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| | Lawn care is vital that most women. Sure, there are some people who have three foot tall grass in their yards, but most people getting a nice looking yard. However in order to take a nice yard, you in order to be care for. It takes an effort for optimal nice lawn, but in order to worth it in the end. In this article I for you to give you some tips on lawn health.<br><br><br><br>In June 2011, I offered up this rose as a 'current favourite' and person who I was trialling throughout my garden. I originally chose it as it was reported to be'.extremely healthy, repeat flowering all summer [if you dead head], almost thornless with clusters of pretty, single flowers starting as soft apricot buds opening to white along with a hint of sentimental lemon and ideal for low hedges or in a mixed border'. A year later I'm able to wholeheartedly recommend this pretty little rose, it's going on my list of 'good doers'.<br><br>Most people like houses with large, bright, cheerful homes. How do you get this looks? If you have a lot of windows, half your is actually solved. Don't keep the blinds closed - open the drapes and encourage the sun shine in! In case the house doesn't receive lots of sunshine, check outside to determine if there region cutting off your light source. The solution can be as simple as clearing overgrown shrubs or just washing the windows.<br><br>But being the disease progressed, he learned to humble himself and more of their receiver than the giver. But still, my mother to be able to be cautious and not do everything for him as it robbed him of his self worthwhile. Being a fighter, he made a decision to make one of the most of nintendo wii thing, and did his best to boost his state of mind until he died at age 79 in 1994. However, it wasn't Parkinson's' ailment that took his life since he had a series of strokes, losing his mobility and eventually lost the battle of his life to pneumonia.<br><br>There will be no better or cheaper way carryout a good first impression compared to buying a new, clean doormat. First impressions are absolutely crucial, and a well used worn out doormat is really a telltale sign of owner ignore. Also, a good doormat will stop mud from being tracked into your home during the show intervals.<br><br>Before you commit completely to another paint color for the exterior of your home, spend the effort driving around and looking at homes which you want the as well as color scheme of. Paint chips are way too hard to envision, so seeing it in person can aid in making a better decision.<br><br>Are you more informed when it appears to redesigning? Do you need to plan functions now? Have your skills improved? Can you now use things effort with your home? Do so no more complaining how effectively install issues? With any luck, the tips above really needs helped you answer these questions.<br><br>If you loved this informative article and you wish to receive details about [http://www.hopesgrovenurseries.co.uk/ hedge] assure visit our own web-site. |
| <div class="thumbinner" style="width:220px;">
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| {| class="wikitable" style="width:220px;"
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| ! {{nobreak|n \ m}} !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 !! 7 !! 8 !! 9 !! 10
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| |-
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| ! 1
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| |bgcolor="#FFFF00"| 1 || || || || || || || || || ||
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| |-
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| ! 3
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| |bgcolor="#FFFF00"| 0
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| |bgcolor="#FFFF00"| 1 || -1 || || || || || || || ||
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| |-
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| ! 5
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| |bgcolor="#FFFF00"| 0
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| |bgcolor="#FFFF00"| 1 || -1 || -1
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| |bgcolor="#FFFF00"| 1 || || || || || ||
| |
| |-
| |
| ! 7
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| |bgcolor="#FFFF00"| 0
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| |bgcolor="#FFFF00"| 1
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| |bgcolor="#FFFF00"| 1 || -1
| |
| |bgcolor="#FFFF00"| 1 || -1 || -1 || || || ||
| |
| |-
| |
| ! 9
| |
| |bgcolor="#FFFF00"| 0
| |
| |bgcolor="#FFFF00"| 1 || 1 || 0
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| |bgcolor="#FFFF00"| 1 || 1 || 0
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| |bgcolor="#FFFF00"| 1 || 1 || ||
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| |-
| |
| ! 11
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| |bgcolor="#FFFF00"| 0
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| |bgcolor="#FFFF00"| 1 || -1
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| |bgcolor="#FFFF00"| 1
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| |bgcolor="#FFFF00"| 1
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| |bgcolor="#FFFF00"| 1 || -1 || -1 || -1
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| |bgcolor="#FFFF00"| 1 || -1
| |
| |}
| |
| <div class="thumbcaption">
| |
| Jacobi symbol (m/n) for various ''m'' (along top) and ''n'' (along left side). Only 0 ≤ ''m'' < ''n'' are shown, since due to rule (2) below any other ''m'' can be reduced modulo ''n''. [[Quadratic residue]]s are highlighted in yellow — note that no entry with a Jacobi symbol of -1 is a quadratic residue, and every quadratic residue with ''m'' ≢ 0 (mod ''n'') has a Jacobi symbol of 1, but some entries with a Jacobi symbol of 1 (in the ''n''=9 row) are not quadratic residues. Notice also that when either ''n'' or ''m'' is a square, all values are 0 or 1.
