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  • example needing two Lagrange multipliers | planetmath.org
    Primary tabs View active tab Coauthors PDF Source Edit example needing two Lagrange multipliers Find the semi axes of the ellipse of intersection formed when the plane z x y z x y z x y intersects the ellipsoid x 2 4 y 2 5 z 2 25 1 superscript x 2 4 superscript y 2 5 superscript z 2 25 1 frac x 2 4 frac y 2 5 frac z 2 25 1 Let x y z x y z x y z be any point of the ellipsoid The square x 2 y 2 z 2 superscript x 2 superscript y 2 superscript z 2 x 2 y 2 z 2 of the distance of this point from the midpoint 0 0 0 0 0 0 0 0 0 has under the constraints g x 2 4 y 2 5 z 2 25 1 0 h x y z 0 g x24y25z2251 0h xyz 0 displaystyle begin cases g frac x 2 4 frac y 2 5 frac z 2 25 1 0 h x y z 0 end cases 1 the minimum and maximum values at the end points of the semi axes of the ellipse Since we have two constraints we must take equally many Lagrange multipliers λ λ lambda and μ μ mu A necessary condition of the extremums of f x 2 y 2 z 2 assign f superscript x 2 superscript y 2 superscript z 2 f x 2 y 2 z 2 is that in addition to 1 also the equations f x λ g x μ h x 2 x 1 2 x λ μ 0 f y λ g y μ h y 2 y 2 5 y λ μ 0 f z λ g z μ h z 2 z 2 25 z λ μ 0 f x λ g x μ h x 2x 12xλμ 0 f y λ g y μ h y 2y 25yλμ 0 f z λ g z μ h z 2z 225zλμ 0 displaystyle begin cases frac partial f partial x lambda frac partial g partial x mu frac partial h partial x 2x frac 1 2 x lambda mu 0 frac partial f partial y lambda frac partial g partial y mu frac partial h partial y 2y frac 2 5 y lambda mu 0 frac partial f partial z lambda frac partial g partial z mu frac partial h partial z 2z frac 2 25 z lambda mu 0 end cases 2 are satisfied I e we have five equations 1 2 and five unknowns λ λ lambda μ μ mu x x x y y y z z z The equations 2 give x 2 μ λ 4 y 5 μ 2 λ 10 z 25 μ 2 λ 50 formulae sequence x 2 μ λ 4 formulae sequence y 5 μ 2 λ 10 z 25 μ 2 λ 50 x frac 2 mu lambda 4 quad y

    Original URL path: http://www.planetmath.org/exampleneedingtwolagrangemultipliers (2016-04-25)
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  • example of analytic continuation | planetmath.org
    modules sparql sparql module User 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 3A26B10 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 3A26B10 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 example of analytic continuation The function defined by f z n 0 z n 1 z z 2 assign f z superscript subscript n 0 superscript z n 1 z superscript z 2 normal f z sum n 0 infty z n 1 z z 2 ldots is as a sum of power series analytic in the disc of convergence D z ℂ z 1 fragments D fragments normal z C normal normal z normal 1 normal D z in mathbb C vdots z 1 The function g z 1 1 i n 0 z i 1 i n 1 1 i z i 1 i 2 z i 2 1 i 3 assign g z 1 1 i superscript subscript n 0 superscript z i 1 i n 1 1 i z i superscript 1 i 2 superscript z i 2 superscript 1 i 3 normal g z frac 1 1 i sum n 0 infty left frac z i 1 i right n frac 1 1 i frac z i 1 i 2 frac z i 2 1 i 3 ldots similarly is analytic in the bigger disc E z ℂ z i 2 fragments E fragments normal z C normal normal z i normal 2 normal E z in mathbb C vdots z i sqrt 2 But we have f z 1

