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  • one year ago

Does this sum have a closed form?

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  1. Empty
    • one year ago
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    \[\sum_{n=0}^\infty (2n+1) e^{-k(n^2+n)} \] where k > 0 is a real number.

  2. amistre64
    • one year ago
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    when i take the integral, i get something close to the closed form.

  3. Empty
    • one year ago
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    Hmmm, how could I go about finding the error of the approximation? I am fine with this idea this sounds pretty good.

  4. amistre64
    • one year ago
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    \[c_n~e^{-k(n^2+n)}\] gives me the 1/e stuff, but my coefficients are not proper ...

  5. amistre64
    • one year ago
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    i wouldnt know off hand how to find the error stuff. its been a long day

  6. amistre64
    • one year ago
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    k=1; 3,10,21,36,55,78,... k=2; 3,10,21,36,55,78,... k=1/2; same progresion

  7. amistre64
    • one year ago
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    3,10,21,36,55,78 7 11 15 19 23 4 4 4 4 3+7n+4n(n-1)/2 seems to cover that ... but thats starting at n=0 so a shift is needed

  8. amistre64
    • one year ago
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    heh, 2n^2+n is the simplified shift ..

  9. Empty
    • one year ago
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    Yeah that's totally fine, I was thinking more or less of solving them independently to within a constant or something I'm not too concerned this is for my own interest in looking for the closed form of the molecular partition function of a linear rigid rotor. I noticed it doesn't have a closed form and the book I'm reading says to evaluate it numerically but I thought that might be kind of lame and wanted a closed form haha.

  10. amistre64
    • one year ago
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    with a little help from the wolf tech ... \[\sum_{n=0}^{\infty} (2n+1)e^{-k(n^2+n)} =(2n^2+n) e^{-k(n^2+n)}\] it works for a couple of test values for k

  11. amistre64
    • one year ago
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    http://www.wolframalpha.com/input/?i=table [%282n^2%2Bn%29e^%28-2%28n^2%2Bn%29%29%2C{n%2C10}] http://www.wolframalpha.com/input/?i=table [sum%28i%3D1+to+n%29+%282n%2B1%29e^%28-%28n^2%2Bn%29%2F2%29%2C{n%2C10}]

  12. amistre64
    • one year ago
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    firefox doesnt like to paste the links correctly ..

  13. amistre64
    • one year ago
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    i notice the shift in the polynomial part, like a usual difference equation. \[(1+2n) +0n^2+0n^3+...\color{red}{\implies} 0+(1n+2n^2)+0n^3+...\]

  14. amistre64
    • one year ago
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    not sure how useful my prattling is, but it looked fun to try out is all :)

  15. amistre64
    • one year ago
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    the setup doesnt work for n=0 tho ...

  16. Empty
    • one year ago
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    That's fine I think this helped actually haha thanks xD

  17. thomas5267
    • one year ago
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    Mathematica returned nothing useful. Looks like it doesn't have a close form.

  18. amistre64
    • one year ago
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    i noticed an error in my plans ... my table indexed i, from 0 to n .... but i forgot to use i in the process :/ oh well. it still looks fun tho

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