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  • Euler phi at a product | planetmath.org
    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 If the positive greatest common divisor of the integers a a a and b b b is d d d then φ a b φ a φ b d φ d φ a b φ a φ b d φ d varphi ab frac varphi a varphi b d varphi d Proof Using the positive prime factors p p p the right hand side of the asserted equation is d a p a p 1 p b p b p 1 p d p a p b p 1 p normal d a subscript product fragments p normal a normal p 1 p b subscript product fragments p normal b p 1 p d subscript product fragments p normal a normal p normal b p 1 p displaystyle frac d cdot a prod p mid a frac p 1 p cdot b prod p mid b frac p 1 p d prod p mid a p mid b frac p 1 p a b p a p b p 1 p p a p b p 1 p p b p a p 1 p p b p a p 1 p p a p b p 1 p normal a b subscript product fragments p normal a normal p not divides b normal p 1 p subscript product fragments p normal a normal p normal b normal p 1 p subscript product fragments p normal b normal p not divides a normal p 1 p subscript product fragments p normal b normal p normal a p 1 p subscript product

    Original URL path: http://www.planetmath.org/eulerphiataproduct (2016-04-25)
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  • Euler polynomial | planetmath.org
    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 3A11 00 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 Euler polynomial The Euler polynomials E 0 x E 1 x E 2 x subscript E 0 x subscript E 1 x subscript E 2 x normal E 0 x E 1 x E 2 x ldots are certain polynomials of the indeterminate x x x with rational coefficients whose denominators may only be powers 1 2 4 8 1 2 4 8 normal 1 2 4 8 ldots of 2 The Euler polynomials may be defined by means of the generating function such that 2 e x t e t 1 n 0 E n x t n n 2 superscript e x t superscript e t 1 superscript subscript n 0 subscript E n x superscript t n n frac 2e xt e t 1 sum n 0 infty E n x frac t n n i e one can get them by dividing the Taylor series 2 2 x t x 2 t 2 1 3 x 3 t 3 2 2 x t superscript x 2 superscript t 2 1 3 superscript x 3 superscript t 3 normal 2 2xt x 2 t 2 frac 1 3 x 3 t 3 ldots by the Taylor series 2 t 1 2 t 2 1 6 t 3 2 t 1 2 superscript t 2 1 6 superscript t 3 normal 2 t frac 1 2 t 2 frac 1 6 t 3 ldots There are also explicit formulae for the polynomials e g E n x k 0 n n k E k 2 k x 1 2 n k subscript E n x superscript subscript k 0 n binomial n k subscript E k superscript 2 k superscript x 1 2 n k E n x sum k 0 n n choose k frac E k 2 k left x frac 1 2 right n k via the Euler numbers E k subscript E k E k Conversely the Euler numbers are expressed with the Euler polynomials through E k 2 k E k 1 2 subscript E k superscript 2 k subscript E k 1 2 E k 2 k E k left frac 1 2 right The first seven Euler polynomials are E 0 x 1 subscript E 0 x 1

