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  • example of finding catacaustic | 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 3A26B15 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 finding catacaustic Let us find the catacaustic when the family of vertical rays x t x t x t coming from above reflects from the curve y e x y superscript e x y e x The line x t x t x t meets the curve in the point t e t t superscript e t t e t where the slope angle α α alpha of the normal line of the curve satisfies tan α e t α superscript e t tan alpha e t and the slope angle of the reflected ray is 2 α π 2 2 α π 2 2 alpha frac pi 2 so its slope is tan 2 α π 2 1 tan 2 α tan 2 α 1 2 tan α e 2 t 1 2 e t e t e t 2 sinh t 2 α π 2 1 2 α superscript 2 α 1 2 α superscript e 2 t 1 2 superscript e t superscript e t superscript e t 2 t tan 2 alpha frac pi 2 frac 1 tan 2 alpha frac tan 2 alpha 1 2 tan alpha frac e 2t 1 2e t frac e t e t 2 sinh t N B tan 2 α π 2 2 α π 2 tan 2 alpha frac pi 2 cannot be determined with the addition formula of tangent but by using the complement formula tan 2 α π 2 tan π 2 2 α cot 2 α 1 tan 2 α 2 α π 2 π 2 2 α 2 α 1 2 α tan 2 alpha frac pi 2 tan frac pi 2 2 alpha cot 2 alpha frac 1 tan 2 alpha Thus the equation of the reflected ray is y e t sinh t x t y superscript e t t x t y e t sinh t x t i e F x y t x t sinh t y e t 0 assign F x y t x t t y superscript e t 0 displaystyle F x y t x t sinh t y e t 0 1 We have the partial derivative F

    Original URL path: http://www.planetmath.org/exampleoffindingcatacaustic (2016-04-25)
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  • example of gcd | planetmath.org
    sites all modules sparql sparql module Primary tabs View active tab Coauthors PDF Source Edit example of gcd If one calculates values of the polynomial f x x 4 x 2 1 assign f x superscript x 4 superscript x 2 1 f x x 4 x 2 1 on the successive positive integers n n n one may observe an interesting thing when factoring the values f 1 3 f 1 3 f 1 3 f 2 21 3 7 f 2 21 normal 3 7 f 2 21 3 cdot 7 f 3 91 7 13 f 3 91 normal 7 13 f 3 91 7 cdot 13 f 4 273 3 7 13 f 4 273 normal 3 7 13 f 4 273 3 cdot 7 cdot 13 f 5 651 3 7 31 f 5 651 normal 3 7 31 f 5 651 3 cdot 7 cdot 31 f 6 1333 31 43 f 6 1333 normal 31 43 f 6 1333 31 cdot 43 f 7 2451 3 19 43 f 7 2451 normal 3 19 43 f 7 2451 3 cdot 19 cdot 43 f 8 4161 3 19 73 f 8 4161 normal 3 19 73 f 8 4161 3 cdot 19 cdot 73 f 9 6643 7 13 73 f 9 6643 normal 7 13 73 f 9 6643 7 cdot 13 cdot 73 f 10 10101 3 7 13 37 f 10 10101 normal 3 7 13 37 f 10 10101 3 cdot 7 cdot 13 cdot 37 f 11 14763 3 7 19 37 f 11 14763 normal 3 7 19 37 f 11 14763 3 cdot 7 cdot 19 cdot 37 It seems as if two consecutive values always have at least one common odd prime factor i e they have the greatest common divisor 2 absent 2 2 It is indeed true The reason of this fact is not particularly deep It is easily understood if we can factorize this polynomial f x x 2 x 1 x 2 x 1 f x superscript x 2 x 1 superscript x 2 x 1 f x x 2 x 1 x 2 x 1 Then f n n 2 n 1 n 2 n 1 f n superscript n 2 n 1 superscript n 2 n 1 displaystyle f n n 2 n 1 n 2 n 1 1 and the next value is f n 1 n 1 2 n 1 1 n 1 2 n 1 1 n 2 3 n 3 n 2 n 1 f n 1 superscript n 1 2 n 1 1 superscript n 1 2 n 1 1 superscript n 2 3 n 3 superscript n 2 n 1 displaystyle f n 1 n 1 2 n 1 1 n 1 2 n 1 1 n 2 3n 3 n 2 n 1 2 Thus f n f n f n and f n 1 f n 1 f n 1 have

