read the question carefully you will see that each $x_i$ is itself a vector. – Zhen Lin Feb 04 '15 at 22:42
  • @Mnifldz so I can write $||x^i x_i ||^2=x^i x_i$ ? – ParaH2 Feb 04 '15 at 22:56
  • @Shadock I edited my post above so please let me know if it is clearer now. In your previous comment, we do NOT write that. The convention implies $x^ix_i = x_1^2 + x_2^2 + \ldots + x_k^2$ where the enumerated values are the components of some vector $x$. We need only write $x^ix_i$ indicate the norm. More concisely, $x^ix_i = ||x||^2$. – Mnifldz Feb 04 '15 at 23:30
  • Well I don't know what is co or contravariant vectors but yes it's better thank you :) – ParaH2 Feb 05 '15 at 00:07
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    Einstein invented the notation to make general relativity a little easier to understand, and the idea of covariance and contravariance are ideas borrowed from differential geometry. They're not needed for the work you're doing here, but in Einstein notation it's important to write inner products with one upper and one lower index. Glad to help :) – Mnifldz Feb 05 '15 at 00:11
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If your vectors are in an euclidean space then you could also write your identity as:

\begin{equation} (x_i)_a(x_j)_a = \frac{1}{n} \delta_{ij} (x_k)_a (x_k)_a \end{equation}

Your second expression can't really be put in ENS, there are no repeated indices there. You could consider, as above, that the inner product is eculidean, and hence I think the best you could do for your second expression would be:

\begin{equation} \sum_{i,j =1}^{n} (x_i)_a (x_j)_a \end{equation}

Rogelio Molina
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