Wikipedia:Reference desk/Mathematics: Difference between revisions

::Hence, writing <math>\sqrt2</math> by abuse of notation for <math>\omega(1-\omega^2),</math> we have:

:::<math>\begin{alignat}{2}7&=(3+\sqrt2)(3-\sqrt2)\\&=(2\sqrt2+1)(2\sqrt2-1)\\&=(5+3\sqrt2)(5-3\sqrt2)\\&=(4\sqrt2+5)(4\sqrt2-5).\end{alignat}</math>

:::So which primes <math>p</math> are still primes in the ring <math>Z[e^{\frac{\pi i}{4}}]</math>? How about <math>Z[e^{\frac{\pi i}{5}}]</math> and <math>Z[e^{\frac{\pi i}{6}}]</math>? [[Special:Contributions/220.132.216.52|220.132.216.52]] ([[User talk:220.132.216.52|talk]]) 06:32, 9 December 2024 (UTC)

::::As I wrote, this is only a minuscule contribution. We do not do research on command; in fact, we are actually not supposed to do any original research here. &nbsp;--[[User talk:Lambiam#top|Lambiam]] 09:23, 9 December 2024 (UTC)

 

: I'll state things a little more generally, in the cyclotomic field <math>\mathbb Q[e^{2\pi i /n}]</math>. (Your n is twice mine.) A prime q factors as <math>q = (q_1\cdots q_r)^{e_r}</math>, where each <math>q_i</math> is a prime ideal of the same degree <math>f</math>, which is the least positive integer such that <math>q^f \equiv 1\pmod n</math>. (We have assumed that q does not divide n, because if it did, then it would ramify and not be prime. Also note that we have to use ideals, because the cyclotomic ring is not a UFD.) In particular, <math>q</math> stays prime if and only if <math>q</math> generates the group of units modulo <math>n</math>. When n is a power of two times an odd composite, the group of units is not cyclic, and so the answer is ''never''. When n is a prime or twice a prime, the answer is when q is a primitive root mod n. If n is 4 times a power of two times a prime, the answer is ''never''. [[User:Tito Omburo|Tito Omburo]] ([[User talk:Tito Omburo|talk]]) 11:08, 8 December 2024 (UTC)

:::This equation is just the definition of function ''g''. For instance if function ''f'' has the inverse function ''f''⁻¹ then we have ''g(x)=x''. [[User:Ruslik0|Ruslik]]_[[User Talk:Ruslik0|<span style="color:red">Zero</span>]] 20:23, 8 December 2024 (UTC)

::: If f is the temperature, and g is the evolution of an ensemble of particles in thermal equilibrium (taken at a single time, say one second later), then because temperature is a function of state, one has <math>f(x)=f(g(x))</math> for all ensembles x. Another example from physics is when <math>g</math> is a Hamiltonian evolution. Then the functions <math>f</math> with this property (subject to smoothness) are those that (Poisson) commute with the Hamiltonian, i.e. "constants of the motion". [[User:Tito Omburo|Tito Omburo]] ([[User talk:Tito Omburo|talk]]) 20:33, 8 December 2024 (UTC)

:Let <math>f</math> be a function from <math>X</math> to <math>Y</math> and <math>g</math> a function from <math>X</math> to <math>X.</math> Using the notation for [[function composition]], the property under discussion can concisely be expressed as <math>f\circ g=f.</math> An equivalent but verbose way of saying the same is that the [[preimage]] of any set <math>B \subseteq Y</math> under <math>f</math> is [[Closure (mathematics)|closed]] under the application of <math>g.</math> &nbsp;--[[User talk:Lambiam#top|Lambiam]] 08:54, 9 December 2024 (UTC)

 

== IEEE Xplore paper claim to acheive exponentiation inversion suitable for pairing in polynomial time. Is it untrustworthy ? ==

 

= December 9 =