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  • Fermat$-$Torricelli theorem | planetmath.org
    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 3A11A41 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 3A11A41 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 Fermat Torricelli theorem Theorem Fermat Torricelli Let all angles of a triangle A B C A B C ABC be at most 120 superscript 120 120 circ Then the inner point F F F of the triangle which makes the sum A F B F C F A F B F C F AF BF CF as little as possible is the point from which the angle of view of every side is 120 superscript 120 120 circ Proof Let s perform the rotation of 60 superscript 60 60 circ about the point A A A When P P P is the image of the point C C C the triangle A C P A C P ACP is equilateral and its angles are 60 superscript 60 60 circ Let F F F be any inner point of the triangle A B C A B C ABC and Q Q Q its image in the rotation We infer that if the sides of the triangle A B C A B C ABC are all seen from F F F in the angle 120 superscript 120 120 circ then the points B B B F F F Q Q Q P P P lie on the same line A A A B B B C C C P P P F F F Q Q Q Generally the triangles A P Q A P Q APQ and A C F A C F ACF are congruent whence C F Q P C F Q P CF QP From the equilateral triangles we obtain A

    Original URL path: http://www.planetmath.org/fermattorricellitheorem (2016-04-25)
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  • field adjunction | planetmath.org
    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 Let K K K be a field and E E E an extension field of K K K If α E α E alpha in E then the smallest subfield of E E E that contains K K K and α α alpha is denoted by K α K α K alpha We say that K α K α K alpha is obtained from the field K K K by adjoining the element α α alpha to K K K via field adjunction Theorem K α K α K alpha is identical with the quotient field Q Q Q of K α K α K alpha Proof 1 Because K α K α K alpha is an integral domain as a subring of the field E E E all possible quotients of the elements of K α K α K alpha belong to E E E So we have K α K α Q E K α K α Q E K cup alpha subseteq K alpha subseteq Q subseteq E and because K α K α K alpha was the smallest then K α Q K α Q K alpha subseteq Q 2 K α K α K alpha is a subring of E E E containing K K K and α α alpha therefore also the whole ring K α K α K alpha that is K α K α K α K α K alpha subseteq K alpha And because K α K α K alpha is a field it must contain all possible quotients of the elements of K α K α K alpha i e Q K α Q K α Q subseteq K alpha In addition to the adjunction of one single element we can adjoin to K K K an arbitrary subset S S S of E E E the resulting field K S K S K S is the smallest of such subfields of E E E i e the intersection of such subfields of E E E that contain both

    Original URL path: http://www.planetmath.org/fieldadjunction (2016-04-25)
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  • field of algebraic numbers | planetmath.org
    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 3A12F99 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 As special cases of the theorem of the parent polynomial equation with algebraic coefficients of this entry one obtains the Corollary If α 𝛼 and β 𝛽 are algebraic numbers then also α β 𝛼 𝛽 α β 𝛼 𝛽 α β 𝛼 𝛽 and α β 𝛼 𝛽 provided β 0 𝛽 0 are algebraic numbers If α 𝛼 and β 𝛽 are algebraic integers then also α β 𝛼 𝛽 α β 𝛼 𝛽 and α β 𝛼 𝛽 are algebraic integers The case of α β 𝛼 𝛽 needs an additional consideration If x m b 1 x m 1 b m 1 x b m superscript 𝑥 𝑚 subscript 𝑏 1 superscript 𝑥 𝑚 1 subscript 𝑏 𝑚 1 𝑥 subscript 𝑏 𝑚 is the minimal polynomial of β 𝛽 the equation β m b 1 β m 1 b m 1 β b m 0 superscript 𝛽 𝑚 subscript 𝑏 1 superscript 𝛽 𝑚 1 subscript 𝑏 𝑚 1 𝛽 subscript 𝑏 𝑚 0 implies 1 β m b m 1 b m 1 β m 1 b 1 b m 1 β 1 b m 0 superscript 1 𝛽 𝑚 subscript 𝑏 𝑚 1 subscript 𝑏 𝑚 superscript 1 𝛽 𝑚 1 subscript 𝑏 1 subscript 𝑏 𝑚 1 𝛽 1 subscript 𝑏 𝑚 0 Hence 1 β 1 𝛽 is

