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Everything posted by Manifold

  1. Hi! I've got a question concerning neighbourhoods of points in 2- and 3-dimensional space. How can we explicitly show, using only the definition of a given metric, that the according neighbourhood is some figure? For example the three metrics are given: 1) [math]d(x,y)=\sqrt{\sum_{i=1}^{n}{(x_i-y_i)^2}}[/math]; 2) [math]d(x,y)=\displaystyle\max_i|x_i-y_i|[/math]; 3) [math]d(x,y)=\sum_{i=1}^{n}|x_i-y_i|[/math] If [math]a\in{\mathbb{R}}^2[/math] and the metric is 1), then it's clear: [math]d(a,x)=\sqrt{(a_1-x_1)^2+(a_2-x_2)^2}<\epsilon[/math] and it's clear that it's a circle...because one can rewrite:[math](a_1-x_1)^2+(a_2-x_2)^2<{\epsilon}^2[/math] But I have problems to see a picture when looking at the other metrics: [math]d(a,x)=\displaystyle\max_2\{|a_1-x_1|,|a_2-x_2|\}<\epsilon[/math]; [math]d(a,x)=|a_1-x_1|+|a_2-x_2|<\epsilon[/math] tell me nothing at the moment about what the according neighbourhood might look like. Could you please enlighten me on this case? :smile:
  2. Ok, thanks Matt Grime! Wow, Jesus, this was our first class on induction at university and our exercise group all solved the task the way I did, so it appears that we all have a wrong solution! Very funny! From your post I could not get whether a) was alright... but anyway... First I find it now confusing that "Assume that A(n) is true" is a wrong sentence to start the induction step, since our prof says it exactly in the same way. Second, didn't I show that (n+1)^2<=2n^2 with the string of inequalities?... In my first attempt for b) for example I used A(n): n^2<=2^n is equivalent n^2+2n+1<=2^n+2n+1, which is true according to one of the order axioms. I actually use them throughout,... in the secon case in b) I used the fact that n^2<=n^2 and 2n+1<=3n-1 and again the order axiom: from a<=b and c<=d follows a+c<=b+d. The same thing with a). Here if we assume A(n) i.e. 2n+1<=2^n we can write 2n+1+2<=2^n+2 (again order axioms) and 2^n+2<=2^n+2^n since 2<=2^n. I can't see why this argumentation should be wrong.
  3. Manifold


    Hello! We've got the following to prove by induction: a) [math]2n+1\le{2^n}[/math] b) [math]n^2\le{2^n}[/math] (It is assumed that 0 is a natural number) a) This inequality is not valid for [math]n=1,2[/math], so to prove the inequality one has to show its validity for all [math]n\ge{3}[/math]: 1) [math]A(3):7\le{8}[/math] (true) 2) Assume that [latex]A(n)[/latex] is true. [math]A(n+1): 2(n+1)+1\le{2^{n+1}}[/math] [math]2(n+1)+1=(2n+1)+2\le{2^n+2}\le{2^{n+1}}[/math], since for [math]2^n+2\le{2^{n+1}}[/math] we have [math]n\ge{1}[/math] Is there a difference if I make the induction step this way?: 2)[math]A(n+1): 2(2n+1)\le{2^{n+1}}[/math], so to prove A(n) we must show that [math]2n+3\le{4n+2}[/math], which is also true for [math]n\ge{1}[/math] The thing that disturbs me is that if we assume 0 to be a natural number the last inequalities in 2) in both cases do not hold...but I think this case doesn't play a role since we are to prove it for n greater 3...does it? b) This inequality is invalid for all [math]n=3[/math], so to show that the inequality is valid we must show that it is valid for all [math]n\ge{4}[/math] 1) [math]A(4)=16\le{16}[/math] (true) 2) Assume that [math]A(n)[/math] is true. [math]A(n+1): (n+1)^2\le{2^{n+1}}[/math] [math](n+1)^2=n^2+2n+1\le{2^n+2n+1}\le{2^{n+1}}[/math]. The last inequality in this string is equivalent to [math]2n+1\le{2^n}[/math], which was proved in a). Here, equally, the same question...Is there a difference if I do it this way?: 2) [math]A(n+1): 2n^2\le{2^{n+1}}[/math] [math](n+1)^2=n^2+2n+1\le{n^2+3n-3}\le{n^2+n^2-3}\le{2n^2}[/math]
  4. OK, forget the last thing you don't understand ...I now see myself that it's a bit of nonsense... I got what you meant and have a clear picture of the matter now. Thanks a lot!
  5. Well, I understood why we took the 11 smallest numbers, but I don't quite understood what is actually meant by this. The 11 smallest of what...of which sets/subsets? Or do you simply mean that in the product of 100 numbers one can pick 11 whose product is fewer than the the product of the rest?...which follows from: if ab>1 then either a>1 and b>1 or a>1 and b<1 or a<1 and b>1...? Aren't the expressions in terms of this and in terms of "smallest" self-excluding? (Although if so, this doesn't make the solution wrong!)