    Original URL path: http://www.planetmath.org/exampleofgcd (2016-04-25)
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  • example of improper integral | planetmath.org
    x 2 dx 1 is undefined both at the lower and the upper limit However the value of the improper integral exists and may be found via the more general integral I y 0 1 arctan x y x 1 x 2 d x I y superscript subscript 0 1 x y x 1 superscript x 2 d x displaystyle I y int 0 1 frac arctan xy x sqrt 1 x 2 dx 2 Denote the integrand of 2 by f x y f x y f x y For any fixed real value y y y f x y O 1 as x 0 f x y O 1 1 x 2 as x 1 formulae sequence f x y O 1 as x normal 0 f x y O 1 1 superscript x 2 as x normal 1 f x y in O 1 mbox as x to 0 quad f x y in O frac 1 sqrt 1 x 2 mbox as x to 1 where the Landau big ordo notation has been used Accordingly the integral 2 converges for every y y y The inequality f x y y 1 1 x 2 y 2 1 x 2 1 1 x 2 f x y y 1 1 superscript x 2 superscript y 2 1 superscript x 2 1 1 superscript x 2 left frac partial f x y partial y right frac 1 1 x 2 y 2 sqrt 1 x 2 leqq frac 1 sqrt 1 x 2 and the convergence of the integral 0 1 d x 1 x 2 π 2 superscript subscript 0 1 d x 1 superscript x 2 π 2 int 0 1 frac dx sqrt 1 x 2 frac pi 2 imply that the integral 0 1 f x y y d x superscript subscript 0 1 f x y y d x displaystyle int 0 1 frac partial f x y partial y dx 3 converges uniformly on the whole y y y axis and equals I y superscript I normal y I prime y For expressing this derivative in a closed form one may utilise the changes of variable x cos φ tan φ t formulae sequence assign x φ assign φ t x cos varphi quad tan varphi t which yield I y superscript I normal y displaystyle I prime y 0 1 d x 1 x 2 y 2 1 x 2 0 π 2 d φ 1 y 2 cos 2 φ normal superscript subscript 0 1 d x 1 superscript x 2 superscript y 2 1 superscript x 2 superscript subscript 0 π 2 d φ 1 superscript y 2 superscript 2 φ displaystyle int 0 1 frac dx 1 x 2 y 2 sqrt 1 x 2 int 0 frac pi 2 frac d varphi 1 y 2 cos 2 varphi 0 d t 1 y 2 t 2 sijoitus t 0 1 1 y 2 arctan t 1 y

    Original URL path: http://www.planetmath.org/exampleofimproperintegral (2016-04-25)
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  • example of isogonal trajectory | planetmath.org
    x 2 y 2 R 2 is 2 x d x 2 y d y 0 2 x d x 2 y d y 0 2x dx 2y dy 0 i e x y d y d x 0 x y d y d x 0 frac x y frac dy dx 0 Thus by the model 2 of the parent entry the differential equation of the isogonal trajectory reads x y y tan π 4 1 y tan π 4 0 x y superscript y normal π 4 1 superscript y normal π 4 0 displaystyle frac x y frac y prime tan frac pi 4 1 y prime tan frac pi 4 0 1 which can be rewritten as y y x y x y x 1 y x 1 superscript y normal y x y x y x 1 y x 1 y prime frac y x y x frac frac y x 1 frac y x 1 Here one may take y x t assign y x t frac y x t as a new variable see ODE types reductible to the variables separable case when y x t y d y d x t x d t d x formulae sequence y x t superscript y normal d y d x t x d t d x y xt quad y prime frac dy dx t x frac dt dx and in the resulting equation t x d t d x t 1 t 1 t x d t d x t 1 t 1 t x frac dt dx frac t 1 t 1 one can separate the variables 1 t 1 t 2 d t d x x 1 t 1 superscript t 2 d t d x x frac 1 t 1

