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November 1

What does it mean to be mathematically educated?

Since there does not seem to be anyone asking questions today, I shall propose this one (for which I myself do not know the answer). What does it mean to be mathematically educated? I should make it clear that I am not wondering about the ability required to do mathematics, but rather the breadth of knowledge necessary (within reasonable bounds!) that should allow one to claim that he/she "knows" mathematics. Succintly, for one particular aspect of the question - which fields of mathematics approximately encapsulate all ways of thinking within mathematics (fundamentally) and which such fields arise in "many areas"? By "all ways of thinking", I mean "sorts of thinking". Perhaps more concretely, thinking geometrically/topologically is an important skill, but so is thinking algebraically, or thinking like a number theorist. However, the possibility exists that these ways of thinking are actually the same, but in a different guise.

In simpler terms, consider the examples of finite group theory, field theory and ring theory. Connections underpinned by Galois theory show that certain finite degree field extensions may be analysed via finite group theory. But general field theory, such as the theory of totally transcendental extensions, must be tackled differently; at least in some respects. Is the thinking involved simply that of ring theory, lattice theory or category theory in a different guise? For an analytic example, consider set-theoretic topology and low-dimensional topology. Of course, set-theoretic topology deals with more abstract spaces than the latter (and thus bears deep relations to axiomatic set theory), but nevertheless, low-dimensional topology may have similarities with set-theoretic topology, at least in terms of the thinking involved (geometric intuition - "moving around a manifold" and using this to construct embeddings is an example (keeping in mind the algebraic topology)). Can one claim, in terms of ways of thinking, that one of these fields is a subfield of the later?

Are the fields of number theory (algebraic, analytic et cetera), analysis and algebra exhaustive in terms of encapsulating ways of thinking? Do other fields merely disguise "ways of thinking", but still have the same ways of thinking? Is familiarity with these fields enough to guarantee "mathematical education"? Note that I have of course not considered all fields within mathematics; rather I have given examples which may be modified to answer (or at least provide insight) into these (rather philosophical) questions. --PST 12:08, 1 November 2009 (UTC)

This may have nothing to do with the question but to do number theory, specifically right now I'm reading a textbook in modular forms (no research yet), I have to know abstract algebra, linear algebra, complex analysis, at least undergrad real analysis, point-set topology, algebraic topology, hyperbolic geometry, differential geometry, number theory itself, and maybe stuff I don't know I need to know yet. I'm not saying I do know all these. But, I need to. My friends doing work in graph theory just need to know graph theory. My friends doing work in Moufang loops are just working with loops. I am not suggesting that someone must be a number theorist to be able to be called a mathematician. It's just that such a person needs to know a lot more just to work on one problem it seems to me. StatisticsMan (talk) 14:22, 1 November 2009 (UTC)

Number Theory is the most interdisciplinary field of maths I know. It includes bits of pretty much every other field of pure maths, although you don't need to learn about those fields in as much depth as a specialist would (if you did, nobody would be able to do number theory). --Tango (talk) 15:57, 1 November 2009 (UTC)

Trying to partition mathematics into different fields isn't particularly useful. All the commonly mentioned fields overlap in various ways. I don't think it is possible for anyone to know all of mathematics (it may have been a couple of hundred years ago, but so much more research has been done since then). I think the key feature of a good mathematical education is to develop "mathematical maturity". By learning and doing lots of maths (it doesn't really matter what maths) you get a feel for how the subject works and how to go about learning it. Once you have that you can learn any necessary maths to do whatever it is you are trying to do. --Tango (talk) 15:57, 1 November 2009 (UTC)

I agree that number theory is one of the most diverse fields of mathematics. It demonstrates that the better you understand the techniques used in fields prerequisite to number theory, the better your ability will be to research number theory (an enormous amount of field theory or algebraic topology may not be needed, but I feel that since a huge number of people have already approached number theory from a number-theoretic perspective, people should start attempting to develop new insights). I also partially agree with Tango that "mathematical maturity" is quite an important skill. However, my question is about the sorts of ways of thinking different fields encapsulate.

For example, in your mathematical experience, do you believe that fields like category theory, measure theory, or functional analysis, have any connections with number theory (of course they do!)? Would a number theorist be aided by appreciating such fields (that is, do such fields encapsulate different ways of thinking)? Or, are these fields redundant, in the sense that although interesting questions arise, and although they are wonderful subjects, can an algebraist have an intuitive feel for these fields (that is, do these fields have the same sorts of thinking to algebra, involved)?

I am of course not trying to partition different fields of mathematics, but rather trying to understand various points of view. Specialists abound in numerous disciplines, and many (if not all) have the qualities Tango suggests. But would not broader education help a specialist in his field (at least, aid him/her with new ways of thinking)? For instance, consider the famous Burnside problem. Analyzing the problem from a group-theoretic point of view is useful, but connecting the problem to graph theory may yield new insights (as researchers have already learnt). In effect, a group theorist should have some appreciation of graph theory (a quite well known link). But suppose someone else arrives with a topological viewpoint. It is quite possible that he may provide new insight, and this would imply that topology encapsulates different ways of thinking. Is this the case, in your experience; namely that although fields overlap in terms of content, they actually encapsulate the same ways of thinking? In the simplest terms possible, which fields encapsulate all sorts of thinking (or if there are none, why is this the case?)? --PST 01:53, 2 November 2009 (UTC)

November 2

Cauchy integral theorem

Dear collegues, article Cauchy's integral theorem says that "The Cauchy integral theorem is valid in a slightly stronger form than given above. Suppose U is an open simply connected subset of C whose boundary is the image of the rectifiable path γ. If f is a function which is holomorphic on U and continuous on the closure of U, then ∫_γf = 0".

I'm disappointed by that there is no any reference about it. Maybe the trick can be found in the proof of Goursat-lemma. If T is a triangle, int(T) ≠ 0 and f is regular at each point of int(T), then the sequence of the mid triangles (T_n) converges to a point of the int(T). Therefore, it is sufficient to require the differentiability of f at the points of int(T) (and of course the integrability of f along ∂T). Or, is it more difficult? Thanks, Mozó (talk) 11:56, 2 November 2009 (UTC).

To be precise it's not quite true if the hypotheses are stated that way. The obvious objection is that if U is an open simply connected subset of C, its closure $\scriptstyle {\overline {U}}$ need not be simply connected. And the boundary of U may be the image of a smooth curve γ that is not contractible in $\scriptstyle {\overline {U}}$. For example take $U:=B(1,2)\setminus {\overline {B(0,1)}}$ where $B(x,r)$ denotes the open disk of center x and radius r. There is a parametrization γ of the boundary of U which has index 2 to wrto 0, so f(z)=1/z has ∫_γf ≠ 0. Of course with the right hypotheses the generalization is true and quite easy to prove. --pma (talk) 14:28, 2 November 2009 (UTC)

Wow, fascinating counterexample! Thanks! I think it's not about precisity, it's about true or false :) Then we should correct the article.