| |
| </div> | |
| </div> | |
| </div> | |
| The '''Jacobi symbol''' is a generalization of the [[Legendre symbol]]. Introduced by [[Carl Gustav Jakob Jacobi|Jacobi]] in 1837,<ref>C.G.J.Jacobi "Uber die Kreisteilung und ihre Anwendung auf die Zahlentheorie", ''Bericht Ak. Wiss. Berlin'' (1837) pp 127-136.</ref> it is of theoretical interest in [[modular arithmetic]] and other branches of [[number theory]], but its main use is in [[computational number theory]], especially [[primality testing]] and [[integer factorization]]; these in turn are important in [[cryptography]].
| |
| | |
| ==Definition==
| |
| For any integer <math>a</math> and any positive odd integer <math>n</math> the Jacobi symbol is defined as the product of the [[Legendre symbol]]s corresponding to the prime factors of <math>n</math>:
| |
| | |
| :<math>\Bigg(\frac{a}{n}\Bigg) = \left(\frac{a}{p_1}\right)^{\alpha_1}\left(\frac{a}{p_2}\right)^{\alpha_2}\cdots \left(\frac{a}{p_k}\right)^{\alpha_k}\mbox{ where } n=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}</math>
| |
| | |
| <br/> | |
| <math>\left(\tfrac{a}{p}\right)</math> represents the Legendre symbol, defined for all integers <math>a</math> and all odd primes <math>p</math> by
| |
| | |
| :<math>
| |
| \left(\frac{a}{p}\right) = \begin{cases}
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| \;\;\,0\mbox{ if } a \equiv 0 \pmod{p}
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| \\+1\mbox{ if }a \not\equiv 0\pmod{p} \mbox{ and for some integer }x, \;a\equiv x^2\pmod{p}
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| \\-1\mbox{ if there is no such } x. \end{cases}</math>
| |
| | |
| Following the normal convention for the empty product, <math>\left(\tfrac{a}{1}\right) = 1.</math> The Legendre and Jacobi symbols are indistinguishable exactly when the lower argument is an odd prime, in which case they have the same value.
| |
| | |
| ==Properties==
| |
| | |
| The following facts, even the reciprocity laws, are straightforward deductions from the definition of the Jacobi symbol and the corresponding properties of the Legendre symbol.<ref>Almost any textbook on elementary or algebraic number theory, e.g. Ireland & Rosen pp. 56–57 or Lemmermeyer p. 10</ref> | |
| | |
| The Jacobi symbol is defined only when the upper argument ("numerator") is an integer and the lower argument ("denominator") is a positive odd integer.
| |
| | |
| :1) If <math>n</math> is (an odd) prime, then the Jacobi symbol <math>\Bigg(\frac{a}{n}\Bigg)</math> is equal to (and written the same as) the corresponding Legendre symbol.
| |
| | |
| :2) If <math>a \equiv b \pmod{n}</math> then <math>\Bigg(\frac{a}{n}\Bigg) = \left(\frac{b}{n}\right)</math>
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| | |
| :3) <math>\left(\frac{a}{n}\right) =
| |
| \begin{cases}
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| \;\;\,0\mbox{ if } \gcd(a,n) \ne 1
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| | |
| \\\pm1\mbox{ if } \gcd(a,n) = 1\end{cases}
| |
| </math>
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| If either the top or bottom argument is fixed, the Jacobi symbol is a [[completely multiplicative function]] in the remaining argument:
| |
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| :4) <math>\left(\frac{ab}{n}\right) = \Bigg(\frac{a}{n}\Bigg)\left(\frac{b}{n}\right)</math>, so <math>\left(\frac{a^2}{n}\right) = 1 \textrm{\ or\ } 0</math>
| |
| | |
| :5) <math>\left(\frac{a}{mn}\right)=\left(\frac{a}{m}\right)\left(\frac{a}{n}\right)</math>, so <math>\left(\frac{a}{n^2}\right) = 1 \textrm{\ or\ } 0</math>
| |
| | |
| The [[law of quadratic reciprocity]]: if ''m'' and ''n'' are odd positive coprime integers, then
