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  1. #11
    Senior Member Urchin's Avatar
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    When writing a proof, I look at the assumptions and try to figure out what they imply. I then look at the conjecture and figure out what would imply it. I try to step inward from either end until I figure out what the linchpin of the proof will be (or the crux as ygolo said). Once I've done that, I pick a method, work from assumptions to linchpin, and then the rest sort of topples into place (hopefully). It's the first parts that are difficult. Sometimes I play a bit with the assumptions, but I try to keep it abstract and avoid playing with specific cases.
    "Having is not such a pleasing thing as wanting. It is not logical, but it is often true." --Spock

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  2. #12
    Glowy Goopy Goodness The_Liquid_Laser's Avatar
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    Generally when I prove something first I have to believe it is true (working examples and whatnot). Once I convince myself then I try to understand what the statement is saying as completely and simply as possible. I think of the first step and the last step, and then the middle steps tend to come to me all at once (or at least in large chunks).
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  3. #13

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    How could I neglect linear algebra?

    I am absolutely fascinated by Singular Value Decomposition.

    Eigenvalue decomposition was interesting, but nothing compared to the power and universality of the SVD.

    It is absolutely vital for signal processing and information recovery.

    Any favorite theorems in particular, for analysis, topology, or various other branches of math? LaGrange's Theorem (which Urchin mentioned) is my favorite group theory theorem.

    Stone-Weierstrass theorem may be my favorite result from analysis, but I haven't studied enough to have a deep understanding of it.

    I suppose, by now, you can tell I am more an applied math type.

    I haven't studied enough topology to pick a favorite but Baire category theorem fascinates me, though I have close to zero understanding of why that is true.

    nemo, do you have any clear picture/understanding of Stone-Weierstrass or the Baire Category theorems?

    Accept the past. Live for the present. Look forward to the future.
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    "[A] scientist looking at nonscientific problems is just as dumb as the next guy." Richard Feynman
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  4. #14
    Senior Member Urchin's Avatar
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    This is part of the magic of my childhood. I know it's long, but I promise that it's worth watching.
    "Having is not such a pleasing thing as wanting. It is not logical, but it is often true." --Spock

    MBTI: INTP
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  5. #15
    Enigma Nadir's Avatar
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    Quote Originally Posted by Urchin View Post
    This is part of the magic of my childhood. I know it's long, but I promise that it's worth watching.
    Thanks for sharing, it is nice.
    I like math, but it is always less interesting and engaging to me during the actual, step-by-step functional analysis. (Though I should clarify that I haven't played around with this sphere-warping, topology stuff yet) It's probably because the numbers in any given exercise all exist in a vacuum, and do not really have a point beyond the exercise itself (wonders of high school, I guess). I'm looking forward to a time where they have some other, more tangible significance.
    And before anyone asks, yeah, I'm not perfect at maths, though I get by. Blame this disinterest!
    Last edited by Nadir; 02-06-2008 at 01:08 PM. Reason: Some clarifications
    Not really.

  6. #16
    no clinkz 'til brooklyn Nocapszy's Avatar
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    Here's a question. I couldn't think of any place more appropriate than here.

    I'm not sure of the answer (partly why I'm asking here) but I think I know.

    If you can use the numbers 1 2 and 3 in any combination you want, and you're allowed to use any of the 3 as many times as you want, how many sets of three can you get?

    I think it's just 3^3 = 27, but I'm not sure. I can't think right now.
    we fukin won boys

  7. #17
    Senior Member Urchin's Avatar
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    Quote Originally Posted by Nocapszy View Post
    Here's a question. I couldn't think of any place more appropriate than here.

    I'm not sure of the answer (partly why I'm asking here) but I think I know.

    If you can use the numbers 1 2 and 3 in any combination you want, and you're allowed to use any of the 3 as many times as you want, how many sets of three can you get?

    I think it's just 3^3 = 27, but I'm not sure. I can't think right now.
    Does order matter? If so, it's 27. If not, it's 10. The way I did it was kind of a hack, because there are so few possibilities. I'm sure there's a better way to do it than simply counting them, but I'm not well-versed in probability, and this gets the job done. I guess you could make a tree diagram. That's kind of what I did in my head.
    "Having is not such a pleasing thing as wanting. It is not logical, but it is often true." --Spock

    MBTI: INTP
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  8. #18
    no clinkz 'til brooklyn Nocapszy's Avatar
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    Order does not matter. It can be 123, or 333 or 312 or 223... doesn't matter. You can use any of the three numbers as many times as you want to create as many sets of 3 as possible.

    The way I did it was, I thought, to create an actual cube.

    Bottom later would look like this
    Code:
    111
    222
    333
    Next layer should have the similar digits along the opposite axis
    Code:
    321
    321
    321
    then the top layer would have a sort of a rolling thing going
    Code:
    123
    312
    231
    And then you can take that, and anywhere you find yourself with 3 in a row, it's a possibility.

    I'll see if I can actually put together a cube like this (in paint or IRL or something) and see if it works. Seems like it would 'cause there are 3 variables.
    we fukin won boys

  9. #19
    no clinkz 'til brooklyn Nocapszy's Avatar
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    Hmm.... I'm thinking that's not right now. I'm sure it can be turned into a cube, but I don't think that the model I gave is any good.
    we fukin won boys

  10. #20

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    If I'm reading you correctly, it is a matter of combinatorics. I may have interpreted you incorrectly but I agree with Urchin.

    Think of this way, you have 3 options for a shirt (first slot), 3 options for pants (second slot), and three options for a tie (third slot).

    You have three independent choices and three things to choose from for each choice. So yes, it is 3^3.

    But the order-independent version can be thought to be equivalent to the following:

    You have a giant(infinite) vat of ping-pong balls numbered 1 to 3, and choose 3 ping pong balls from there. You subsequently forgot which order you chose them. How many distinguishable possibilities are there?

    It is 10, like Urchin mentioned, but how do you do it more generally? With choosing r balls from n types of balls? That is an interesting question, imo.

    The short answer is it is C(n+r-1,r), where C is the choose function.

    I tried typing the "star" and "bar" explanation but I didn't like the formating.

    So, here is a link.

    Nocapszy, if you like these types of questions you'll like Discrete Math (which is the mathematics behind computer science).

    The following source has explanations of the basics.
    Discrete Math Project - Permutations with Repetition

    Also note, that these basics come in four flavors:
    1)permutations without replacement
    2)combinations without replacement
    3)permutations with replacement
    4)combinations with replacement

    You'll find that transforming your counting problem into one of these is the most likely source of answers. That along with generating functions will fulfill most people's combinatorics type questions.

    Accept the past. Live for the present. Look forward to the future.
    Robot Fusion
    "As our island of knowledge grows, so does the shore of our ignorance." John Wheeler
    "[A] scientist looking at nonscientific problems is just as dumb as the next guy." Richard Feynman
    "[P]etabytes of [] data is not the same thing as understanding emergent mechanisms and structures." Jim Crutchfield

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