    Original URL path: http://www.planetmath.org/exampleofisogonaltrajectory (2016-04-25)
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  • example of jump discontinuity | planetmath.org
    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 3A34A09 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 3A34A09 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 jump discontinuity The elementary real function f x 1 1 e 1 x normal f maps to x 1 1 superscript e 1 x f colon x mapsto frac 1 1 e frac 1 x has a jump discontinuity at the origin since lim x 0 f x 1 and lim x 0 f x 0 formulae sequence subscript normal x limit from 0 f x 1 and subscript normal x limit from 0 f x 0 lim x to 0 f x 1 quad mathrm and quad lim x to 0 f x 0 Indeed if x 0 normal x limit from 0 x to 0 then 1 x normal 1 x displaystyle frac 1 x to infty e 1 x 0 normal superscript e 1 x 0 displaystyle e frac 1 x to 0 1 1 e 1 x 1 normal 1 1 superscript e 1 x 1 displaystyle frac 1 1 e frac 1 x to 1 if x 0 normal x limit from 0 x to 0 then 1 x normal 1 x displaystyle frac 1 x to infty e 1 x normal superscript e 1 x displaystyle e frac 1 x to infty 1 1 e 1 x 0 normal 1 1 superscript e 1 x 0 displaystyle frac 1 1 e frac 1 x to 0 These results can be seen also from the series expansions of the function gotten by performing the divisions for x 0 x 0 x 0 we obtain the converging alternating series 1 1 e 1 x k 0 1 k e k x 1 e 1 x e 2 x e 3 x fragments 1 normal fragments

    Original URL path: http://www.planetmath.org/exampleofjumpdiscontinuity (2016-04-25)
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  • example of Riemann double integral | planetmath.org
    left frac y sqrt 1 x 2 right 2 right 2 0 1 1 1 x 2 2 1 x 2 2 y 0 1 x arctan y 1 x 2 y 1 x 2 1 y 2 1 x 2 d x absent superscript subscript 0 1 normal 1 superscript 1 superscript x 2 2 1 superscript x 2 2 superscript subscript normal y 0 1 x y 1 superscript x 2 y 1 superscript x 2 1 superscript y 2 1 superscript x 2 d x displaystyle int 0 1 frac 1 1 x 2 2 cdot frac sqrt 1 x 2 2 operatornamewithlimits Big y 0 quad 1 x left arctan frac y sqrt 1 x 2 frac frac y sqrt 1 x 2 1 frac y 2 1 x 2 right dx 0 1 1 2 1 x 2 3 2 arctan 1 x 1 x 2 1 x 1 x x 2 1 x 2 d x absent superscript subscript 0 1 1 2 superscript 1 superscript x 2 3 2 1 x 1 superscript x 2 1 x 1 x superscript x 2 1 superscript x 2 d x displaystyle int 0 1 left frac 1 2 1 x 2 frac 3 2 arctan frac 1 x sqrt 1 x 2 frac 1 x 1 x x 2 1 x 2 right dx The last expression seems quite difficult to calculate to a closed form Some appropriate substitution x x u v y y u v formulae sequence assign x x u v assign y y u v x x u v quad y y u v directly to the form 1 could offer a better way of calculation The general formula for such substitutions is D f x y d x d y Δ f x u v y u v x y u v d u d v subscript double integral D f x y d x d y subscript double integral normal Δ f x u v y u v x y u v d u d v displaystyle iint D f x y dx dy iint Delta f x u v y u v left frac partial x y partial u v right du dv 2 What kind a change of variables would be good One idea were to use some natural substitution i e such one that would give constant limits For example the equations x y u y x v formulae sequence assign x y u assign y x v x y u quad frac y x v map the triangular domain D D D to the rectangle Δ 0 u 1 0 v fragments Δ normal 0 u 1 normal 0 v normal Delta 0 leqq u leqq 1 quad 0 leqq v infty Then we need the Jacobian x y u v u v 2 v 1 3 x y u v u superscript v 2 superscript v 1 3 frac partial x y partial u

