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  • conditional congruences | planetmath.org
    error Socket error Could not connect to http planetmath org 8890 sparql query 0APREFIX msc 3A 3Chttp 3A 2F 2Fmsc2010 org 2Fresources 2FMSC 2F2010 2F 3E PREFIX skos 3A 3Chttp 3A 2F 2Fwww w3 org 2F2004 2F02 2Fskos 2Fcore 23 3E PREFIX dct 3A 3Chttp 3A 2F 2Fpurl org 2Fdc 2Fterms 2F 3E PREFIX local 3A 3Chttp 3A 2F 2Flocal virt 2F 3E SELECT 3Flabel WHERE 7B GRAPH 3Chttp 3A 2F 2Flocalhost 3A8890 2FDAV 2Fhome 2Fpm 2Frdf sink 23this 3E 7B msc 3A51F20 skos 3AprefLabel 3Flabel FILTER langMatches 28 lang 28 3Flabel 29 2C 22en 22 29 7D 7D proxy 0 Connection refused in ARC2 Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module User error missing stream in getFormat via ARC2 Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module User error missing stream in readStream http planetmath org 8890 sparql query 0APREFIX msc 3A 3Chttp 3A 2F 2Fmsc2010 org 2Fresources 2FMSC 2F2010 2F 3E PREFIX skos 3A 3Chttp 3A 2F 2Fwww w3 org 2F2004 2F02 2Fskos 2Fcore 23 3E PREFIX dct 3A 3Chttp 3A 2F 2Fpurl org 2Fdc 2Fterms 2F 3E PREFIX local 3A 3Chttp 3A 2F 2Flocal virt 2F 3E SELECT 3Flabel WHERE 7B GRAPH 3Chttp 3A 2F 2Flocalhost 3A8890 2FDAV 2Fhome 2Fpm 2Frdf sink 23this 3E 7B msc 3A51F20 skos 3AprefLabel 3Flabel FILTER langMatches 28 lang 28 3Flabel 29 2C 22en 22 29 7D 7D via ARC2 Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module Primary tabs View active tab Coauthors PDF Source Edit conditional congruences Consider congruences of the form f x a n x n a n 1 x n 1 a 0 0 mod m assign f x subscript a n superscript x n subscript a n 1 superscript x n 1 normal subscript a 0 annotated 0 pmod m displaystyle f x a n x n a n 1 x n 1 ldots a 0 equiv 0 mathop rm mod m 1 where the coefficients a i subscript a i a i and m m m are rational integers Solving the congruence means finding all the integer values of x x x which satisfy 1 If a i 0 mod m subscript a i annotated 0 pmod m a i equiv 0 mathop rm mod m for all i i i s the congruence is satisfied by each integer in which case the congruence is identical cf the formal congruence Therefore one can assume that at least a n 0 mod m not equivalent to subscript a n annotated 0 pmod m a n not equiv 0 mathop rm mod m since one would otherwise have a n x n 0 mod m subscript a n superscript x n annotated 0 pmod m a n x n equiv 0 mathop rm mod m and the first term could be left out of 1 Now we say that the degree of

    Original URL path: http://www.planetmath.org/conditionalcongruences (2016-04-25)
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  • conditionally convergent real series | planetmath.org
    org 8890 sparql query 0APREFIX msc 3A 3Chttp 3A 2F 2Fmsc2010 org 2Fresources 2FMSC 2F2010 2F 3E PREFIX skos 3A 3Chttp 3A 2F 2Fwww w3 org 2F2004 2F02 2Fskos 2Fcore 23 3E PREFIX dct 3A 3Chttp 3A 2F 2Fpurl org 2Fdc 2Fterms 2F 3E PREFIX local 3A 3Chttp 3A 2F 2Flocal virt 2F 3E SELECT 3Flabel WHERE 7B GRAPH 3Chttp 3A 2F 2Flocalhost 3A8890 2FDAV 2Fhome 2Fpm 2Frdf sink 23this 3E 7B msc 3A11A05 skos 3AprefLabel 3Flabel FILTER langMatches 28 lang 28 3Flabel 29 2C 22en 22 29 7D 7D proxy 0 Connection refused in ARC2 Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module User error missing stream in getFormat via ARC2 Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module User error missing stream in readStream http planetmath org 8890 sparql query 0APREFIX msc 3A 3Chttp 3A 2F 2Fmsc2010 org 2Fresources 2FMSC 2F2010 2F 3E PREFIX skos 3A 3Chttp 3A 2F 2Fwww w3 org 2F2004 2F02 2Fskos 2Fcore 23 3E PREFIX dct 3A 3Chttp 3A 2F 2Fpurl org 2Fdc 2Fterms 2F 3E PREFIX local 3A 3Chttp 3A 2F 2Flocal virt 2F 3E SELECT 3Flabel WHERE 7B GRAPH 3Chttp 3A 2F 2Flocalhost 3A8890 2FDAV 2Fhome 2Fpm 2Frdf sink 23this 3E 7B msc 3A11A05 skos 3AprefLabel 3Flabel FILTER langMatches 28 lang 28 3Flabel 29 2C 