    Original URL path: http://www.planetmath.org/eulerpolynomial (2016-04-25)
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  • Euler product formula | planetmath.org
    1 p 2 s 1 p 2 2 s ν 2 0 1 p 2 ν 2 s 1 1 1 superscript subscript p 2 s 1 1 superscript subscript p 2 s 1 superscript subscript p 2 2 s normal superscript subscript subscript ν 2 0 1 superscript subscript p 2 subscript ν 2 s frac 1 1 frac 1 p 2 s 1 frac 1 p 2 s frac 1 p 2 2s ldots sum nu 2 0 infty frac 1 p 2 nu 2 s Since these series are absolutely convergent their product see multiplication of series may be written as 1 1 1 p 1 s 1 1 1 p 2 s ν 1 ν 2 0 1 p 1 ν 1 s 1 p 2 ν 2 s ν 1 ν 2 0 1 p 1 ν 1 p 2 ν 2 s normal 1 1 1 superscript subscript p 1 s 1 1 1 superscript subscript p 2 s superscript subscript subscript ν 1 subscript ν 2 0 normal 1 superscript subscript p 1 subscript ν 1 s 1 superscript subscript p 2 subscript ν 2 s superscript subscript subscript ν 1 subscript ν 2 0 1 superscript superscript subscript p 1 subscript ν 1 superscript subscript p 2 subscript ν 2 s frac 1 1 frac 1 p 1 s cdot frac 1 1 frac 1 p 2 s sum nu 1 nu 2 0 infty frac 1 p 1 nu 1 s cdot frac 1 p 2 nu 2 s sum nu 1 nu 2 0 infty frac 1 left p 1 nu 1 p 2 nu 2 right s where ν 1 subscript ν 1 nu 1 and ν 2 subscript ν 2 nu 2 independently on each other run all nonnegative integers This equation can be generalised by induction to ν 1 k 1 1 1 p ν s ν 1 ν 2 ν k 0 1 p 1 ν 1 p 2 ν 2 p k ν k s superscript subscript product ν 1 k 1 1 1 superscript subscript p ν s superscript subscript subscript ν 1 subscript ν 2 normal subscript ν k 0 1 superscript superscript subscript p 1 subscript ν 1 superscript subscript p 2 subscript ν 2 normal superscript subscript p k subscript ν k s displaystyle prod nu 1 k frac 1 1 frac 1 p nu s sum nu 1 nu 2 ldots nu k 0 infty frac 1 left p 1 nu 1 p 2 nu 2 cdots p k nu k right s 3 for s 0 s 0 s 0 and for arbitrarily great k k k the exponents ν 1 ν 2 ν k subscript ν 1 subscript ν 2 normal subscript ν k nu 1 nu 2 ldots nu k run independently all nonnegative integers Because the prime factorization of positive integers is unique we can rewrite 3 as ν 1 k 1 1 1 p ν

    Original URL path: http://www.planetmath.org/eulerproductformula (2016-04-25)
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  • Euler's derivation of the quartic formula | planetmath.org
    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 Euler s derivation of the quartic formula Let us consider the quartic equation y 4 p y 2 q y r 0 superscript y 4 p superscript y 2 q y r 0 displaystyle y 4 py 2 qy r 0 1 where p q r p q r p q r are arbitrary known complex numbers We substitute in the equation y u v w assign y u v w displaystyle y u v w 2 We get firstly y 2 u 2 v 2 w 2 2 v w w u u v superscript y 2 superscript u 2 superscript v 2 superscript w 2 2 v w w u u v y 2 u 2 v 2 w 2 2 vw wu uv y 4 u 2 v 2 w 2 2 4 u 2 v 2 w 2 v w w u u v 4 v 2 w 2 w 2 u 2 u 2 v 2 8 u v w u v w superscript y 4 superscript superscript u 2 superscript v 2 superscript w 2 2 4 superscript u 2 superscript v 2 superscript w 2 v w w u u v 4 superscript v 2 superscript w 2 superscript w 2 superscript u 2 superscript u 2 superscript v 2 8 u v w u v w y 4 u 2 v 2 w 2 2 4 u 2 v 2 w 2 vw wu uv 4 v 2 w 2 w 2 u 2 u 2 v 2 8uvw u v w Thus 1 attains the form 4 v 2 w 2 w 2 u 2 u 2 v 2 u 2 v 2 w 2 2 p u 2 v 2 w 2 r 4 superscript v 2 superscript w 2 superscript w 2 superscript u 2 superscript u 2 superscript v 2 superscript superscript u 2 superscript v 2 superscript w 2 2 p superscript u 2 superscript v 2 superscript w 2 r 4 v 2 w 2 w 2 u 2 u 2 v 2 u 2 v 2 w 2 2 p u 2 v 2 w 2 r qquad v w w u u v 4 u 2 v 2 w 2 2 p u v w 8 u v w q 0 v w w u u v 4 superscript u 2 superscript v 2 superscript w 2 2 p u v w 8 u v w q 0 vw wu uv 4 u 2 v 2 w 2 2p u v w 8uvw q 0 When u v w u v w u v w are determined so