"quite easy to prove" typical mathematician. :) The generalization is not contained in basic complex analysis books. So, would you please offer us a reference? Or how can I extend my proof to non-triangular shapes (if it is correct)? Mozó (talk) 16:43, 2 November 2009 (UTC)

You're right about "easy to prove", sorry! But in fact what I meant is the trivial remark (and typical mathematician too) that, under suitable hypotheses, every statement becomes true and easy to prove ;-) For instance, we may just change the assumption on γ in the original statement. Assume: (a): γ is a loop in $\scriptstyle {\overline {U}}$ which is limit in the C¹ sense (that is, uniformly with the derivative) of a sequence of C¹ loops γ_j in U . Then 0 = ∫_γj f converges to ∫_γf, and we are done. In fact (having defined conveniently the path integral for BV curves) it would be sufficient: (b): γ is uniform limit of BV loops in U, with bounded length. However I understand that the spirit of that statement is giving a condition on U only. In this case, (a) is certainly true for any smooth loop in $\scriptstyle {\overline {U}}$ if $\scriptstyle {\overline {U}}$ is a C ¹ sub-manifold with boundary of R². I am not sure at the moment whether it is sufficient to assume a plain: "both U and $\scriptstyle {\overline {U}}$ are simply connected" though, or maybe just "$\scriptstyle {\overline {U}}$ is simply connected". Maybe there is a simple answer that I don't see now. I'll look for references, but I'm not optimistic. --pma (talk) 19:21, 2 November 2009 (UTC)

"spirit of that statement is giving a condition on U only" Well, complex analysts are esoteric guys, cause while we all know what is "sub-manifold with boundary", they don't use such clear concepts. They're always talking about "moving" contours, paths along the boundary, ... without any discussion on convergency or existence. By the way, Cauchy's integral formula is a highly surprising and inexplicable result, so they must be esoteric :) Mozó (talk) 21:02, 2 November 2009 (UTC)

Now, I finally see what you mean on "with the right hypotheses the generalization is true and quite easy to prove". The generalization is an immediate consequence of the Cauchy Theorem if we consider some well-known differential topological argument using the right hypotheses. Mozó (talk) 08:24, 3 November 2009 (UTC)

The answer is: Kodaira, Complex Analysis, Camb. Univ. Press 2007, Theorem 2.3. Since, he proved his celebrated vanishing theorem, I trust in him and indeed, he has nontrivial assumption about the boundary. Mozó (talk) 09:16, 3 November 2009 (UTC)

What I had in mind as easy case was essentially what I mentioned, an open simply connected domain U with a smooth boundary (or just Lipschitz). I do not see clear the full generality of "$\scriptstyle {\overline {U}}$ is simply connected", even if I'm possibly lacking something. Do you see a more general situation? What is the precise statement of Kodaira's result in the context we are talking of, that is U=a domain of C and its boundary? --pma (talki) 09:28, 3 November 2009 (UTC)

Kodaira's way: (1) If the boundary of a compact, connected set D consists of finite numbers of "disjoint", smooth Jordan curves then D has a cellular decomposition (the well-known "squares" in pictures of books) (2) He proves the generalization for rectangles by a construction of a uniformly convergent curve sequence, just like you. (3) He states the generalization for a ∂D in (1).

He does not require simply connectedness, since in his construction the integrals along the "inner" and "outer" loops balance each others. Mozó (talk) 18:52, 3 November 2009 (UTC)

Thanks, very interesting. In the meanwhile, I had some further thoughts that I hope may be of some interest for you:

1. Let U be a bounded domain in C bounded by a simple closed rectifiable loop γ (here by rectifiable loop I mean: of bounded total variation, and of course continuous, $\textstyle \gamma :\partial D\to \mathbb {C}$ )

A classical addendum to the Riemann mapping theorem says that in this situation the biholomorphic Riemann diffeomorphism $\scriptstyle h:D\to U$ extends to a continuous

$h:{\overline {D}}\to {\overline {U}}$

such that $h_{|\partial {D}}$ is a weakly monotone reparametrization of $\gamma$; moreover, h has finite energy (it's a minimizer for the energy indeed. It's a special case of Rado's solution of the Plateau's problem, i.e. for plane curves!).

2. Actually we may think $h_{|\partial {D}}=\gamma$ with no loss of generality, because the path integral is certainly invariant by reparametrization. The preceding fact (1) then allows to conclude that ∫_γf = 0 for any f holomorphic on U and continuous up to $\scriptstyle {\overline {U}}$, just because the analogous fact is true for the unit disk D, and you can change variable with h (the fact that h has finite energy is relevant to me, because it allows to approximate uniformly $\gamma (\theta )=h(\theta ,1)$ with a sequence $\gamma _{n}(\theta ):=h(\theta ,\rho _{n})$ with $\rho _{n}\to 1$ (here I'm using polar coordinates) in such a way that $\gamma _{n}$ have bounded length. So one can pass to the limit in the sequence of the path integrals (which are all zero).

3. Using the above fact (2), I think one could make a proof for this: assume U is any open set of C; that f is holomorphic on U and continuous up to $\scriptstyle {\overline {U}}$; that γ is any rectifiable loop in $\scriptstyle \partial {U}.$ Assume further ind(γ,z)=0 for any z in U (this was exactly the missing condition in the counterexample). Then ∫_γf = 0. The idea should be decomposing γ into a countable family of simple loops γ_k that meet the conditions in (2), and in such a way that we can write ∫_γf as a series of the ∫_{γ k}f.

I think I see quite clearly the decomposition lemma, but it would be some effort to put it down here in all details. I think one needs no finiteness assumptions however. I guess this should be more or less equivalent to Kodaira's argument, with the difference that all the Jordan thing and the approximation scheme is already included in the known result (1). So, what do you like more, mine or Kodaira's? forget about the fact that mine is only sketched ;-) --pma (talk) 11:24, 4 November 2009 (UTC)

True or False? (Game Theory)

"In game theory, a player is supposed to choose only one strategy from her strategy set."

===========================True or False???

"To define a game, we must specify every player's payoff at every possible strategy profile."

===========================True or False???

"Among the games below, choose all games that have no pure-strategy Nash equilibrium." 1. Matching Penny; 2. Prisoners' dilemma; 3. Battle of sexes; 4. (Pure) Coordination games.

Thanks a lot! —Preceding unsigned comment added by Aaa50211 (talk • contribs) 14:06, 2 November 2009 (UTC)

Please do your own homework.

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. You might find our article Game Theory to be helpful. --Sean 14:12, 2 November 2009 (UTC)

Have you tried out that search box at the top left of the page? just put in a term at a time like 'battle of sexes' and have a look at the results. Dmcq (talk) 19:33, 2 November 2009 (UTC)

Euler-Lagrange equations and (possibly!) Lagrange multipliers

Hi all,

I'm trying to solve the following problem:

Write down the Euler-Lagrange equation for the functional

$I[u]=\int _{-\infty }^{+\infty }{\frac {1}{2}}(u')^{2}+(1-\cos u){\text{dx}}=\int _{-\infty }^{+\infty }f$

and find all solutions which satisfy $\lim _{x\to -\infty }u(x)=0$ and $\lim _{x\to +\infty }u(x)=2\pi$.

The first bit is just

${\frac {\partial {f}}{\partial {u}}}-{\frac {d}{dx}}{\frac {\partial {f}}{\partial {u'}}}=0=\sin u-u''$, right?

Now how on earth do I solve that?