22en 22 29 7D 7D via ARC2 Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module Primary tabs View active tab Coauthors PDF Source Edit conditionally convergent real series Theorem If the series u 1 u 2 u 3 subscript u 1 subscript u 2 subscript u 3 normal displaystyle u 1 u 2 u 3 ldots 1 with real terms u i subscript u i u i is conditionally convergent i e converges but u 1 u 2 u 3 subscript u 1 subscript u 2 subscript u 3 normal u 1 u 2 u 3 cdots diverges then the both series a 1 a 2 a 3 and b 1 b 2 b 3 subscript a 1 subscript a 2 subscript a 3 normal and subscript b 1 subscript b 2 subscript b 3 normal displaystyle a 1 a 2 a 3 ldots quad mbox and quad b 1 b 2 b 3 ldots 2 consisting of the positive and negative terms of 1 are divergent more accurately lim n i 1 n a n and lim n i 1 n b n formulae sequence subscript normal n superscript subscript i 1 n subscript a n and subscript normal n superscript subscript i 1 n subscript b n lim n to infty sum i 1 n a n infty quad mbox and quad lim n to infty sum i 1 n b n infty Proof If both of the series 2 were convergent having the sums A A A and B B B then we had 0 u 1 u 2 u

    Original URL path: http://www.planetmath.org/conditionallyconvergentrealseries (2016-04-25)
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  • cone in $\mathbb{R}^3$ | planetmath.org
    S such that any ruling ℓ normal ℓ ell on S S S passes through O O O By definition it is readily seen that this point O O O is unique for otherwise all rulings of S S S that pass through O O O as well as another point O superscript O normal O prime must all coincide concluding that S S S must be nothing more than a straight line contradicting the fact that S S S is a surface The point O O O is commonly known as the apex of the conical surface S S S Figure 1 A circular conical surface Remarks No plane can be a conical surface because one can always find a line ruling on the plane not passing through O O O A conical surface is strictly speaking not a regular surface in the sense of a differentiable manifold This is because of the differentiability at the apex breaks down In fact no neighborhood of the apex is diffeomorphic to ℝ 2 superscript ℝ 2 mathbb R 2 Nevertheless it is easy to see that any two points on a conical surface can be joined by a simple continuous curve Any plane passing through the apex O O O of a conical surface S S S must contain at least two lines ℓ m normal ℓ m ell m through the apex such that ℓ S m S O normal ℓ S m S O ell cap S m cap S O In fact it can be shown that there is a plane with the above property such that ℓ m perpendicular to normal ℓ m ell perp m Given a conical surface S S S if there is a plane π π pi in ℝ 3 superscript ℝ 3 mathbb R 3 such that π S π S pi cap S varnothing then it can be shown that S S S is planar that is S S S lies on a plane This shows that if S S S is a non planar surface it must have non empty intersection with any plane in ℝ 3 superscript ℝ 3 mathbb R 3 Given a plane π π pi not passing through the apex O O O of a non planar conical surface S S S the intersection of π π pi and S S S is non empty as guaranteed by the previous remark Let c c c be the intersection of π π pi and S S S It is not hard to see that c c c is necessarily a curve The curve may be bounded or unbounded and it may have disjoint components If there is a plane not passing through the apex of a conical surface S S S such that its intersection with S S S is a bounded connected closed loop then we call this surface S S S a closed cone or cone for short Intuitively it can be pictured as a

    Original URL path: http://www.planetmath.org/coneinmathbbr3 (2016-04-25)
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  • conformality of stereographic projection | planetmath.org
