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Math Help - Strange Cal II questions

  1. #1
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    Strange Cal II questions

    These are some rather odd questions that I got presented with at my university today. I'd really like to get them solved and I know just about what I need to solve them, just not sure how to go about putting it into practice.

    The first has to do with probabilities and infinite series, I know that much.

    A board game consists of four positions labeled A, B, C, and D. Whenever you reach position A or B, you roll some dice whose results determine what position you will move to next. Ah, but these dice are loaded.

    From position A, there is a 1/4 chance of moving to position B, a 1/12 chance of moving to position C, and a 1/6 chance of moving to D.


    From position B, there is a 1/3 chance of staying at B, there is a 1/3 chance of moving to position A, a 1/6 chance of moving to position C, and a 1/6 chance of moving to D.

    Whenever you reach position C or D the game is over and you win some cake.

    Suppose you begin the game at position A, what is the probability you end the game at position C
    I tried taking the probabilities:

    P(Moving from A to C): 1/12
    P(Moving from A to B to C): 1/4 x 1/6 = 1/24Here is where I get stuck. Normally I would just keep drawing out the probabilities to the situation I want. Not sure how to do it since, you could possibly move back to either A or stay at B after B. If anyone has any tips about how I can go about uncovering the correct infinite sequence for this I would be much obliged.



    Also it is a strange question but I got this question too:

    Find two functions f(x) and g(x) such that the integration of the product (f(x) x g(x)) is equal to the product of the integration of f(x) and the integration of g(x).


    Integration(f(x) x g(x)) = Integration(f(x)) x Integration(g(x))
    The zero functions f(x) = 0, and g(x) = 0, while correct in this context are not the desired answers.

    This seems like an easy questions but the more I got to thinking about it, I could not off the top of my head come up with two functions for which this is true.
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    Find two functions f(x) and g(x) such that the integration of the product (f(x) x g(x)) is equal to the product of the integration of f(x) and the integration of g(x).

    Integration(f(x) x g(x)) = Integration(f(x)) x Integration(g(x))
    Try some exponentials, f(x) = e^{\alpha x},\ g(x) = e^{\beta x} for suitable \alpha,\ \beta.

    A board game consists of four positions labeled A, B, C, and D. Whenever you reach position A or B, you roll some dice whose results determine what position you will move to next. Ah, but these dice are loaded.
    .
    .
    .
    Suppose you begin the game at position A, what is the probability you end the game at position C?
    Are you supposed to know something about the theory of stochastic matrices? It will be hard to do this problem otherwise.
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    No, nothing to do with matrices I believe. I think that we are supposed to eliminate the probabilities of staying at A (from A) or staying at B (from B) by multiplying and combining them with the other probabilties so that it is simpler. And then draw it up into an infinite series. Just not sure exactly how to go about that, there are so many different probabilties going on at the same time.

    As for the other. The exponential function alone doesn't work. I cannot seem to think about what alpha and beta might work with what you have there.

    (Finally figured out these math tags. Sorry for the horrible depiction of the equation in my OP)

    (\int f(x) dx)(\int g(x) dx) = \int f(x)g(x) dx

    For what functions f(x) and g(x) is the above true...?

    Well if we use the exponential function alone, the left side of the above equation is simply e^{2x} I believe, and the right would be (1/2)e^{2x}

    Thought it might have something to do with using inverse functions together, but I'm still stumped....
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    Quote Originally Posted by CoryG89 View Post
    No, nothing to do with matrices I believe. I think that we are supposed to eliminate the probabilities of staying at A (from A) or staying at B (from B) by multiplying and combining them with the other probabilities so that it is simpler. And then draw it up into an infinite series. Just not sure exactly how to go about that, there are so many different probabilities going on at the same time.
    That is exactly the difficulty. There are so many different routes for getting from A to C. You can shuttle back and forth between A and B arbitrarily often, and you can stay for an indefinite number of moves at A or at B. The matrix method takes all these possibilities into account and provides a clean method for getting the answer (namely a probability 7/18 of ending at C if you started at A). If you can get that answer without using matrices, I'll be impressed!

    Quote Originally Posted by CoryG89 View Post
    As for the other. The exponential function alone doesn't work. I cannot seem to think about what alpha and beta might work with what you have there.

    (Finally figured out these math tags. Sorry for the horrible depiction of the equation in my OP)

    (\int f(x) dx)(\int g(x) dx) = \int f(x)g(x) dx

    For what functions f(x) and g(x) is the above true...?

    Well if we use the exponential function alone, the left side of the above equation is simply e^{2x} I believe, and the right would be (1/2)e^{2x}
    Try f(x) = g(x) = e^{2x}. The objection I have to this problem is that it takes no account of the constant of integration. The indefinite integral of e^{2x} is \frac12e^{2x} + C, and you have to take the constant C to be 0 in order to get \int(e^{2x}*e^{2x})\,dx = \int e^{2x}\,dx\int e^{2x}\,dx = \frac14e^{4x}.
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