I have a feeling I may also be meant to incorporate the constraint "find all solutions which satisfy..." using Lagrange multipliers - the conditions imply $\int _{-\infty }^{+\infty }u'=2\pi$ so am I meant to say $I[u]+\lambda \int _{-\infty }^{+\infty }u'=0$ for some $\lambda$? If so, I tried using the fact that the funtion f has no explicit dependence on x, i.e. $f=f(u,u')$, and then following up with the Beltrami Identity [1] which gave me

$k-\cos u=u'^{2}$, if I did the math correctly. I'm not sure which route to pursue (they probably both lead the same way), or whether I'm doing the right thing: attempting a series solution seems like a terrible idea and other than that I'm out of suggestions!

Thanks for the help :) Spalton232 (talk) 21:56, 2 November 2009 (UTC)

Write $\textstyle u''={\frac {d(u')}{dx}}={\frac {d(u')}{du}}{\frac {du}{dx}}={\frac {1}{2}}{\frac {d\left(u'^{2}\right)}{du}}$ and separate the variables $\textstyle u'^{2}$ and $\textstyle u$. After integration, you'll have a first-order equation to separate and integrate again. The general solution is not an elementary function, but the demanded one may be (haven't checked). — Pt _(T) 04:18, 3 November 2009 (UTC)

Few more words. Yours is the nonlinear simple pendulum equation. You'll find it more commonly written $v''=-\sin(v)$ (corresponding to u=v-π in your equation). The limit conditions at $\scriptstyle \pm \infty$ describe a homoclinic orbit to the unstable equilibrium (the pendulum is vertical with the mass in the highest position at $\scriptstyle -\infty$, then falls in infinite time and reaches again the unstable equilibrium for $\scriptstyle t\to +\infty .$) The differential equation is solved by means of an elliptic integral and studied qualitatively, starting as shown by Pt_(T), by looking at the phase space (u,u'). (according to the value of the energy, you have periodic or monotone solutions &c). In this context, the role of the action functional I(u) is classically bounded to describe some properties of the solutions (that you have found independently, by means of ODE techniques); so you do not have to bother too much about the domain of I(u) &c. Things are different if you consider more complicated (still variational) equations: then, the corresponding action functional becomes a key tool to prove existence and to get information on the solutions, which are found as critical points of the functional (there's a huge amount of calculus of variations and global analysis to do that). Note however that in your problem the limit conditions at $\pm \infty$ may be encoded in the domain of the functional, and you do not have to use Lagrange multipliers. Take as a domain the affine space of all u of the form $u=u_{0}+w$ where $u_{0}$ is a fixed smooth function with $u_{0}(t)=0$ for t<0 and $u_{0}(t)=2\pi$ for t>1, and where $w$ varies among all functions with $\scriptstyle \int _{\mathbb {R} }(w'^{2}+w^{2})dt<\infty$ (say of class C¹, but if you want a Hilbert space, the right choice is the Sobolev space $\scriptstyle W^{1,2}(\mathbb {R} ).$ You may also check that if $\scriptstyle u:\mathbb {R} \to \mathbb {R}$ is any weakly differentiable function, and $\scriptstyle I(u)<\infty$, then u is a (Hölder) continuous function with limits at $\scriptstyle \pm \infty$ that are integer multiples of 2π). --pma (talk) 08:14, 3 November 2009 (UTC)

November 3

Probability of getting 3 of a kind in a 5 card poker hand.

Yes this is a hw problem for me but I've tried it and keep getting stuck. The way I think it should be solved is 49 choose 2 / 52 choose 5. The 49 choose 2 represents the favorable outcomes --- there are 49 other cards besides the 3 of a kind and you want to select 2 out of these. The total number of hands would be 52 choose 5. I'm thinking that the 49 choose 2 doesn't completly make sense because there are 4 of each kind though. The answer I've been given is .2. If I solve it the way I described above I get an answer thats wrong by a factor of about 50. I'm thinking I should multiply the answer by 52 -- I get a solution that's close, but I can't figured why I would do that. 66.133.196.152 (talk) 00:09, 3 November 2009 (UTC)

This is how I'd do it. First compute how many different three-ace hands there are, considering both that there are several ways to choose the three aces from the four in the deck and that there are several ways to choose the two remaining cards from the 48 non-aces. Then simply multiply that by 13 to get the total number of three-of-a-kind hands. —JAO • T • C 00:35, 3 November 2009 (UTC)

Keep in mind you probably only want the probability of three of a kind, excluding any better hands (like a full house, or four of a kind). You can first pick the rank (13 ways), then 3 suits for that rank (4 ways), and then you have to pick the other two cards to ensure you don't get a full house or four of a kind. Particularly, those last two cards had better have ranks different from each other and from the rank chosen for the three of a kind. An answer of .2 is definitely wrong; if you've played poker for any amount of time, you'd notice that. The answer would be closer to .02, though you should try to get an exact probability.Nm420 (talk) 03:08, 3 November 2009 (UTC)

Consider the hand with three aces and two other cards. We split the pack into two piles: the four aces and the 48 non-aces. There are (4 × 3 × 2)/3! = 4 ways of choosing three aces from four, and (48 × 47)/2! = 1,128 ways of choosing two cards from the 48 non-aces. That gives a total of 4 × 1,128 = 4,512 five card hands made up of three aces and two non-aces. To avoid a full house the two non-aces cannot make a pair. Since there are four 2s in the non-ace pile there are (4 × 3)/2! = 6 ways of making a pair of 2s. Likewise, there are 6 ways of making a pair of 3s, etc. So there are 6 × 12 = 72 ways of making a pair with the two cards drawn from the non-ace pile. Therefore, there are 4,512 − 72 = 4,440 five card hands with exactly trip As. We can do the same for trip 3s, trip 4s, …, trip Ks. This makes 13 × 4,440 = 57,720 hands with exactly three of a kind. There are (52 × 51 × 50 × 49 × 48)/5! = 2,598,960 five card hands. The odds of getting exactly three of a kind from a five card hand are then 1,629-to-37 against (about 44-to-1 against).
The probability is then 37/1,666 ≈ 0.02. ~~ Dr Dec (Talk) ~~ 12:47, 3 November 2009 (UTC)

The above solution is close, but not quite. The subtraction is being made at the wrong place. In general, how I like to solve these types of problems, is to use only multiplication of binomial coefficients (when possible of course), and of course state the "real-world" choice being made with each. Nm420 (talk) 14:45, 3 November 2009 (UTC)

Oh yeah, silly me. We have to take out the possibility of the pairs right from the beginning. There are 4 ways of choosing three aces from the aces pile, and there are 1,128 − 72 = 1,056 choices of two non-paired cards from the non-ace pile. That gives a total of 4 × 1,056 = 4,224 five card hands made up of three aces and two non-paired, non-ace cards. Doing this for three 2s, three 3s, …, three Ks gives 13 × 4,224 = 54,912 hands with exactly three of a kind. There are still 2,598,960 five card hands. This gives revised odds of 4,077-to-88 against (about 46-to-1 against). The probability is then 88/4165 ≈ 0.02. The Poker Probability article gives a full list of odds. ~~ Dr Dec (Talk) ~~ 15:08, 3 November 2009 (UTC)

Mensuration and geometry

Could you please explain me the difference between mensuration and geometry with examplesKasiraoj (talk) 11:49, 3 November 2009 (UTC)