    Original URL path: http://www.planetmath.org/exampleofanalyticcontinuation (2016-04-25)
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  • example of Cauchy multiplication rule | planetmath.org
    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 3A30A99 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 example of Cauchy multiplication rule Let us form the Taylor expansion of e x sin y superscript e x y e x sin y starting from the known Taylor expansions e x 1 x x 2 2 x 3 3 superscript e x 1 x superscript x 2 2 superscript x 3 3 normal e x 1 x frac x 2 2 frac x 3 3 ldots sin y y y 3 3 y 5 5 y 7 7 fragments y y superscript y 3 3 superscript y 5 5 superscript y 7 7 normal sin y y frac y 3 3 frac y 5 5 frac y 7 7 ldots and multiplying these series with Cauchy multiplication rule As power series both series are absolutely convergent for all real and complex values of x x x and y y y The rule gives immediately the series y y 3 3 x y y 5 5 x y 3 3 x 2 y 2 y 7 7 x y 5 5 x 2 y 3 2 3 x 3 y 3 y 9 9 x y 7 7 x 2 y 5 2 5 x 3 y 3 3 3 x 4 y 4 y superscript y 3 3 x y superscript y 5 5 x superscript y 3 3 superscript x 2 y 2 superscript y 7 7 x superscript y 5 5 superscript x 2 superscript y 3 2 3 superscript x 3 y 3 superscript y 9 9 x superscript y 7 7 superscript x 2 superscript y 5 2 5 superscript x 3 superscript y 3 3 3 superscript x 4 y 4 normal displaystyle y frac y 3 3 xy frac y 5 5 frac xy 3 3 frac x 2 y 2 frac y 7 7 frac xy 5 5 frac x 2 y 3 2 3 frac x 3 y 3 frac y 9 9 frac xy 7 7 frac x 2 y 5 2 5 frac x 3 y 3 3 3 frac x 4 y 4 1 The parenthesis expressions here seem a bit irregular but we can regroup and rearrange the terms in new parentheses e

    Original URL path: http://www.planetmath.org/exampleofcauchymultiplicationrule (2016-04-25)
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  • example of changing variable | planetmath.org
    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 3A30B10 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 3A30B10 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 User 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 3A26A24 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 3A26A24 skos 3AprefLabel 3Flabel FILTER langMatches 28 lang 28 3Flabel 29 2C 22en 22 29 7D

    Original URL path: http://www.planetmath.org/exampleofchangingvariable (2016-04-25)
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  • example of contractive sequence | planetmath.org
    n 1 normal 2 normal 3 normal normal normal normal displaystyle a 1 1 qquad a n 1 sqrt 5 2a n quad n 1 2 3 ldots 1 We see by induction that the radicand in 1 cannot become negative in fact we justify that 1 a n 3 1 subscript a n 3 displaystyle 1 leqq a n leqq sqrt 3 2 for every n n n It s clear when n 1 n 1 n 1 If it is true for an a n subscript a n a n it implies that 1 5 2 a n 3 1 5 2 subscript a n 3 1 5 2a n leqq 3 i e 1 a n 1 3 1 subscript a n 1 3 1 a n 1 leqq sqrt 3 As for the convergence of the sequence which is not monotonic one could think to show that it is a Cauchy sequence Unfortunately it is almost impossible to directly express and estimate the needed absolute value of a m a n subscript a m subscript a n a m a n Fortunately the recursive definition 1 allows quite easily to estimate a n a n 1 subscript a n subscript a n 1 a n a n 1 Then it turns out that it s a question of a contractive sequence whence it is by the parent entry a Cauchy sequence We form the difference a n a n 1 subscript a n subscript a n 1 displaystyle a n a n 1 5 2 a n 1 5 2 a n 5 2 a n 1 5 2 a n 5 2 a n 1 5 2 a n absent 5 2 subscript a n 1 5 2 subscript a n 5 2 subscript a n 1 5 2 subscript a n 5 2 subscript a n 1 5 2 subscript a n displaystyle frac sqrt 5 2a n 1 sqrt 5 2a n sqrt 5 2a n 1 sqrt 5 2a n sqrt 5 2a n 1 sqrt 5 2a n 2 a n 1 a n 5 2 a n 1 5 2 a n absent 2 subscript a n 1 subscript a n 5 2 subscript a n 1 5 2 subscript a n displaystyle frac 2 a n 1 a n sqrt 5 2a n 1 sqrt 5 2a n where n 1 n 1 n 1 Thus we can estimate its absolute value by using 2 a n a n 1 2 a n 1 a n 5 2 a n 1 5 2 a n 2 a n 1 a n 5 2 3 5 2 3 a n 1 a n 5 2 3 subscript a n subscript a n 1 2 subscript a n 1 subscript a n 5 2 subscript a n 1 5 2 subscript a n 2 subscript a n 1 subscript a n 5 2 3 5 2 3 subscript a n 1

