Timeline for Find exponential generating function from the first few terms
Current License: CC BY-SA 3.0
9 events
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| May 20, 2015 at 22:30 | comment | added | J. M.'s missing motivation♦ |
I've found that sometimes, you have to take the long route and use FindSequenceFunction[] on your sequence first, and then apply GeneratingFunction[] to the sequence function thus generated.
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| May 20, 2015 at 18:11 | vote | accept | Manolito Pérez | ||
| May 20, 2015 at 11:34 | comment | added | Manolito Pérez | @LLIAMnYP: Oh, I see, you have created the function findExponentialGeneratingFunction. | |
| May 20, 2015 at 11:32 | comment | added | Manolito Pérez | Thanks a lot to you all for your answers and comments. In fact, I had tried FindGeneratingFunction with the terms divided by the corresponding factorials, but it did not work. Apparently, it needed more terms. That was the reason for my second question. As for the function "findExponentialGeneratingFunction", where is it? I can't find it. Please excuse my ignorance :-) | |
| May 20, 2015 at 11:11 | comment | added | LLlAMnYP |
@Mr.Wizard findExponentialGeneratingFunction[{1, 2, 5, 16, 65, 326},x] returns the correct result. Sometimes you want less terms, it seems.
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| May 20, 2015 at 11:01 | comment | added | Mr.Wizard |
By the way Factorial is listable: Range[0, Length@foo - 1] ! -- or Array[Factorial, Length@foo, 0]
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| May 20, 2015 at 10:59 | comment | added | LLlAMnYP |
Indeed, FindGeneratingFunction needs a sequence of nine 1s to return 1/(1-x). It's just as much as Mathematica can do.
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| May 20, 2015 at 10:56 | comment | added | Mr.Wizard |
I confirm this works in a number of cases. One that is less than ideal is {1,2,5,16,65,326,1957,13700,109601} which should be E^x/(1 - x) but gives something far more complex. That's not your doing of course but perhaps it could be improved nevertheless.
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| May 20, 2015 at 10:49 | history | answered | LLlAMnYP | CC BY-SA 3.0 |