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  • example of solving a cubic equation | planetmath.org
    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 3A30C20 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 us use Cardano s formulae for solving algebraically the cubic equation x 3 3 x 2 1 0 superscript 𝑥 3 3 superscript 𝑥 2 1 0 1 First apply the Tschirnhaus transformation x y 1 assign 𝑥 𝑦 1 for removing the quadratic term from y 1 3 3 y 1 2 1 0 superscript 𝑦 1 3 3 superscript 𝑦 1 2 1 0 we get the simplified equation y 3 3 y 2 0 superscript 𝑦 3 3 𝑦 2 0 2 We now suppose that y u v assign 𝑦 𝑢 𝑣 Substituting this into 2 and rewriting the equation in the form u 3 v 3 2 3 u v 1 u v 0 superscript 𝑢 3 superscript 𝑣 3 2 3 𝑢 𝑣 1 𝑢 𝑣 0 one can determine u 𝑢 and v 𝑣 such that u 3 v 3 2 0 superscript 𝑢 3 superscript 𝑣 3 2 0 and u v 1 0 𝑢 𝑣 1 0 i e u 3 v 3 2 u 3 v 3 1 cases superscript 𝑢 3 superscript 𝑣 3 2 otherwise superscript 𝑢 3 superscript 𝑣 3 1 otherwise Using the properties of quadratic equation we infer that u 3 superscript 𝑢 3 and v 3 superscript 𝑣 3 are the roots of the resolvent equation z 2 2 z 1 0 superscript 𝑧 2 2 𝑧 1 0 Therefore u 𝑢 and v 𝑣 satisfy the binomial equations u 3 1 2 v 3 1 2 formulae sequence superscript 𝑢 3 1 2 superscript 𝑣 3 1 2 3 respectively If we choose the real radicals u u 0 1 2 3 𝑢 subscript 𝑢 0 3 1 2 and v v 0 1 2 3 𝑣 subscript 𝑣 0 3 1 2 the other solutions u v 𝑢 𝑣 of 3 are ζ u 0 ζ 2 v 0 ζ 2 u 0 ζ v 0 𝜁 subscript 𝑢 0 superscript 𝜁 2 subscript 𝑣 0 superscript 𝜁 2 subscript 𝑢 0 𝜁 subscript 𝑣 0 4 where ζ 1 i 3 2 ζ 2 1 i 3 2 formulae sequence 𝜁 1 𝑖 3 2 superscript 𝜁 2 1 𝑖 3 2 are the primitive third roots of unity One must combine the pairs u v 𝑢 𝑣 of 4 so

    Original URL path: http://www.planetmath.org/exampleofsolvingacubicequation (2016-04-25)
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  • example of solving a functional equation | planetmath.org
    f y f superscript y 2 f y f y f y 2 whence f y f y f y f y f y f y So f f f is an odd function We differentiate both sides of 1 with respect to y y y and the result with respect to x x x f x y f x y f x y f x y 2 f y f y superscript f normal x y f x y superscript f normal x y f x y 2 f y superscript f normal y f prime x y f x y f prime x y f x y 2f y f prime y f x y f x y f x y f x y f x y f x y f x y f x y 0 superscript f x y f x y superscript f normal x y superscript f normal x y superscript f x y f x y superscript f normal x y superscript f normal x y 0 f prime prime x y f x y f prime x y f prime x y f prime prime x y f x y f prime x y f prime x y 0 The result is simplified to f x y f x y f x y f x y superscript f x y f x y superscript f x y f x y f prime prime x y f x y f prime prime x y f x y i e f x y f x y f x y f x y superscript f x y f x y superscript f x y f x y f prime prime x y f x y f prime prime x y f x y Denoting x y u assign x y u x y u x y v assign x y v x y v we obtain the equation f u f u f v f v superscript f u f u superscript f v f v frac f prime prime u f u frac f prime prime v f v for all real values of u u u and v v v This is not possible unless the proportion f u f u superscript f u f u frac f prime prime u f u has a constant value independent on u u u Thus the second order homogeneous linear differential equation f t f t k 2 superscript f t f t plus or minus superscript k 2 f prime prime t f t pm k 2 or f t k 2 f t superscript f t plus or minus superscript k 2 f t f prime prime t pm k 2 f t with k k k some constant is valid There are three cases 1 k 0 k 0 k 0 Now f t 0 superscript f t 0 f prime prime t equiv 0 and consequently f t C t f