"Mensuration" just means measurement. You could be measuring anything - time, temperature, electric current, luminosity etc. - so many measurment techniques have nothing to do with geometry. On the other hand, if you are measuring a length, an area, a volume, or related concepts such as density, speed or acceleration, then you may be using geometric techniques. For example, to measure the width of a river, you can use surveying techniques, which are an application of geometry and trigonometry. Gandalf61 (talk) 12:01, 3 November 2009 (UTC)

November 4

Complete vector spaces under the uniform norm

Hi all:

I'm wondering whether the following spaces are complete under the uniform norm $\|f(x)\|_{\infty }=\sup _{x}(|f(x)|)$, and I'm a little concerned my arguments are too rough and I'm neglecting things which could lead me to be wrong because i haven't thought it through well enough, I'd appreciate it if someone could point out areas where I'm going wrong, which I have no doubt there will be!

i) The space $C_{b}(\mathbb {R} )$ of bounded continuous functions $f:\,\mathbb {R} \to \mathbb {R}$.

ii) The space $C_{0}(\mathbb {R} )$ of continuous functions $f:\,\mathbb {R} \to \mathbb {R}$ with $f(x)\to 0$ as $|x|\to \infty$

iii) The space $C_{c}(\mathbb {R} )$ of continuous functions $f:\,\mathbb {R} \to \mathbb {R}$ with $f(x)=0$ for $|x|$ sufficiently large.

My thoughts:

i) I know that on the bounded interval ($f:\,[0,1]\to \mathbb {R} )$, the space is complete, and so presumably the same is true of any interval [a,b] by stretching and translating. The functions f_n (Cauchy sequence) converge to their pointwise limit on these intervals, and since the interval is arbitrary, the functions have a limit f which is continuous on any interval [a,b] and thus bounded on it too. Thus, the only manner in which f can be unbounded is as x tends to (± infinity): however, if this is the case (WLOG take + infinity), then since f is the pointwise limit of the f_n, and the sequence is Cauchy which implies that, if as n tends to infinity the f_n have arbitrarily increasing upper bounds (pointwise convergence to an unbounded function) and we can make the f_n arbitrarily close for large enough n > some N by the Cauchy property, the f_n all tend to infinity as x tends to infinity, since they must remain as close to each other as we require by choice of N. Thus the space is complete.

ii) The functions are continuous and by their definition must be bounded, so by (i) they converge to a bounded continuous function in the limit, and because this function is in fact their pointwise limit, if it does not tend to 0 as |x| tends to 0, then the f_n do not either, and so by a contradiction argument, f too must tend to 0, and so the space is complete again.

iii) I believe I've constructed a series of functions which are 'truncations' of f(x)=1 if |x|<1, 1/x if |x|>1 and which in the limit tend to this f(x) which is not equal to 0 for sufficiently large |x|, but the sequence is Cauchy because as m,n tend to infinity, f_m and f_n agree with each other precisely, for larger and larger values of |x|, and thus $\|\cdot \|_{\infty }$ tends to 0, so the seq. is Cauchy and hence this is not a complete space.

Sorry if any of that doesn't make sense, please try to muddle through :) I do intend to refine this argument more when I write the question up in full, but since it's quite long I'd rather sort things out in rough before I dive in fully rigorously - I have, no doubt, made some stupid mistakes, and your help would be greatly appreciated! (And thanks for reading this far, too.) Typeships17 (talk) 02:50, 4 November 2009 (UTC)

i) This looks right but I think you can tighten it up a bit. To show the pointwise limit f is bounded, pick any ε > 0. There exists an N such that ||f_N - f_n|| < ε for all n > N. Suppose f_N is bounded by M, then you can show f is bounded by M + ε.

ii) It's not true in general that the pointwise limit of a sequence of functions that all go to 0 as x goes to infinity must go to 0 as x goes to infinity. For example let g_n(x) = 1 for |x| < n and g_n(x) = 0 otherwise. However for any ε > 0 you can find N as before, and then some B such that |f_N(x)| < ε for all |x| > B, and then you should be able to argue that |f(x)| < 2ε for all |x| > B.

iii) This looks good.

Rckrone (talk) 04:09, 4 November 2009 (UTC)

Wow, I wasn't expecting an answer so quickly, thanks very much! :D I'll get to writing them up properly then! Typeships17 (talk) 04:16, 4 November 2009 (UTC)

As a further remark, note that the space (iii) is dense in the space (ii), and different from it, so it's not complete. Both (i) and (ii) are closed subspaces of the space B(R) of all bounded functions R→R, so (i) and (ii) are complete since the latter is complete. For analogous cases, you may find it useful the general result (same proof) that for any set S and any Banach E the space B(S,E) with the sup norm is still a Banach space (this way you get at once the completeness of a lot of norms based on the sup norm: dual and operator norms, &c). Although (iii) is not complete with the sup norm, it has another natural structure of TVS as inductive limit (in the category TVS) of the inclusions C(K)→C(R), K compact. This topology is locally convex, non metrizable, sequentially complete. --pma (talk) 08:28, 4 November 2009 (UTC)

Two parallelogram areas theorem

When two parallelograms lie between two parallels and same base, their areas are equal. I have got problem with the proof. When the figures are like this, it can be done by proving the triangles congruent (Side-angle-angle) but when the figures are like this, how do we do it? Srinivas 15:32, 4 November 2009 (UTC)

What about adding congruent triangles to both, so as to obtain congruent trapezoids. --pma (talk) 15:53, 4 November 2009 (UTC)

(e/c) If all else fails, you can introduce sufficiently many other parallelograms in between so that the first case applies to each pair of neighbours, and use transitivity of equality. — Emil J. 15:57, 4 November 2009 (UTC)

You can also make the same congruent triangle argument with triangles ACE and BDF. You just need to do some subtraction to get to the areas you want. Rckrone (talk) 18:56, 4 November 2009 (UTC)

Incompleteness theorem

The incompleteness theorem says that in any mathematical system, there are statements which can be neither proved nor disproved. Is there any way to prove that a statement is unprovable by this theorem? --75.50.49.177 (talk) 23:05, 4 November 2009 (UTC)

I'm afraid I don't know what you're really asking. You'll probably need to phrase it more precisely before anyone can help. In the meantime you might take a look at our article on Gödel's incompleteness theorems and see if it addresses your question. --Trovatore (talk) 23:12, 4 November 2009 (UTC)

The incompleteness theorem states that some statements are neither provable nor disprovable. However, is there any way of determining which statements those are? --75.50.49.177 (talk) 23:14, 4 November 2009 (UTC)

For a given theory T satisfying the hypotheses, the (modified) proof of the first incompleteness theorem gives an effective way of finding a sentence G_T such that, if T is consistent, then G_T is true, yet neither provable nor refutable from T. Sorry I can't make it less complicated than that. All the stipulations are necessary in order to have a correct statement of the situation, so you'll just have to work your way through them. -Trovatore (talk) 23:21, 4 November 2009 (UTC)

That says how to construct a statement which is unprovable. However, given a statement (e.g. the Riemann Hypothesis), is it possible to determine whether or not that statement is provable? --70.250.215.137 (talk) 02:41, 5 November 2009 (UTC)

Not in general, no. Well, at least not by any fixed mechanical procedure. In technical terms, the set of theorems of T is computably enumerable but not computable. --Trovatore (talk) 02:44, 5 November 2009 (UTC)