    jcorneli beta sites all modules sparql sparql module User error missing stream in getFormat via ARC2 Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module User error missing stream in readStream http planetmath org 8890 sparql query 0APREFIX msc 3A 3Chttp 3A 2F 2Fmsc2010 org 2Fresources 2FMSC 2F2010 2F 3E PREFIX skos 3A 3Chttp 3A 2F 2Fwww w3 org 2F2004 2F02 2Fskos 2Fcore 23 3E PREFIX dct 3A 3Chttp 3A 2F 2Fpurl org 2Fdc 2Fterms 2F 3E PREFIX local 3A 3Chttp 3A 2F 2Flocal virt 2F 3E SELECT 3Flabel WHERE 7B GRAPH 3Chttp 3A 2F 2Flocalhost 3A8890 2FDAV 2Fhome 2Fpm 2Frdf sink 23this 3E 7B msc 3A51M04 skos 3AprefLabel 3Flabel FILTER langMatches 28 lang 28 3Flabel 29 2C 22en 22 29 7D 7D via ARC2 Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module Primary tabs View active tab Coauthors PDF Source Edit conformality of stereographic projection N N N P P P P superscript P normal P prime m m m psplot linecolor blue 0 603 33 0 36 x x mul sub sqrt mul psplot linecolor blue 0 60 3 33 0 36 x x mul sub sqrt mul psplot linestyle dotted 00 63 33 0 36 x x mul sub sqrt mul psplot linestyle dotted 00 6 3 33 0 36 x x mul sub sqrt mul psplot linecolor blue 22 0 3 4 x x mul sub sqrt mul psplot linestyle dotted 220 3 4 x x mul sub sqrt mul l l l l 1 subscript l 1 l 1 We shall see that the stereographic projection P P maps to P superscript P normal displaystyle P mapsto P prime 1 mapping the points P P P of the closed complex plane ℂ ℂ mathbb C cup infty bijectively onto the points P superscript P normal P prime of the Riemann sphere preserves the angles between two curves i e it is conformal Therefore it is used as a map projection Let l l l be a line in the complex plane If it passes through the origin then 1 maps it onto a meridian as m m m of the sphere If l l l don t pass through the origin we can think all lines passing through the North Pole N N N and a point of l l l these lines form a plane π π pi which intersects the sphere along a circle c c c Thus 1 maps any line onto a circle which passes through the point N N N Moreover the tangent line t t t of the circle c c c at N N N is parallel to the line l l l because it is the intersection line of the plane π π pi and the tangent plane of the Riemann sphere set at N N N and the tangent plane is parallel to the complex plane Think then two lines l

    Original URL path: http://www.planetmath.org/conformalityofstereographicprojection (2016-04-25)
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  • congruence in algebraic number field | planetmath.org
    modules sparql sparql module User error missing stream in getFormat via ARC2 Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module User error missing stream in readStream http planetmath org 8890 sparql query 0APREFIX msc 3A 3Chttp 3A 2F 2Fmsc2010 org 2Fresources 2FMSC 2F2010 2F 3E PREFIX skos 3A 3Chttp 3A 2F 2Fwww w3 org 2F2004 2F02 2Fskos 2Fcore 23 3E PREFIX dct 3A 3Chttp 3A 2F 2Fpurl org 2Fdc 2Fterms 2F 3E PREFIX local 3A 3Chttp 3A 2F 2Flocal virt 2F 3E SELECT 3Flabel WHERE 7B GRAPH 3Chttp 3A 2F 2Flocalhost 3A8890 2FDAV 2Fhome 2Fpm 2Frdf sink 23this 3E 7B msc 3A32C15 skos 3AprefLabel 3Flabel FILTER langMatches 28 lang 28 3Flabel 29 2C 22en 22 29 7D 7D via ARC2 Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module Primary tabs View active tab Coauthors PDF Source Edit congruence in algebraic number field Definition Let α α alpha β β beta and κ κ kappa be integers of an algebraic number field K K K and κ 0 κ 0 kappa neq 0 One defines α β mod κ α annotated β pmod κ displaystyle alpha equiv beta mathop rm mod kappa 1 if and only if κ α β fragments κ normal α β kappa mid alpha beta i e iff there is an integer λ λ lambda of K K K with α β λ κ α β λ κ alpha beta lambda kappa Theorem The congruence equiv modulo κ κ kappa defined above is an equivalence relation in the maximal order of K K K There are only a finite amount of the equivalence classes the residue classes modulo κ κ kappa Proof