EPT
Happy 410th Birthday, Fermat!
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EsX_Raptor
United States2801 Posts
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ComaDose
Canada10357 Posts
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EsX_Raptor
United States2801 Posts
![[image loading]](http://i56.tinypic.com/wc0rno.png) Whoever solves that differential equation can ask the next question. Edit: This rooms needs some abstract decoration, too: ![[image loading]](http://i52.tinypic.com/10pua6f.png) | ||
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Iranon
United States983 Posts
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ComaDose
Canada10357 Posts
y(t) = (const)/t^2 + 1/12(3*t^2 - 4t + 6) next question: to what extent was Albert Einstein involved in the creation of the atomic bomb. | ||
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NeoLearner
Belgium1847 Posts
On August 17 2011 22:55 ComaDose wrote: Ans: y(t) = (const)/t^2 + 1/12(3*t^2 - 4t + 6) next question: to what extent was Albert Einstein involved in the creation of the atomic bomb. Well, E=mc2 shows the huge amount of energy that could be released from the small mass that disappears during nuclear fission. Also, he wrote a letter urging the government to develop an A-bomb before the Germans. Was that the answer you were looking for? If so, prove that e^(pi i) = -1 | ||
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ComaDose
Canada10357 Posts
On August 17 2011 23:01 NeoLearner wrote: Well, E=mc2 shows the huge amount of energy that could be released from the small mass that disappears during nuclear fission. Also, he wrote a letter urging the government to develop an A-bomb before the Germans. Was that the answer you were looking for? If so, prove that e^(pi i) = -1 I will accept that answer. I was looking more toward pointing out the lack of desire to physically take part in the construction and testing leading up to it despite being pressured to by the American physicists/government. He spent that time sailing and contemplating other theories. Should I let someone else write out Euler's theorem? EDIT: made me think of + Show Spoiler + ![[image loading]](http://www.smbc-comics.com/comics/20110518.gif) | ||
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EsX_Raptor
United States2801 Posts
e^(pi * i) = -1, it is necessary to understand what raising a number to an imaginary power means. However, according to de Moivre's formula, we four out e^(i * x) = cos(x) + i * sin(x) for all x. This illustrates how closely related the exponential function is to the trigonometric functions. It thus follows that e^(pi * i) = cos(pi) + i * sin(pi) = -1 + i * 0 = -1. Do write out the Euler theorem! | ||
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ComaDose
Canada10357 Posts
And you didn't ask another question ;p | ||
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EsX_Raptor
United States2801 Posts
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Iranon
United States983 Posts
On August 17 2011 23:15 EsX_Raptor wrote: To understand the equation e^(pi * i) = -1, it is necessary to understand what raising a number to an imaginary power means. However, according to de Moivre's formula, we four out e^(i * x) = cos(x) + i * sin(x) for all x. This illustrates how closely related the exponential function is to the trigonometric functions. It thus follows that e^(pi * i) = cos(pi) + i * sin(pi) = -1 + i * 0 = -1. Do write out the Euler theorem! For a neat bit of insight, look at the Taylor series for e^x. Replace x with iz in the e^z series. Now look at the Taylor series for cos x and sin x... On August 17 2011 23:24 EsX_Raptor wrote: In laymen terms, what is the Riemann hypothesis all about? Not hard, but not that helpful either. There's this function on the complex numbers called the Riemann zeta function. It's easy to see that the function is zero at the negative integers, but there are other zeros too. We think they all lie on the vertical line which hits the real axis at 1/2, but nobody knows how to prove it. If it turned out that this were true (Riemann's hypothesis), it would imply all sorts of useful results, most notably in number theory. | ||
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ComaDose
Canada10357 Posts
On August 17 2011 23:24 EsX_Raptor wrote: In laymen terms, what is the Riemann hypothesis all about? can you put that in laymens terms lol. should we start by describing what a 0 is ;p EDIT: i guess he^ did *the negative even integers* -correction i would have added that its somehow related to the location of prime numbers too. but you didn't ask another question. | ||
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EsX_Raptor
United States2801 Posts
Iranon, you forgot to ask your question! | ||
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Nemesis
Canada2568 Posts
fun^10 x int^40 = Ir2 | ||
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Archas
United States6531 Posts
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EsX_Raptor
United States2801 Posts
![[image loading]](http://i54.tinypic.com/2rojbco.png) http://www.youtube.com/watch?v=ATbMw6X3T40 | ||
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Iranon
United States983 Posts