As Trovatore says, not in general, no. But there are some specific examples where undecidability can be proved: one of the most famous problems of the early 20th century was to prove or disprove the continuum hypothesis. This turns out to be both unproveable and un-disprovable (also known as independence) if you start from the usual axioms of set theory. The proof of independence was done in the 1960's by Paul Cohen and doesn't use the incompleteness theorem directly (it uses a complicated technique called forcing). That proof was considered a very major result. For other examples, see the article about independence that I linked to. 69.228.171.150 (talk) 05:39, 5 November 2009 (UTC)

There are plenty of mathematical systems in which every statement can be either proved or disproved. Examples include the theory of real closed ordered fields and the theory of dense linear orderings without endpoints. There are several hypotheses in Gödel's incompleteness theorem and they are all important for the result. — Carl (CBM · talk) 03:28, 5 November 2009 (UTC)

Hmm, it occurs to me that I gave the answer sbout the theorems being c.e. but not computable without a proof immediately in mind, and when I thought of one, it requires T to be Sigma-1 sound. (The proof is, you can reduce the halting problem to the theorems of T -- given a Turing machine that you want to know whether it halts, just decide whether T proves that it halts. The usual hypotheses of the Gödel theorems suffice to show that, if the TM halts, then T proves that it halts, but for the converse you need that T is Sigma-1 sound. Does anyone see how to eliminate this hypothesis? I flat don't believe that the theorems of, say, PA+~Con(PA) form a computable set, even though that theory proves that certain Turing machines halt that actually don't halt.) --Trovatore (talk) 05:24, 5 November 2009 (UTC)

As you're saying, no consistent completion of PA is computable, whether that completion is sound or not. Syntactically, you get rid of the soundness assumption by using Rosser's trick. But there is also a completely computational proof, related to the concept of a PA degree. We can prove that given any completion T of PA and any nonempty $\Pi _{1}^{0}$ class C in Cantor space 2^ω, there is an element of C computable from T. So if you take T to be a $\Pi _{1}^{0}$ class with no computable elements, you obtain a corollary that there is no computable completion of PA. — Carl (CBM · talk) 11:25, 5 November 2009 (UTC)

Well, unless I'm missing something, that's a little different question. I wasn't asking about completions. The question is, given a consistent r.e. theory extending PA and closed under logical consequence, can you show that that theory is not computable? --Trovatore (talk) 11:34, 5 November 2009 (UTC)

Yes, and asking it to extend Q rather than PA is more than enough. — Emil J. 11:39, 5 November 2009 (UTC)

Rosser's trick is one way to prove it, and the other argument by Carl works too: you just have to observe that every decidable theory has a decidable completion. The usual way to prove the theorem along the lines of your proof above is to take a recursively inseparable pair of r.e. sets instead of the halting problem. — Emil J. 11:44, 5 November 2009 (UTC)

Why does every decidable theory have a decidable completion? --Trovatore (talk) 20:02, 5 November 2009 (UTC)

You can do this explicitly. The language is countable, so fix an enumeration of all sentences. Then go along the list, and for each sentence φ use your decision procedure to determine which of φ and ~φ is provable from your theory so far. If one of them is, add it to the theory, if neither is, add φ. This gives a completion which is decidable by construction (since you can decide a sentence just by repeating the construction) Thus all you have to show is that if T is decidable, then T union {φ} is decidable, and this is obvious (to see if θ follows from T u {φ}, check if φ→θ follows from T). Algebraist 21:39, 5 November 2009 (UTC)

Nice; thanks. --Trovatore (talk) 22:19, 5 November 2009 (UTC)

Exactly. I think I did misread the original question. For PA (or any consistent extension of Q) the set A of formulas φ such that ${\mbox{PA}}\vdash \varphi$ and the set B of formulas θ such that ${\mbox{PA}}\vdash \lnot \theta$ are already recursively inseparable r.e. sets, that is, there is no computable set C including A and disjoint from B. Any consistent decidable extension of PA would give such a set, so there is no such extension. So PA (and Q) is "essentially undecidable". — Carl (CBM · talk) 12:48, 5 November 2009 (UTC)

November 5

Provable or Disprovable or Independent?

The above discussion about Godel's incompleteness theorem prompted a question I've had for a while: Is it true that any statement can either be proved from a set of axioms, disproved, or shown to be independent of those axioms?

Here's my reasoning: I've heard people suppose that, in addition to those possibilities, it might be undecidable whether or not something is undecidable, or undecidable whether or not it's undecidable that it's undecidable, and so on ad infinitum. However, if something is (undecidable)^n, then clearly there exists no proof of that statement, and there also exists no proof of its negation. But that means it is simply undecidable: it is independent of the axioms. It seems, then, that there is no possibility for a statement's undecidability to be undecidable-- given any problem, there are three options: we can prove it, disprove it, or give a proof of its independence. Is this valid? Do I need any conditions on a theory for this argument to apply? 140.114.81.68 (talk) 09:20, 5 November 2009 (UTC)—Preceding unsigned comment added by 140.114.81.68 (talk) 09:19, 5 November 2009 (UTC)

If you do things this way, you have to keep careful track of where you're proving things. For example, the Goedel sentence G_PA for Peano arithmetic (PA), cannot be proved or refuted in PA. Also, PA cannot prove that G_PA is independent of PA.

However the slightly stronger theory PA+"PA is consistent" can prove that G_PA is independent of PA, and also that G_PA is outright true.

The thing to keep in mind is that "independent" is not a truth value. It's a statement about whether something is provable or refutable from a specific formal theory, and the theory always needs to be specified. Truth sempliciter on the other hand is a much more robust notion; the Goedel sentence of a consistent theory, though not provable in that theory, is always true, and you don't have to specify the theory that makes it true: It's just true, period. Remembering this is a good anchor when learning the Goedel stuff; it helps keep you from floating off into confusion. --Trovatore (talk) 09:35, 5 November 2009 (UTC)

Also within any given consistent system of sufficient power there will be statements that we can not prove, nor prove the negation, nor give a proof that neither of the previous proofs exists. Taemyr (talk) 10:11, 5 November 2009 (UTC)

If something is (undecidable)^n in T, as you say, then it is in particular undecidable in T (because if something is decidable in T, it is also provable decidable). However, the conclusion does not follow: there is no reason why we should be able to prove in T that the statement is (undecidable)^n (indeed, we never can do that, because undecidability of anything implies consistency of T). — Emil J. 11:52, 5 November 2009 (UTC)

In a certain sense, it is true that a particular sentence φ will be either provable from PA, disprovable from PA, or independent of PA. This is just tautologous. But it does not mean that "we can prove it, disprove it, or give a proof of its independence". There are many statements of interest where we can do none of those three things at present. Also, it will usually be true that the proof of independence is carried out in a metatheory rather than the theory at hand, as others have pointed out above. — Carl (CBM · talk) 13:46, 5 November 2009 (UTC)

Relation of "contiguity space" and "proximity space"

Are the usual meanings of "contiguity space" and "proximity space" essentially the same? At Talk:Contiguity space we have a suggestion that they are the same, mainly on the basis of a google result that included "proximity space or contiguity space". Essentially I am asking if the proposed redirect and renaming would conflict with established usage, or with what should be on Wikipedia. An article on proximity space already exists. If "contiguity space" is a separate idea we could still move out the stuff related to probability measures, but that would leave a very poor article. Melcombe (talk) 11:47, 5 November 2009 (UTC)

