E-Government The X-Factor For Government: Fill & Download for Free

This rhetoric that we have nothing nice to say about England needs to stop. It’s embarrassing, and not to independence supporters.A for Austen (And Attenborough)B for Bowie (and Beatles, Black Sabbath and so on)C for Carroll (and Cheddar and Cornwall and so on)D for Dr Who (and Dickens and Dawkins and Darwin, et cetera)E for Idris Elba (I think he’d be an excellent James Bond… there’s another for the B’s)F for Football (although not personally)G for GaskellH for Hawking and Higgs and Hot FuzzI for Iron MaidenJ for Judas PriestK for Keswick (And Kew Gardens)L for Leeds CastleM for Ian McKellen (And Shane Meadows)N for NewtonO for Gary OldmanP for Princess Diana (and Paddington Bear)Q for QueenR for RowlingS for Maggie Smith (Shaun of the Dead, Jenny Saville and Shakespeare… I really love Maggie Smith though)T for Turner (and Turing and This is England)U for UB40V for VondelparkW for White Cliffs of Dover (and Whitesnake)X for X Factor (I hate these sort of shows, irrespective of what country they’re from but a lot of people do like it)Y for YorkZ for Zayn Malik (again not personally a fan but he’s popular and The Zutons)So there you have it. Artists, authors, scientists, areas of natural beauty, sports, structures etc and that’s just off the top of my head.No ones denying any of that. If you want me to be positive about consecutive Tory governments, Boris Johnson and Brexit, then you’re asking too much. That’s not anti English, I’d still hate those things if Scotland asked for it but we didn’t and that’s the issue.g

