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psi20

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Everything posted by psi20

  1. psi20

    x to the xtreme

    Ah cool. Is the method to find x described in my initial post a valid one?
  2. Putting powers upon powers aren't taught in school. But this problem is interesting. The problem is to find x. x^x^x^x^x^x... = 2. Find x. This was in the book The Art of the Infinite. Now the question is whether or not there is such a number x. If there was, it'd have to be greater than 1. But it'd have to be less than 2. The idea is that because x^x^x^x^x... = 2, it is the same as x^x^x^x... = 2 and so you can have x^2 = 2, and x is the square root of 2. Now the steps seemed logical to me, but I was skeptical so I tried to check it on the calculator. But x can't be the square root of 2 because this diverges. I tried it on the calculator different ways, pressing Ans ^ Ans and then going step by step, but it still showed that it diverged. So the question still remains. What is x? I played on the calculator a bit and saw something interesting. sqr rt 2 to sqrrt 2 to sqr rt 2. cube root of 3 of to the cube root of 3 to the cube root of 3 to the cube root of 3. and so on. Interesting to look at, but I didn't make much sense of it. Oh well.
  3. You can't solve this by hand (perhaps you could do it in your head). You can use Newton's Method, which requires some calculus and takes a long time to do. http://tutorial.math.lamar.edu/AllBrowsers/2413/NewtonsMethod.asp is a link to that. BhavinB used matlab to solve it. I don't know how that program works. I used a graphing calculator, graphed the function and pressed a few buttons to find when the graph becomes 0. The function above is in y. This can get confusing, because you need to put "Y^4 - 12Y^3 - 328Y^2 + 4800Y - 14400 = 0" as "X^4 - 12X^3 - 328X^2 + 4800X - 14400 = 0" into the calculator and solve for the zeros of the function. There are several roots but only one of the roots makes sense. The equation should really be (Y^4 - 12Y^3 - 328Y^2 + 4800Y - 14400) / (y - 6)^2 = 0 . The roots are the same, but the graphs are different.
  4. It'd be the other way around. X ~ 9.04 ft and Y ~ 17.84 ft. The top of the ladder would be 9.04 ft from the ground. If you need to calculate the opposite and adjacent sides, you need to have an angle as a reference. But anyways, you got the numbers. Or do you need to find the opposite and adjacent angles?
  5. I didn't write the puzzle! Don't blame me!
  6. [hide] Imagine the the box flattened so that the four 30' x 12' walls line up, connected by the 30' sides. One 12' x 12' wall will be connected to one of the four 30' x 12'. The last wall would be connected on the side two walls down and on the opposite side of the 30' x 12' wall. I'd draw a picture but I don't think that a spoiler would work on that. But imagine folding the box up again and placing the spider and fly where they're supposed to be. When you plug the numbers in, use the Pythagorean theorem, the answer is 40'. One leg of the triangle is 32 (1' + 1' + 30') and the other leg is 24 (12' + 6' + 6')[/hide]
  7. Hehe. This is supposed to be a more mathematical puzzle than a logic one. I didn't figure this one out either. It was in a book. I thought that the method to find the answer was really interesting, so I thought I'd post the puzzle here. I'd post the answer, but someone may want to figure it out first. How do you do the spoiler thing and black out the answer?
  8. Good question. I don't think so. Dudeney never knew about those.
  9. That's a good answer. Now suppose that the only way the spider can move is if it crawls. Fragile little thing can't jump for fear of breaking a leg.
  10. This puzzle was from a famous puzzle maker Dudeney. Don't give it away if you know the answer In a rectangular room, 30' x 12' x 12', a spider is at the middle of an end wall (one with dimensions 12' x 12'), one foot from the ceiling. The fly is at the middle of the opposite end wall, one foot above the floor. The fly is so scared it can't move. What is the shortest distance the spider must crawl in the room to capture the fly? No web spinning. Hint: It is less than 42 '.
  11. What would you teach, how would you teach it, and why these ways?
  12. Ah yeah, now it makes sense, thanks.
  13. I have little background in number theory, groups, conditions, and stuff like that. So I got a book called Teach Yourself Mathematical Groups and this is one of the examples. Prove that a necessary and sufficient condition for a number N expressed in denary notation to be divisible by 3 is that the sum of the digits of N is divisible by 3. I see the proof in the book, but I can't get it. It shows what denary notation is, decimal notation written out like 1x10^4 + 2x10^3 ... Let N= a 10^n + b 10^(n-1) + ... + z (The book uses subscripts instead of different letters for the digits a, b, ..., z) The proof says: If 3 divides N, then 3 divides a 10^n + b 10^(n-1) + ... + z . The part I don't understand is 'For all powers of 10, dividing by 3 gives a remainder of 1. So the remainder when N is divided by 3 is a + b + ... z.' How did that work?