On August 17 2011 23:38 EsX_Raptor wrote: It is also believed that a proof to the Riemann hypothesis would compromise internet security. Iranon, you forgot to ask your question! Oh, right! And it sure would -- for number theoretic reasons. It would allow for a fast factoring algorithm, which compromises RSA. I'll keep on the topic of fun math that a lot of people know about but relatively few actually follow. Can you turn a sphere inside out without poking holes in it or making any sharp creases? The sphere's surface can pass through itself, and you can stretch and rotate parts of it as much as you like, but you can't break it and glue it back together. If you know what the terms mean, I'm asking for a diffeomorphism between two spheres that reverses the orientation. | ||
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ComaDose
Canada10357 Posts
On August 17 2011 23:49 Iranon wrote: Oh, right! And it sure would -- for number theoretic reasons. It would allow for a fast factoring algorithm, which compromises RSA. I'll keep on the topic of fun math that a lot of people know about but relatively few actually follow. Can you turn a sphere inside out without poking holes in it or making any sharp creases? The sphere's surface can pass through itself, and you can stretch and rotate parts of it as much as you like, but you can't break it and glue it back together. If you know what the terms mean, I'm asking for a diffeomorphism between two spheres that reverses the orientation. This is a good one. I have no idea. I'm reading about Riemann and internet security and I'm getting very motivated to become involved in pure math ;p | ||
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EsX_Raptor
United States2801 Posts
On August 17 2011 23:49 Iranon wrote: Oh, right! And it sure would -- for number theoretic reasons. It would allow for a fast factoring algorithm, which compromises RSA. I'll keep on the topic of fun math that a lot of people know about but relatively few actually follow. Can you turn a sphere inside out without poking holes in it or making any sharp creases? The sphere's surface can pass through itself, and you can stretch and rotate parts of it as much as you like, but you can't break it and glue it back together. If you know what the terms mean, I'm asking for a diffeomorphism between two spheres that reverses the orientation. You can! I found a rather amazing animation of this possibility (min. 1:24): http://www.youtube.com/watch?v=R_w4HYXuo9M&t=1m25s | ||
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ComaDose
Canada10357 Posts
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intotheheart
Canada33091 Posts
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EsX_Raptor
United States2801 Posts
Assume that we have the following string: MI and that we can perform operations on it based strictly off the following rules: I. Add a U to the end of any string ending in I. For example: MI to MIU. II. Double any string after the M (that is, change Mx, to Mxx). For example: MIU to MIUIU. III. Replace any III with a U. For example: MUIIIU to MUUU. IV. Remove any UU. For example: MUUU to MU. Is it possible to obtain MU? If you find an answer to this, please post it in spoilers. I'm still working on it! Edit: New page: More decoration: ![[image loading]](http://www.mathfail.com/optical-illusion2.jpg) | ||
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SecondChance
Australia603 Posts
 I understand to an extent, but it looks like this: <|understood|-----lolwut---------------------------------------------------------------------------------------------------------|> | ||
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ComaDose
Canada10357 Posts
On August 18 2011 00:37 SecondChance wrote: + Show Spoiler + All those who didn't go so well in highshcool and have no fucking idea what is going on in this thread yet wished they understood say "I". I understand to an extent, but it looks like this: <|understood|-----lolwut---------------------------------------------------------------------------------------------------------|> You wont learn this in any high-school I've heard of. I spoiled myself with a solution to the MI string from google lol . I want more decoration! | ||
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EtherealDeath
United States8366 Posts
On August 18 2011 00:35 EsX_Raptor wrote: Let's do the good-old MU Puzzle: Assume that we have the following string: MI and that we can perform operations on it based strictly off the following rules: I. Add a U to the end of any string ending in I. For example: MI to MIU. II. Double any string after the M (that is, change Mx, to Mxx). For example: MIU to MIUIU. III. Replace any III with a U. For example: MUIIIU to MUUU. IV. Remove any UU. For example: MUUU to MU. Is it possible to obtain MU? If you find an answer to this, please post it in spoilers. I'm still working on it! Edit: New page: More decoration: + Show Spoiler [solution] + We approach this algebraically. Suppose we have n Is, and we want to reduce them to a single U. We approach this instead as k+3 I's, since the first 3 I's we can reduce to a U. The question is then, given what k can we completely erase the string. The only way we can remove strings is to remove UU. UU is produced by k=6. Thus, in order for the string subsequent to M to be reducible to U, it must consist of 3+6n I's, for some n. At the same time, we can only generate powers of 2 for the number of I's. Thus 3+6n = 2^m for n,m, which is so say we want an m such that 2^m - 3 ≅ 0 mod 6. 2^1 ≅ 1 mod 6 2^2 ≅ 4 mod 6 2^3 ≅ 2 mod 6 2^4 ≅ 4 mod 6 etc we have hit a cycle Therefore it is impossible to produce MU. Other cases ignored because it is clearly impossible to reduce them. edit - Aw fuck it according to wiki I could have just used invariants LOL so much simpler fml. | ||
