Re: Proof of ordered powerset

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 03/21/05 

Date: Mon, 21 Mar 2005 16:43:26 -0600

On Mon, 21 Mar 2005 09:44:59 -0600, "Jon Slaughter"
<Jon_Slaughter@Hotmail.com> wrote:

>
>"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message
>news:oejt31911k04uvc71000e03qtlffc8ltsa@4ax.com...
>> On Sun, 20 Mar 2005 19:54:04 -0600, "Jon Slaughter"
>> <Jon_Slaughter@Hotmail.com> wrote:
>>
>>>
>>>"David C. Ullrich" <ullrich@math.okstate.edu> wrote in message
>>>news:or4r319uuhvc63onlr4dg7jcvniuq9tbi5@4ax.com...
>>>> On Sat, 19 Mar 2005 09:35:00 -0600, "Jon Slaughter"
>>>> <Jon_Slaughter@Hotmail.com> wrote:
>>>>
>>>>>
>>>>>"M.J.T. Guy" <mjtg@cus.cam.ac.uk> wrote in message
>>>>>news:d1ha3p$6n3$1@gemini.csx.cam.ac.uk...
>>>>>> Jon Slaughter <Jon_Slaughter@Hotmail.com> wrote:
>>>>>>>Anyone know of an elegant proof of \sum(k!*nCk,k=0..n) = floor(e*n!),
>>>>>>>that
>>>>>>>involves only arithematic?
>>>>>>
>>>>>> Not sure what you mean by "involves only arithematic", but try
>>>>>> comparing
>>>>>> the sum on the left with the first n terms of the series expansion
>>>>>> of e*n!.
>>>>>>
>>>>>>
>>>>>> Mike Guy
>>>>>
>>>>>Well, what I ment was an elementary proof(one that doesn't use
>>>>>"advanced"
>>>>>concepts like integration, the gamma function or hypergeometric series,
>>>>>etc... actually, anything that doesn't really involve calculus)).
>>>>
>>>> Looking below it seems you don't want any sort of infinite anything
>>>> involved. I don't think you can even _define_ e using the sort
>>>> of thing you're allowing.
>>>>
>>>> Or maybe you can: What definition of e did you have in mind
>>>> when you asked the question?
>>>>
>>>
>>>yeah, I'm not sure either ;/
>>
>> If you can't tell me what _definition_ of e you have in mind then
>> your question simply makes no sense at all - before we can prove
>> anything about e we need to know what e _is_.
>>
>>
>>
>> ************************
>>
>> David C. Ullrich
>
>heh, I don't know what e is ;/ How else can you define what e is without
>some limiting process? Does the necessarily mean that you cannot find the
>solution to the original series without using some infinite limiting
>process?

Yes, of course it does! Anything you prove about e has to involve
the definition of e at some point.

************************

David C. Ullrich

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