cosine

I'm taking a Transformations & Geometries course right now, where the professor is exploring Euclidean geometry, and then is going to build towards hyperbolic geometry. He eludes frequently to a "point of infinity" in hyperbolic geometry where parallel lines intersect. Can this be thought of as giving something a value of infinity?

Sorry, even though the answer is correct, there is a mistake in the work. [math] \cos(xy)=y [/math] [math]-\sin(xy)(y)dx - \sin(xy)(x)dy = dy [/math] [math]-\sin(xy)(y)dx = dy(1 + \sin(xy)(x))[/math] [math]\frac{dy}{dx} = \frac{-\sin(xy)(y)}{1 + \sin(xy)(x)}[/math] That is the derivitive, though for this particular question it just happens to give you the same answer. Hope this helps.

Another way (at least the way I'm looking at it) to show why you can't do this, is because your going from something with 4 dimensions (4 variables) to 3 dimensions (3 variables) which you can do, but then trying to go backwards. I would just like to let you realize this intuitively by imagining projecting every point on a plane onto a line. This shouldn't be too hard, but what about projecting every point on a line onto a plane? The projection could only make a line. To make more you would be inserting new information. Hope this helps.

Yeah its really nifty. I couldn't give you the proof off hand, but you can verify that it seems to work. For a proof you could check Euclid's number theory chapters in the elements. Internet search Euclid's Elements, then somewhere from books 7-10 I think. Or you may be able to search "Euclidean algorithm" for a proof. But in any case it works.

have you heard of the Euclidean algorithm? That would give you the greatest common factor of two numbers a and b. The algorithm basically says to subtract the lesser of the two numbers from the bigger number. Compare this difference to the lesser of the original numbers, subtract the lesser of those two from the larger, etc. Just don't subtract so much that you wind up with a 0 or negative number. So once you've gotten this number, that is the greatest common factor. Example, find the GCF of 23 and 81: 1) 81-23 = 58 2) 58-23 = 35 3) 35-23 = 12 4) 23-12 = 11 5) 12-11 = 1 6) 11-1 = 10 7) 10-1 = 9 ... 15) 2-1 = 1 16) 1-1 = 0 So the greatest common factor is 1. Here is an example of numbers that aren't coprime: 16 and 56 1) 56-16 = 40 2) 40-16 = 24 3) 24-16 = 8 4) 16-8 = 8 5) 8-8 = 0 So the greatest common factor is 8. Hope this helps.

I'm pretty sure the op meant is there any perfect square with a 2, 3, 7, or 8 in the units place a.k.a. the ones place. As for the question itself, imagine doing the calculation by the "long multiplication" method. Consider doing it with both numbers being equal. We can agree that the units digit of the result must be equal to the units digit of the product of the units digit of the number being squared. Since there are only 10 possible digits in the units place of the number to be squared, there are a finite amount of possiblities to consider. Namely, the squares of the 10 digits, namely: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81. The units digits of these numbers are only 0, 1, 4, 5, 6, and 9. Thus showing that there are no perfect squares with a units digit of 2, 3, 7, or 8. Hope this helps.

Haha, you mean the mathematical formula shouldn't just assume that you're a point?

Um....... I'm trying to resist why........soo...ahhhhhhhh....ummmm....errrrrr... for what reason?

If you assume that the rain is randomly distributed, then I would expect that your chances of being rained on don't change whether you are moving or still. I mean this is assumming random distribution though, otherwise if some places have less rain on then than others, then the trivial solution would be to go and stay where the least rain is. If the place of least rain is changing, but the position of the rain is moving to discrete points and not moving along a continuous line, I think that would be a pretty interesting problem.

