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{{DISPLAYTITLE:Root of unity modulo ''n''}}
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In [[mathematics]], a '''''k''-th root of unity modulo ''n''''' for positive [[integer]]s ''k'', ''n''&nbsp;≥&nbsp;2, is a solution ''x'' to the [[congruence]] <math>x^k \equiv 1 \pmod{n} </math>.  If ''k'' is the smallest such exponent for ''x'', then ''x'' is called a '''primitive ''k''-th root of unity modulo ''n'''''.<ref>
{{Cite journal
  | last1 = Finch
  | first1 = Stephen
  | last2 = Martin
  | first2 = Greg
  | last3 = Sebah
  | first3 = And Pascal
  | title = Roots of unity and nullity modulo ''n''
  | journal = Proceedings of the American Mathematical Society
  | volume = 138
  | issue = 8
  | pages = 2729–2743
  | publisher = American Mathematical Society
  | url = http://www.math.ubc.ca/~gerg/papers/downloads/RUNM.pdf
  | accessdate = 2011-02-20}}
</ref>
 
Do not confuse this with a [[Primitive root modulo n|primitive element modulo ''n'']], where the primitive element must generate all [[Unit (ring theory)|units]] of the [[residue class ring]] <math>\mathbb{Z}/n\mathbb{Z}</math> by exponentiation.
That is, there are always roots and primitive roots of unity modulo ''n'' for ''n''&nbsp;≥&nbsp;2, but for some ''n'' there is no primitive element modulo ''n''.  Being a root or a primitive root modulo ''n'' always depends on the exponent ''k'' and the modulus ''n'', whereas being a primitive element modulo ''n'' only depends on the modulus ''n'' — the exponent is automatically <math>\varphi(n)</math>.
 
== Roots of unity ==
 
=== Properties ===
 
* If ''x'' is a ''k''-th root of unity, then ''x'' is a unit (invertible) whose inverse is <math>x^{k-1}</math>. That is, ''x'' and ''n'' are [[coprime]].
* If ''x'' is a unit, then it is a (primitive) ''k''-th root of unity modulo ''n'', where ''k'' is the [[multiplicative order]] of ''x'' modulo ''n''.
* If ''x'' is a ''k''-th root of unity and <math>x-1</math> is not a [[zero divisor]], then <math>\sum_{j=0}^{k-1} x^j \equiv 0 \pmod{n}</math>, because
:: <math>(x-1)\cdot\sum_{j=0}^{k-1} x^j \equiv x^k-1 \equiv 0 \pmod{n}.</math>
 
=== Number of ''k''-th roots ===
 
For the lack of a widely accepted symbol, we denote the number of ''k''-th roots of unity modulo ''n'' by <math>f(n,k)</math>.
It satisfies a number of properties:
 
* <math>f(n,1) = 1</math> for <math>n\ge2</math>
* <math>f(n,\lambda(n)) = \varphi(n)</math> where &lambda; denotes the [[Carmichael function]] and <math>\varphi</math> denotes [[Euler's totient function]]
* <math>n \mapsto f(n,k)</math> is a [[multiplicative function]]
* <math>k\mid l \implies f(n,k)\mid f(n,l)</math> where the bar denotes [[divisibility]]
* <math>f(n,\mathrm{lcm}(a,b)) = \mathrm{lcm}(f(n,a),f(n,b))</math> where <math>\mathrm{lcm}</math> denotes the [[least common multiple]]
* For [[prime]] <math>p</math>, <math>\forall i\in\mathbb{N}\ \exists j\in\mathbb{N}\ f(n,p^i) = p^j</math>. The precise mapping from <math>i</math> to <math>j</math> is not known. If it would be known, then together with the previous law it would yield a way to evaluate <math>f</math> quickly.
 
== Primitive roots of unity ==
 
=== Properties ===
 
* The maximum possible radix exponent for primitive roots modulo <math>n</math> is <math>\lambda(n)</math>, where &lambda; denotes the [[Carmichael function]].
* A radix exponent for a primitive root of unity is a [[divisor]] of <math>\lambda(n)</math>.
* Every divisor <math>k</math> of <math>\lambda(n)</math> yields a primitive <math>k</math>-th root of unity. You can obtain one by choosing a <math>\lambda(n)</math>-th primitive root of unity (that must exist by definition of &lambda;), named <math>x</math> and compute the power <math>x^{\lambda(n)/k}</math>.
* If ''x'' is a primitive ''k''-th root of unity and also a (not necessarily primitive) ℓ-th root of unity, then ''k'' is a divisor of ℓ. This is true, because [[Bézout's identity]] yields an integer [[linear combination]] of ''k'' and ℓ equal to <math>\mathrm{gcd}(k,\ell)</math>. Since ''k'' is minimal, it must be <math>k = \mathrm{gcd}(k,\ell)</math> and <math>\mathrm{gcd}(k,\ell)</math> is a divisor of ℓ.
<!--
* If <math>x</math> is a primitive <math>k</math>-th root of unity and for another exponent <math>l</math> it holds <math>x^l \equiv 1 \pmod{n}</math>, then <math>k</math> is a divisor of <math>l</math>.
-->
 