Lie Brackets

I put a question about Lie brackets here. Basically, I don't understand how they are useful and so why someone would have come up with them. —Ben FrantzDale (talk) 12:16, 5 November 2009 (UTC)

The Lie bracket of vector fields has a nice geometric interpretation. Let M be a smooth Riemannian manifold and consider the C^∞(M,R)-module of smooth vector fields over M. Let X and Y be smooth vector fields on M. We can consider the new vector field [X,Y] given by [X,Y] = XY − YX. If we flow along X a distance of ε and then along Y a distance of ε we reach a point p ∈ M, then if we flow first along Y a distance of ε and then along X a distance of ε we reach a point q ∈ M. The difference between p and q is given by ε²[X,Y]. More formally: "The Lie bracket [X,Y] of two vector fields X and Y measures the O(ε²) gap in an incomplete quadrilateral of O(ε) arrows made alternatively from εX and εY." So it's measuring how movement is commutative. In our real world experience if we move two paces east and then two paces north we end up at the same point as if we had taken first two paces north and then two paces east. This wouldn't be the case for more exotic manifolds and more exotic vector fields. ~~ Dr Dec (Talk) ~~ 15:22, 5 November 2009 (UTC)

Thanks. What I don't understand is how that is useful, beyond characterizing the local non-flatness of the manifold. When would I use it to solve a problem? Would it just be to answer questions about the manifold itself? Perhaps I'm just over my head and need to fill in more understanding to ask meaningful questions :-) —Ben FrantzDale (talk) 19:34, 6 November 2009 (UTC)

Well, you've answered your own question there. The torsion and curvature tensors both use the Lie bracket in their definitions. They're both very fundamental and interesting objects in differential geometry. Asking where the Lie bracket might be useful or when you might use it to solve a problem is kind of like, to a much lesser extent, asking how addition is useful or how you might use addition to solve a problem: they're both mathematical constructions with their own uses. I know that addition is far more useful and commonplace than the Lie bracket, but I hope you get my point. ~~ Dr Dec (Talk) ~~ 19:54, 6 November 2009 (UTC)

Transporting tensors on Lie groups

I have a question about parallel transport of tensors on Lie groups. Thanks. —Ben FrantzDale (talk) 18:02, 5 November 2009 (UTC)

The exponential map gives a mapping from a Lie algebra to its Lie group. For the connexion between the two you might like to read this section of this article. Moreover, given your last question about Lie brackets, you might like to read this section of this article. Also, you might like to read this article. ~~ Dr Dec (Talk) ~~ 18:36, 5 November 2009 (UTC)

November 6

Counterexample

Could anyone suggest a nice example of a continuously differentiable function on R^2 which has only 1 stationary point - a local minimum (or maximum), but which is not a global minimum (or maximum)?

I keep looking for functions with minimum at 0,0 but I can't seem to find anything which increases locally in all 4 directions, i.e. along both axes, and yet doesn't produce any more stationary points when it eventually decreases somewhere. I found one in a paper by Ash and Sexton but it was quite unpleasant and I was hoping to find something nice and neat. Any ideas?

Thanks. 82.6.96.22 (talk) 00:50, 6 November 2009 (UTC)

I think (x³ - x)e^y + e^2y works. Rckrone (talk) 01:31, 6 November 2009 (UTC)

As an aside, increasing along both axes doesn't guarantee a local min. For example x⁴ + y⁴ - (x+y)⁴/4 - (x-y)⁴/4 has only a saddle point at (0, 0). Rckrone (talk) 02:02, 6 November 2009 (UTC)

In polar coordinates: unless I misunderstand your question or am not thinking straight, try

$f(r,\theta )=r^{2}\,\cos \,r$

That should have a local minimum at the origin but oscillate unboundedly as you get further away. 69.228.171.150 (talk) 06:10, 6 November 2009 (UTC)

A further note. Consider a diffeo h of Rⁿ with an open subset A of Rⁿ itself, for instance a ball, h:Rⁿ→A. If f: Rⁿ→R is any differentiable function, the composition g(x):=f(h(x)) is a differentiable function whose critical set has the same structure (same number, index, nullity, level) of the critical points of f in A. Of course, the latter may be whatever. If the global max and min points of f belongs to the boundary of A, then g has also the same sup/inf as f. So for instance there is a function on Rⁿ with critical points of prescribed level, index, and cardinality, and the interval f(Rⁿ) may also be prescribed. --pma (talk) 09:35, 6 November 2009 (UTC)

With regards to :$f(r,\theta )=r^{2}\,\cos \,r$, would that not have more than 1 stationary point though? 82.6.96.22 (talk) 14:23, 6 November 2009 (UTC)

Oh yes, I did misunderstand. Hmm, I'll think about whether I can fix it by suitably twisting up the shape while keeping the equations reasonably neat. 69.228.171.150 (talk) 22:50, 6 November 2009 (UTC)

Try this section of the maxima and minima article. It says that ƒ : R² → R given by ƒ(x,y) = x² + y²(1 − x)³ has its only critical point at x = y = 0 but that this isn't a global minimum because ƒ(4,1) = −11 < ƒ(0,0) = 0. ~~ Dr Dec (Talk) ~~ 15:44, 6 November 2009 (UTC)

Polynomials

When you say R[x]/<x^2+1> is isomorphic to the complex numbers, do you mean that because [x]^2+[1]=[0], where the brackets denotes the respective equivalence classes, and x is the polynomial f(x)=x? Is the following correct: [x]^2=[x^2] and hence [x]^2+[1] = [x^2+1] = [0], so [x] is like the imaginary unit? Najor Melson (talk) 05:38, 6 November 2009 (UTC)

If R is a ring and if I is an ideal of R, the map sending r to the coset r + I is a surjective ring homomorphism. In particular, using your notation, [x]² = [x²] is true, as well as your other assertions. Consider the surjective homomorphism defined on R[x] that sends a polynomial f to its image at i (the imaginary unit). Since R[x] is a principal ideal domain, and since x² + 1 is irreducible over the real numbers, the kernel of this homomorphism is generated by x² + 1. In particular, $\mathbb {C} \simeq \mathbb {R} [X]/(X^{2}+1).$. Furthermore, since the image of x under this isomorphism is i, x does indeed "act like i" in the given quotient ring (ismorphisms preserve the behaviour of elements). Hope this helps. --PST 05:53, 6 November 2009 (UTC)

It means not only that [x]^2+[1]=[0], but more generally there is a one-to-one correspondence between these equivalence classes and the complex numbers, such that multiplication and addition of equivalence classes correspond exactly to multiplication and addition of complex numbers. Michael Hardy (talk) 16:58, 6 November 2009 (UTC)

ordinal strength

The ordinal strength of PA is ε₀. How does one figure it out for PA+Con(PA)? Thanks. 69.228.171.150 (talk) 08:36, 6 November 2009 (UTC)

It depends on the particular way you choose to measure ordinal strength. There is a trivial answer for measures based on computable functions, such as the smallest ordinal α such that every provably total computable function is β-recursive for some β < α: since the set of p.t.c.f. of a theory is unaffected by addition of true $\Pi _{1}^{0}$-sentences, PA + Con(PA) has the same ordinal as PA, ε₀, in this measure. I'm afraid the answer is more complicated for measures sensitive to the $\Pi _{1}^{0}$-fragment of the theory. For example, the $\Pi _{1}^{0}$-ordinal measure by Beklemishev[2][3] gives ε₀·2 for PA + Con(PA). — Emil J. 16:03, 7 November 2009 (UTC)

Birth and Death Chain - Markov Processes

Hi all :)

I asked this a while ago but never got an answer, hopefully someone might be able to help me this time. For a fairly simple birth-death chain (1D, with probabilities $p_{i}=({\frac {i+1}{i}})^{k}q_{i}$, $p_{01}=1$, $p_{i}+q_{i}=1\forall \,i$ of moving in each direction at node i), what's the easiest way to find (*) P(X_n-> infinity as n -> infinity) - e.g. should I be looking at the expected time of something as n tends to infinity, or P(X_n<k) as n tends to infinity, or the transition matrix P^n etc? Hopefully once I know a sensible way to show (*), the actual question will be fairly simple.

