Shadow

W/A is built upon the Mathematica language, in which # denotes the slot symbol (a placeholder if you will). From the way I interpret the syntax (which is almost certainly going to be incorrect), the # should be replaced by a one (because of the &, 1 at the end). However, that would make no sense at all, so your guess is as good as mine. I do remember having similar output from the root function in Maple, and somehow managing to expand it into something meaningful, but I still have no clue as to what it actually means. Another page you might find useful is the documentation for Root.

Very interesting, but this method only work for numbers of the form [math]10^n + x*10^{n-2}[/math] (ie. 108, 1017, 1048796...the first digit always has to be one, the second always has to be zero and the third mustn't be a zero). To extend this method to work for (almost) all numbers, we've got to modify the procedure a bit. Given a number [math] A = x*10^n + y*10^k[/math]: (Step zero: check if [math]n-k = 1[/math]; if yes, this procedure becomes somewhat lengthy and thus ineffective, as explained below. In other words, if the second digit isn't a zero, don't bother.) Step one: Add [math]A + y[/math]. Step two: Multiply the number you got in step one by [math]x[/math], and write it down. Step three: Add (as in concatenate) [math]n-k-2[/math] zero's to the end of the number you wrote down in step two. If [math]n-k-2 = -1[/math], the procedure requires modifications of an increasingly tedious nature, and quite frankly I think simple multiplication by hand would be faster at this point. Step four: Add (as in concatenate) [math]y^2[/math] to the end of the number you wrote down in step three. Example: [math]8005^2 = 64080025[/math] Step one: [math]8005 + 5 = 8010[/math] Step two: [math]8*8010 = 64080[/math] Step three: Instead of computing the equivalent of [math]n-k-2[/math] in the example above, I just look at how many zeroes there are between the digits 8 and 5, subtract one, and that's the number of zeroes I have to add (if we were dealing with the number 8050, I wouldn't add any zeroes, if with 800 005 I would add three zeroes and so on). So we get [math]640800[/math]. Step four: We add (as in concatenate) [math]5^2 = 25[/math] to 640800, which gives us [math]64080025 = 8005^2[/math]. I didn't spend that much time checking whether this procedure really works for all numbers (except the ones described in step zero), so I apologize if I overlooked something. Nevertheless, I think this is a very interesting method with practical applications, so thank you Klaplunk

Sorry, that's just a typo, the calculations were performed using the correct equation for a line. Corrected. Merged post follows: Consecutive posts mergedSorry for the bump, but this has been bugging me for a while and I'd be extremely grateful for any help.

Let h be a hyperbole given by the equation: [math]h : \frac{(x-m)^2}{a^2} - \frac{(y-n)^2}{b^2} = 1[/math] Its asymptotes are then: [math] a_{1, 2}: y-n= \pm \frac{b}{a} (x-m)[/math] We were told in our math class that every line parallel to the asymptotes of a hyperbola intersects the hyperbola in only one point. So let us define a line l which is parallel to one of the asymptotes: [math]l: y=\frac{b}{a}x + c[/math] From what I've been told, this line should intersect the hyperbola only once for every real value of c excluding the ones for which [math] l = a_{1, 2}[/math]. Let's pretend we don't know that, and try to calculate c so that the resulting line intersects the hyperbola in only one point. I'm not sure if this is the method that's generally used to calculate c, but this is what they taught us. If we substitute y in the equation of h for [math]\frac{b}{a}x + c[/math]: [math]\frac{(x-m)^2}{a^2} - \frac{(\frac{b}{a}x + c-n)^2}{b^2} - 1 = 0[/math] we are effectively finding the intersection points of h and l. If we expand the left side of the above equation, we (should) get a parametrized quadratic equation with the parameter c. Since we want the line to only have one intersection point with the hyperbola, we want the parameter to be such that there will only be one (multiple) solution, ie. the discriminant of this quadratic polynomial must be zero. So we take the discriminant D with respect to x, and solve the equation [math]D=0[/math] Now, in accordance with the fact that every line parallel to one of the asymptotes only has one intersection point with the hyperbola, the above equation should have an infinite number of solutions, excluding the one for which the line is the asymptote. Unfortunately, what happens is that the terms with [math]x^2[/math] in them cancel out, and what should have been a parametrized quadratic equation becomes a parametrized linear equation. The discriminant of a linear polynomial is equal to 1, and this is my first question: why is the discriminant of a linear equation equal to one? Regardless, the equation [math]1=0[/math] has no solutions, which basically means that there exists no c for which the line [math]y=\frac{b}{a}x + c[/math] intersects the hyperbola only once. So my question is, why doesn't this method of calculating c work when dealing with lines parallel to the asymptotes? Thanks

The title pretty much sums it up. Also, I wasn't sure where to post this, so apologies if it's not in the section it's supposed to be.

The title pretty much sums it up. If you rub a balloon against your jeans, it'll stick to a wall, but eventually fall off. Can the same thing be said about magnets?