  6. Hi, I've just picked up the following problem and got stuck...although seems easy. You have 100 positive real numbers. The product of each 11 of them (11 different) is greater than 1. Show that the product of all hundred numbers is greater than 1. Please, give me a hint first. Thanks
  7. I think you could also show it this way: 1. An electron is attracted towards the nucleus by the force: [math]\vec{F}=\frac{q_e{q_p}}{4\pi{\epsilon_0}r^3}\vec{r}[/math] where [math]\vec{F}\cdot\vec{r}=1[/math]; 2. On the other side [math]\frac{q_e{q_p}}{4\pi{\epsilon_0}r^3}\vec{r}=m\vec{a}[/math], since the electric field is non-homogeneous, and [math]\frac{q_e{q_p}}{4\pi{\epsilon_0}mr}=\vec{a}\vec{r}\neq{0}[/math], the electron gets in the electric field of the nucleus an acceleration with the normal component directed towards the nucleus, which is in the extreme case has a zero angle with the radius-vector (radial acceleration); 3. It is now clear that the trajectory of the electron is anyway curvilinear. Since it moves in the non-homogeneous electric field with increasing force gradient directed towards the nucleus, it gets angular acceleration, moment of inertia and hence orbital moment of force [math]L=rmv[/math]. This is that classical model, which allows the electron to fall on the nucleus.
  8. Back to your question again, Swansot. The electric field is mainly generated by the transformator, which works as plasma heater. Because there are always fluctuations of charge in plasma current, the electric field is in places where differences in charge occur. There is another thing I don't understand. It says that spiral magnetic field lines, which are set by both toroidal and poloidal fields, are the best protection from the cross drift. It also says, that they go along on interleaving toroidal surfaces. At the same time, it reads that a particle, traveling along such a magnetic field line, moves through magnetic field of varying intensity, that is through all layers with different magnetic field strength. Now how can that be? The toroidal surfaces have constant cross-section radius, and hence the B-field spirals themselves. A particle moves along only one such spiral, a B-field line with constant intensity value. It cannot change its trajectory can it?... so why should it travel through B-field of different strength?
  9. Thanks Mezarashi! The link is just great! The fact that the Lorentz force causes spiral-like motion of a particle along the magnetic field axis is clear to me. It reads on the page you've given me that there is a gradient of magnetic field - along the inner axis of the torus the magnetic field is at strongest. Ok, but since the field is weaker also to the right and to the left for example, why is there no drift in these directions, but only the cross drift? I find it confusing, that the ions should move upward and electrons downward. My source claims that the converse is true (Soros Educational Journal). I also can't understand, why we are dealing with homogeneous magnetic field here then. How can there be a gradient???...Oh sorry, have just noticed it...In a toroid the field is non-homogeneous!!!...I'll make corrections above. Swansot, you asked exactly the same question I asked myself. I read it in other sources again and again...it's only magnetic field involved, although I can't get it either...plasmas are also (very good) conductors, and as far as I know there is also electric field around any conductor segment. I'm totally confused now!
  10. I've got an understanding problem. It's about the behaviour of charged particles in plasma inside of a tokamak. It says that a particle in the non-homogeneous magnetic field (toroidal field), moving on the spiral trajectory all along the magnetic field lines, tends to drift downward (for ions) or upward (for electrons), together with the axis of the Larmor-spiral, which makes plasma highly unstable and damages the chamber walls. That's the reason why the second field (poloidal field) is necessary, which is provided by plasma current itself. Which force is it, that causes the particles to drift (toroidal drift) downward/upward?
  11. Hello! What I'd like to ask is, whether there is a mathematical theory of consciousness in development or any promising serious research projects in describing conscious process mathematically or physically with sufficient mathematical background
  12. I think it's inappropriate to make such an example in this situation. I read a number of biographies about prominent mathematicians, who invented problems of their own when they were young, in their pre-university years. They had succeeded in a number of olympiads themselves, so they were good problem solvers. But somehow they managed to make a transition from problem-solvers to problem-inventors. You can't put it down to their assumed genius, can you? I really think there is more to this than their mathematical ingenuity.
  13. hm well I think it will at most help training those problem solving strategies...I've got enough books to do that , although you can assume that similar methods would be applicable to few of the problems. I can imagine refining an already solved problem, investigating conditions, constraints, looking for special cases and so on. But the basis for this will be still the problem you already solved. I wonder however if it is the only method of invention in mathematics. What if I go straight on the "body" of mathematics and try to invent a new problem out of the material I'm working on. I can imagine asking various questions about each theorem, lemma, corollary or whatever. So far it has led me to clarification of the whole picture of how things work in a particular situation, but not to new problems in traditional sense.