For justifying the transitivity of equiv suppose 1 and β γ mod κ β annotated γ pmod κ beta equiv gamma mathop rm mod kappa then there are the integers λ λ lambda and μ μ mu of K K K such that α β λ κ α β λ κ alpha beta lambda kappa β γ μ κ β γ μ κ beta gamma mu kappa Adding these equations we see that α γ λ μ κ α γ λ μ κ alpha gamma lambda mu kappa with the integer λ μ λ μ lambda mu of K K K Accordingly α γ mod κ α annotated γ pmod κ alpha equiv gamma mathop rm mod kappa Let ω ω omega be an arbitrary integer of K K K and ω 1 ω 2 ω n subscript ω 1 subscript ω 2 normal subscript ω n omega 1 omega 2 ldots omega n a minimal basis of the field Then we can write ω a 1 ω 1 a 2 ω 2 a n ω n ω subscript a 1 subscript ω 1 subscript a 2 subscript ω 2 normal subscript a n subscript ω n omega a 1 omega 1 a 2 omega 2 ldots a n

    Original URL path: http://www.planetmath.org/congruenceinalgebraicnumberfield (2016-04-25)
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  • congruence of arbitrary degree | planetmath.org
    the congruence f x a n x n a n 1 x n 1 a 0 0 mod p assign f x subscript a n superscript x n subscript a n 1 superscript x n 1 normal subscript a 0 annotated 0 pmod p displaystyle f x a n x n a n 1 x n 1 ldots a 0 equiv 0 mathop rm mod p 1 where p a n not divides p subscript a n p nmid a n has at least n n n incongruent roots x 1 x 2 x n subscript x 1 subscript x 2 normal subscript x n x 1 x 2 ldots x n Form the congruence f x a n x x 1 x x 2 x x n mod p f x annotated subscript a n x subscript x 1 x subscript x 2 normal x subscript x n pmod p displaystyle f x equiv a n x x 1 x x 2 cdots x x n mathop rm mod p 2 Both sides have the same term a n x n subscript a n superscript x n a n x n of the highest degree whence they may be cancelled from the congruence and the degree of 2 has a lower degree than n n n Because 2 however clearly has n n n incongruent roots x 1 x 2 x n subscript x 1 subscript x 2 normal subscript x n x 1 x 2 ldots x n it must by the induction hypothesis be simplifiable to the form 0 0 mod p 0 annotated 0 pmod p 0 equiv 0 mathop rm mod p and thus be an identical congruence Now if the congruence 1 had an additional incongruent root x n 1 subscript x n 1 x n 1 i e P x n 1 0 mod p P subscript x n 1 annotated 0 pmod p P x n 1 equiv 0 mathop rm mod p then the identical congruence 2 would imply a n x n 1 x 1 x n 1 x 2 x n 1 x n 0 mod p subscript a n subscript x n 1 subscript x 1 subscript x n 1 subscript x 2 normal subscript x n 1 subscript x n annotated 0 pmod p a n x n 1 x 1 x n 1 x 2 cdots x n 1 x n equiv 0 mathop rm mod p Yet this is impossible since no one of the factors of the left hand side is divisible by p p p This settles the induction proof Cf SpringerLink Example When f x x 5 x 1 0 mod 7 assign f x superscript x 5 x 1 annotated 0 pmod 7 f x x 5 x 1 equiv 0 mathop rm mod 7 we have f 0 1 mod 7 f 0 annotated 1 pmod 7 f 0 equiv 1 mathop rm mod 7 f 1 3 mod 7

    Original URL path: http://www.planetmath.org/congruenceofarbitrarydegree (2016-04-25)
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  • conjugate diameters of ellipse | planetmath.org
    superscript a 2 superscript m 2 superscript b 2 x 0 frac 2a 2 mk a 2 m 2 b 2 2 frac a 2 mk a 2 m 2 b 2 The ordinate y 0 subscript y 0 y 0 of the midpoint of the chord satisfies y 0 m x 0 k subscript y 0 m subscript x 0 k y 0 mx 0 k Eliminating the parameter k k k from the two last equations gives us y 0 b 2 a 2 m x 0 subscript y 0 superscript b 2 superscript a 2 m subscript x 0 displaystyle y 0 frac b 2 a 2 m x 0 3 This equation says that the point x 0 y 0 subscript x 0 subscript y 0 x 0 y 0 is situated on the line y b 2 a 2 m x y superscript b 2 superscript a 2 m x y frac b 2 a 2 m x which passes through the origin the centre of the ellipse So we have the Theorem 1 The diameter of ellipse i e the bisector of the parallel chords