Hello, over the summer I was checking out this infinite series, the alternating harmonic series, and was able to reach two contradictory answers. Let the sum of the series be S: [math]S = \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + - +...[/math] Group the terms in to the odd and even fractions: [math]S = \left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} +...\right)[/math] Now by the distributive property, pull out a half: [math]S = \left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) - \frac{1}{2}\left(\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} +...\right)[/math] Group the subtracted group now into odds and evens: [math]S = \left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) - \frac{1}{2}\left[\left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) + \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} +...\right)\right][/math] Now seperate those groups: [math]S = \left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) - \frac{1}{2}\left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) - \frac{1}{2}\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} +...\right)[/math] Adding the odds together, you get: [math]S = \frac{1}{2}\left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) - \frac{1}{2}\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} +...\right)[/math] Multiply both sides of the equation by 2: [math]2S = \left(\frac{1}{1} + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} +...\right) - \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \frac{1}{8} +...\right)[/math] Combining this equation and the first equation, we deduce that: [math]2S = S[/math] [math]S = 0[/math] However, there are arguements that the sum of the alternating harmonic series is [math]ln(2)[/math]. If you know taylor series, you can verify that. Here is another arguement: [math]f(x) = \frac{x^{1}}{1} - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \frac{x^{5}}{5} -+...[/math] {1} If we can find the value for [math]f(1)[/math] then our problem will be solved. Differentiate [math]f(x)[/math] so: [math]f'(x) = 1 - x^{1} + x^{2} - x^{3} + x^{4} - x^{5} +-...[/math] {2} Find [math]xf'(x)[/math]: [math]xf'(x) = x^{1} - x^{2} + x^{3} - x^{4} + x^{5} -+...[/math] {3} Adding {2} and {3} together, all but the first 1 in {2} negate each other: [math]f'(x) + xf'(x) = (1+x)f'(x) = 1[/math] {4} Dividing both sides of {4} by the [math](1+x)[/math] term: [math]f'(x) = 1/(1+x)[/math] {5} Now integrate both sides of equation {5}: [math]f(x) = ln(1+x) + C[/math] {6} You can deduce that [math]C = 0[/math] by stating [math]f(0) = 0[/math], as known from {1}, thus: [math]f(x) = ln(1+x)[/math] {7} Thus [math]f(1) = ln(1+1) = ln(2)[/math] Though there are several proofs to show that the series adds up to [math]ln2[/math], why would the first proof I showed you that it equals 0 be wrong?

Here's a good source for competition problems: http://www.claymath.org/millennium/ (Haha, just kidding, I'll try to message you some.)

Not to contradict Mot, but I would like to suggest that you think about exploring some non-analysis math too. This isn't the order that I've fallen in love with math, but since you're already fairly deep into it, I would suggest you check out some things like elementary number theory, or combinatorics. There are many things in these topics you can teach yourself. Also, it may sounds like a dry suggestion, it gets fun to sort of read your way through Euclid's Elements. Its especially fun if you're reading one with a good commentary, like Sir Thomas Heath's translation. There are also translations on the interenet, with commentary. I suggest you search for "Euclid's Elements" and I'm sure you'll find something. Otherwise I would also suggest Arther T. Benjamin's "Proofs That Really Count." I saw him talk this summer at Math Fest, it was amazing. I would also suggest any of William Dunham's books. This should provide you with many things that you will definitely find interesting, as well as leading you to new topics. Good luck, and remember, have fun!

I think this is an interesting question, at the very least for aesthetic purposes. Geometric proofs can be very pretty... "proofs without words." The only problems with them is that they are really only good when working up to powers of 2. At least Plane geometry. Sure you can get higher dimensional geometry, but looking at an n-dimensional proof gets less intuitive after a while. So while geometric proofs are cool, you'll probably have to rely on some symbolic representation after a while, which is cool in itself. One of the most interesting things I like to look at is the "Geometric Algebra" outlined throughout Euclid's elements. Such that algebraic expressions were represented and solved using geometrical methods. The Greeks mostly preferred to represent their expressions geometrically after Hippocrates demostrated the irrationality of the squareroot of 2. This was seen as a difficiency in algebra and arithmetic, and it was figured that geometry was superior because it had no problem representing the squareroot of 2 as the hypoteneuse of an isoseles right triangle.

<op> Haha wow, that is nifty. Yeah now that Fermat's Last Theorem is proved, I suppose we can just start using it now...

