Let S = K[r,s,t,u,v,w,x,y} and I the ideal of S generated by f = sy-ux , g = rw-tv , h = rt-su ,Show that there exist no monomial order < o S such that {f,g,h } I a Grobner basis of I with respect to < .

— Preceding unsigned comment added by 182.187.77.255 (talk • contribs) 04:56, 5 July 2013‎

If you want some hints, you should present the context more clearly.

I suppose that you intend K to be a field, and K[r,s,t,u,v,w,x,y} [SIC!] the commutative polynomial ring over that field. Is that correct?

In what kind of coursis or other context did you get the question? Specifically, have you access to tools to decide whether or not S is a Koszul algebra? (Note, that if the "denominator" ideal for a homogeneous ring of this type has a Gröbner basis with only quadratic (homogeneous) elements, then it is necessarily Koszul.) JoergenB (talk) 21:39, 5 July 2013 (UTC)[reply]

${\displaystyle \int _{0}^{\infty }{b(x) \over B(x)}\ dx\ =\ ?\ ,\qquad \qquad b(x)=\sum _{n=1}^{\infty }{n^{x} \over n^{n}}\ ,\qquad \qquad B(x)=\sum _{n=1}^{\infty }{n^{x} \over n!}}$

Observations:

1. B(x) is e times a Bell number.

2. If the numerator of each function would be xⁿ instead of n^x, and each sum would start at 0 instead of 1, then the value of our integral would become ${\displaystyle \sum _{n=0}^{\infty }{n! \over n^{n}}\ ,}$ where lim nⁿ → 1 when n → 0.

— 79.113.210.135 (talk) 20:45, 6 July 2013 (UTC)[reply]

You could try to use Ramanujan's master theorem. Count Iblis (talk) 21:40, 6 July 2013 (UTC)[reply]

In that case, s = 1 and ${\displaystyle f(x)={\frac {b(x)}{B(x)}}\ ,}$ whose Taylor expansion is absolutely catastrophic... 79.113.210.135 (talk) 23:11, 6 July 2013 (UTC)[reply]

To make matters even more complicated than they already are, integration by parts would seem to suggest that the integral is divergent if any of the two sums starts from n = 1, despite numerical data indicating the contrary... The question now would be whether starting at n = 2 would really make that much of a difference anyway... 79.113.225.68 (talk) 13:58, 7 July 2013 (UTC)[reply]

Drop it. Looie496 (talk) 18:21, 7 July 2013 (UTC)[reply]

Yeah, drop it. It makes no sense. —SeekingAnswers (reply) 12:59, 8 July 2013 (UTC)[reply]

As it stands, what you've written doesn't make sense. However, if you'd like to discuss your ideas and try and clarify them down to what you are wanting to express, I'd be happy to help on my talk page. Even if this particular conception doesn't work out to anything, I'm sure there are some great articles and books we can find on the topics that lead you to it (and this could help you refine what you want to say into actual mathematical language). Just drop me a line on my talk page:-)Phoenixia1177 (talk) 06:23, 9 July 2013 (UTC)[reply]

Like Very charitable of you! SemanticMantis (talk) 14:09, 9 July 2013 (UTC)[reply]

Given, ${\displaystyle x+1/x+y+1/y=6}$: Using this condition, there are two related but different questions:

i) ${\displaystyle x/y+y/x=6}$, then ${\displaystyle x-1/x+y-1/y=?}$

ii) ${\displaystyle xy+1/xy=3}$, then ${\displaystyle x-y+1/y-1/x=?}$

Another question from the same book, this one also I can't make it through: If ${\displaystyle x^{2}+2=2x;x^{4}-x^{3}+x^{2}+1=?}$

Now if you take out the values of x,y that's cheating, you have to find out the answers by manipulating the equations, not the long way. These are the questions of my book which I can't solve. Please help. 117.197.64.60 (talk) 13:20, 9 July 2013 (UTC)[reply]

I added <math> formatting for ease of interpretation – I hope you do not object. — Quondum 13:41, 9 July 2013 (UTC)[reply]

These questions seem to be aimed at building the ability to spot a way forward intuitively. In you last example, you should be able to spot that subexpressions in the latter part can be simplified using the former. We would generally try to get rid of variables, and a useful approach is often reduce the degree of a polynomial. A rather straightforward approach is to rewrite the first as ${\displaystyle x^{2}-2x+2=0}$, and then to add any multiple of the LHS to the unknown expression that will cancel its highest power. At worst you'll simplify it to a first-degree polynomial.

