text: Solving 2-Can, 1-Cup Problem: f(n)
  
  • JanEnClaesen
  In summary, the conversation discusses a setup where two cans containing different but miscible liquids are used to transfer and mix the liquids multiple times. The resulting mixtures have equal volume but different concentrations. The problem is then posed as determining the concentration of one of the liquids after a certain number of transfers. The conversation also mentions a similar brain teaser involving milk and water.
  
  JanEnClaesen
  
  Given are two equally full cans containing two different but miscible fluids and one smaller, empty cup.
  Take one cup of one of the liquids and pour its contents into the can containing the other liquid, similarly, take one cup of the resulting mixture and pour it back into the other can. This results in two mixtures of equal volume but of different concentrations.
  
  Set this transfer equal to n = 1.
  
  (You might have encountered this set-up in the form of a brain teaser, the tease being the relationship between the transferred liquids. The answer is equality.)
  
  Now consider f(n), the concentration of one of the liquids in one of the cans after n transfers.
  What is f(n)?
  
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  Perhaps the problem is badly posed: it's basically a continuation and quantification of this brain teaser:
  
  There are two equally filled jars, one contains milk, the other water. A teaspoon of milk goes into the water and is stirred. Then a teaspoon of the mixture goes back into the milk. Is there more water in the milk or milk in the water?
  
  I don't see the additional challenge if you can solve it for n=1.
  It is just a geometric progression towards 1/2.
  
  I think this better fits into our https://www.physicsforums.com/forumdisplay.php?f=33 forum.
  
  1 person
  So you are saying, concretely:
  
  ( only the even powers of x in the expansion of (x+1)^n if n is not even and vice versa ) / (x+1)^n where x is as the can is to the cup
  
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  I have no idea how you got that formula.
  
  Put a fraction x of container A (with content a) to container B (with content b), then container B has a fraction x/(1+x) of a and 1/(1+x) of b. Put x back, and you get x/(1+x) of b in container a and (by symmetry) the same of a in container b. The concentration difference to 1/2 reduces from 1/2 to 1/(1+x)-1/2, or by a factor of 2/(1+x)-1. This will be the same for all steps, so the concentration after n steps is 1/2*(1+(2/(1+x)-1)^n).
  
  1 person
  Like this:
  
  x times the cup equals the can. After one step the concentration of liquid a is x/(1+x) in can A and 1/(1+x) in can B, the second step combines x/(1+x) of a from A and x*1/(1+x) of a from B in B. f(2) = 2x/(1+x)^2 The third step combines (x^2 + 1)/(1+x) of a from A and x*2x/(1+x)^2 of a from B in B. So f(3) = (3x^2 + 1)/(1+x)^3 ...
  
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