    Original URL path: http://www.planetmath.org/eulersderivationofthequarticformula (2016-04-25)
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  • Euler's substitutions for integration | planetmath.org
    2 c 2 t 2 d t x 2 c t 2 c 2 t t t 2 c 2 t formulae sequence x superscript t 2 c 2 t formulae sequence d x superscript t 2 c 2 superscript t 2 d t superscript x 2 c superscript t 2 c 2 t t superscript t 2 c 2 t x frac t 2 c 2t quad dx frac t 2 c 2t 2 dt quad sqrt x 2 c frac t 2 c 2t t frac t 2 c 2t Accordingly we obtain d x x 2 c t 2 c 2 t 2 d t t 2 c 2 t d t t ln t C ln x x 2 c C d x superscript x 2 c superscript t 2 c 2 superscript t 2 d t superscript t 2 c 2 t d t t t C x superscript x 2 c C int frac dx sqrt x 2 c int frac frac t 2 c 2t 2 dt frac t 2 c 2t int frac dt t ln t C ln x sqrt x 2 c C Especially the cases c 1 c plus or minus 1 c pm 1 give the formulas d x x 2 1 arsinh x C d x x 2 1 arcosh x C x 1 fragments d x superscript x 2 1 arsinh x C normal d x superscript x 2 1 arcosh x C fragments normal x 1 normal normal int frac dx sqrt x 2 1 arsinh x C quad int frac dx sqrt x 2 1 arcosh x C x 1 2 The integral c 2 x 2 x d x superscript c 2 superscript x 2 x d x displaystyle int frac sqrt c 2 x 2 x dx is needed in deriving the equation of the tractrix We use for integrating the second substitution c 2 x 2 x t c superscript c 2 superscript x 2 x t c sqrt c 2 x 2 xt c then c 2 x 2 x 2 t 2 2 c x t c 2 superscript c 2 superscript x 2 superscript x 2 superscript t 2 2 c x t superscript c 2 c 2 x 2 x 2 t 2 2cxt c 2 which implies x 2 c t t 2 1 d x 2 c 1 t 2 d t 1 t 2 2 c 2 x 2 2 c t 2 t 2 1 c c t 2 1 t 2 1 formulae sequence x 2 c t superscript t 2 1 formulae sequence d x 2 c 1 superscript t 2 d t superscript 1 superscript t 2 2 superscript c 2 superscript x 2 2 c superscript t 2 superscript t 2 1 c c superscript t 2 1 superscript t 2 1 x frac 2ct t 2 1 quad dx frac 2c 1 t 2 dt 1 t 2

    Original URL path: http://www.planetmath.org/eulerssubstitutionsforintegration (2016-04-25)
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  • evolute | planetmath.org
    normal 2 superscript y normal superscript y η y 1 superscript y normal 2 superscript y xi x frac 1 y prime 2 y prime y prime prime qquad eta y frac 1 y prime 2 y prime prime For examining the properties of the evolute we choose for parameter the arc length s s s measured from a certain point of the curve then in 1 the quantities x y ϱ α x y ϱ α x y varrho alpha and thus ξ ξ xi and η η eta are functions of s s s We assume that all needed derivatives exist and are continuous Differentiating 1 with respect to s s s we obtain d ξ d s d x d s ϱ d α d s cos α d ϱ d s sin α d η d s d y d s ϱ d α d s sin α d ϱ d s cos α formulae sequence d ξ d s d x d s ϱ d α d s α d ϱ d s α d η d s d y d s ϱ d α d s α d ϱ d s α frac d xi ds frac dx ds varrho frac d alpha ds cos alpha frac d varrho ds sin alpha qquad frac d eta ds frac dy ds varrho frac d alpha ds sin alpha frac d varrho ds cos alpha and recalling that d x d s cos α d x d s α frac dx ds cos alpha d y d s sin α d y d s α frac dy ds sin alpha and ϱ d α d s 1 ϱ d α d s 1 varrho frac d alpha ds 1 it yields d ξ d s d ϱ d s sin α d η d s d ϱ d s cos α formulae sequence d ξ d s d ϱ d s α d η d s d ϱ d s α displaystyle frac d xi ds frac d varrho ds sin alpha qquad frac d eta ds frac d varrho ds cos alpha 2 If d ϱ d s 0 d ϱ d s 0 frac d varrho ds neq 0 in the point x y x y x y of γ γ gamma the derivatives d ξ d s d ξ d s frac d xi ds and d η d s d η d s frac d eta ds do not vanish simultaneously and so the evolute has in the corresponding point ξ η ξ η xi eta a tangent line with the slope d η d s d ξ d s 1 tan α normal d η d s d ξ d s 1 α frac d eta ds frac d xi ds frac 1 tan alpha Since the right side of this is the slope of the normal line of the given curve γ γ gamma we have the Theorem 1 The

