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Crazily Looping Machine

Discussion in 'Mathematics' started by AVDBA, Sep 24, 2019.

  1. AVDBA

    AVDBA New commenter

    A machine of a casino seems unconventional. The machine can be used only once and then replaced. The machine asks for a deposit of $2 and for the number of rounds the game is going to be played. Each round the game is played the machine saves the previously saved amount plus the square of the inverse of the number of times the game is going to be played. Finished the final round, the machine multiplies all of the saved amounts and pays this quantity to the player. Being allowed the infinity mode (which is a mode that plays infinitely many times), knowing that the player uses just one machine and that the machine saves $1 after the initial round - how much is going the player to collect by playing the infinity mode?
     
  2. adamcreen

    adamcreen Occasional commenter

    By "inverse", do you mean reciprocal?
     
  3. AVDBA

    AVDBA New commenter

    Hi adamcreen! Yes, considering x, I mean 1/x as the inverse (x not equal to 0). So, I am going to restate. Thanks a lot.
     
  4. AVDBA

    AVDBA New commenter

    Restatement:

    A machine of a casino seems unconventional. The machine can be used only once and then replaced. The machine asks for a deposit of $2 and for the number of rounds the game is going to be played. Each round the game is played the machine saves the previously saved amount plus the square of the inverse (reciprocal) of the number of times the game is going to be played. Finished the final round, the machine multiplies all of the saved amounts and pays this quantity to the player. Being allowed the infinity mode (which is a mode that plays infinitely many times), knowing that the player uses just one machine and that the machine saves $1 after the initial round - how much is going the player to collect by playing the infinity mode?
     
  5. AVDBA

    AVDBA New commenter

    I would like to say adamcreen has raised an important point, since, rigorously, the inverse of, e.g.,

    y = 7x => x = y/7

    by renaming the variables:

    y = x/7

    which is the inverse function. Sometimes, here, we call the inverse as meaning the reciprocal, but, rigorously speaking, adamcreen is correct. So, I have restated the problem.
     
  6. gainly

    gainly Lead commenter

    The reciprocal could also be referred to as the multiplicative inverse but not just the inverse.

    I still can't understand this problem.
     
  7. adamcreen

    adamcreen Occasional commenter

    I won't be happy until this thread is longer than the X Bar
     
  8. AVDBA

    AVDBA New commenter

    Thanks gainly.

    Restatement:

    A machine of a casino seems unconventional. The machine can be used only once and then replaced. The machine asks for a deposit of $2 and for the number of rounds the game is going to be played. Each round the game is played the machine saves the previously saved amount plus the square of the reciprocal of the number of times the game is going to be played. Finished the final round, the machine multiplies all of the saved amounts and pays this quantity to the player. Being allowed the infinity mode (which is a mode that plays infinitely many times), knowing that the player uses just one machine and that the machine saves $1 after the initial round - how much is going the player to collect by playing the infinity mode?
     
  9. AVDBA

    AVDBA New commenter

    I would like to provide the following hints:

    * Let n be the number of rounds the game is going to be played

    * Finished the 1st round: Saved amount: 1

    * Finished the 2nd round: Saved amount: 1 + [(1/n)^2]

    * Finished the 3rd round: Saved amount: {1 + [(1/n)^2]} + (1/n)^2 = 1 + [2/(n^2)]
    ...

    * Finished the nth round: Saved amount: 1 + [(n - 1)/(n^2)]
     
  10. AVDBA

    AVDBA New commenter

    This problem deals with a product that becomes a summation by applying a property of logarithms. Furthermore, the problem turns out to require an analysis related to the logarithm as well. I would like to suggest the following solution. I would like to suggest to TES, for the future, the possibility of implementing Latex here, since this would make life easier for writing equations here, so that I am going to define some notations [e.g., lim(n->infinity) denoting the limit] for writing the equations.

