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Everything posted by Genady
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You are not mistaken. It didn't touch Shannon entropy yet, and I don't expect it to do so.
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Ukraine using drones to 'great effect' on Russian forces: Pentagon updates (yahoo.com)
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Can this diarrhea be stopped?
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MIT OCW has recently uploaded a new complete set of video lectures from this class: Theory of Computation | Mathematics | MIT OpenCourseWare I am 25% through and enjoying it.
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Et tu, Brute? The very last time we watched TV was the first Obama inauguration.
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In this case, what is a proper label for that kind of social media, the one with the algorithms?
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Me too.
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The vertical force he exert is equal to the weight. However, the weight itself, i.e. the heavy wheel doesn't stay above his head. If we look only at the vertical displacement of the wheel, it goes up and then down.
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I'm quite positive that the OP is talking about this video: No, it doesn't suggest that the weight changes. It is a pretty good demonstration and is followed by explanation.
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There is a summation in the formula (2) for A(), but in your calculation there is no summation. So, you have calculated only one components of A(3,2), rather than the value A(3,2). In fact, A(3,2) = 6. For example, if your objects are a, b, and c, then there are these 6 ways to distribute them in 2 distinguishable cells with no empty cells: a | bc b | ac c | ab bc | a ac | b ab | c
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The general form is OK, but the variables are not. First, do it the way in which the eq (1) is given. I.e. use A(..., n) as given and derive A(r, n+1). It Should look like A(r, n+1) = .... A(r-k, n). Second, provide explanation for each component, i.e. what is the coefficient in front of A(r-k, n) and what values of k are you summing for.
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Russian officials' false statements in chronological order leading to the war: Oh, How They Lied. The Many Times Russia Denied Ukraine Invasion Plans (polygraph.info)
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1. You need to derive the equations for generic variables r and n. Working with specific values, such as r=4 and n=2, will not give you a general derivation. 2. The upper limit of summation is still wrong. 3. I don't know where the second part of these two simple sums came from. From nowhere? 4. You still did not derive equation (1). I've given you a fat hint for that. If you derive it, you will see where your summation limit is wrong.
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Yes, both equations (1) and (2) relate to distinguishable objects and distinguishable cells. Eq (1) is recursive. That is, assume that there are A(x, n) ways to distribute x distinguishable objects between n distinguishable cells, with no empty cells. Knowing this for any number of distinguishable objects x > n, we want to find A(r, n+1), number of ways to distribute r distinguishable objects between n+1 distinguishable cells, with no empty cells. Start thinking about it in the following way. Take one of the n+1 cells aside. Let's call it a "new" cell. You have one new cell and n old cells. You have to put some number of objects, k>0, in the new cell. The rest r-k objects will go into the n old cells. This can be done in A(r-k, n) ways, the number that we assume we know. Can you continue from here so we get the expression for A(r, n+1), which is the eq (1)?
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In the eq (1) here your upper limit of summation is k=r. Are you sure about it? I don't think it is correct. The case (2) doesn't make sense. You cannot distribute 2 objects in 3 cells such that there are no empty cells. I think, you have to have at least as many objects as there are cells for these formulas to make sense. That's why I think that the upper limit of summation in eq (1) is incorrect. Did you derive the eq (1) by a combinatorial argument, as requested?