Thanks a lot, Otherlobby17 (talk) 16:28, 6 November 2009 (UTC)

Could you describe the problem in more words? For example, what is k? 69.228.171.150 (talk) 23:00, 6 November 2009 (UTC)

I want to find it for values of k in $(0,\infty )$ - is there a general formula of any sort? Initially I want to show that the probability is 1 for k=2, but then I've been asked to find the formula for a general k in the positive reals - why, is that not feasible? Other than the $p_{i}=({\frac {i+1}{i}})^{k}q_{i}$, $p_{01}=1$, $p_{i}+q_{i}=1\forall \,i$ formulae, I don't really have anything else to explain unfortunately, I mean it's a markov chain along a line (natural numbers, say) with probabilities p_ij of moving from i to j. Sorry but that's pretty much the whole problem! Otherlobby17 (talk) 15:06, 7 November 2009 (UTC)

Help with Logic Problem

I think i've made progress, but I'm stuck--I was hoping for a nudge in the right direction

I'm only allowed to use the 18 valid argument forms.

Here's what I have so far

1. A≣B (premise)
2. ~(A ⋅ ~R) ⊃ (A ⋅ S) (premise) /∴ ~(B ⋅ S) ⊃ ~(A ⋅ R)
3. (A ⊃ B) ⋅ (B ⊃ A) 1 EQUIV
4. A ⊃ B 3 SIMP
5. (A ⋅ ~R) v (A ⋅ S) 2 IMP
6. A ⋅ (~R v S) 5 DIST
7. A 6 SIMP
8. B 7, 4 MP
9. ~R v S 6 SIMP

but now i'm completely stuck! I have a sneaking suspicion that ADD might help......UGH, any thoughts?

sorry for the double post, i fixed the formatting209.6.48.226 (talk) 18:41, 6 November 2009 (UTC)

Well, "the 18 valid argument forms" doesn't really ring a bell — you should know that which inference rules are taken as basic is something that varies by presentation. There's a good chance that the specific list of inference rules and their names/acronyms is particular to, perhaps, even just the one professor from whom you're taking the course. It's also possible that this is some more widely recognized list that someone else can help you with, just not me. But it's very far from universal, and you should be aware of that. --Trovatore (talk) 20:24, 6 November 2009 (UTC)

I hadn't heard of 18 valid argument forms either, but there are a few ghits. 69.228.171.150 (talk) 20:40, 6 November 2009 (UTC)

November 7

Probability question

This is nothing to do with Homework, just an interesting problem that I came across. Assume there are three possible states for a man : He can be healthy, sick, or dead. The probability that he is sick the next day given that he is healthy today is 0.2 (the values don't really matter much here). The probability that he is dead the next day given he is sick today is 0.25. The probability that he is dead the next day given he is healthy today is zero. The probability that he is healthy the next day given he is sick is 0.4. Given he is healthy today, what is the probability that he will never die ? My instinct tells me that the answer is zero, he has to get sick somewhere down the line, and eventually die. Rkr1991 ^{(Wanna chat?)} 06:32, 7 November 2009 (UTC)

Yes, it's 0. Assuming the independence you may observe that the probability to be dead within 2 days is in any case at least p=0.05, so the probability to be still alive after 2n days is not larger than (1-p)ⁿ, that goes to 0 exponentially. Yours is an example of a Markov chain. --pma (talk) 07:59, 7 November 2009 (UTC)

OK. So how do you find the answer to the slightly harder question, what is the probability that he is alive after n days given he is alive today ? Rkr1991 ^{(Wanna chat?)} 10:55, 7 November 2009 (UTC)

(I assume you mean "given he is healthy today" ). Just write the transition matrix relative to the states 1=H, 2=S, 3=D, which is

P:=${\begin{bmatrix}0.8&0.2&0\\0.4&0.35&0.25\\0&0&1\end{bmatrix}},$

where p_ij is the probability to pass from the state i to the state j. Then, in Pⁿ the coefficient p⁽ⁿ⁾_1,3 is the probability to be in 3 at time n starting from 1 at time 0. To write Pⁿ efficiently you should diagonalize P first (you can: it has 3 simple eigenvalues: 1, 0.213, and 0.936), P=LDL^-1, Pⁿ=LDⁿL^-1. Check Markov chain for details.--pma (talk) 12:13, 7 November 2009 (UTC)

Great, thanks. Rkr1991 ^{(Wanna chat?)} 13:47, 7 November 2009 (UTC)

Moment of inertia of a trapezoidal prism

I'm trying to put together a little tool to calculate a first approximation to the centre of gravity and moments of inertia of a conventional airliner. Most of it is pretty simple: I'll model the fuselage as a solid cylinder (though I'm tempted to try to make it more a combination of the structural mass as a cylindrical shell and the payload mass as a cylinder inside it), the engines also as cylinders, and the moments of inertia around the aircraft CG of the tail parts are dominated by the parallel axis theorem so I can neglect their contributions about their own CGs.

The challenge I'm finding is the wing. It is not small and it is close to the aircraft CG so I need it's own moment of inertia. I would like to take into account the effects of wing sweep, dihedral, and taper (that is, the difference in chord between the wing tip and the wing root). I don't need anything more complex like elliptical wings or multiple taper ratios. Sweep and dihedral I'll deal with by just rotating the moment of inertia matrix once I have it. So what I need is the moment of inertia of a trapezoidal prism. But I can't find any equations for that anywhere.

From the Moment of inertia article:

$\mathbf {I} =\iiint _{V}\rho (x,y,z)\left(\|\mathbf {r} \|^{2}\mathbf {E} _{3}-\mathbf {r} \otimes \mathbf {r} \right)\,dx\,dy\,dz,$

The convention I am using is that x is rearward, y is out the right wing, and z is up. So the wing is a trapezoid when projected on the x-y plane, with a constant (relateively small) thickness in the z direction. I am happy keeping the density (rho) constant. I expect that the answer will be in the forms of formulae for Ixx, Iyy, Izz about the principal axes (the diagonal of the matrix), with Ixz, Iyz, and Ixy zero and I will have to rotate the matrix to work in my coordinate system.

I would also like some guidance on the appropriate thickness to choose as the z-dimension. Clearly it is airfoil shaped, but I was going to approximate it as rectangular. Given the maximum airfoil thickness, what would be a rough guide to the right prism thickness? Actually, now that I think about it, in truth the height of the airfoil is proportional to the chord as well, that is it decreases from root to tip. But I am happy to neglect that. An alternative to modelling it as a rectangular section would be a diamond section, i.e. a quadrilateral with two sets of two equal sides, and the user of my tool would have to specify the maximum airfoil thickness and the location of the maximum airfoil thickness.