    The problem states that, finished the finite final round n, the machine pays to the player:

    C(n) = (1) x [1 + 1/(n^2)] x [1 + 2/(n^2)] x ... x [1 + (n - 1)/(n^2)]

    The function ln(x + 1), satisfies the following property:

    x - (x^2)/2 < ln(x + 1) < x, for all x in (0, infinity)

    This can be seen by defining the following functions:

    l(x) = ln(x + 1) - x + (x^2)/2
    r(x) = ln(x + 1) - x

    by virtue of:

    l(0) = r(0) = 0

    and by virtue of the differentiation sign within the interval (0, infinity):

    l'(x) = 1/(x + 1) - 1 + x = (1 - x - 1 + x^2 + x)/(x + 1) = (x^2)/(x + 1) > 0 for all x in (0, infinity)
    r'(x) = 1/(x + 1) - 1 = (1 - x - 1)/(x + 1) = -[x/(x + 1)] < 0 for all x in (0, infinity)

    Hence:

    r(x) < 0 and l(x) > 0 => ln(x + 1) - x < 0 and ln(x + 1) - x + (x^2)/2 > 0

    therefore:

    x - (x^2)/2 < ln(x + 1) < x for all x in (0, infinity)

    We are interested in the infinity mode, i.e.:

    lim(n->infinity) C(n) = lim(n->infinity) {(1) x [1 + 1/(n^2)] x [1 + 2/(n^2)] x ... x [1 + (n - 1)/(n^2)]}

    Taking the logarithm, we have [now we will drop the notation (n->infinity) for simplicity]:

    ln[lim C(n)] = ln{lim {(1) x [1 + 1/(n^2)] x [1 + 2/(n^2)] x ... x [1 + (n - 1)/(n^2)]}},

    which can be interchanged by the continuity within the considered interval:

    lim ln[C(n)] = lim {0 + ln[1 + 1/(n^2)] + ln[1 + 2/(n^2)] + ... + ln[1 + (n - 1)/(n^2)]}

    therefore:

    lim ln[C(n)] = lim Sum[k, 1, n - 1]{ln[1 + k/(n^2)]}

    and, by the previous property:

    lim Sum[k, 1, n - 1]{k/(n^2) - (k^2)/[2(n^4)]} < lim Sum[k, 1, n - 1]{ln[1 + k/(n^2)]} < lim Sum[k, 1, n - 1][k/(n^2)]

    Naming the LHS of the previous inequality:

    L = lim Sum[k, 1, n - 1]{k/(n^2) - (k^2)/[2(n^4)]}

    inferring, for the summation in k, that we have a summation of terms in arithmetic progression minus a sum of squares of natural numbers {this latter having degree 3, which can be obtained by induction = [2(n - 1)^3 + 3(n - 1)^2 + (n - 1)]/6, albeit a known result for the summation of the squares of natural numbers}, gives L = 1/2 [the term that survives from the summation over k giving the terms in arithmetic progression leading to (1 + (n - 1))(n - 1)/2, having degree 2 that cancels the denominator n^2 by taking the limit].

    Naming the RHS of the previous inequality:

    R = lim Sum[k, 1, n - 1][k/(n^2)], by virtue of the fact this summation is in k, we have a summation of terms in an arithmetic progression, so the degree is the same as n^2, surviving hence L = 1/2 for the taken limit.

    Henceforth:

    lim Sum[k, 1, n - 1]{ln[1 + k/(n^2)]} = 1/2

    and we conclude that:

    lim ln[C(n)] = 1/2 => lim C(n) = e^(1/2) ~ 1.649

    and the player collects e^(1/2) by playing infinity mode, i.e., the player looses 2 - 1.649 = 0.351 by playing the infinity mode.
     
  11. adamcreen

    adamcreen Occasional commenter

    *loses* not looses
     
  12. AVDBA

    AVDBA New commenter

    Thanks adamcreen. Sorry for incorrectly misspelling "loses". Would it be correct: he looses money from his pocket?
     

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