Yes, it is correct that only perfect squares will be on. The reasoning follows this way. We can agree that if a bulb has been flipped an even number of times, it will be off. And If a bulb has been flipped an odd number of times, it will be on. Now whenever starting from a bulb that is a Factor of the Nth bulb, the Nth bulb will be flipped that turn. So the problem is equivalent to finding what numbers have an even number of factors, and which have an odd number of factors. Consider a number N, if a is a factor of N, then there exists a number b such that ab = N, by definition of a factor. So that means that for every factor, there exists another factor in the set of factors of N. Consider N such that that for all a that are factors of N, the b for which ab = N is not a. Thus for every factor a, there is another factor. Thus, there is an even number of factors. Consider N such that there exists an a in the set of of factors of N, such that ab = N where b is a, or in other words that a^2 = N. For now, assume that there is only one such a (to be proven later). Now if you remember from the previous case, the number of all factors of N besides a is even. Thus including a would lead to an odd number of factors. Now, if our assumption is true, then we have proved what we wanted to. To proove there is at most one a such that a*a = N, assume the opposite. That is, assume there is distinct A and B such that there squares are equal to N, with A being the greater. Let B be represented by A - C. Then A*A = (A - C)(A - C). Take the square root of both sides, A = A - C. But this is true only if C = 0. But then B = A, contradictory to hypothesis. Thus there is no more than 1 a such that a*a = N. So all of this demonstrates mathematically why only the square numbers are on.

Darn, I just wrote a long post and accidentily went back a page, and the post cleared. I have a few important points to say. 1) I'm sorry about the long list of numbers. I'm not here to be a jerk. I was hoping that getting through more numbers would get us to numbers in which new techniques would need to be found. 2) God knows what I was thinking last night in my tired stupor that 4^3 = 128. So that eliminates many of my answers. I'm sure they could be easily found by people here smarter than me. 3) F_n questionable? Its a mathematically defined function. Should the rules only allow continuous functions? 4) I'd like to say again I apologize for offendable behavior, thats not my intention whatsoever. That said, it can't help for me to linger on it. So I'll move on. 5) Since we are questioning functions, we need a lot more rigourous definition of what is acceptable. For instance, sqrt(x) is 4^(1/2) so there is an implied 1 and 2. Is cuberoot acceptable? Quadroot? Quintroot? Etc.

sin^-1[cos(sqrt(4))] + 4! + 4 + F_4 = 119 where F_n is the nth Fibonacci number. (4 + 4/4)*4! = 120 4^F_4 - sqrt(4)*F_4 = 121 sin^-1[cos(sqrt(4))] + 4! + sqrt(4)*F_4 = 122 (4 + 4/4)! + F_4 = 123 does that 3 post rule also apply to posts that are answers? (4 + 4/4)! + 4 = 124 4^F_4 - F_(sqrt(4)*sqrt(4)) = 125 4^F_4 - 4/sqrt(4) = 126 4^F_4 - 4/4 = 127 (sqrt(4)*sqrt(4))^F_(sqrt(4)*sqrt(4)) = 128 4^F_4 + 4/4 = 129 4^F_4 + 4/sqrt(4) = 130 (sqrt(4)*sqrt(4))^F_4 + F_4 = 131 (sqrt(4)*sqrt(4))^F_4 + 4 = 132 4^F_4 + sqrt(4) + F_4 = 133 4^F_4 + sqrt(4)*F_4 = 134 4^F_4 + 4 + F_4 = 135 4^F_4 + 4 + 4 = 136 4^F_4 + (F_4)*F_4 = 137 sqrt(4) * [F_4 * 4! - F_4] = 138 F_(F_4 * 4) - sqrt(4) - F_4 = 139 F_(F_4 * 4) - sqrt(4)*sqrt(4) = 140 F_(F_4 * 4) - F_(sqrt(4) + sqrt(4)) = 141 F_(F_4 * 4) - 4/sqrt(4) = 142 ...yeah, its kinda late, I'll chill for a bit. Goodnight!