The first question presumably is aimed at a similar simple manipulation of the given equations to allow substitutions of subexpressions, with the result of simplifying the expression. This approach is often simpler than solving for one of the unknowns. — Quondum 14:06, 9 July 2013 (UTC)[reply]

The whole idea is to multiply all equations with the product xy. And then to factorize them so as to bring out s = x + y and p = xy. Like this:

x + ${\displaystyle {\tfrac {1}{x}}}$ + y + ${\displaystyle {\tfrac {1}{y}}}$ = 6 | * xy

x²y + y + xy² + x = 6 xy

(x + y)(xy + 1) = 6 xy

s (p + 1) = 6p

s = ${\displaystyle {\tfrac {6p}{p+1}}\quad {\color {white}.}}$ (1)

i)

${\displaystyle {\tfrac {x}{y}}}$ + ${\displaystyle {\tfrac {y}{x}}}$ = 6 | * xy

x² + y² = 6 xy | + 2 xy

(x + y)² = 8 xy

s² = 8p

p = ± ${\displaystyle {\tfrac {s}{2{\sqrt {2}}}}\quad {\color {white}.}}$ (2)

By replacing the value of p obtained from (2) into (1) we deduce the value of s, and then that of p.

x - ${\displaystyle {\tfrac {1}{x}}}$ + y - ${\displaystyle {\tfrac {1}{y}}}$ = z | * xy

x²y - y + xy² - x = z xy

(x + y)(xy - 1) = z xy

s (p - 1) = z * p

z = s ${\displaystyle {\tfrac {p-1}{p}}}$ = ...

Similarly for case ii). — 79.118.170.186 (talk) 16:37, 9 July 2013 (UTC)[reply]

What does the term "complete" mean in this context and how does it contrast with "consistent"? — Melab±1 ☎ 04:37, 11 July 2013 (UTC)[reply]

Consistent means that a contradiction can't be derived. Complete means that every statement that can be formed in the system can be proven to be either true or false. Bubba73 ^{You talkin' to me?} 04:45, 11 July 2013 (UTC)[reply]

See complete theory and consistency. Gandalf61 (talk) 12:15, 11 July 2013 (UTC)[reply]

Are there any findings about when the sum of kth powers of three or more positive integers can equal to another positive integer's kth power?like that in this book

Do Fermat's Last Theorem also apply to complex numbers?--128.237.207.243 (talk) 21:29, 11 July 2013 (UTC)[reply]

Are you referring to complex numbers of the form Z², whose real and imaginary parts are both integers ? — 79.113.215.215 (talk) 23:11, 11 July 2013 (UTC)[reply]

Also known as Gaussian integers.--80.109.106.49 (talk) 23:12, 11 July 2013 (UTC)[reply]

Yes, it should be --128.237.137.221 (talk) 23:33, 11 July 2013 (UTC)[reply]

See

Euler's sum of powers conjecture#Generalizations

Beal's conjecture

Jacobi–Madden equation

Prouhet–Tarry–Escott problem

Taxicab number

and these relevant Google results for its extension to Gaussian integers

— 79.113.210.229 (talk) 01:37, 12 July 2013 (UTC)[reply]

Also Pythagorean quadruple. Duoduoduo (talk) 18:56, 14 July 2013 (UTC)[reply]

If cosecά-sinα = a³ and secα-cosα = b³ then prove that a²b²(a² + b²)=1 — Preceding unsigned comment added by ManishRaichand (talk • contribs) 20:29, 12 July 2013 (UTC)[reply]

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. Dolphin (t) 12:12, 13 July 2013 (UTC)[reply]

As a hint, putting t=tan(α/2) will lead to (ab)^3=sinαcosα, a/b=cosα/sinα, from which it is easy to obtain the expression which is to equal unity.86.140.134.100 (talk) 21:49, 13 July 2013 (UTC)[reply]

Suppose I normally use a quick algorithm for my RNG generator in software. I think it's very good. I have a random hardware generator that's given to me by the NSA and told "trust us, there's no backdoor in here." I also have a lava lamp.

So, now I will combine these streams by xor'ing the next bit from all three to get my next random bit.

Isn't it true now true that if any ONE of the three are secure, the stream is truly random?

Meaning, the fact htat I've included all three can *only* increase the entropy, and cannot possibly decrease it? Even if it were so broken that one of the three only always returns 1? Even if one of them is explicitly included to try to break the randomness of hte stream?

The argument being that xor is commutative, so that you can arrange any of the three items in the xor stream to come "last". At that point it MUST act like a OTP (one time pad) and completely hide the 'brokenness' of the other two, with the key immediately being discarded since it is not saved anywhere. The brokenness HAS to disappear, otherwise it would in some statistical or other sense be present in the 'ciphertext' that is the output of the OTP performed by the 'good' source in the chain.