    Original URL path: http://www.planetmath.org/evolute (2016-04-25)
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  • evolute of cycloid | planetmath.org
    rolling angle of the circle with radius a a a forming the cycloid The parametric equations of the evolute of a curve x x u y y u formulae sequence x x u y y u x x u y y u are ξ x ϱ sin α η y ϱ cos α formulae sequence ξ x ϱ α η y ϱ α displaystyle xi x varrho sin alpha qquad eta y varrho cos alpha 2 with α α alpha the slope angle of the tangent and ϱ ϱ varrho the radius of curvature of the given curve in the point x y x y x y ϱ x 2 y 2 3 2 x y x y ϱ superscript superscript x normal 2 superscript y normal 2 3 2 superscript x normal superscript y superscript x superscript y normal varrho frac x prime 2 y prime 2 3 2 x prime y prime prime x prime prime y prime In the case of the cycloid 1 we have x a 1 cos u y a sin u x a sin u y a cos u formulae sequence superscript x normal a 1 u formulae sequence superscript y normal a u formulae sequence superscript x a u superscript y a u x prime a 1 cos u qquad y prime a sin u qquad x prime prime a sin u qquad y prime prime a cos u Now we get x 2 y 2 2 a 2 1 cos u 4 a 2 sin 2 u 2 superscript x normal 2 superscript y normal 2 2 superscript a 2 1 u 4 superscript a 2 superscript 2 u 2 x prime 2 y prime 2 2a 2 1 cos u 4a 2 sin 2 frac u 2 x y x y a 2 cos u 1 2 a 2 sin 2 u 2 superscript x normal superscript y superscript x superscript y normal superscript a 2 u 1 2 superscript a 2 superscript 2 u 2 x prime y prime prime x prime prime y prime a 2 cos u 1 2a 2 sin 2 frac u 2 and thus the radius of curvature red in the diagram is ϱ 4 a sin u 2 ϱ 4 a u 2 displaystyle varrho 4a sin frac u 2 3 We utilised the identity 1 cos u 2 sin 2 u 2 1 u 2 superscript 2 u 2 displaystyle 1 cos u 2 sin 2 frac u 2 4 see the half angle formula of sine in the goniometric formulas It is easy to show that the point where the circle touches the x x x axis bisects the radius of curvature which lies on the normal line at the point x y x y x y of the cycloid Using then the derivative for parametric form we obtain tan α d y d x y x sin u 1 cos u α d y d x superscript y normal superscript

    Original URL path: http://www.planetmath.org/evoluteofcycloid (2016-04-25)
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  • exact differential equation | planetmath.org
    ℝ 2 mathbb R 2 and let the functions X R ℝ normal X normal R ℝ X R to mathbb R Y R ℝ normal Y normal R ℝ Y R to mathbb R have continuous partial derivatives in R R R The first order differential equation X x y Y x y d y d x 0 X x y Y x y d y d x 0 X x y Y x y frac dy dx 0 or X x y d x Y x y d y 0 X x y d x Y x y d y 0 displaystyle X x y dx Y x y dy 0 1 is called an exact differential equation if the condition X y Y x X y Y x displaystyle frac partial X partial y frac partial Y partial x 2 is true in R R R By 2 the left hand side of 1 is the total differential of a function there is a function f R ℝ normal f normal R ℝ f R to mathbb R such that the equation 1 reads d f x y 0 d f x y 0 d f x y 0 whence its general integral is f x y C f x y C f x y C The solution function f f f can be calculated as the line integral f x y P 0 P X x y d x Y x y d y assign f x y superscript subscript subscript P 0 P X x y d x Y x y d y displaystyle f x y int P 0 P X x y dx Y x y dy 3 along any curve γ γ gamma connecting an arbitrarily chosen point P 0 x 0 y 0 subscript P 0 subscript x 0 subscript y 0 P 0 x 0 y 0 and the point P x y P x y P x y in the region R R R the integrating factor is now 1 absent 1 equiv 1 Example Solve the differential equation 2 x y 3 d x y 2 3 x 2 y 4 d y 0 2 x superscript y 3 d x superscript y 2 3 superscript x 2 superscript y 4 d y 0 frac 2x y 3 dx frac y 2 3x 2 y 4 dy 0 This equation is exact since y 2 x y 3 6 x y 4 x y 2 3 x 2 y 4 y 2 x superscript y 3 6 x superscript y 4 x superscript y 2 3 superscript x 2 superscript y 4 frac partial partial y frac 2x y 3 frac 6x y 4 frac partial partial x frac y 2 3x 2 y 4 If we use as the integrating way the broken line from 0 1 0 1 0 1 to x 1 x 1 x 1 and from this to x y x y x

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