=== Number of primitive ''k''-th roots ===
 
For the lack of a widely accepted symbol, we denote the number of primitive ''k''-th roots of unity modulo ''n'' by <math>g(n,k)</math>.
It satisfies the following properties:
 
* <math>g(n,k) = \begin{cases} >0 &\text{if } k\mid\lambda(n), \\ 0 & \text{otherwise}. \end{cases}</math>
* Consequently the function <math>k \mapsto g(n,k)</math> has <math>d(\lambda(n))</math> values different from zero, where <math>d</math> computes the [[number of divisors]].
* <math>g(n,1) = 1</math>
* <math>g(4,2) = 1</math>
* <math>g(2^n,2) = 3</math> for <math>n \ge 3</math>, since -1 is always a [[square root]] of 1.
* <math>g(2^n,2^k) = 2^k</math> for <math>k \in [2,n-1)</math>
* <math>g(n,2) = 1</math> for <math>n \ge 3</math> and <math>n</math> in {{OEIS|A033948}}
* <math>\sum_{k\in\mathbb{N}} g(n,k) = f(n,\lambda(n)) = \varphi(n)</math> with <math>\varphi</math> being [[Euler's totient function]]
* The connection between <math>f</math> and <math>g</math> can be written in an elegant way using a [[Dirichlet convolution]]:
:: <math>f = \bold{1} * g</math>, i.e. <math>f(n,k) = \sum_{d\mid k} g(n,d)</math>
: You can compute values of <math>g</math> recursively from <math>f</math> using this formula, which is equivalent to the [[Möbius inversion formula]].
 
=== Testing whether ''x'' is a primitive ''k''-th root of unity modulo ''n'' ===
 
By [[Exponentiation by squaring|fast exponentiation]] you can check that <math>x^k \equiv 1 \pmod{n}</math>.
If this is true, ''x'' is a ''k''-th root of unity modulo ''n'' but not necessarily primitive.
If it is not a primitive root, then there would be some divisor ℓ of ''k'', with <math>x^{\ell} \equiv 1 \pmod{n}</math>.
In order to exclude this possibility, one has only to check for a few ℓ's equal ''k'' divided by a prime.
That is, what needs to be checked is:
:<math>\forall p \text{ prime dividing}\ k,\quad x^{k/p} \not\equiv 1 \pmod{n}.</math>
 
=== Finding a primitive ''k''-th root of unity modulo ''n'' ===
 
Among the primitive ''k''-th roots of unity, the primitive <math>\lambda(n)</math>-th roots are most frequent.
It is thus recommended to try some integers for being a primitive <math>\lambda(n)</math>-th root, what will succeed quickly.
For a primitive <math>\lambda(n)</math>-th root ''x'', the number <math>x^{\lambda(n)/k}</math> is a primitive <math>k</math>-th root of unity.
If ''k'' does not divide <math>\lambda(n)</math>, then there will be no ''k''-th roots of unity, at all.
 
=== Finding multiple primitive ''k''-th roots modulo ''n'' ===
 
Once you have a primitive ''k''-th root of unity ''x'', every power <math>x^l</math> is a <math>k</math>-th root of unity, but not necessarily a primitive one. The power <math>x^l</math> is a primitive <math>k</math>-th root of unity if and only if <math>k</math> and <math>l</math> are [[coprime]]. The proof is as follows: If <math>x^l</math> is not primitive, then there exists a divisor <math>m</math> of <math>k</math> with <math>(x^l)^m \equiv 1 \pmod{n}</math>, and since <math>k</math> and <math>l</math> are coprime, there exists an inverse <math>l^{-1}</math> of <math>l</math> modulo <math>k</math>. This yields <math>1 \equiv ((x^l)^m)^{l^{-l}} \equiv x^m \pmod{n}</math>, which means that <math>x</math> is not a primitive <math>k</math>-th root of unity because there is the smaller exponent <math>m</math>.
 
That is, by exponentiating ''x'' one can obtain <math>\varphi(k)</math> different primitive ''k''-th roots of unity, but these may not be all such roots. However, finding all of them is not so easy.
 
=== Finding an ''n'' with a primitive ''k''-th root of unity modulo ''n'' ===
 
You may want to know, in what integer [[residue class ring]]s you have a primitive ''k''-th root of unity.
You need it for instance if you want to compute a [[Discrete Fourier Transform]] (more precisely a [[Number theoretic transform]]) of a <math>k</math>-dimensional integer [[Vector (mathematics and physics)|vector]].
In order to perform the inverse transform, you also need to divide by <math>k</math>, that is, ''k'' shall also be a unit modulo <math>n</math>.
 