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EsX_Raptor
United States2801 Posts
On August 18 2011 01:57 EtherealDeath wrote: + Show Spoiler [solution] + We approach this algebraically. Suppose we have n Is, and we want to reduce them to a single U. We approach this instead as k+3 I's, since the first 3 I's we can reduce to a U. The question is then, given what k can we completely erase the string. The only way we can remove strings is to remove UU. UU is produced by k=6. Thus, in order for the string subsequent to M to be reducible to U, it must consist of 3+6n I's, for some n. At the same time, we can only generate powers of 2 for the number of I's. Thus 3+6n = 2^m for n,m, which is so say we want an m such that 2^m - 3 ≅ 0 mod 6. 2^1 ≅ 1 mod 6 2^2 ≅ 4 mod 6 2^3 ≅ 2 mod 6 2^4 ≅ 4 mod 6 etc we have hit a cycle Therefore it is impossible to produce MU. Other cases ignored because it is clearly impossible to reduce them. That's what I thought. Whoever made that puzzle made me spend quite some time trying to solve the unsolvable. lol By the way, it's your turn to ask us a question! | ||
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EtherealDeath
United States8366 Posts
On August 18 2011 02:05 EsX_Raptor wrote: That's what I thought. Whoever made that puzzle made me spend quite some time trying to solve the unsolvable. lol By the way, it's your turn to ask us a question! Hmm... Explain why the following scenario is true. Suppose we have two entangled particles, let's say their state is a|1>|0>+b|0>|1> to be simple, where the values of a and b are not particularly important save that their squares sum to 1, but let's use one of the Bell States, that is a=b=1/sqrt(2). Now we measure one of the particles to determine it's actual state. We know that instantaneously, the state of the other particle is set. However, it is impossible to determine any classical information from this instantaneous effect - i.e., cannot gain any physical information. Why? Or rather, what information do you think you could gain that you didn't have before? | ||
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TL AntiHack
39 Posts
On August 18 2011 00:21 IntoTheheart wrote: Best blog ever. Just sayin'. | ||
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EtherealDeath
United States8366 Posts
On August 18 2011 02:17 TL AntiHack wrote: 5/5, you sir deserve applause. Thank you for making my day better. hax. | ||
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Pengtoss
207 Posts
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Aletheia27
United States267 Posts
On August 17 2011 23:49 Iranon wrote: Oh, right! And it sure would -- for number theoretic reasons. It would allow for a fast factoring algorithm, which compromises RSA. I'll keep on the topic of fun math that a lot of people know about but relatively few actually follow. Can you turn a sphere inside out without poking holes in it or making any sharp creases? The sphere's surface can pass through itself, and you can stretch and rotate parts of it as much as you like, but you can't break it and glue it back together. If you know what the terms mean, I'm asking for a diffeomorphism between two spheres that reverses the orientation. I believe you can although it hinges on the axiom of choice... EDIT: I don't remember exactly why though... Guess I wasn't meant to be a math major | ||
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Primadog
United States4411 Posts
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Ruyguy
Canada988 Posts
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Kukaracha
France1954 Posts
But then I grew up and found out that I missing out a whole beautiful world... I don't know if I'll ever dedicate myself to it but I'd like to give a try oince more, by myself this time. | ||
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ComaDose
Canada10357 Posts
I only took first year chemistry and promptly forgot everything involved. I only mention chemistry becuase i only vaguely recognize the syntax involved in "a|1>|0>+b|0>|1>" as from that class moar math and physics! I propose a new question: "whoes 410th birthday is it and what is he famous for?" should be easy enough. | ||
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Aletheia27
United States267 Posts
On August 18 2011 04:06 ComaDose wrote: EtherealDeath's question was too hard! I only took first year chemistry and promptly forgot everything involved. I only mention chemistry becuase i only vaguely recognize the syntax involved in "a|1>|0>+b|0>|1>" as from that class moar math and physics! I propose a new question: "whoes 410th birthday is it and what is he famous for?" should be easy enough. It's a quantum physics problem actually.... but i don't want to answer this cause I don't havea good question  | ||
