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March 10

How do you prove that $x$ generates the multiplicative group of $GF(p)[x]/f(x)$ if $f(x)$ is a primitive polynomial?

How do you prove that $x$ ~~(and any element in $GF(p)[x]/f(x)$ other than the additive identity)~~ generates the multiplicative group of $GF(p)[x]/f(x)$ if $f(x)$ is a primitive polynomial?

If it is fairly easy to prove, I'd like to give it a try, but could use some hints. On the other hands, if it is difficult, could someone point me to a complete proof? Thanks. --98.114.146.242 (talk) 02:57, 10 March 2010 (UTC)[reply]

Well, x generates the field because it is a root of a primitive polynomial... However, for example, if the multiplicative group of the field has size 8 and z is a generator, then z⁴ cannot possibly be a generator, because z⁴ will have order 2. So it will not be true that "any element in

GF(p)[x]/f(x)

other than the additive identity" will generate the multiplicative group. In general, a cyclic group of size n has only φ(n) generators. — Carl (CBM · talk) 03:04, 10 March 2010 (UTC)[reply]

You're right. I confused a special case with the general case. --98.114.146.242 (talk) 03:28, 10 March 2010 (UTC)[reply]

Project Euler problem

In this problem, how many simulations of random walks would you expect to have to do in order to get a result accurate to 6 decimal places? (and yes, I am aware this isn't a good way to solve the problem). 149.169.164.152 (talk) 07:11, 10 March 2010 (UTC)[reply]

That's easier to estimate once you've done a few simulations. You'll get very rough estimates of the mean

\mu

and standard deviation

\sigma

. If you then do n simulations and take their average, the result will be distributed normally with expectation

\mu

and standard error

\sigma /{\sqrt {n}}

. Let

\mu =m\cdot 10^{e}

,

1<m<10

. To be highly confident the result is accurate to 6 s.f., you'll want the SE to be about

10^{e-6}

, so

n=\sigma ^{2}10^{12-2e}\approx {\frac {\sigma ^{2}}{\mu ^{2}}}10^{13}\approx {\frac {{\hat {\sigma }}^{2}}{{\hat {\mu }}^{2}}}10^{13}

. If the distribution of the experiment result is roughly exponential, expect the required number to be in the trillions. -- Meni Rosenfeld (talk) 08:27, 10 March 2010 (UTC)[reply]

Oh, decimal places, not s.f.. Then what you really want is an SE of

10^{-7}

, which means

n\approx {\hat {\sigma }}^{2}10^{14}

, which is much higher (order of

10^{20}

, maybe?) -- Meni Rosenfeld (talk) 08:35, 10 March 2010 (UTC)[reply]

Is C^∞ dense between the measure preserving transformations?

My problem is:

Suppose we have a diffeomorphism T which preserves the Lebesgue measure and is C³. Is it possible to find a T' which is C³-close to T, preserves the Lebesgue measure and is C^∞? Usually density arguments make use of convolution but I guess this would destroy the property of preserving the measure. How can I do?--Pokipsy76 (talk) 10:39, 10 March 2010 (UTC)[reply]

At least in the case of a C³ diffeo T:M→M on a C^∞ compact connected differentiable manifold M with a volume form α, I think this follows quite immediately by a celebrated theorem by Jürgen Moser [1], that states that if α and β are two volume forms with the same total mass on such an M, there exists a C^∞ diffeo f:M→M such that β=f^*(α), and in fact f can be chosen arbitrarily C^∞ close to the identity provided α and β are suitably close. Now, take a C^∞ diffeo S which is C³ close to T. Then S^*(α) is a volume form close to α and has the same total mass. By Moser theorem with β=S^*(α) there exists a C^∞ diffeo f:M→M close to the identity such that S^*(α)=f^*(α), so (Sf ^-1)^*(α)=α, which mans that Sf ^-1 is a C^∞ diffeo C³ close to T which preserves α, as you wanted. I strongly suggest you to read Moser's beautiful proof, that may give you several hints. It has been generalized to various cases (manifold with boundary, non-compact, &c). You should work out the details of what I sketched, but I think it's ok and I hope it's of help. It is possible that there is a more direct proof, but I guess in any case one has to pass close to some of Moser's proof's lemmas. --pm a 21:15, 11 March 2010 (UTC)[reply]

Thank you very much, I think this article will be useful.--Pokipsy76 (talk) 13:10, 12 March 2010 (UTC)[reply]

Another thing. Let F:R×R^m→R^m be a time-dependent field with with vanishing divergence (wrto the x variable), and such that its integral flow

\scriptstyle \partial _{t}g(t,x)=F(t,g(t,x))

\scriptstyle g(0,x)=x

is defined at time 1 for any initial value x.

