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Proper usage of Einstein sum notation

  1. Apr 21, 2017 #1
    1. The problem statement, all variables and given/known data

    I'm dealing with some pretty complex derivatives of a kernel function; long story short, there's a lot of summations going on, so I'm trying to write it down using the Einstein notation, for shortness and hopefully reduction of errors (also for the sake of a paper in which I have to write all this stuff down and possibly do it without blowing past the page's margins). Right now I was testing something that's relatively simple, but I'm not sure I'm using this correctly.

    2. Relevant equations

    My test example was a relatively simple derivative. For reference, these are the symbols I am using:

    $$ P_{ij} = exp[-(x_i-x_j)^2]
    \qquad
    P_{ij}' = \frac{dP_{ij}}{dx_i} = -\frac{dP_{ij}}{dx_j}
    \qquad
    P_i = P_{ij}\delta_{jj}
    \qquad
    P_i' = P_{ij}'\delta_{jj}
    $$

    I'm already unsure about the use of ##\delta_{jj}## there, but then comes the problem. As a first exercise I'm trying an example of a derivative, with an additional index ##n##:

    $$\frac{d(P_iP_i)}{dx_n}$$

    3. The attempt at a solution

    Here's my solution:

    $$\frac{d(P_iP_i)}{dx_n} = 2P_i\frac{dP_i}{dx_n} = 2P_i\left[\frac{dP_i}{dx_i}\delta_{in}-\frac{dP_i}{dx_j}\delta_{jn}\right] = 2P_nP_n' - 2P_iP_{in}'
    $$

    Which actually works (tested numerically), but seems ugly and wrong to me due to those repeated ##n## indices which seem to imply a summation that isn't really there. Did I do something wrong? Is there some other symbol I'm disregarding or some rule I don't know? Thanks!
     
  2. jcsd
  3. Apr 21, 2017 #2

    vela

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    One error is that the same index shouldn't show up more than twice in a term, so ##P_i = P_{ij}\delta_{jj}## doesn't make sense because ##j## appears three times. It's not clear to me what you're trying to do there. What is ##P_i## supposed to be equal to in normal summation notation?
     
  4. Apr 21, 2017 #3
    In regular notation,

    $$P_i = \sum_j P_{ij} $$

    I suppose I could get the same result by multiplying by an array of ones with a single index, I just don't know if there's a conventional symbol for that.
     
  5. Apr 21, 2017 #4

    vela

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    That's probably the most straightforward way. You can define a vector of ones, say ##e = (1, 1, \dots, 1)##, then ##P_i = P_{ij}e_j##.

    You also need to clean up the notation for the derivative. The chain rule gives you (with no implied summation here)
    $$\frac{d}{dx_n} P_{ij} = \frac{\partial P_{ij}}{\partial x_i}\frac{dx_i}{dx_n} + \frac{\partial P_{ij}}{\partial x_j}\frac{dx_j}{dx_n}.$$
     
  6. Apr 21, 2017 #5
    Yes, right, I'll fix that. The minus sign came from me knowing it appears in the end but it's not correct there.

    EDIT: apparently I can't edit the first post in the thread? Sorry for that.
     
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