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EsX_Raptor
United States2801 Posts
By the way, I have been practicing writing trivial proofs and decided to give proving the MU puzzle a try: ![[image loading]](http://i56.tinypic.com/20kf7m0.png) I most likely (and as per-usual) left out something critical. | ||
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ComaDose
Canada10357 Posts
On August 18 2011 04:13 Aletheia27 wrote: It's a quantum physics problem actually.... but i don't want to answer this cause I don't havea good question You should answer and give a lame question then.... now I'm googling cause I feel ignorant. Its pretty much your calling to continue this thread. EDIT: Your proof is true but a little.... shallow, like i mean just missing some whys and therefore... pretty much exactly like a professional would do it good job ;p randall from xkcd helped me understand it better ;p linky EDIT: can your question be rewritten as this? + Show Spoiler + Suppose we have two entangled particles. Now we measure one of the particles to determine it's actual state. We know that instantaneously, the state of the other particle is set. However, it is impossible to determine any classical information from this instantaneous effect - i.e., cannot gain any physical information. Why? Or rather, what information do you think you could gain that you didn't have before? i.e. is the answer your looking for true for all entangled particles? | ||
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EsX_Raptor
United States2801 Posts
From now on, I will try a mathematical approach to every puzzle I encounter to save myself the potential time-loss of trying one with no solution. ;p ![[image loading]](http://i56.tinypic.com/2ccxcfc.png) M.C. Escher And someone still needs to answer ComaDose's question! On August 18 2011 04:06 ComaDose wrote: I propose a new question: "whoes 410th birthday is it and what is he famous for?" | ||
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EtherealDeath
United States8366 Posts
You know that if one measures Particle A to be in state 1, then B is in 0. If A is measured to be in state 2, then B is in 1, and conversely if you measure B, then A is immediately set if it has not already been set by measurement. Some have thought this could be used for FTL (faster than light) communication, since the "information transfer" is instantaneous. However, this is not the case. In short, what does measuring your particle tell you that you didn't already know? | ||
]343[
United States10328 Posts
On August 17 2011 23:15 EsX_Raptor wrote: However, according to de Moivre's formula, we four out e^(i * x) = cos(x) + i * sin(x) for all x. This illustrates how closely related the exponential function is to the trigonometric functions. rawrgh that's not De Moivre De Moivre can be proven elementarily, and states that (cos n*x + i sin n*x) = (cos x + i sin x)^n for positive integer n. The fact that cos x + i sin x = e^(i*x) is deeper and requires that you do the Taylor series thing. Since I haven't had enough quantum yet to think about entanglement, here are a few cute combinatorics problems: 1. For n a positive integer, let A_1, A_2, ... A_{n+1} be distinct subsets of {1, 2, ..., n}, each containing exactly 3 elements. Show that some two of these subsets have exactly one common element. 2. Show that any n points, not all collinear, determine at least n distinct lines. + Show Spoiler [hint] + linear algebra! edit: http://www.teamliquid.net/forum/viewmessage.php?topic_id=246103¤tpage=4#61 on entanglement  | ||
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EtherealDeath
United States8366 Posts
The key is that if you already knew the state, it is impossible to determine whether you did the first measurement or they did. You don't actually gain any information, classically speaking. edit - How do you do the n points problem by linear algebra? I can think of a proof by induction, which tbh was the immediately obvious method for me. Not a clue atm how to proceed with linear algebra though, maybe cause I haven't used it in a while lol. | ||
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Foolishness
United States3044 Posts
On August 18 2011 11:48 EtherealDeath wrote: Well, Nawyria does answer the question in that thread you linked lol. The key is that if you already knew the state, it is impossible to determine whether you did the first measurement or they did. You don't actually gain any information, classically speaking. edit - How do you do the n points problem by linear algebra? I can think of a proof by induction, which tbh was the immediately obvious method for me. Not a clue atm how to proceed with linear algebra though, maybe cause I haven't used it in a while lol. Likewise I think I can do the proof using graph theory and k-partite sets (proof by induction seems much simpler though ), but I can't think of a linear algebra and now I'm probably going to lose sleep over it. | ||
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Antifate
United States415 Posts
