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  1. #41
    Ridill
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    Mmm I see latex. Delicious, as my mathematical modeling professor would say, "the equation look so good you want to lick it!"

  2. #42
    Title: "HUBBLE GOTCHU!" (without the quotes, of course [and without "(without the quotes, of course)", of course], etc)
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    My LaTex trolls me by putting my lines over the margin in ways that are hard to fix.



    If I bring that expression down to the next line, then the preceding line stops way too early. If I stretch the proceeding line by the appropriate amount the letters end up spaced really far apart. I can't figure out how to make stuff like this not look ugly without completely rewording sentences (and a lot of times it's hard to figure out how to reword it in a way that actually fixes the problem).

  3. #43
    Ridill
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    What packages are you using? Also, I ended up getting enough of a working piece-mesh of code to finish my project. Submitted it yesterday. Whatever. I hate this semester, lol.

  4. #44
    Title: "HUBBLE GOTCHU!" (without the quotes, of course [and without "(without the quotes, of course)", of course], etc)
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    \usepackage{amssymb,latexsym,amsmath}

    Is there a package I should (or shouldn't) be using that would help counter this problem?

  5. #45
    Ridill
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    I always use wrapfig and graphicx also, but yeah, idk. sometimes when shit is wonky I'll just purposefully // or set something as a $$ $$ to break up a paragraph if I know it's gonna format weirdly.

  6. #46
    2600klub

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    1.
    log(9x -2) =1 + log(x-4) hint log10 =1

    2.
    3^x = 9^x+5

  7. #47
    An exploitable mess of a card game
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    Question or answer to an earlier problem? If question (And I haven't taken any courses in like... 3 yrs):
    Spoiler: show

    1. x = 38
    2. x = (-10)

    I assume question (2) was 9^(x+5)

    For Question 1
    1. 10^(log(9x-2)) = 10^(1 + log(x-4))
    2. The right half breaks down into (10^1)*(10^(log(x-4)) using the rule (10^A)*(10^B)=10^(A+B)
    3. The right half then simplifies into (10)*(x-4) => 10x-40
    4. The left half simplifies to 9x-2
    5. 9x-2=10x-40
    6. x = 38

    For Question 2
    1. 9^(x+5) can change into the same base as 3^x using the rule (3^2 = 9); this gives 3^(2*(x+5))
    2. The formula from (1) simplifies to 3^(2x+10)
    3. Same base on both so you can set x = 2x + 10
    4. x = -10

    Remember that you can't do 3^(-10) on the calculator; you change that to a fraction:

    3^(-10) => 1/(3^10)
    9^(-5) => 3^(2*(-5)) => 3^(-10) => 1/(3^10)

    Again, many years since doing this; hope I'm not fucking that up.

  8. #48
    Title: "HUBBLE GOTCHU!" (without the quotes, of course [and without "(without the quotes, of course)", of course], etc)
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    Another way to do part 2 is to immediately take the log of both sides and then solve for x. You should get the same answer as Yugl got, but you'll have to use some properties of logarithms to actually get that answer. Yugl's way is better though.

  9. #49
    2600klub

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    ;_; thanks guys

  10. #50
    Title: "HUBBLE GOTCHU!" (without the quotes, of course [and without "(without the quotes, of course)", of course], etc)
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    We talk a lot about the "War On Women" and the "War on Drugs", or the wars our soldiers are fighting overseas. But there's a much bigger threat facing us today that we don't talk enough about. And that's the war that math is waging on our minds. And that's the topic of a rant I just wrote on facebook.

    It's a wall of text. Don't say I didn't warn you.

    Spoiler: show
    The unfortunate thing about math is that pathological and counter intuitive stuff is actually the most common type of stuff in our mathematics, we just have a hard time finding them because we're wired to think of objects with nice properties. Convergence is just another one of the many casualties of this war that pathological objects and paradoxes like to wage on our weak human minds. First they took away our rational numbers, which pissed Pythagoras off. Then they took away our countable sets, with Cantor's diagonal argument, which was fine for a while, except now we have Skolem's paradox, which asserts that all sets from ZF set theory ARE countable. Oh, by the way, the word "paradox" is a misnormer; all sets *really are* countable, including the reals, and this in no way contradicts the fact that our model of set theory has the uncountability of the reals as a theorem. And if you reject the axiom of choice, the real numbers is actually a countable union of countable sets. We also have weird stuff like the Weirstrauss function which is continuous everywhere but differentiable nowhere. And it turns out, "most" continuous functions aren't differentiable at all whatsoever. Our brains just aren't wired to find such pathological functions. And don't even get me started on the Cantor set, which has measure zero but is somehow uncountable. And, on the topics of measure theory and set theory, we have the whole Axiom of Choice dilemma. If we accept it, we get the Banach-Tarski paradox where you can basically chop up a pea and reassemble it so that it's bigger than the entire milky way if you wanted to. But if we reject it, we lose soooooooooooo many nice things, like every vector space having a basis (and every basis having the same cardinality. How do you even define the dimension of a vector space without those two results?). We'd also lose the Hahn-Banach theorem. These last two things would impact hilbert spaces too, so any discipline that uses Hilbert Spaces would be affected (actually, I'm not too sure if physicists would care or even how much this would affect them, but still...). The axiom of choice, by the way, implies that the real numbers are well ordered. In other words, there's some ordering in which every bounded subset has a minimum and maximum, not just an infimum and supremum. With our standard, intuitive idea of ordering numbers (i.e. the way we decide which numbers are bigger than other numbers) this is impossible, so this means our idea of what it means for one number to be bigger than another is inferior in a sense. There's a better way to order than the one we use. What does that say about humans when the way we decide if one number is bigger than the other isn't actually the best way to do it? That's such an intuitive idea that we're all born with, and we may actually be wrong about that. Which reminds me, depending on how arithmetic are formulated, there may be an infinite amount of "non standard models". It's kind of like how, for example, there are 5 groups of order 8. Well, there are multiple systems of natural numbers (an infinite amount, to be exact) if you try to formulate it in terms of first order logic (the same kind of logic that tells us that the reals are countable). So not only are there an infinite number of models of numbers other than the one we use, but our theory of numbers can't even prove its own consistency because Godel hated everyone and wanted to explode our brains. Speaking of which, there's Godel's theorem, where our whole idea of "truth" may be shaken. In some systems, if they are consistent, they must have undecidable statements. Philosophers argue about whether that's a big deal in the grand scheme of things, but if this does affect epistemology in any way, it's going to be in a negative way for sure. The fact is, math has been waging war on our intuition for as long as we've been able to think. I say it's time we fight back! I'm not sure how we wage war on abstract concepts and theories, but we should at least try! lol