A simple way to find such an ''n'' is to check for primitive ''k''-th roots with respect to the moduli in the [[arithmetic progression]] <math>k+1, 2k+1, 3k+1, \dots</math>.
All of these moduli are coprime to ''k'' and thus ''k'' is a unit. According to [[Dirichlet's theorem on arithmetic progressions]] there are infinitely many primes in the progression,
and for a prime <math>p</math> it holds <math>\lambda(p) = p-1</math>.
Thus if <math>mk+1</math> is prime then <math>\lambda(mk+1) = mk</math> and thus you have primitive ''k''-th roots of unity.
But the test for primes is too strong, you may find other appropriate moduli.
 
=== Finding an ''n'' with multiple primitive roots of unity modulo ''n'' ===
 
If you want to have a modulus <math>n</math> such that there are primitive <math>k_1</math>-th, <math>k_2</math>-th, ..., <math>k_m</math>-th roots of unity modulo <math>n</math>, the following theorem reduces the problem to a simpler one:
 
: For given <math>n</math> there are primitive <math>k_1</math>-th, ..., <math>k_m</math>-th roots of unity modulo <math>n</math> if and only if there is a primitive <math>\mathrm{lcm}(k_1, ..., k_m)</math>-th root of unity modulo ''n''.
 
; Proof
 
Backward direction:
If there is a primitive <math>\mathrm{lcm}(k_1, ..., k_m)</math>-th root of unity modulo <math>n</math> called <math>x</math>, then <math>x^{\mathrm{lcm}(k_1, \dots, k_m)/k_l}</math> is a <math>k_l</math>-th root of unity modulo <math>n</math>.
 
Forward direction:
If there are primitive <math>k_1</math>-th, ..., <math>k_m</math>-th roots of unity modulo <math>n</math>, then all exponents <math>k_1, \dots, k_m</math> are divisors of <math>\lambda(n)</math>. This implies <math>\mathrm{lcm}(k_1, \dots, k_m) \mid \lambda(n)</math> and this in turn means there is a primitive <math>\mathrm{lcm}(k_1, ..., k_m)</math>-th root of unity modulo <math>n</math>.
 
== See also ==
* [[Root of unity]]
* [[Primitive root modulo n|Primitive root modulo ''n'']]
 
== References ==
{{reflist}}
 
[[Category:Modular arithmetic]]

Latest revision as of 22:31, 3 September 2014

A couple of weeks ago I bought a Gerber Bear Grylls survival knife to try out. To be truthful, I'd never observed the guy's show, so I watched element of an episode on Youtube to see what it was about. If you loved this article and you would like to receive more info pertaining to the pocket knife i implore you to visit the web page. If he actually did half the crazy stuff in a genuine emergency survival scenario that he does on his show he'd be dead inside a week. Not that he doesn't know his stuff, but the issues these shows do for high ratings would get most individuals killed.



Image your self in a survival scenario. Possibly you got injured on a solo hunting trip or perhaps you became disoriented even though walking via a forest and got lost. Regardless of how you finish up in a survival predicament, a knife is 1 of the couple of tools that will give you a fighting likelihood at making it out alive. I now discover that I use this knife as an axe, and I routinely attain for it as an alternative of my camp axe. It is an excellent for any survival pack. Nevertheless, I would not take into account it an everyday carry knife A fixed blade knife is extra durable and trustworthy than a folding knife. Whilst I like a superior folder for Each Day Carry (EDC), a fixed blade has the upper hand when it comes to meeting the demands a survival predicament could possibly present.

My name is Josh and I am a fan of the outdoors. On the weekends you will obtain me hiking or attempting to total any of the many adventures I have on my list. I live in Southern California and really like the deversity that permits me to have in my each day life. I am part of the OutdoorPros Adventure Group. I also really like photography and you can obtain me blogging on my California Travel Blog or Our Exceptional Globe Photo Blog Bear Grylls Survival Knife Review , 8.5 out of 10 primarily based on 1 rating

The black sections on the manage are a rubber sleeve/coating that provide traction when holding. The orange sections are a hard grivory-like material, primarily ABS plastic. The knife seems to be fairly indestructible, at least based on my experience. The Bear Grylls Ultimate Knife, a new solution from Gerber, has two holes in its handle, allowing a user to lash it to a stake and create a spear. The finish of the manage is textured and blunt, created for pounding. Integrated into the sheath, a little magnesium rod lets you draw the blade and strike metal to develop hot sparks and begin a fire. At seven inches extended, this blade is a great deal also unwieldy to use for whittling. Nevertheless, with this knife being the heaviest it created for easy function chopping at wood.