[R] Infinite Series
Don McKenzie
dmck at u.washington.edu
Sat Jul 25 06:14:15 CEST 2015
but you get a different hide with sum vs. cumsum. David is right if you want the sum of n terms.
> sum(1/(1:100)^2) - pi^2/6
[1] -0.009950167
> sum(1/(1:1000)^2) - pi^2/6
[1] -0.0009995002
etc.
> On Jul 24, 2015, at 7:24 PM, Janh Anni <annijanh at gmail.com> wrote:
>
> Wow! So many (simpler) ways to skin a cat. Thanks!
>
> On Fri, Jul 24, 2015 at 8:07 PM, David Winsemius <dwinsemius at comcast.net>
> wrote:
>
>>
>> On Jul 24, 2015, at 4:37 PM, Janh Anni wrote:
>>
>>> Hello Jeff,
>>>
>>> Thanks a lot. I tried it and see that it prints out the entire 100
>> partial
>>> sums, so I can take the last value as the partial sum for the first 100
>>> terms. Would there be any way cumsum can print only the nth partial sum,
>>> i.e. the last value in the array, instead of printing the entire array?
>>> Thanks again.
>>
>> Wouldn't that just mean using sum instead of cumsum?????
>>
>> Can even check the error from the analytical limit.
>>
>>> sum(1/(1:100)^2) - pi^2/6
>> [1] -0.009950167
>>
>>
>>>
>>> Joseph
>>>
>>> On Fri, Jul 24, 2015 at 2:02 PM, Jeff Newmiller <
>> jdnewmil at dcn.davis.ca.us>
>>> wrote:
>>>
>>>> Please reply-all so the mailing list stays in the loop.
>>>>
>>>> cumsum(1/(1:100)^2)
>>>>
>>>> gives you the partial sums up through i=100.
>>>>
>> ---------------------------------------------------------------------------
>>>> Jeff Newmiller The ..... ..... Go
>> Live...
>>>> DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live
>>>> Go...
>>>> Live: OO#.. Dead: OO#.. Playing
>>>> Research Engineer (Solar/Batteries O.O#. #.O#. with
>>>> /Software/Embedded Controllers) .OO#. .OO#.
>> rocks...1k
>>>>
>> ---------------------------------------------------------------------------
>>>> Sent from my phone. Please excuse my brevity.
>>>>
>>>> On July 24, 2015 10:30:09 AM PDT, Janh Anni <annijanh at gmail.com> wrote:
>>>>> Hello Jeff,
>>>>>
>>>>> Thank you so much for the suggestion, I searched cumsum as suggested
>>>>> but
>>>>> not sure it is what I had in mind. For instance if I had the infinite
>>>>> series: [image: Inline image 1]
>>>>>
>>>>> and want to compute the sum of the, say, first 100 terms, how could I
>>>>> use
>>>>> cusum to do that?
>>>>>
>>>>> Thanks again,
>>>>>
>>>>> Janh
>>>>>
>>>>>
>>>>> On Thu, Jul 23, 2015 at 11:51 PM, Jeff Newmiller
>>>>> <jdnewmil at dcn.davis.ca.us>
>>>>> wrote:
>>>>>
>>>>>> ?cumsum
>>>>>>
>>>>
>>>>>
>> ---------------------------------------------------------------------------
>>>>>> Jeff Newmiller The ..... ..... Go
>>>>> Live...
>>>>>> DCN:<jdnewmil at dcn.davis.ca.us> Basics: ##.#. ##.#. Live
>>>>>> Go...
>>>>>> Live: OO#.. Dead: OO#..
>>>>> Playing
>>>>>> Research Engineer (Solar/Batteries O.O#. #.O#. with
>>>>>> /Software/Embedded Controllers) .OO#. .OO#.
>>>>> rocks...1k
>>>>>>
>>>>
>>>>>
>> ---------------------------------------------------------------------------
>>>>>> Sent from my phone. Please excuse my brevity.
>>>>>>
>>>>>> On July 23, 2015 8:23:39 PM PDT, Janh Anni <annijanh at gmail.com>
>>>>> wrote:
>>>>>>> Dear All,
>>>>>>>
>>>>>>> Does anyone know of any R functions that compute partial sums of
>>>>>>> series?
>>>>>>>
>>>>>>> Thanks in advance!
>>>>>>>
>>>>>>> Janh
>>>>>>>
>>>>>>> [[alternative HTML version deleted]]
>>>>>>>
>>>>>>> ______________________________________________
>>>>>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
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>>>>>>
>>>>>>
>>>>
>>>>
>>>
>>> [[alternative HTML version deleted]]
>>>
>>> ______________________________________________
>>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
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>>> PLEASE do read the posting guide
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>>> and provide commented, minimal, self-contained, reproducible code.
>>
>> David Winsemius
>> Alameda, CA, USA
>>
>>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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