| |
| | |
| :6) <math>\left(\frac{m}{n}\right)
| |
| = \left(\frac{n}{m}\right)(-1)^{\tfrac{m-1}{2}\tfrac{n-1}{2}} =
| |
| \begin{cases}
| |
| \;\;\;\left(\frac{n}{m}\right) & \text{if }n \equiv 1 \pmod 4 \text{ or } m \equiv 1 \pmod 4 \\
| |
| -\left(\frac{n}{m}\right) & \text{if }n\equiv m \equiv 3 \pmod 4
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| \end{cases}
| |
| </math>
| |
| | |
| and its supplements | |
| | |
| :7) <math>
| |
| \left(\frac{-1}{n}\right)
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| = (-1)^\tfrac{n-1}{2}
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| = \begin{cases} \;\;\,1 & \text{if }n \equiv 1 \pmod 4\\ -1 &\text{if }n \equiv 3 \pmod 4\end{cases}
| |
| </math>
| |
| | |
| :8) <math>
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| \left(\frac{2}{n}\right)
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| = (-1)^\tfrac{n^2-1}{8}
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| = \begin{cases} \;\;\,1 & \text{if }n \equiv 1,7 \pmod 8\\ -1 &\text{if }n \equiv 3,5\pmod 8\end{cases}
| |
| </math>
| |
| | |
| Like the Legendre symbol,
| |
| | |
| :If <math>\left(\frac{a}{n}\right) = -1</math> then <math>a</math> is a quadratic nonresidue <math>\pmod{n}</math>
| |
| | |
| :If <math>a</math> is a [[quadratic residue]] <math>\pmod{n}</math> and <math>a \not\equiv 0\pmod n</math>, then <math>\left(\frac{a}{n}\right) = 1</math>
| |
| | |
| But, unlike the Legendre symbol
| |
| | |
| :If <math>\left(\frac{a}{n}\right) = 1</math> then <math>a</math> may or may not be a quadratic residue <math>\pmod{n}</math>.
| |
| | |
| This is because for ''a'' to be a residue (mod ''n'') it has to be a residue modulo ''every'' prime that divides ''n'', but the Jacobi symbol will equal one if for example ''a'' is a non-residue for exactly two of the primes which divide ''n''.
| |
| | |
| Although the Jacobi symbol can't be uniformly interpreted in terms of squares and non-squares, it can be uniformly interpreted as the sign of a permutation by [[Zolotarev's lemma]].
| |
| | |
| The Jacobi symbol <math>(\tfrac{a}{n})</math> is a [[Dirichlet character]] to the modulus ''n''.
| |
| | |
| ==Calculating the Jacobi symbol==
| |
| | |
| The above formulas lead to an efficient O((log ''a'')(log ''b''))<ref>Cohen, pp. 29–31</ref> algorithm for calculating the Jacobi symbol, analogous to the [[Euclidean algorithm]] for finding the GCD of two numbers. (This should not be surprising in light of rule 3).
| |
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| # Reduce the "numerator" modulo the "denominator" using rule 2.
| |
| # Extract any factors of 2 from the "numerator" using rule 4 and evaluate them using rule 8.
| |
| # If the "numerator" is 1, rules 3 and 4 give a result of 1. If the "numerator" and "denominator" are not coprime, rule 3 gives a result of 0.
| |
| # Otherwise, the "numerator" and "denominator" are now odd positive coprime integers, so we can flip the symbol using rule 6, then return to step 1.
| |
| | |
| ==Example of calculations==
| |
| | |
| The Legendre symbol <math>(\tfrac{a}{p})</math> is only defined for odd primes ''p''. It obeys the same rules as the Jacobi symbol (i.e., reciprocity and the supplementary formulas for <math>(\tfrac{-1}{p})</math> and <math>(\tfrac{2}{p})</math> and multiplicativity of the "numerator".)
| |
| | |
| Problem: Given that 9907 is prime, calculate <math>\left(\frac{1001}{9907}\right).</math>
| |
| | |
| ===Using the Legendre symbol===
| |
| | |
| :<math>
| |
| \left(\frac{1001}{9907}\right)
| |
| =\left(\frac{7}{9907}\right) \left(\frac{11}{9907}\right) \left(\frac{13}{9907}\right).
