
#1
Apr713, 05:19 AM

P: 190

Considering e is the limit>+oo of (1+1/n)^n, then is e "what you get if you wait for the least gain, by waiting for the most amount of time"? Something like "e is the patience number".




#2
Apr713, 05:32 AM

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P: 12,016

Hmm..No.




#3
Apr713, 05:57 AM

P: 190





#4
Apr713, 06:09 AM

HW Helper
P: 3,434

Is this 'e' interpretation correct? 



#5
Apr713, 07:11 AM

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P: 12,016

which, for every particular choice of "n" will have a less gain waited for for an even greater period of time than the one you happende to pick. 



#6
Apr813, 08:46 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,877

What explanation could be given when you haven't said what you mean by "gain" or "waiting".



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