So: weak software rng -> xor -> lava lamp that actually always ends up returning 1 due to my bug -> xor -> NSA provided backdoor hardware rng

By including hte hardware RNG in this stream, I could *ONLY* make the stream stronger. But I could not make it broken if it's not already broken; it could 'only' save the day. (For example if there's some regularity in my weak software rng), but by some miracle the NSA provided box really is truly random.

what do you think? 178.48.114.143 (talk) 23:14, 13 July 2013 (UTC)[reply]

If one is good and the other two aren't, you should be fine. But in principle you could go wrong if two are good independently but are correlated with each other. If you can rule that out, I think you should be okay. Looie496 (talk) 00:46, 14 July 2013 (UTC)[reply]

Surely if you have a real valued function which is differentiable but not smooth, for example

${\displaystyle f(x)={\begin{cases}-x^{2},&{\mbox{if }}x<0\\x^{2},&{\mbox{if }}x\geq 0\end{cases}}}$

and you define a complex valued function simply by

${\displaystyle g(z)=f(\operatorname {Re} (z))}$

g is a complex differentiable function which is not complex smooth. 130.56.84.56 (talk) 08:07, 14 July 2013 (UTC)[reply]

Actually, g wouldn't even be differentiable; unless f is constant. If you look at our article, Cauchy–Riemann equations, you'll see that for f(x + yi) = u(x, y) + iv(x, y) and u, v differentiable; then f is diff. iff u, v satisfy the equations. In your case, both partial derivs. of v are 0 since v is 0. Moreover, u only depends on x and has deriv. wrt x of 0, so it must be constant. If you look at our articles, Smooth function, Holomorphic function, and the stub Looman–Menchoff theorem; complex differentiable, basically, implies complex analytic and smooth.Phoenixia1177 (talk) 08:34, 14 July 2013 (UTC)[reply]

Of course. Differentiable in ${\displaystyle \mathbb {R} ^{2}}$ does not mean differentiable in ${\displaystyle \mathbb {C} }$. 130.56.84.56 (talk) 08:50, 14 July 2013 (UTC)[reply]

I don't understand your response, where you agreeing or objecting? (I'm autistic and that trips me up sometimes)Phoenixia1177 (talk) 09:31, 14 July 2013 (UTC)[reply]

I was agreeing. The limit ${\displaystyle \lim _{z\rightarrow 0}{\frac {g(z)}{|z|}}=0}$ exists, but the limit ${\displaystyle \lim _{z\rightarrow 0}{\frac {g(z)}{z}}}$ does not. 130.56.84.56 (talk) 11:13, 14 July 2013 (UTC)[reply]

The page at http://www.flexprojector.com/gallery/gallery.html shows some map projections created by Flex Projector. They only give the coefficient tables that define the projection, and no formula is given to compute the x-y coordinates from long-lat spherical coordinates given the tables supplied. I already know how to compute for pseudocylindrical projections, but what do the parallel curvature value do? Czech is Cyrillized (talk) 15:27, 14 July 2013 (UTC)[reply]

The article mentioned in the text can be found at [1]. That might help. It looks like they divide the latitude into 5° increments and there is one value of Length of Parallels (l), Distance of Parallels from Equator (d), Bending (b) and Meridians Distribution (m). Calculate these for your given latitude, let θ be the longitude. Ignoring bending and Meridians Distribution for a moment the coords will be x=l*θ and y=d. It looks like the bending follows a sine or cosine curve, so you may have y=d+b*cos(θ). Not quite sure how the Meridian distribution works.--Salix (talk): 16:55, 14 July 2013 (UTC)[reply]

What about cubic and quadratic bending? Czech is Cyrillized (talk) 00:08, 15 July 2013 (UTC)[reply]

Then I guess you have a cubic or quadratic equation y=d + α θ + β θ² + γ θ³.--Salix (talk): 04:04, 15 July 2013 (UTC)[reply]

Hello. Suppose I have ten counters to distribute as I wish among five squares, numbered 1,2,3,4,5. Then a random number 1-5 is called repeatedly, with known probabilities p1 to p5 (p1+...+p5=1) and if there is a counter on the corresponding square I remove it.