of an ellipse is a line passing through the centre of the ellipse We see from the last equation that the slope m superscript m normal m prime of the diameter which bisects the chords with the slope m m m is m b 2 a 2 m superscript m normal superscript b 2 superscript a 2 m m prime frac b 2 a 2 m accordingly one has the symmetric equation m m b 2 a 2 m superscript m normal superscript b 2 superscript a 2 mm prime frac b 2 a 2 between m m m and m superscript m normal m prime From it one can infer that conversely the diameter with the slope m m m bisects all chords which have the slope m superscript m normal m prime If we have two diameters of the ellipse each one bisecting the chords parallel to the other then these chords are conjugate diameters of each other Apparently the major axis and the minor axis are a pair of conjugate diameters If the line 1 especially is a tangent of the ellipse then the points P 1 subscript P 1 P 1 and P 2 subscript P 2 P 2 and the midpoint x 0 y 0 subscript x 0 subscript y 0 x 0 y 0 coincide and thus the equation 3 gives the slope of the tangent m t b 2 x 0 a 2 y 0 subscript m t superscript b 2 subscript x 0 superscript a 2 subscript y 0 m t frac b 2 x 0 a 2 y 0 So we may write the equation of the tangent y y 0 b 2 x 0 a 2 y 0 x x 0 y subscript y 0 superscript b 2 subscript x 0 superscript a 2 subscript y 0 x subscript

    Original URL path: http://www.planetmath.org/conjugatediametersofellipse (2016-04-25)
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  • conjugate fields | planetmath.org
    7D via ARC2 Reader in sparql request line 92 of home jcorneli beta sites all modules sparql sparql module Primary tabs View active tab Coauthors PDF Source Edit conjugate fields If ϑ 1 ϑ 2 ϑ n subscript ϑ 1 subscript ϑ 2 normal subscript ϑ n vartheta 1 vartheta 2 ldots vartheta n are the algebraic conjugates of the algebraic number ϑ 1 subscript ϑ 1 vartheta 1 then the algebraic number fields ℚ ϑ 1 ℚ ϑ 2 ℚ ϑ n ℚ subscript ϑ 1 ℚ subscript ϑ 2 normal ℚ subscript ϑ n mathbb Q vartheta 1 mathbb Q vartheta 2 ldots mathbb Q vartheta n are the conjugate fields of ℚ ϑ 1 ℚ subscript ϑ 1 mathbb Q vartheta 1 Notice that the conjugate fields of ℚ ϑ 1 ℚ subscript ϑ 1 mathbb Q vartheta 1 are always isomorphic but not necessarily distinct All conjugate fields are equal i e ℚ ϑ 1 ℚ ϑ 2 ℚ ϑ n ℚ subscript ϑ 1 ℚ subscript ϑ 2 normal ℚ subscript ϑ n mathbb Q vartheta 1 mathbb Q vartheta 2 ldots mathbb Q vartheta n or equivalently ϑ 1 ϑ n subscript ϑ 1 normal subscript ϑ n vartheta 1 ldots vartheta n belong to ℚ ϑ 1 ℚ subscript ϑ 1 mathbb Q vartheta 1 if and only if the extension ℚ ϑ 1 ℚ ℚ subscript ϑ 1 ℚ mathbb Q vartheta 1 mathbb Q is a Galois extension of fields The reason for this is that if ϑ 1 subscript ϑ 1 vartheta 1 is an algebraic number and m x m x m x is the minimal polynomial of ϑ 1 subscript ϑ 1 vartheta 1 then the roots of m x m x m x are precisely the algebraic conjugates of ϑ 1 subscript ϑ 1 vartheta 1 For example let ϑ 1 2 subscript ϑ 1 2 vartheta 1 sqrt 2 Then its only conjugate is ϑ 2 2 subscript ϑ 2 2 vartheta 2 sqrt 2 and ℚ 2 ℚ 2 mathbb Q sqrt 2 is Galois and contains both ϑ 1 subscript ϑ 1 vartheta 1 and ϑ 2 subscript ϑ 2 vartheta 2 Similarly let p p p be a prime and let ϑ 1 ζ subscript ϑ 1 ζ vartheta 1 zeta be a primitive p p p th root of unity Then the algebraic conjugates of ζ ζ zeta are ζ 2 ζ p 1 superscript ζ 2 normal superscript ζ p 1 zeta 2 ldots zeta p 1 and so all conjugate fields are equal to ℚ ζ ℚ ζ mathbb Q zeta and the extension ℚ ζ ℚ ℚ ζ ℚ mathbb Q zeta mathbb Q is Galois It is a cyclotomic extension of ℚ ℚ mathbb Q Now let ϑ 1 2 3 subscript ϑ 1 3 2 vartheta 1 sqrt 3 2 and let ζ ζ zeta be a primitive 3 3 3 rd root of unity i e

    Original URL path: http://www.planetmath.org/conjugatefields (2016-04-25)
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