    Original URL path: http://www.planetmath.org/fieldofalgebraicnumbers (2016-04-25)
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  • finding another particular solution of linear ODE | planetmath.org
    0 of 1 it s possible to derive from it via two quadratures another solution y 2 x subscript y 2 x y 2 x linearly independent on y 1 x subscript y 1 x y 1 x thus one can write the general solution y C 1 y 1 x C 2 y 2 x y subscript C 1 subscript y 1 x subscript C 2 subscript y 2 x y C 1 y 1 x C 2 y 2 x of that homogeneous differential equation We will now show the derivation procedure We put y u v y u v displaystyle y uv 2 which renders 1 to v P v Q v u 2 v P v u u v 0 superscript v P superscript v normal Q v u 2 superscript v normal P v superscript u normal superscript u v 0 displaystyle v prime prime Pv prime Qv u 2v prime Pv u prime u prime prime v 0 3 Here one can choose v y 1 x assign v subscript y 1 x v y 1 x whence the first addend vanishes and 3 gets the form 2 y 1 P y 1 u y 1 u 0 2 superscript subscript y 1 normal P subscript y 1 superscript u normal subscript y 1 superscript u 0 displaystyle 2y 1 prime Py 1 u prime y 1 u prime prime 0 4 This equation may be written as u u 2 y 1 y 1 P superscript u superscript u normal 2 superscript subscript y 1 normal subscript y 1 P frac u prime prime u prime 2 frac y 1 prime y 1 P which is integrated to ln d u d x ln 1 y 1 2 P d x constant d u d x 1 superscript subscript y 1 2 P d x constant ln left frac du dx right ln frac 1 y 1 2 int P dx mbox constant i e d u d x C y 1 2 e P d x d u d x C superscript subscript y 1 2 superscript e P d x frac du dx frac C y 1 2 e int P dx A new integration results from this the general solution of 4 u C e P d x y 1 2 d x C u C superscript e P d x superscript subscript y 1 2 d x superscript C normal u C int frac e int P dx y 1 2 dx C prime Thus by 2 we have obtained the wanted other solution y 2 x y 1 x e P d x y 1 2 d x subscript y 2 x subscript y 1 x superscript e P d x superscript subscript y 1 2 d x y 2 x y 1 x int frac e int P dx y 1 2 dx which is clearly linearly independent on y 1 x Consequently we can express the

    Original URL path: http://www.planetmath.org/findinganotherparticularsolutionoflinearode (2016-04-25)
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  • finite changes in convergent series | planetmath.org
    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 finite changes in convergent series The following theorem means that at the beginning of a convergent series one can remove or attach a finite amount of terms without influencing on the convergence of the series the convergence is determined alone by the infinitely long tail of the series Consequently one can also freely change the order of a finite amount of terms Theorem Let k k k be a natural number A series n 1 a n superscript subscript n 1 subscript a n displaystyle sum n 1 infty a n converges iff the series n k 1 a n superscript subscript n k 1 subscript a n displaystyle sum n k 1 infty a n converges Then the sums of both series are connected with n k 1 a n n 1 a n n 1 k a n superscript subscript n k 1 subscript a n superscript subscript n 1 subscript a n superscript subscript n 1 k subscript a n displaystyle sum n k 1 infty a n sum n 1 infty a n sum n 1 k a n 1 Proof Denote the k k k th partial sum of n 1 a n superscript subscript n 1 subscript a n sum n 1 infty a n by S k subscript S k S k and the n n n th partial sum of n k 1 a n superscript subscript n k 1 subscript a n sum n k 1 infty a n by S n superscript subscript S n normal S n prime Then we have S n n k 1 k n a n S k n S k superscript subscript S n normal superscript subscript n k 1 k n subscript a n subscript S k n subscript S k displaystyle S n prime sum n k 1 k n a n S k n S k 2 1 superscript 1 1 circ If n 1 a n superscript subscript n 1 subscript a n sum n 1 infty a n converges i e lim n S n S assign subscript normal n subscript S n S lim n to infty S n S exists as a finite number then 2 implies lim n S n lim n S k n lim n S k S S k subscript normal n superscript subscript S n normal subscript normal n subscript S k n subscript normal n subscript S k S subscript S k lim n to infty S n prime lim n to infty S k n lim n to infty S k S S k Thus n k 1 a n superscript subscript n k 1 subscript a n sum n k 1 infty a n converges and 1 is true 2 superscript 2 2 circ If we suppose n k 1 a n superscript