You are probably familiar with the fact that any polynomial [math]p(X)[/math] with complex coefficients can be written down in the form[math]\displaystyle p(X) = a(X - x_0)(X - x_1)\ldots(X - x_k) \tag*{}[/math]where [math]x_0, x_1, x_2, \ldots x_k[/math] are its roots and [math]a[/math] is a non-zero constant. The Weierstrass factorization theorem tells you how to extend this result to entire functions, which might have infinitely many roots.What is an entire function? It is a complex-valued function [math]f(z)[/math] that is defined on the entire complex plane, such that it has a complex-derivative at every point—that is, for every complex number [math]z_0[/math], the limit[math]\displaystyle \lim_{z \rightarrow z_0} \frac{f(z) - f(z_0)}{z - z_0} \tag*{}[/math]converges to some complex number. Common examples of entire functions include polynomials, [math]e^z[/math], [math]\sin(z)[/math], [math]\cos(z)[/math], and compositions, sums, and products thereof.Using polynomials for motivation, one might think that any entire function [math]f(z)[/math] can be written down in the form[math]\displaystyle f(z) = a \prod_k (z - z_k) \tag*{}[/math]where the [math]z_k[/math] are the roots of [math]f(z)[/math]—there might be infinitely many of those, in general—and [math]a[/math] is a non-zero constant. However, this is certainly wrong, because [math]e^z[/math] doesn’t have any zeros in the complex plane, and it is definitely not constant. In fact, for any non-constant entire function [math]g(z)[/math], [math]e^{g(z)}[/math] will not have any zeros in the complex plane and will not be constant. So, as a minimum, we should change our proposed decomposition to[math]\displaystyle f(z) = e^{g(z)} \prod_k (z - z_k). \tag*{}[/math]However, this is still wrong. The problem is that if the product is infinite, it won’t converge. So, we need to find some nice standard functions that have the desired zeros, but when we multiply an infinite number of them together, we still get something that converges to a complex number. We call these functions the elementary factors, and we define them as follows:[math]\displaystyle E_n(z) = \begin{cases} 1 - z & \text{if } n = 0 \\ (1 - z)e^{z + \frac{z^2}{2} + \frac{z^3}{3} + \ldots + \frac{z^n}{n}} & \text{otherwise.} \end{cases} \tag*{}[/math]The idea is that [math]E_n(z)[/math] has only one root, at [math]z = 1[/math]. The parameter [math]n[/math] is called the degree, and it is part of what governs the growth rate. And so, while I won’t prove it here, the Weierstrass factorization theorem tells you that for any entire function [math]f(z)[/math] with:a root of multiplicity [math]m[/math] at [math]z = 0[/math] (we allow [math]m = 0[/math], if it doesn’t have a root there),non-zero roots [math]z_0, z_1, z_2, \ldots[/math]there exists some entire function [math]g(z)[/math] and non-negative integers [math]k_i[/math] such that[math]\displaystyle f(z) = z^m e^{g(z)} \prod_i E_{k_i}\left(\frac{z}{z_i}\right). \tag*{}[/math]This is already useful, but there is a stronger form that applies if we know something about how quickly [math]f(z)[/math] grows, known as the Hadamard factorization theorem. Suppose there is some finite, positive real number [math]r[/math] such that[math]\displaystyle |f(z)| \leq e^{C|z|^r} \tag*{}[/math]for some constant [math]C > 0[/math] for all complex numbers [math]z[/math] such that [math]|z|[/math] is large enough. We call the infimum of all such [math]r[/math] the order of growth of [math]f(z)[/math]. So, suppose that we have a function [math]f(z)[/math] satisfying the conditions of the Weierstrass factorization theorem, and let’s suppose that [math]f(z)[/math] has order of growth [math]\rho[/math]. Then[math]\displaystyle f(z) = z^m e^{P(z)} \prod_i E_T \left(\frac{z}{z_i}\right), \tag*{}[/math]where [math]T[/math] is the smallest integer greater or equal to [math]\rho[/math], and [math]P(z)[/math] is a polynomial of degree no more than [math]T[/math]!Here is a classic application of this theorem. The function [math]\sin(\pi z)[/math] is entire, with zeros at [math]z = k[/math] for every integer [math]k[/math]. Since[math]\displaystyle \sin(\pi z) = \frac{e^{\pi i z} - e^{-\pi i z}}{2i}, \tag*{}[/math]we see that [math]\sin(\pi z)[/math] has order of growth [math]1[/math]. Therefore, by the Hadamard factorization theorem, for some linear polynomial [math]P(z)[/math],[math]\begin{align*} \sin(\pi z) &= z e^{P(z)} \prod_{k \neq 0} \left(1 - \frac{z}{k}\right) \\ &= z e^{P(z)} \prod_{k = 1}^\infty \left(1 - \left(\frac{z}{k}\right)^2\right) \\ &= a z e^{bz} \prod_{k = 1}^\infty \left(1 - \left(\frac{z}{k}\right)^2\right), \end{align*} \tag*{}[/math]for some complex numbers [math]a,b[/math]. It remains to work out what [math]a[/math] and [math]b[/math] are. We can figure out what [math]a[/math] is by looking at the value of the first derivative at [math]z = 0[/math]:[math]\begin{align*} \pi &= \left.\frac{d}{dz}(\sin(\pi z))\right|_{z = 0} \\ &= \left. a e^{b z} \prod_{k = 1}^\infty \left(1 - \left(\frac{z}{k}\right)^2\right)\right|_{z = 0} \\ &= a. \end{align*} \tag*{}[/math]We can determine what [math]b[/math] is by noting that [math]\sin(-\pi z) = -\sin(\pi z)[/math]—but this is only possible if [math]b = 0[/math]. Therefore, we have proven the famous product formula for [math]\sin(z)[/math]:[math]\displaystyle \sin(\pi z) = \pi z \prod_{k = 1}^\infty \left(1 - \left(\frac{z}{k}\right)^2\right). \tag*{}[/math]This formula was a key component of Euler’s resolution of the Basel problem, which was non-rigorous when he originally wrote it down, but was proven rigorously by later generations of mathematicians.

Different things to different people.A lady in New Hampshire called Trump classy. Personally I don't see it.I don't think Trump has the gravitas to be a head of state. Head of Government? Maybe? But not head state and in this country e combine the two.Some people cannot accept Obama as President. While not the FDR figure I hoped for , he has proven to be cool and confident without being overbearing in difficult times.I don't see that in Trump.I thought Carter getting out of his Limo and walking down Pennsylvania Avenue gave him a "man of the people" quality. His detractors thought it "unpresidential".I thought Reagan was Symbolism over substance. His supporters thought he was Presidential.It is a X factor thing and what works in one era doesn't in another.