    Original URL path: http://www.planetmath.org/exampleofsolvingafunctionalequation (2016-04-25)
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  • example of solving the heat equation | planetmath.org
    and thus 1 gets the form X Y X Y 0 superscript X Y X superscript Y 0 displaystyle X prime prime Y XY prime prime 0 2 and the boundary conditions X 0 X π 0 X x C Y π Y 0 0 formulae sequence X 0 X π 0 formulae sequence X x C Y π superscript Y normal 0 0 X 0 X pi 0 quad X x frac C Y pi quad Y prime 0 0 We separate the variables in 2 X X Y Y superscript X X superscript Y Y frac X prime prime X frac Y prime prime Y This equation is not possible unless both sides are equal to a same negative constant k 2 superscript k 2 k 2 which implies for X k 2 X superscript X superscript k 2 X X prime prime k 2 X the solution X C 1 cos k x C 2 sin k x assign X subscript C 1 k x subscript C 2 k x X C 1 cos kx C 2 sin kx and for Y k 2 Y superscript Y superscript k 2 Y Y prime prime k 2 Y the solution Y D 1 cosh k y D 2 sinh k y assign Y subscript D 1 k y subscript D 2 k y Y D 1 cosh ky D 2 sinh ky The two first boundary conditions give 0 X 0 C 1 0 X 0 subscript C 1 0 X 0 C 1 0 X π 0 C 2 sin k π 0 X π 0 subscript C 2 k π 0 X pi 0 C 2 sin k pi and since C 2 0 subscript C 2 0 C 2 neq 0 we must have sin k π 0 k π 0 sin k pi 0 i e 0 k n 1 2 3 formulae sequence 0 k assign n 1 2 3 normal 0 k n 1 2 3 ldots Therefore X x C 2 sin n x Y y n D 1 sinh n y n D 2 cosh n y formulae sequence assign X x subscript C 2 n x superscript Y normal y n subscript D 1 n y n subscript D 2 n y X x C 2 sin nx quad Y prime y equiv nD 1 sinh ny nD 2 cosh ny The fourth boundary condition now yields that 0 Y 0 n D 2 0 superscript Y normal 0 n subscript D 2 0 Y prime 0 nD 2 thus D 2 0 subscript D 2 0 D 2 0 and Y y D 1 cosh n y assign Y y subscript D 1 n y Y y D 1 cosh ny So 1 has infinitely many solutions u n C 2 D 1 sin n x cosh n y A n sin n x cosh n y assign subscript u n subscript C 2 subscript D 1 n

