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Empty
 one year ago
Does this sum have a closed form?
Empty
 one year ago
Does this sum have a closed form?

This Question is Closed

Empty
 one year ago
Best ResponseYou've already chosen the best response.1\[\sum_{n=0}^\infty (2n+1) e^{k(n^2+n)} \] where k > 0 is a real number.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1when i take the integral, i get something close to the closed form.

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Hmmm, how could I go about finding the error of the approximation? I am fine with this idea this sounds pretty good.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1\[c_n~e^{k(n^2+n)}\] gives me the 1/e stuff, but my coefficients are not proper ...

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i wouldnt know off hand how to find the error stuff. its been a long day

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1k=1; 3,10,21,36,55,78,... k=2; 3,10,21,36,55,78,... k=1/2; same progresion

amistre64
 one year ago
Best ResponseYou've already chosen the best response.13,10,21,36,55,78 7 11 15 19 23 4 4 4 4 3+7n+4n(n1)/2 seems to cover that ... but thats starting at n=0 so a shift is needed

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1heh, 2n^2+n is the simplified shift ..

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Yeah 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.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1with 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

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1http://www.wolframalpha.com/input/?i=table [%282n^2%2Bn%29e^%282%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}]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1firefox doesnt like to paste the links correctly ..

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i notice the shift in the polynomial part, like a usual difference equation. \[(1+2n) +0n^2+0n^3+...\color{red}{\implies} 0+(1n+2n^2)+0n^3+...\]

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1not sure how useful my prattling is, but it looked fun to try out is all :)

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1the setup doesnt work for n=0 tho ...

Empty
 one year ago
Best ResponseYou've already chosen the best response.1That's fine I think this helped actually haha thanks xD

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0Mathematica returned nothing useful. Looks like it doesn't have a close form.

amistre64
 one year ago
Best ResponseYou've already chosen the best response.1i 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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