  11. #51
    Title: "HUBBLE GOTCHU!" (without the quotes, of course [and without "(without the quotes, of course)", of course], etc)
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    Seriously, screw math and logic. I'm only in it to learn their weaknesses so I can strike when they least suspect it.

  12. #52
    Renegade Philosopher
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    That's the exciting part. If all of it was intuitive, there'd be no sense of excitement and discovery when some paradox or seemingly unintuitive result gets resolved or reinterpreted in a way that makes it seem like another small part of the universe has come into the human understanding.

  13. #53
    Sea Torques
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    Need help solving this please!

    EDIT: Nevermind, got it.

  14. #54
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    Lets say I have five independent binomial ratios with different sample sizes.
    * How do I determine if they my sample size is large enough for them to be worth doing statistics on and reporting?
    * How do I statistically determine how to cluster them? - Let say group 1 and 2 are not significantly different and group 2 and 3 are not statistically different, but group 1 and 3 are statistically different.
    * How does this change if they are no longer binomial and are also not normal?

  15. #55
    Siralin
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    Any tips on getting back in the swing of things with math? Took calc1 spring semester and ended with a 98 average...skip to last Thursday taking a pretest in calc 2 and I can't remember shit. Each time I'd start a problem I'd be like "Oh yeah I need to do that!"...but then whatever "that" was never totally formed in my head and I'd be left with a dead end...

  16. #56
    United States of Smash!
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    Hey guys I am taking Intro to Analysis 1 this semester and I always have a tough time with proofs so I might be asking for help. I am stuck on this totally stupid question right now and it seems so simple but I cannot figure out how I am looking at the problem the wrong way. It asks me to solve (x/(x-1))<1

    In the back of the book they give the answer as -inifnity < x < 1 but I have no idea how they got there.

    Thanks for the help.

  17. #57

    Quote Originally Posted by zoobernut View Post
    Hey guys I am taking Intro to Analysis 1 this semester and I always have a tough time with proofs so I might be asking for help. I am stuck on this totally stupid question right now and it seems so simple but I cannot figure out how I am looking at the problem the wrong way. It asks me to solve (x/(x-1))<1

    In the back of the book they give the answer as -inifnity < x < 1 but I have no idea how they got there.

    Thanks for the help.
    Are you allowed to do proofs by induction? Base case: x = 2, trivial to prove the inequality is true. inductive hypothesis: the inequality is true for x > 2. prove it's true for x + 1 using algebra. Should work out pretty simply.

    That's only half the answer, obviously. This inequality is true over the Reals, except at x = 1, so you'd have to show it holds using induction from x > 1 and from x < 1. It's the "hard" way, and I'm sure there's an easier way, but a proof is a proof (tautological proofs are superior).

  18. #58
    United States of Smash!
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    Thanks Blubbartron. That was helpful. I was thinking of just plugging values in and building a picture of the inequality that way but I figured there was a way to solve the inequality for x without plugging in values for x.

  19. #59
    Title: "HUBBLE GOTCHU!" (without the quotes, of course [and without "(without the quotes, of course)", of course], etc)
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    I'm not sure if induction is a legitimate way to solve this. The set of numbers for which our property is true, (-infinity,1), is uncountable. The properties of natural numbers that allow us to use induction do not apply here (unless you know something that I don't. But if that is the case, I'm pretty sure it would be well beyond the scope of a Real Analysis class).

    The easiest way to do this would be to solve it directly. Solve it the same way you solved equations and inequalities in algebra. Of course, you have to keep track of whether or not your operations will change the direction of the inequality (i.e. if you multiply both sides by (x-1), this will flip the inequality if x-1 <0, but not if x-1>0. So you're going to have to do two cases: x <1 and x > 1. The x=1 case is not necessary (you're going to want to state the reason for this in your answer).

    Edit: After working it out, it looks something like this

    Case 1: x < 1
    x/(x-1) < 1 iff ... iff ... ... ... iff <obviously false expression>. Therefore, this never holds for any x < 1.
    Case 2: x > 1
    x/x-1 < 1 iff ... iff ... ... ... <obviously true expression>. Therefore, this always holds for any x > 1.
    EDIT: I just realized I reversed these. Case 1 is "always holds" and Case 2 is "never holds".

    The key to realizing that this was the way to go was to notice that if you were to try to solve it the usual way (multiply both sides by x-1) you'd get different answers depending on whether or not that expression is positive or negative (because this would determine whether or not you have to flip the inequality).

  20. #60
    Title: "HUBBLE GOTCHU!" (without the quotes, of course [and without "(without the quotes, of course)", of course], etc)
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    Also, is there a way to set it up so that I get an email or something when someone posts in this thread? lol, I'm not on BG much anymore because...well, I don't know why.

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