    Original URL path: http://www.planetmath.org/exampleofsolvingtheheatequation (2016-04-25)
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  • example of summation by parts | planetmath.org
    by parts we obtain n N N P cos n φ n p 0 P 1 N p cos N p φ p 0 P 1 1 N p 1 N p 1 S N N p 1 N P S N N P superscript subscript n N N P n φ n superscript subscript p 0 P 1 N p N p φ superscript subscript p 0 P 1 1 N p 1 N p 1 subscript S N N p 1 N P subscript S N N P absent left sum n N N P frac cos n varphi n right left sum p 0 P frac 1 N p cos N p varphi right left sum p 0 P 1 left frac 1 N p frac 1 N p 1 right S N N p frac 1 N P S N N P right leqq p 0 P 1 1 N p 1 N p 1 S N N P 1 N P S N N P fragments superscript subscript p 0 P 1 fragments normal 1 N p 1 N p 1 normal normal subscript S N N P normal 1 N P normal subscript S N N P normal leqq sum p 0 P 1 left frac 1 N p frac 1 N p 1 right S N N P frac 1 N P S N N P p 0 P 1 1 N p 1 N p 1 2 K φ 1 N P 2 K φ 1 N 2 K φ fragments superscript subscript p 0 P 1 fragments normal 1 N p 1 N p 1 normal normal 2 subscript K φ 1 N P normal 2 subscript K φ 1 N normal 2 subscript K φ normal sum p 0 P 1 left frac 1 N p frac 1 N p 1 right cdot 2K varphi frac 1 N P cdot 2K varphi frac 1 N cdot 2K varphi the last form is gotten by telescoping the preceding sum and before that by using the identity S N N p cos φ cos 2 φ cos N p φ cos φ cos 2 φ cos N 1 φ subscript S N N p φ 2 φ normal N p φ φ 2 φ normal N 1 φ S N N p cos varphi cos 2 varphi ldots cos N p varphi cos varphi cos 2 varphi ldots cos N 1 varphi Thus we see that n N N P cos n φ n 2 K φ N ε superscript subscript n N N P n φ n 2 subscript K φ N ε left sum n N N P frac cos n varphi n right frac 2K varphi N varepsilon for all natural numbers P P P as soon as N 2 K φ ε N 2 subscript K φ ε N frac 2K varphi varepsilon According to the Cauchy criterion the latter series is convergent for the mentioned values of

    Original URL path: http://www.planetmath.org/exampleofsummationbyparts (2016-04-25)
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  • example of telescoping sum | planetmath.org
    superscript subscript k 1 n k α sum k 1 n sin k alpha may be telescoped if the terms are first edited by a suitable goniometric formula product formula E g we may write k 1 n cos k α 1 sin α 2 k 1 n cos k α sin α 2 superscript subscript k 1 n k α 1 α 2 superscript subscript k 1 n k α α 2 sum k 1 n cos k alpha frac 1 sin frac alpha 2 sum k 1 n cos k alpha sin frac alpha 2 The product formula cos x sin y 1 2 sin x y sin x y x y 1 2 x y x y cos x sin y frac 1 2 sin x y sin x y alters this to k 1 n cos k α 1 2 sin α 2 k 1 n sin 2 k 1 α 2 sin 2 k 1 α 2 superscript subscript k 1 n k α 1 2 α 2 superscript subscript k 1 n 2 k 1 α 2 2 k 1 α 2 sum k 1 n cos k alpha frac 1 2 sin frac alpha 2 sum k 1 n left sin frac 2k 1 alpha 2 sin frac 2k 1 alpha 2 right or k 1 n cos k α 1 2 sin α 2 sin 3 α 2 sin α 2 sin 5 α 2 sin 3 α 2 sin 2 n 1 α 2 sin 2 n 1 α 2 fragments superscript subscript k 1 n k α 1 2 α 2 fragments normal 3 α 2 α 2 5 α 2 3 α 2 normal 2 n 1 α 2 2 n 1 α 2 normal normal sum k 1

    Original URL path: http://www.planetmath.org/exampleoftelescopingsum (2016-04-25)
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  • example of using Eisenstein criterion | 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 3A40A05 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 using Eisenstein criterion For showing the irreducibility of the polynomial P x x 5 5 x 11 assign P x superscript x 5 5 x 11 P x x 5 5x 11 one would need a prime number dividing its other coefficients except the first one but there is no such prime However a suitable substitution x y a assign x y a x y a may change the situation Since the binomial coefficients of y 1 5 superscript y 1 5 y 1 5 except the first and the last one are divisible by 5 and 11 1 mod 5 11 annotated 1 pmod 5 11 equiv 1 mathop rm mod 5 we try x y 1 assign x y 1 x y 1 Then P y 1 y 5 5 y 4 10 y 3 10 y 2 10 y 5 P y 1 superscript y 5 5 superscript y 4 10 superscript y 3 10 superscript y 2 10 y 5 P y 1 y 5 5y 4 10y 3 10y 2 10y 5 Thus the prime 5 divides