  14. Hey guys! (I posted the same thread in physicsforums. To grasp a wider audience I decided to post it here too.) They say the best way to develop and train problem solving skills is by creating one's own problems. I'm still a "passive" problem solver but I'd like to learn solving problems in an "active" way. Could you give me some advice, name some guidelines on how to create new, own problems? Thanks in advance.
  15. There is a method, including no angle calculations, and needing only the information about the inner diagonal FC of the cube. The tetrahedron is nothing but the solid cut off the cube and placed on its rest, so I will use only the left figure and its symbols. 1. We will find the diagonal FC: [math]FC=\sqrt{(FG^2+GH^2)+HC^2}=a\sqrt{3}[/math] ; 2. Next we prove that F' lies on the segment FC: Take the plane FACH. The points F and C both lie in this plane, then the straight line and, hence, the segment FC lie in this plane. Take the segment MH. M an H both lie in FACH, then the segment MH lies in this same plane. Since F and C lie in the opposite halfplanes in respect to MH, then FC crosses MH. Because of the fact that MH lies in both FACH and BDH, it is the only intersection line, so any line of FACH crosses BDH at a point lying on MH. The point of intersection is exactly F'. 3. Construct the height from M to the base FGHE, let it be N. Then the parallel segments NA and MH divide the segment FC in proption 1:1:1, that is in three equal parts. Since the two parallel lines NA and MH divide AC in propotion [math]AM:MC=1:1[/math] then the same lines divide FC in the same propotion (see Fales theorem). The same is valid for NA,MH and the lines FH and FC. Since we operate with equal segments FN=CM=AM=HN, we can conclude that FC is split in three equal segment too. So FF':F'C=2:1 and it follows [math]FF'=\frac{2}{3}FC=\frac{2}{3}a\sqrt{3}[/math] That's all. There is actually very little maths needed. Cheers!
  16. Well, I worked at this problem for a while and I believe to possess the (right I hope) solution. I'm not at all enthusiastic about the prospect for my message, containing only the answer without the according steps, to be deleted ...It already happened to me once in another forum...that's why this question: Do you wish the answer first, so that we can compare our results, or should I write it all down and place the answer at the end, or should I give a hint?
  17. Well, despite my opinion expressed above, what I think is that you're lucky to have such a father, who cares about his daughter and doesn't let her to be brought up by the bad side of this world and become something like a crazy party girl...and your father, in turn, is lucky to have such a daughter...big respect for your family!
  18. I agree, but then the prospect of being stuck indoors and denying any contact with a representator of the global community is by no means attractive, at least for me. One should be careful, that's all. You learn about a person in a forum like this, and you can drop it anytime all as soon as you notice something suspicious about him while chatting. I still can't get what the problem actually is. Forums are much more passive, if it comes to exchanging ideas, and espacially scientific ones, that's why I wouldn't fully turn down live conversations. Both things are interchangeable. In my opinion, the best way to learn more about this world is by learning people from different cultures, with different views and so on. Internet gives all the necessary tools, fantastic tools! By talking live to somebody you get the emotional side of one's personality, but you cannot do it in the forum...smilies are by far not enough. So I'll stick to my opinion that Bettina's father is wrong.
  19. Well, I didn't mean chatrooms. Skype is internet-telephony...one-to-one call. I'm not sure, but there is a limit to the number of users that can take part in a conference at a time...I think it's 5 or so...so it's not a public party but a private conversation. As you perhaps heard MSN closed down all its chatrooms due to obscene behaviour of the participants, so it became more like a telephone service too. If you get to know the other person (me e.g.) well enough to trust him, I can't see any reason why you should not use the chance to chat with him. I can also prove your father that I'm a good guy! Cheers!
  20. Hi World!!! If you happen to use SKYPE or MSN-messenger, feel free to skype or msn me...both audio and video chat is possible! I'm a Russian living in Germany. I'd like to practice English with people from all over the globe...and you can also pick up some Russian and German if you wish to! My interests: science, mathematics, sports, world cultures, languages and computers. Hope to hear from you soon! Cheers, Manifold Skype-name: symplectic_manifold; MSN-account: symplectic_manifold@msn.com
  21. Manifold

    help please

    I'm not quite familiar with (in)finite sums and products...but I think it has something to do with it...and I guess one might be able to apply them to this problem...
  22. This reminds me of an article in the August issue of Scientific American "Time Before the Big Bang"...if you adopt the string theory as the theory which describes the world best, there is a limit for size (or length), the so called "Length-Quantum", and it's 10^-34 metres...but it's one of special consequences of String Theory...not more...
  23. ...well...the actual solution... how to perform the bijection in a)...and how to justify the proof in b)...I am myself confused about all that...
  24. I created a new thread on this topic..."Infinite Sets and the Schröder-Bernstein Theorem"...take a look at it...I think it all needs even more discussion...
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