    Original URL path: http://www.planetmath.org/finitechangesinconvergentseries (2016-04-25)
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  • finite limit implying uniform continuity | planetmath.org
    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 finite limit implying uniform continuity Theorem If the real function f f f is continuous on the interval 0 0 0 infty and the limit lim x f x subscript normal x f x displaystyle lim x to infty f x exists as a finite number a a a then f f f is uniformly continuous on that interval Proof Let ε 0 ε 0 varepsilon 0 According to the limit condition there is a positive number M M M such that f x a ε 2 x M formulae sequence f x a ε 2 for all x M displaystyle f x a frac varepsilon 2 quad forall x M 1 The function is continuous on the finite interval 0 M 1 0 M 1 0 M 1 hence f f f is also uniformly continuous on this compact interval Consequently there is a positive number δ 1 δ 1 delta 1 such that f x 1 f x 2 ε x 1 x 2 0 M 1 with x 1 x 2 δ formulae sequence f subscript x 1 f subscript x 2 ε for all subscript x 1 subscript x 2 0 M 1 with subscript x 1 subscript x 2 δ displaystyle f x 1 f x 2 varepsilon quad forall x 1 x 2 in 0 M 1 mbox with x 1 x 2 delta 2 Let x 1 x 2 subscript x 1 subscript x 2 x 1 x 2 be nonnegative numbers and x 1 x 2 δ subscript x 1 subscript x 2 δ

    Original URL path: http://www.planetmath.org/finitelimitimplyinguniformcontinuity (2016-04-25)
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  • finite ring has no proper overrings | planetmath.org
    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 3A26A15 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 finite ring has no proper overrings The regular elements of a finite commutative ring R R R are the units of the ring see the parent of this entry Generally the largest overring of R R R the total ring of fractions T T T is obtained by forming S 1 R superscript S 1 R S 1 R the extension by localization using as the multiplicative set S S S the set of all regular elements which in this case is the unit group of R R R The ring R R R may be considered as a subring of T T T which consists formally of the fractions a s a s 1 a s a superscript s 1 frac a s as 1 with a R a R a in R and s S s S s in S Since every s s s has its own group inverse s 1 superscript s 1 s 1 in S S S and so in R R R it s evident that T T T contains no other elements than the

    Original URL path: http://www.planetmath.org/finiteringhasnoproperoverrings (2016-04-25)
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  • finite subgroup | planetmath.org
    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 3A13G05 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 finite subgroup Theorem A non empty finite subset K K K of a group G G G is a subgroup of G G G if and only if x y K for all x y K formulae sequence x y K for all x y K displaystyle xy in K quad mbox for all quad x y in K 1 Proof The condition 1 is apparently true if K K K is a subgroup Conversely suppose that a nonempty finite subset K K K of the group G G G satisfies 1 Let a a a and b b b be arbitrary elements of K K K By 1 all positive powers of b b b belong to K K K Because of the finiteness of K K K there exist positive integers r s r s r s such that b r b s r s 1 formulae sequence superscript b r superscript b s r s 1 b r b s quad r s 1 By 1 K b r s 1 b r s b 1 e b 1 b 1 contains K superscript b r s 1 superscript b r s superscript b 1 e superscript b

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