  14. I tried a search on the forums and didn't find anything. So what kind of jobs are there? Just as a brainstorm. I'm not particularly looking at math applications in physics or geology, but more math-focused. Also, when you find discover something in math, what do you do?
  15. Now, I remember some time within the last three years that one of my teachers in high school said that the number of atoms in the universe is about 10^? . I've forgotten what the number was. Is there an absolute number of atoms in the universe? I thought atoms decay and split apart. Although I suppose if a fission occurred for every fusion, there would be an absolute number. I'm also curious to know how this giant number could be obtained. I was reading Archimedes' The Sand Reckoner and it reminded me of the day my teacher told us the number of atoms in the universe.
  16. That was an interesting problem. I tried RyanJ's way and some interesting things came up. If you try that for 3, the last digit is 1 for exponents in the form 4n (where n is an integer greater than or equal to 0), 3 for (4n +1), 9 for (4n + 2), and 7 for (4n + 3). For 7, they are 1, 7, 9, and 3. It seems like for positive integer powers, the last digit always repeats itself after 4 or less terms upon increasing the power by 1. The reason why they always repeat is because all you're doing is multiplying over and over again. There are only 10 digits (in our counting system). So once you multiply one number by 3 for example, some where along the line you will have to multiply by that number again, getting the same result. So the last digits repeat. Why they repeat after 4 or less terms I haven't looked into. Problem Bb is the easiest so we'll start with that. The second to the last digit also repeats itself. For 7, it repeats like this 01 07 49 43 7^0=01 7^1=07 7^2=49 7^3=43 7^4=01 7^5=07 7^6=49 7^7=43 and so on. For the same reason as stated above, they repeat in this manner. So the last two digits of 7^1994 is 49, because 1994 is in the form (4n + 2). For problem A, doing the same thing is harder, but it works. I start with 3^0 and go across 3^1, 3^2, 3^3, then 3^4 is lined up vertically with 3^0 and so on. 01 03 09 27 81 43 29 87 61 83 49 47 41 23 69 07 21 63 89 67 and after this it repeats. So instead of (4n + C) , we're looking for (20n + C) For 3^1994, the last two digits will be 69. From this, problem C will be 18 after 69 + 49 = 118 Problem D will be 80. It becomes A49 - B69 = C0 where A and B are everything in front of the numbers, and C is the second to the last digit. Seeing that 7^1994 is bigger than 3^1994, you can "borrow" from A and you get 80.
  17. I think you can only factor it once with (x-2) and the other factor can't be factored.
  18. [math] \int{(32-x^{5})^{(.2)}dx} [/math] How do you do that and what integration method(s) did you use?
  19. Hey. I can't understand why this is weird. I'm not familiar with the math notation so I'll explain it in words, it's not hard to explain. Take the derivative of ln(5 x^4). You get 4/x If you take the integral of 4/x, you get 4 ln |x| + C or ln (x^4) + C. But this is different than what the original function was, and that's what's confusing me.
  20. Given the ratio of a function to its derivative, is it possible to find the function?
  21. I don't know what you mean. Is a first order equation a linear equation? And is k a fraction between 0 and 1? Edit: Whoops, I realize why I have no idea what you're talking about. I thought this was math, but you're talking about chemistry.
  22. Sorry, my wording is confusing. First you'll need to know what the chain rule is. You might've been taught it a different way than me, although the general concept will be the same. Derivative of the outside times the derivative of the inside was how I was taught it. It seems like you're familiar with it. Secondly, you'll need to know how to differentiate inverse tan(x). The derivative of that is [math]\frac{1}{x^2 +1}[/math] Normally, if a problem required you to differentiate inverse tan(x^2), you'd use the chain rule once. So it'd be [math]\frac{1}{\left(x^2 \right)^2 +1} * 2x[/math] Hehe, let me try out latex. Anyways, you would differentiate inverse tan(x^2) that way only if x was an independent variable. However, this is a related rate problem and it says that x is dependent on time. That means we'll have to differentiate x, which is [math]\frac{dx}{dt}[/math] and multiply it to the derivative up there (chain rule). So in the end, what you end up with is [math]\frac{1}{\left(x^2 \right)^2 +1} * 2x * \frac{dx}{dt}[/math]
  23. Actually, I think all this is fantasizing is improbable. I need to brush up on some basics of the Games Factory and see what I can do. The Games Factory can do 4 players, 2 on joysticks, one on the keyboard, and one on the mouse. Whether or not a laptop can hold the two joysticks is now my concern and how much the 2 button joysticks will cost me.
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