    Original URL path: http://www.planetmath.org/exampleofcontractivesequence (2016-04-25)
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  • example of converging increasing sequence | planetmath.org
    a x n which implies x n q a x n superscript subscript x n q a subscript x n x n q a x n i e x n q x n a 0 superscript subscript x n q subscript x n a 0 displaystyle x n q x n a 0 2 for all n n n We study the polynomial f x x q x a x x q 1 1 1 assign f x superscript x q x a x superscript x q 1 1 1 f x x q x a x x q 1 1 1 From its latter form we see that the function f f f attains negative values when 0 x 1 0 x 1 0 leqq x leqq 1 and that f f f increases monotonically and boundlessly when x x x increases from 1 to infty Because f f f as a polynomial function is also continuous we infer that the equation x q x a 0 superscript x q x a 0 displaystyle x q x a 0 3 has exactly one positive root x M 1 x M 1 x M 1 and that f f f is negative for 0 x 1 0 x 1 0 x 1 and positive for x M x M x M Thus we can conclude by 2 that x n M subscript x n M x n M for all values of n n n The proven facts x 1 x 2 x 3 x n M subscript x 1 subscript x 2 subscript x 3 normal subscript x n normal M x 1 x 2 x 3 ldots x n ldots M settle by the theorem of the parent entry that the sequence x 1 x 2 x 3

    Original URL path: http://www.planetmath.org/exampleofconvergingincreasingsequence (2016-04-25)
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  • example of derivative as parameter | planetmath.org
    2 p superscript y 2 displaystyle x frac y 3p 2py 2 1 with p d y d x p d y d x p frac dy dx according to III in the parent entry we differentiate both sides in regard to y y y getting first 1 p 1 3 p y 3 p 2 2 y 2 d p d y 4 p y 1 p 1 3 p y 3 superscript p 2 2 superscript y 2 d p d y 4 p y frac 1 p frac 1 3p left frac y 3p 2 2y 2 right frac dp dy 4py Removing the denominators we obtain 2 p y 6 p 2 y 2 d p d y 12 p 3 y 0 2 p y 6 superscript p 2 superscript y 2 d p d y 12 superscript p 3 y 0 2p y 6p 2 y 2 frac dp dy 12p 3 y 0 The left hand side can be factored y d p d y 2 p 1 6 p 2 y 0 y d p d y 2 p 1 6 superscript p 2 y 0 displaystyle y frac dp dy 2p 1 6p 2 y 0 2 Now we may use the zero rule of product the first factor of the product in 2 yields y d p d y 2 p y d p d y 2 p y frac dp dy 2p i e 2 d y y d p p ln C 2 d y y d p p C 2 int frac dy y int frac dp p ln C whence y 2 C p superscript y 2 C p y 2 frac C p i e p C y 2 p C superscript y 2 p

    Original URL path: http://www.planetmath.org/exampleofderivativeasparameter (2016-04-25)
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  • example of differentiation under integral sign | planetmath.org
    home jcorneli beta sites all modules sparql sparql module Primary tabs View active tab Coauthors PDF Source Edit example of differentiation under integral sign Differentiation with respect to a parameter under the integral sign may sometimes yield useful formulae One example is given here We know that the equation 0 1 x m d x 1 m 1 superscript subscript 0 1 superscript x m d x 1 m 1 int 0 1 x m dx frac 1 m 1 is valid for all m 1 m 1 m 1 If one differentiates with respect to m m m under the integral sign in succession one gets 0 1 m e m ln x d x 0 1 e m ln x ln x d x 0 1 x m ln x d x 1 m 1 2 superscript subscript 0 1 m superscript e m x d x superscript subscript 0 1 superscript e m x x d x superscript subscript 0 1 superscript x m x d x 1 superscript m 1 2 int 0 1 frac partial partial m e m ln x dx int 0 1 e m ln x ln x dx int 0 1 x m ln x dx frac 1 m 1 2 0 1 m x m ln x d x 0 1 x m ln x 2 d x 1 2 m 1 3 superscript subscript 0 1 m superscript x m x d x superscript subscript 0 1 superscript x m superscript x 2 d x normal 1 2 superscript m 1 3 int 0 1 frac partial partial m x m ln x dx int 0 1 x m ln x 2 dx frac 1 cdot 2 m 1 3 0 1 m x m ln x 2 d

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