HP Prime Infinite Sum Bug?
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11-05-2014, 07:58 AM
(This post was last modified: 11-05-2014 08:06 AM by kcnicho.)
Post: #1
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HP Prime Infinite Sum Bug?
Hi All,
I posted this question over at the HP Calculator site (http://h30499.www3.hp.com/t5/Calculators...-p/6662418), thought I'd ask here as well. HP Prime, in CAS mode, when I enter "sum(1/sqrt(n), n, 1, infinity)", the Prime returns a screen full of warnings (see attachments.) I believe the correct answer should be "+ infinity", as I get from Wolfram Alpha as well as the TI-Nspire CAS. Has anyone else seen this? Can anyone else duplicate this behavior on their own machine? I verified this on the Prime Virtual Calculator, Version 2014 3 31, Rev 6031 (latest firmware, as far as I know.) Thanks, - Kevin |
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11-05-2014, 12:34 PM
Post: #2
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RE: HP Prime Infinite Sum Bug?
Sorry, convergence of series is not yet implemented in giac. I plan to do that as part of the new step by step feature.
sum tries to find a closed form for the discrete antiderivative (that's why you see all these messages), it fails, hence returns the sum unevaluated. |
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11-05-2014, 07:17 PM
Post: #3
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RE: HP Prime Infinite Sum Bug?
(11-05-2014 12:34 PM)parisse Wrote: part of the new step by step feature.Oh, that's exciting (Coming to KhiCAS too ? ) TI-Planet.org co-administrator |
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11-06-2014, 07:05 AM
Post: #4
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RE: HP Prime Infinite Sum Bug?
If I can find a way to display intermediate computations on the nspire screen.
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11-06-2014, 07:55 AM
Post: #5
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RE: HP Prime Infinite Sum Bug?
Thanks for the (definitive) answer!
Happy to hear that things are in the works to address these types of problems in future firmware releases. - Kevin |
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01-28-2015, 02:03 PM
Post: #6
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RE: HP Prime Infinite Sum Bug?
(11-05-2014 12:34 PM)parisse Wrote: Sorry, convergence of series is not yet implemented in giac. I plan to do that as part of the new step by step feature. I hope you could do it soon in a next firmware upgrade, thank you! It would be very interesting... Salvo |
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01-28-2015, 06:32 PM
Post: #7
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RE: HP Prime Infinite Sum Bug?
...another Series I would se calculated in Prime:
sum((-1)^(k+1)/K^2, k, 1, infinity) It should give π^2/12 Thank you Salvo |
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