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Is this 'e' interpretation correct?

  1. Apr 7, 2013 #1
    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".
     
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  3. Apr 7, 2013 #2

    arildno

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    Hmm..No.
     
  4. Apr 7, 2013 #3
    No explanation?
     
  5. Apr 7, 2013 #4

    Mentallic

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    Is this in regards to compound interest returns?
     
  6. Apr 7, 2013 #5

    arildno

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    You might wish to look at, for example, the actual limit of, for example, (1+1/(3n))^(2n)
    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.
     
  7. Apr 8, 2013 #6

    HallsofIvy

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