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Zeno's Coffee Shop

  1. Jul 12, 2003 #1
    You enter a diner and sit down at a booth. The diner is classic Americana... like something from the '50s... but nothing fancy. You read the morning paper and have a smoke as you wait for the waitress dressed in a not-so-clean, pink-and-white-striped, short sleave dress and white leather tennis shoes to bring you a coffee. You like your coffee with a lot of cream... but no sugar.

    Sugar is for kids, you think to yourself.

    The ratio of cream to coffee that you prefer is represented by the fraction c, which is of course less than unity. However, everytime you drink your coffee to the point where your cup is exactly half empty (you are a pessimist... I might mention), the waitress comes by and fills your cup to the brim with steaming hot, black coffee.

    Question: How many times must she do this before you are drinking a cup of black coffee?

    Let the ratio of "cream molecule" size to coffee cup volume be represented by the fraction m... also less than unity.

    Last edited by a moderator: Jul 12, 2003
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  3. Jul 12, 2003 #2


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    So the first time the fluid in your cup is c cream and (1-c) coffee.

    And when she fills you up again you have

    .5 coffee from the fill up
    .5c cream from what was left
    .5(1-c) coffee ditto,

    making .5c cream and .5 + .5(1-c) = (1 - .5c) coffee. It checks so far.

    next time you're down to half of this

    .5*.5c cream and .5*(1-.5c)coffee + .5 coffee
    or .25 c and mmmm, 1-.25c coffee. and so on.

    So you have a geometric series 1, .5, .25, .125, ... proportion of cream and the aliquot proportion of coffee, Of course cream comes in discrete molecules so eventually you'll get down to one molecule, but only probability says you will ever get rid of that one.
  4. Jul 13, 2003 #3
    yes unlike the classic distance zeno's paradox, this problem is quantatative in that the number of cream molecules is measurable. self-adjoint says the rest.
  5. Jul 14, 2003 #4
    Okay, let me clear a few things up.

    1. I created this problem. I know the answer. Don't you think I realize that this problem involves a quantization? Seriously, why do you think I gave you a molecular size to deal with?

    2. No one has yet posted the correct answer. I want an explicit expression for number of times necessary to refill the cup as a function of both c and m.

    3. This isn't a physics problem. It is an idealized math problem. You don't have to worry about the probability of removing that last atom. You don't have to worry about cream atoms adhering to the sides of the cup. Just assume that there is always a uniform distribution of cream in coffee. It's that simple folks.


    P.S. The previous post only strengthens my theory that there is no such thing as smart people in Nebraska.
    Last edited by a moderator: Jul 14, 2003
  6. Jul 14, 2003 #5


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    Well creams made up of several molecules, some of them (for example water) also appear in the coffee too, but that aside:

    (0.5)^n = c, and (0.5 * m[initial])^n = m, but the problem doesn't really work as you get unphysical answers for n+1 when m is odd.
  7. Jul 14, 2003 #6
    There's no need for m...it's just a "trap"...
    Let's do this "the hard way"...
    Let N be the number of times she filled your cup...including the first time when she gave you cream...
    N is also the number of times you have to drink to have a cup of coffee...
    N-c*[(2-1/2^(n-1)]/2>1 is the condition to stop...you had a full cup of coffee...(summing that progression...blabla)...

    c<[(N-2)*2^(N-1)]/2^n-1...and using a calculator we obtain that...
    ...for c=0...N=2...
    ...for c<=4/7...N=3...
    ...for c>4/7...N=4...
    That's my opinion...no use for m...because after a finite number of steps you'll have only one "cream molecule"...so...will you "drink" it or not ?...
  8. Jul 14, 2003 #7
    lets see, you flamed me in the post, you went out of your way to flame me in another post, and yet you dont know me at all... sometimes things dont need to be said entropy, as much as you think you have figured me out, you havent even begun, just as i havent begun to figure you out...

    im sorry im not fantastic at math, attack me for inherent faults, please, im sorry i voiced my opinion, im sorry it was a trivial opinion, im sorry for filling your post with nonsense...

    why do you feel that you need to do these things to other people?
  9. Jul 14, 2003 #8
    Jeez, calm down. I'm just joking with you... except I was a little annoyed by the fact that you acted like I didn't understand my own question. Well, perhaps I'm a bit too harsh at times.

  10. Jul 15, 2003 #9
    Is my solution correct ?
  11. Jul 15, 2003 #10
    My solution in explicit form with explanation....

    We have already stated that c is the preferred ratio of cream to coffee in the cup. This means that:

    c = Cream / Coffee.

    where V, the volume of the cup is:

    V = Cream + Coffee.

    If we allow Cream to represented in discreet quantities, we see:

    Cream = N * CreamMolecule.

    We can rearrange the terms and find:

    c = N * CreamMolecule / (V - N * CreamMolecule).

    Here, N represents a discreet number of cream molecules.

    Now, we remember that the "cream molecule" size to cup volume ratio is:

    m = CreamMolecule / V.

    Notice that both quantities c and m are dimensionless.

    Furthermore, notice the relationship between c and m:

    1/c = 1/(N*m) - 1.

    Rearranging terms again, our new expression for c is:

    c = 1 / (1/(N*m) - 1).

    Here, we remember that in the limit that there is only one cream molecule N = 1. Also, remember that each time you drink half of the coffee, half of the cream molecules vanish. With this information, we form the relationship:

    (1/2)^M * c >= m / (1 - m).

    Solving for M, the number of times the cup is half-emptied, we get:

    M >= - Log2[m / (c*(1 - m))].

    We take to be the nearest integer that satifies this expression. M is also equal to the number of times the waitress refills the cup.

  12. Jul 15, 2003 #11
    Well...if m->0...then M->infinity...
    ...if c->0...M->-infinity...
    ...both situations are quite odd...huh ?
  13. Jul 15, 2003 #12
    What else would you expect?

    If m -> 0, then you have a continuous fluid and the cream can be forever divided into smaller parts; so, it would take M -> infinity to get rid of all the cream.

    Admittedly, the case where c -> 0 should yield M -> 0. This would mean that if you started out with a cup of black coffee, then you wouldn't need any refills to get a cup of black coffee. However, this model only works for as long as you have at least one coffee molecule in your cup from the beginning.


    If c = m / (1 - m), this implies that N = 1... meaning the prefered ratio of cream to coffee is to have just one molecule of cream in the cup.

    Then, we immediately see that:

    M = Log2[ 1 ] = 0.

    This means that it will take zero steps to get just one molecule of coffee in the cup.

    If we wanted to improve this model perhaps we could slightly change the expression to be concerned with the case where zero molecules occur rather than just one.

    So, the limitation of this model is the following:

    m / (1 - m) < c < 1.

    Last edited by a moderator: Jul 15, 2003
  14. Jul 16, 2003 #13
    I thought you said "How many times must she do this before you are drinking a cup of black coffee?"...not until you have a cup full of pure coffee...
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