    Original URL path: http://www.planetmath.org/exampleofriemanndoubleintegral (2016-04-25)
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  • example of Riemann triple integral | planetmath.org
    x y z fragments D assign fragments normal fragments normal x normal y normal z normal superscript ℝ 3 normal x 0 normal y 0 normal z 0 normal superscript fragments normal superscript x 2 superscript y 2 superscript z 2 normal 3 3 superscript a 3 x y z normal normal D x y z in mathbb R 3 vdots x geqq 0 y geqq 0 z geqq 0 x 2 y 2 z 2 3 leqq 3a 3 xyz By the definition D χ D v d v assign D subscript χ D v d v mathbf meas D int chi D v dv in the parent entry the volume in the question is V D 1 d v D d x d y d z V subscript D 1 d v subscript triple integral D d x d y d z displaystyle V int D 1 dv iiint D dx dy dz 1 For calculating the integral 1 we express it by the geographic spherical coordinates through x r cos φ cos λ y r cos φ sin λ z r sin φ x rcosφcosλ y rcosφsinλ z rsinφ displaystyle begin cases x r cos varphi cos lambda y r cos varphi sin lambda z r sin varphi end cases where the latitude angle φ φ varphi of the position vector r normal r vec r is measured from the x y x y xy plane not as the colatitude ϕ ϕ phi from the positive z z z axis λ λ lambda is the longitude For the change of coordinates we need the Jacobian determinant x y z r φ λ x r y r z r x φ y φ z φ x λ y λ z λ cos φ cos λ cos φ sin λ sin φ r sin φ cos λ r sin φ sin λ r cos φ r cos φ sin λ r cos φ cos λ 0 x y z r φ λ x r y r z r x φ y φ z φ x λ y λ z λ cosφcosλ cosφsinλsinφ rsinφcosλ rsinφsinλ rcosφ rcosφsinλ rcosφcosλ0 displaystyle frac partial x y z partial r varphi lambda left begin matrix frac partial x partial r frac partial y partial r frac partial z partial r frac partial x partial varphi frac partial y partial varphi frac partial z partial varphi frac partial x partial lambda frac partial y partial lambda frac partial z partial lambda end matrix right left begin matrix cos varphi cos lambda cos varphi sin lambda sin varphi r sin varphi cos lambda r sin varphi sin lambda r cos varphi r cos varphi sin lambda r cos varphi cos lambda 0 end matrix right which is simplified to r 2 cos φ superscript r 2 φ r 2 cos varphi The equation of the surface attains the form r 6 3 a 3 r 3 cos 2 φ sin φ cos λ sin λ superscript

    Original URL path: http://www.planetmath.org/exampleofriemanntripleintegral (2016-04-25)
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  • example of Schwarz-Christoffel transformation | planetmath.org
    GRAPH 3Chttp 3A 2F 2Flocalhost 3A8890 2FDAV 2Fhome 2Fpm 2Frdf sink 23this 3E 7B msc 3A26A42 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 Schwarz Christoffel transformation Using the Schwarz Christoffel transformation find a function f f f which maps conformally the upper half of the z z z plane onto the domain which located above the red broken line of w w w plane in the below picture such that f 1 i and f 1 0 f 1 i and f 1 0 displaystyle f 1 i mbox and f 1 0 1 psaxes Dx 10 Dy 10 0 0 3 5 1 5 3 5 3 5 y y y x x x l l l i i i 0 0 0 w w w plane The border line of the image domain comes first along l l l from the left and its argument gets in the point w i w i w i the positive increment 3 2 π π k 1 π 3 2 π π subscript k 1 π frac 3 2 pi pi k 1 pi whence k 1 1 2 subscript k 1 1 2 k 1 frac 1 2 Then in the point w 0 w 0 w 0 the argument of the line attains the increment 1 2 π π k 2 π 1 2 π π subscript k 2 π frac 1 2 pi pi k 2 pi whence k 2 1 2 subscript k 2 1 2 k 2 frac 1 2 Thus we have w f z c d

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