I'm shocked something like this even made it to Yahoo!.

http://news.yahoo.com/s/time/20091111/wl_time/08599193737000

Red with a bit of white in it. Seriously though

I see. Thanks ajb

Yup, I'm almost certain that's correct. But if it was meant as an answer to my question, that's not what I asked

Strange...I didn't try putting it through any color filters, maybe I should have...but believe me, the "joke" wasn't intentional. I guess I should verify my friends claims next time

Translation: "Six questions that help identify any hidden psychological diseases a young soldier might have. If the soldier can’t see the number in one of the 6 circles on the test picture, then he may suffer from:" (original version here) Can’t see circle 1: High aggression, proneness to conflict, the recommendation is to add more physical excercise and cold showers. Can’t see circle 2: Possible low than average intellectual abilities, can’t serve with sophisticated equipment. Can’t see circle 3: Possible debauchery, soldier should get increased daily ration, should get more physical activity tasks, should not be connected to food supplies, etc. Can’t see circle 4: Possible inclination to violence, can be assigend as a leader to his unit, as he can preserve discipline. Can’t see circle 5: Possible latent homosexuality. Can be light uncontrolled accesses of attraction to the same sex. Can’t see circle 6: Possible schizophrenic tendency. Required additional inspection. Note, there are no empty circles. Here is a list of numbers in the circles: 1: 25 2: 29 3: 45 4: 56 5: 76 (according to a friend) 6: 8 As hinted above, I can't see 5. So, I guess a change of partners is in order. Not that I actually believe in the results, I was a lot more fascinated by the fact that I actually couldn't see one of the circles. I'm sure we've all stumbled upon hundreds of similar color blindness tests, but I for one have never seen one that, well, I can't see.

Formal as in treat it as I would a variable (ie.: x, y, a, b...)? I'm unfamiliar with the meaning of "formal" in this context.

I wonder if this early snowing will have any radical effect on animals that hibernate, or plant life that's used to 25 °C at these times.

But by itself (ie. not in the context of total derivatives) does it mean anything? ajb, I'm sorry, but I'm not on a level high enough to make much sense of what you wrote. However I'm eager to know the answer. Do you think you could PM me (I don't want to lead this thread off topic) a dumbed down version of what you wrote, or the same thing with quick explanations/references to pretty much everything above high school level?

I was going to put this in another topic, but since you brought it up I'll ask here. I know, or have heard, that multiplying equations by infinitesimals is frowned upon. Can you actually do what you did above? If so, why is it frowned upon? If you can't, why?

Oh I see. Thanks, good to know. EDIT: By the way, it calculates the limit correctly: http://www.wolframalpha.com/input/?i=lim%201%2Fx%20x-%3E%200&t=ff3tb01

I was wondering, is there a proof of some sort that a recursively self-improving AI can be made? If so, can it be created by us (relatively inferior being)?

The limit of 1/x as x -> 0 doesn't exist? I know 1/0 isn't defined, but I though that the limit is infinity...?

Well certainly a possibility, but it won't explain this: http://www.wolframalpha.com/input/?i=1^1%2Fx+for+x%3D0 I think it's a problem on their side. As for the spike part, you're right of course. I keep treating 1/0 as infinity.

I just tried plotting [math]1^{\frac{1}{x-0.5}}[/math] in WolframAlpha, thinking a spike function would result. Instead, I got a straight line which was defined at 0.5. http://www.wolframalpha.com/input/?i=plot+1^1%2F%28x-0.5%29 Is it a plotting error on their side, or is [math]1^{\frac{1}{0}}[/math] defined??

[math]\frac{\mathrm{d} y}{\mathrm{d}x}[/math] means "How does the output of y (a function y(x)) change when x (its only parameter) changes?" [math]\frac{\partial y}{\partial x}[/math] means "How does the output of y (a function y(x,....)) change when x (one of its parameters) changes and all of its other parameters stay the same?" As far as I know, [math]\frac{\mathrm{d} y}{\partial x}[/math] means nothing.

Well, it's snowing here in the Czech Republic. In friggin' October, it's snowing. Not two days ago I was still running around in T-Shirts, and not 10 years ago there wouldn't be any snow for at least another month and a half. Ironic, our president still denies global warming...I hope he's looking out of the window right now. What's it like at your place?

This is one of the problems in our school math contest; I found it exotic, so I though I would share. Note, I have both the solution and the process by which one arrives to the solution, so this is not homework; I can send it to anyone who requests it. I'm not posting the process here (yet) because it's in Czech and I still have to translate it. I expect to be able to post it by the end of today. Here's the problem: For a given prime number [math]p[/math], calculate the number of (all) ordered trios [math](a, b, c)[/math] from the set [math]\left\{1, 2, ..., 2p^2\right\}[/math] that satisfy the relation [math] \frac{LCM(a, c) + LCM(b, c)}{a+b}=\frac{p^2+1}{p^2+2} \cdot c[/math] where [math]LCM(x, y)[/math] is the Least Common Multiple of [math]x, y[/math]. The solution is [math]2p(p-1)[/math]. PS.: The process I'll post isn't a rigorous proof, it's more of a guide with explanations.