Then it's clear that the map at time 1, T(x):=g(1,x) is a diffeomorphism that preserves the Lebesgue measure. Conversely, any measure preserving diffeo is of this form -I don't have a reference here but it is a standard result (not completely elementary; it uses Moser's theorem and a few facts about fibrations); I am pretty sure that you can do it for any T of class C³, like the ones you are treating. Of course, such a map T can be very easily approximated by a C^∞ measure preserving diffeo: just take the integral flow of a regularized field with null divergence (convolution now works well, because it gives a fiels with null divergence). So if you find a ready reference, the approximation should follow very easily. Hope this help. --pm a 22:13, 12 March 2010 (UTC)[reply]

'Extension' of a linear transformation?

Say I have a linear transformation of R², $J$ , and a linear transformation of R⁴, $A=\left({\begin{array}{c|c}J&0\\\hline 0&K\end{array}}\right),$ is it fair to call A an 'extension' of J, since, considering R² a subset of R⁴, J and A bring about the same transformation on this subset? Is there a proper term for this sort of situation? Thanks, Icthyos (talk) 17:03, 10 March 2010 (UTC)[reply]

You could say that J the restriction of A to R². This is term has pretty general application, see Domain of a function#Formal definition. I doubt the specific situation you describe has a name.--RDBury (talk) 03:03, 11 March 2010 (UTC)[reply]

Thanks, that makes sense. Restriction will do nicely! Icthyos (talk) 11:48, 11 March 2010 (UTC)[reply]

Actually A an extension of J (or J a restriction of A, as said above) is fair, exactly with the justification you wrote, and is indeed common language for linear operators also. Note that you can say the same even if there's only one zero-blok in the matrix (which one: NE or SW? ;-) ). If you want to describe the diagonal decomposition, you may also say that J is a factor of A, or that A splits into the direct sum of J and K, and similar expressions. --pm a 16:46, 11 March 2010 (UTC)[reply]

Well, as it stands, A is an extension of both J and K - if only the NE block is zero, it remains an extension of J, but no longer for K, and if only the SW block is zero, it is still an extension of K, but not of J, correct? I suppose I'll just pick my favourite expression and run with it, thanks! Icthyos (talk) 21:57, 11 March 2010 (UTC)[reply]

Complex Analytic Functions

Let's say I have a function f : C --> C, which is infinitely differentiable in a neighbourhood of $z_{0}$ . I define a new sequence of polynomial functions ${f_{n}}$ given by

f_{n}(z):=\sum _{k=0}^{n}{\frac {f^{(k)}(z_{0})}{k!}}(z-z_{0})^{k}.

I want to consider the new sequence of functions ${g_{n}}$ given by $g_{n}(z):=|f(z)-f_{n}(z)|.$ I want to know how to investigate the limit of the sequence ${g_{n}(z_{0})}$ as n tends to infinity. I think this is called the pointwise convergence. Ultimately I'd like to investigate the uniform convergence. But I want some explicit methods. A nice worked example would be great. Let's say $z_{0}=0$ and $f(z)=e^{\sin(z)}$ . I know that $f(z)$ is complex analytic, a.k.a. holomorphic, and so I know that it's equal to it's power series. I just use this as a familiar non-trivial example for the explicit computations. •• Fly by Night (talk) 19:53, 10 March 2010 (UTC)[reply]

it's not quite clear if you are talking of an f defined everywhere on C, but only C^∞ in a nbd of 0. If so, it is a rather strange assumption, and you should be aware that there is no relation at all between f and the f_n in the large (i.e. out of the neighborhood Ω you are talking of). Besides, it is not clear if you mean real or complex differentiability. In the first case, you can have a C^∞ function with f(x)≠0 for all x≠0, and all the f_n identically 0. If you mean infinitely differentiable in the complex sense, then complex differentiability (that is f holomorphic) is sufficient (it's equivalent) ad implies that f_n converges to f uniformly on all closed disks centered at z₀ and contained in Ω. You can't have uniform convergence on C even if f is holomorphic on C (entire) unless f itself is a polynomial (in which case the sequence of the f_n stabilizes and the uniform convergence is trivially true). For instance, the function you wrote is bounded and nonconstant, whereas a nonconstant polynomial is not. So the uniform distance is infinite. --pm a 16:36, 11 March 2010 (UTC) (PS: above I meant: bounded on R, sorry)[reply]

I gave an explicit example:

z_{0}=0

and

f(z)=e^{\sin(z)}

. I also said that I know that

f(z)

is complex analytic, a.k.a. holomorphic, and so I know that it's equal to it's power series. I just used this as a familiar non-trivial example for the explicit computations. •• Fly by Night (talk) 21:47, 12 March 2010 (UTC)[reply]

Ok, then for this particular entire function the polynomials f_n converge to f uniformly on every disk, but do not converge uniformly on C, for the reason I wrote (it is bounded on R whereas the f_n are not. It was not clear to me what you were asking exactly. Is it all clear now? --pm a 22:50, 12 March 2010 (UTC)[reply]