| |
| </math>
| |
| | |
| ::<math>
| |
| \left(\frac{7}{9907}\right)
| |
| =-\left(\frac{9907}{7}\right)
| |
| =-\left(\frac{2}{7}\right)
| |
| =-1
| |
| </math>
| |
| | |
| ::<math>
| |
| \left(\frac{11}{9907}\right)
| |
| =-\left(\frac{9907}{11}\right)
| |
| =-\left(\frac{7}{11}\right)
| |
| =\left(\frac{11}{7}\right)
| |
| =\left(\frac{4}{7}\right)
| |
| =1
| |
| </math>
| |
| | |
| ::<math>
| |
| \left(\frac{13}{9907}\right)
| |
| =\left(\frac{9907}{13}\right)
| |
| =\left(\frac{1}{13}\right)
| |
| =1
| |
| </math>
| |
| | |
| :<math>\left(\frac{1001}{9907}\right) =-1</math>
| |
| | |
| ===Using the Jacobi symbol===
| |
| | |
| :<math>
| |
| \left(\frac{1001}{9907}\right)
| |
| =\left(\frac{9907}{1001}\right)
| |
| =\left(\frac{898}{1001}\right)
| |
| =\left(\frac{2}{1001}\right)\left(\frac{449}{1001}\right)
| |
| =\left(\frac{449}{1001}\right)
| |
| </math>
| |
| | |
| ::<math>
| |
| =\left(\frac{1001}{449}\right)
| |
| =\left(\frac{103}{449}\right)
| |
| =\left(\frac{449}{103}\right)
| |
| =\left(\frac{37}{103}\right)
| |
| =\left(\frac{103}{37}\right)
| |
| </math>
| |
| | |
| ::<math>
| |
| =\left(\frac{29}{37}\right)
| |
| =\left(\frac{37}{29}\right)
| |
| =\left(\frac{8}{29}\right)
| |
| =\left(\frac{2}{29}\right)^3
| |
| =-1.
| |
| </math>
| |
| | |
| The difference between the two calculations is that when the Legendre symbol is used the "numerator" has to be factored into prime powers before the symbol is flipped. This makes the calculation using the Legendre symbol significantly slower than the one using the Jacobi symbol, as there is no known polynomial-time algorithm for factoring integers.<ref>The [[number field sieve]], the fastest known algorithm, requires <math>O\left(e^{(\ln N)^{1/3}(\ln\ln N)^{2/3}(C+o(1))}\right)</math> operations to factor ''N''. See Cohen, p. 495</ref> In fact, this is why Jacobi introduced the symbol.
| |
| | |
| ==Primality testing==
| |
| | |
| There is another way the Jacobi and Legendre symbols differ. If the [[Euler criterion]] formula is used modulo a composite number, the result may or may not be the value of the Jacobi symbol, and in fact may not even be -1 or 1. For example,
| |
| | |
| :<math>\left(\frac{19}{45}\right) = 1\quad\textrm{ and }\quad19^{(45-1)/2} \equiv 1\pmod{45}</math>
| |
| :<math>\left(\frac{8}{21}\right) = -1\quad\textrm{ but }\quad8^{(21-1)/2} \equiv 1\pmod{21}</math>
| |
| :<math>\left(\frac{5}{21}\right) = 1\quad\textrm{ but }\quad5^{(21-1)/2} \equiv 16\pmod{21}</math>
| |
| | |
| So if it's not known whether a number ''n'' is prime or composite, we can pick a random number ''a'', calculate the Jacobi symbol <math>(\tfrac{a}{n})</math> and compare it with Euler's formula; if they differ modulo ''n'', then ''n'' is composite; if they're the same modulo ''n'' for many different values of ''a'', then ''n'' is "probably prime".
| |
| | |
| This is the basis for the probabilistic [[Solovay–Strassen primality test]] and refinements such as the [[Baillie-PSW primality test]] and the [[Miller–Rabin primality test]].
| |
| | |
| ==See also==
| |
| | |
| *The [[Kronecker symbol]] is a generalization of the Jacobi symbol to all integers.
| |
| *The [[power residue symbol]] is a generalization for third, fourth, and higher powers.
| |
| | |
| ==Notes==
| |
| | |
| {{reflist}}
| |
| | |
| ==References==
| |
| | |
| *{{citation
| |
| | last1 = Cohen | first1 = Henri
| |
| | title = A Course in Computational Algebraic Number Theory
| |
| | publisher = [[Springer Science+Business Media|Springer]]
| |
| | location = Berlin
| |
| | date = 1993
| |
| | isbn = 3-540-55640-0}}
| |
| | |
| *{{citation
| |
| | last1 = Ireland | first1 = Kenneth
| |
| | last2 = Rosen | first2 = Michael
| |
| | title = A Classical Introduction to Modern Number Theory (Second edition)
| |
| | publisher = [[Springer Science+Business Media|Springer]]
| |
| | location = New York
| |
| | date = 1990
| |
| | isbn = 0-387-97329-X}}
| |
| | |
| *{{citation
| |
| | last1 = Lemmermeyer | first1 = Franz
| |
| | title = Reciprocity Laws: from Euler to Eisenstein
| |
| | publisher = [[Springer Science+Business Media|Springer]]
| |
| | location = Berlin
| |
| | date = 2000
| |
| | isbn = 3-540-66957-4}}
| |
| | |
| == External links ==
| |
| * [http://www.math.fau.edu/richman/jacobi.htm Calculate Jacobi symbol] shows the steps of the calculation.
| |
| | |
| [[Category:Modular arithmetic]]
| |
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