I know I could have asked this on the Science RD, but I hope I've taken the physics out of the question and it is now actually a math question.

Thanks, moink (talk) 07:13, 7 November 2009 (UTC)

Concatenations with preimages

Given a function f : X → Y, how to prove f(f ⁻¹(B)) ⊆ B and f ⁻¹(f(A)) ⊇ A for all subsets A of X and all subsets B of Y? --88.78.2.122 (talk) 08:52, 7 November 2009 (UTC)

Just use the definitions of the preimage f ⁻¹(B) ("all what goes in B") and of the image f(A) ("where A goes"). So f(f ⁻¹(B)) ⊆ B reads "all what goes in B, goes in B" and f ⁻¹(f(A)) ⊇ A reads "A goes where also goes all what goes where A goes" .

Also notice the useful equivalence, for any subset A of X and any subset B of Y:

f(A) ⊆ B ⇔ A ⊆  f ⁻¹(B),

from which you can deduce both your inclusions starting respectively from f ⁻¹(B) ⊆ f ⁻¹(B) and f(A)) ⊆ f(A). --pma (talk) 09:03, 7 November 2009 (UTC)

Just for talking, you may consider the power set of X and Y, namely ^_℘(X) and ^_℘(Y), as small categories, where objects are subsets, and arrows are inclusions. Then the image map f_*: ^_℘(X)→^_℘(Y) and the preimage map f^*: ^_℘(Y)→^_℘(X) are adjoint functors, and of course the inclusions you wrote are just the co-unity and unity of the adjunction f_*$\scriptstyle \dashv$ f^*. (Check closure operator also).

PS: Talking about Html, does anybody know how to get the ℘ a bit higher, and to get a bit lower the star in f_* ? (I got it in quite a weird way). And why I can't see ⊣ for $\dashv$ ? --pma (talk) 11:25, 7 November 2009 (UTC)

<span style="vertical-align:20%">℘</span> ℘ (but not abusing Weierstrass' p-function symbol for power set is a better option), ''f''<sub style="vertical-align:-80%">*</sub> f_*. Note that fine-tuning like this is very font-dependent, hence you cannot do it in a way which would reliably work for all readers. As for &LeftTee;, there is no such thing among the XHTML 1.0 named character entities[4] (Wikipedia's HTML tidy code will thus remove it even if a particular browser could support it as an extension). The character appears in Unicode on position U+22A3, hence you can get it by &#x22a3; (or &#8867;) in HTML: ⊣⊣ (or simply input the character directly: ⊣). — Emil J. 15:03, 7 November 2009 (UTC)

Thank you! Nice emoticon :⊣) also --pma (talk) 15:37, 7 November 2009 (UTC)

plausibility of mathematical completeness

It's the mid 1920's and you're a top mathematician being recruited to the Hilbert school. Your mission: prove Hilbert's famous contention (from "On the Infinite"):

As an example of the way in which fundamental questions can be treated I would like to choose the thesis that every mathematical problem can be solved. We are all convinced of that. After all, one of the things that attracts us most when we apply ourselves to a mathematical problem is precisely that within us we always hear the call: here is the problem, search for the solution; you can find it by pure thought, for in mathematics there is no ignoramibus.

The subtlety of Gödel's 1931 incompleteness proof may have escaped everyone else, but still, those guys weren't slouches. My question is why could anyone ever have been convinced Hilbert's thesis was true?

Consider a simple enumeration of all the sentences φ₁, φ₂,... over, say, the language of PA. For each sentence φ_k make a red dot on the number line at the point k, and write the formula denoted by φ_k next to the dot. At the end of this process you have an infinite row of red dots, each labelled by a formula. Now go through again and enumerate the theorems t₁, t₂,... and for each t_i, find the dot you made earlier for that formula and flip its color from red to blue. (If it's already blue, don't do anything; this will happen a lot because you will generate all the theorems infinitely often). Also similarly flip the color of the dot for ~t_i to blue. In other words you have colored the decidable formulas' dots blue and left the undecidable ones (if there are any) red. Hilbert says that at the end of this, you will have no red dots left, just blue ones. This seems like a pretty elementary description, certainly accessible and hopefully natural to any logician of that time.

Of course we know today that there will necessarily be red dots at the end, and of course we can reasonably say that until 1931, Hilbert could reasonably harbor some hope that there wouldn't be red dots. But what could make him (or anyone) so sure that there'd be no red dots. The process is just a big combinatorial mess, some weird contraption jumping around all over the place flipping dots from red to blue, and the exact pattern of flipping depends intimately on the exact make-up of the contraption (i.e. the content of the theory being studied). You could construct axioms that resulted in only flipping the prime-numbered dots, or whatever. Was there any mathematically reasonable plausibility argument that the dots would end up all blue for a theory like PA or ZF? Did they expect too much, based on the completeteness of simpler theories like Presburger arithmetic? Or was it just a bogus emotional conviction people had back then, that's alien to us now? 69.228.171.150 (talk) 12:22, 7 November 2009 (UTC)

While waiting for a more technical explanation, my answer to your question is: evidently, at that time it was not at all trivial. Speaking in general, it seems to me that this question reflects a common ahistorical attitude of we contemporary/modern people towards people of the past/ancient people (your post is still correct, I'm just taking the opportunity). We usually say "how could they be so naives to believe something, and not to see what for us is such a triviality". The conclusion is: "we are then smarter than them" (usually only thought), and a certain condescending air. I think I have enough evidence that our brain is not better (experiment: just ask for a formula for the solution of the third degree equation to a mathematician who doesn't know it). If today so many ideas are so easily understandable for so many people of medium intelligence, whereas once they were so difficult and just for a small elite, this seems to me the proof that the people who created these ideas, and those who prepared the ground for them, where giants, and we should have the greatest gratitude and admiration for them. --pma (talk) 14:02, 7 November 2009 (UTC)

Calculus with imaginary constants

Is calculus with imaginary constants, say $\int e^{ix}$, just the same as when the constant is real? 131.111.216.150 (talk) 15:43, 7 November 2009 (UTC)

Broadly. If you want to evaluate an integral like $\int _{a}^{b}f(x)dx$ where f is a complex valued function of a real variable and a and b are reals, then you can just write f as the sum of its real and imaginary parts f = u + iv, and then the integral of f is defined to be the integral of u plus i times the integral of v (as long as the integrals exist). However you do have to be a little careful with things like change of variables. Here's an example: consider the real integral $\int _{\infty }^{\infty }1/(1+x^{4})dx$. Making the substitution y = ix, so dy=i dx doesn't change the limits and, x^4=(y/i)^4=y^4. So $\int _{\infty }^{\infty }1/(1+x^{4})dx=i\int _{\infty }^{\infty }1/(1+y^{4})dy$. It follows the integral is zero....but it obviously isn't. Tinfoilcat (talk) 16:38, 7 November 2009 (UTC)

Mathematical Sequences and inductive reasoning

Use inductive reasoning to write the 10th term in each sequence below. THEN write a formula for the nth term.

a) 4, 13, 22, 31, 40...

b) 3, 5, 9, 17, 33...

c) 0, 6, 24, 60, 120...

d) 5, 10, 17, 28, 47, 82...