I want to place my counters so that they are all removed as quickly as possible (at first I thought that the optimal strategy was to put all your counters on the square with the maximal probability, but my logic was flawed). Actually I want to minimize the expected time to remove the last counter. Is this question a special case of a well-known problem? I thought it might be something like the secretary problem but it isn't. Can anyone advise? Robinh (talk) 20:35, 14 July 2013 (UTC)[reply]

I don't know whether this problem has been named or studied, but it is an interesting question. I think the optimal placing of counters will depend on the probabilities. If the probabilities are more or less equal then placing all your counters on one square is not a good strategy. But if the probabilities are very unequal - say p1 is close to 1 and p2 to p5 are very small - then placing all you counters on one square may indeed be the best strategy. A quick analysis of the 2 squares and 2 counters case (if stuck, consider a simpler case) shows that the expected time to remove the last counter is 2/p if you put both counters on square 1, and (p^2 - p + 1)/p(1-p) if you put one counter on each square (p is the probability of taking a counter from square 1). If p is less than 1/phi then the "one on each square" strategy is better; if p is greater than 1/phi then "both on square 1" is better. Gandalf61 (talk) 10:20, 15 July 2013 (UTC)[reply]

(OP) thanks Gandalf. I hadn't got round to the 2-square problem yet. But I realized last night that the problem was related to the coupon collector's problem but that doesn't have the element of choice at the beginning as we have here. Best wishes, Robinh (talk) 19:45, 15 July 2013 (UTC)[reply]

Why is it flawed logic to put all your counters on the highest probability square? (phrased that badly as it seems like I'm doubting the maths, I just don't get it intuitively) --Iae (talk) 13:16, 16 July 2013 (UTC)[reply]

... and I take n cuts at it. What is the probability that I can form a shape from the pieces? (i.e. the cuts are completely random along the length and I wish to be able to construct a polygon by rotating and translating the line segments resulting from the cut, using all of them). I thought it was ${\displaystyle 1-1/2^{n-1}}$ as after the first cut it would be that to prevent the forming of the polygon would have to fall in the same half as the original cut, but it was pointed out to me that if the first cut fell near the end of the length there would be more than half of the line a cut could fall in and still leave a length greater than half the original length.--Gilderien Chat|List of good deeds 21:08, 14 July 2013 (UTC)[reply]

This thread at MathOverflow is mostly about the triangle case (n=2), but has a reference to an American Mathematical Monthly article, and says the general answer is ${\displaystyle 1-{n}/2^{n-1}}$. AndrewWTaylor (talk) 11:14, 16 July 2013 (UTC)[reply]

Small correction, the AMM article uses n as the number of sides of the polygon, so in the terms used above (n= number of cuts) the answer is ${\displaystyle 1-(n+1)/2^{n}}$. MChesterMC (talk) 12:41, 16 July 2013 (UTC)[reply]

Hello,

I am a graduate student at Old Dominion University in Norfolk, VA. I am trying to run a Monte Carlo simulation that includes the generation of pseudo-random correlated bivariate numbers (or points). The Wikipedia entry for copulas (http://en.wikipedia.org/wiki/Copula_%28probability_theory%29#Empirical_copulas) lists a procedure for generating random samples from a multivariate distribution. The first step is:

1. Draw a sample (U_1^k,...,U_d^k) ~ C (k=1,...,n) of size n from the copula C

I don't how to do this and have unable to find any information anywhere else about it. Could you provide additional information about how to this, or point me to a site that describe how to do this?

Thank you for your help.

Wilfred

173.50.125.48 (talk) 02:42, 16 July 2013 (UTC)[reply]

1,1,4,27,256,3125,46656,823543,8^8,9^9,...if you take enough successive differences, you will get the sequence 0,3,17,169,2079,31261,554483,... I learned today on OEIS that the second sequence is called the binomial transform of the first sequence. Question 1) Analogy with results on symmetric group: The {n!} sequence gets the binomial transform sequence 0,1,2,9,44,.. which are # of permutations in Symm(n) with no fixed points (derangements). So my first question is, as an analogy with permutations, are the numbers, 0,3,17,169,.. likewise an enumeration per n of a distinguished subset of functions on the set {1,2,3,..n}? I've fiddled with various sets of functions, but, for example, I haven't yet found a special 17 element subset of the 27 functions on {1,2,3}. Of course I've looked at functions with no, or at least few, fixed points. Question 2) It is wellknown that the ratio |Symm(n)|/#derangements goes to e ~ 2.718281828 as n goes to infinity. That is, the sequence of ratios 2/1, 6/2, 24/9, 120/44... has e as the limit. So, my second question is, as an anology with the permutation ratio going to e, what the sequence of ratios approach in the functions case? It's about 1.47 and decreasing with n, perhaps it's agm(1,2)? (To be clearer, I'm talking about the ratios 4/3,27/17,256/169,3125/2079,46656/31261,...). So, thanks in advance.Rich (talk) 11:40, 16 July 2013 (UTC)[reply]