On August 18 2011 10:36 ]343[ wrote: rawrgh that's not De Moivre De Moivre can be proven elementarily, and states that (cos n*x + i sin n*x) = (cos x + i sin x)^n for positive integer n. The fact that cos x + i sin x = e^(i*x) is deeper and requires that you do the Taylor series thing. Since I haven't had enough quantum yet to think about entanglement, here are a few cute combinatorics problems: 1. For n a positive integer, let A_1, A_2, ... A_{n+1} be distinct subsets of {1, 2, ..., n}, each containing exactly 3 elements. Show that some two of these subsets have exactly one common element. 2. Show that any n points, not all collinear, determine at least n distinct lines. + Show Spoiler [hint] + linear algebra! edit: http://www.teamliquid.net/forum/viewmessage.php?topic_id=246103¤tpage=4#61 on entanglement Just to comment on Euler's famous equality, one doesn't need to do the Taylor series, just basic derivation works (and the proof is a little sexier). I could be wrong though. Anyway, I've always thought of it like this. f(x) = e^(i*x) Break up the function into its real and imaginary components. f(x) = g(x) + i*h(x) Differentiate f ' (x) = i*e^(ix) = i*f(x) f ' (x) = g ' (x) + i*h ' (x) Equate things f ' (x) = i*f(x) = i*g(x) - h(x) = g ' (x) + i*h ' (x) Equate the real and imaginary parts of f ' (x) g ' (x) = - h(x) h ' (x) = g(x) We end up with two equations about the components functions. h(x)'s derivative is g(x) and g(x)'s derivative is the negative of h(x). So h(x) is the sine function and g(x) is the cosine function. So e^(ix) = cos(x) + isin(x). If x is pi, e^(ix) is negative one (cos(pi) is -1 and sin(pi) is 0). Hopefully this is settled without nasty sigmas! As for your first problem, it's kind of late, so I could be completely wrong, but I'm not getting it. If you want to have multiple subsets, n has to be greater than 3. So let's say n = 4. The subsets are {1, 2, 3}, {1, 2, 4}, {1, 3, 4}, and {2, 3, 4}. There are no two sets here that share only one common element. If n were 5 however, it works because there are 5 unique elements that two sets of 3 can share with one (and only one) in both. So this seems to work for n > 4, or well, any n such that there are 5 unique elements in them? This seems a little simple and I'll probably have to edit this out later in shame. | ||
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NeoLearner
Belgium1847 Posts
On August 17 2011 23:26 Iranon wrote: For a neat bit of insight, look at the Taylor series for e^x. Replace x with iz in the e^z series. Now look at the Taylor series for cos x and sin x... I was looking for the Taylor series actually. Writing it out for e^x, filling Pi*i, re-ordening the terms and getting cos(x) + i * sin(x). I was amazed the first time I did that. EDIT: http://www.math.toronto.edu/mathnet/questionCorner/epii.html | ||
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Jumbled
1543 Posts
On August 18 2011 11:48 EtherealDeath wrote: Well, Nawyria does answer the question in that thread you linked lol. The key is that if you already knew the state, it is impossible to determine whether you did the first measurement or they did. You don't actually gain any information, classically speaking. The problem is with the way you stated your scenario. It's quite correct to say that entanglement cannot be used to instantaneously transmit classical information, but that wasn't what you described, and you didn't even mention spatial separation or a second experimenter in your original post. On August 18 2011 02:14 EtherealDeath wrote: Hmm... Explain why the following scenario is true. Suppose we have two entangled particles, let's say their state is a|1>|0>+b|0>|1> to be simple, where the values of a and b are not particularly important save that their squares sum to 1, but let's use one of the Bell States, that is a=b=1/sqrt(2). Now we measure one of the particles to determine it's actual state. We know that instantaneously, the state of the other particle is set. However, it is impossible to determine any classical information from this instantaneous effect - i.e., cannot gain any physical information. Why? Or rather, what information do you think you could gain that you didn't have before? It is fair to say that no net information is gained in this operation, but only because the particles are in a known, pure state both before and after the measurement. | ||
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ComaDose
Canada10357 Posts
woooooo! + Show Spoiler + Show that any n points, not all collinear, determine at least n distinct lines. People have said that it was possible by induction but have not done so. I interpret the question to mean that a distinct line must only cross 2 points. This is logical becuase the n points are not collinear. At least n lines are possible becuase there is a line from one of the points to all other points. In the case that there are collinear points that lie on one of these lines preventing you from finding n distinct lines immediately, one must simply select these "wasted/overlapped" points as the new "starting point" and revisit the other points to make new lines. I couldn't figure out how to prove a general case with linear algebra aha :S All the reading about twin paradox etc. was fun. | ||
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