    Original URL path: http://www.planetmath.org/exampleofusingeisensteincriterion (2016-04-25)
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  • example of using Lagrange multipliers | planetmath.org
    2Fpm 2Frdf sink 23this 3E 7B msc 3A11C08 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 using Lagrange multipliers One way to determine the perpendicular distance of the parallel planes A x B y C z D 0 and A x B y C z E 0 formulae sequence A x B y C z D 0 and A x B y C z E 0 Ax By Cz D 0 quad mbox and quad Ax By Cz E 0 is to use the Lagrange multiplier method In this case we may to minimise the Euclidean distance of a point x y z x y z x y z of the former plane to a fixed point x 0 y 0 z 0 subscript x 0 subscript y 0 subscript z 0 x 0 y 0 z 0 of the latter plane Thus we have the equation A x 0 B y 0 C z 0 E 0 A subscript x 0 B subscript y 0 C subscript z 0 E 0 Ax 0 By 0 Cz 0 E 0 which we can subtract from the first plane equation getting g A x x 0 B y y 0 C z x 0 D E 0 assign g A x subscript x 0 B y subscript y 0 C z subscript x 0 D E 0 displaystyle g A x x 0 B y y 0 C z x 0 D E 0 1 This is the only constraint equation for minimising the square f x x 0 2 y y 0 2 z x 0 2 assign f superscript x subscript x 0 2 superscript y subscript y 0 2 superscript z subscript x 0 2 displaystyle f x x 0 2 y y 0 2 z x 0 2 2 of the distance of the points The polynomial functions f f f and g g g satisfy the differentiability requirements Accordingly we can find the minimising point x y z x y z x y z by considering the system of equations formed by 1 and f x λ g x 2 x x 0 λ A 0 f y λ g y 2 y y 0 λ B 0 f z λ g z 2 z z 0 λ C 0 f x λ g x 2 xx0 λA 0 f y λ g y 2 yy0 λB 0 f z λ g z 2 zz0 λC 0 displaystyle begin cases frac partial f partial x lambda frac partial g partial x equiv 2 x x 0 lambda A 0 frac partial f partial y lambda frac partial g partial y equiv 2 y y 0 lambda B 0 frac partial f partial z lambda frac partial g partial z

    Original URL path: http://www.planetmath.org/exampleofusinglagrangemultipliers (2016-04-25)
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  • example of using residue theorem | planetmath.org
    residue theorem We take an example of applying the Cauchy residue theorem in evaluating usual real improper integrals We shall calculate the integral I 0 cos k x 1 x 2 d x assign I superscript subscript 0 k x 1 superscript x 2 d x I int 0 infty frac cos kx 1 x 2 dx where k k k is any real number One may prove that the integrand has no antiderivative among the elementary functions if k 0 k 0 k neq 0 Since the integrand is an even and x sin k x 1 x 2 maps to x k x 1 superscript x 2 x mapsto frac sin kx 1 x 2 an odd function we may write I 1 2 cos k x i sin k x 1 x 2 d x 1 2 e i k x 1 x 2 d x I 1 2 superscript subscript k x i k x 1 superscript x 2 d x 1 2 superscript subscript superscript e i k x 1 superscript x 2 d x I frac 1 2 int infty infty frac cos kx i sin kx 1 x 2 dx frac 1 2 int infty infty frac e ikx 1 x 2 dx using also Euler s formula Let s consider the contour integral J γ e i k z 1 z 2 d z assign J subscript contour integral γ superscript e i k z 1 superscript z 2 d z J oint gamma frac e ikz 1 z 2 dz where γ γ gamma is the perimeter of the semicircle consisting of the line segment from R 0 R 0 R 0 to R 0 R 0 R 0 and the semi circular arc c c c connecting these points in the upper half plane R 1 R 1 R 1 The integrand is analytic on and inside of γ γ gamma except in the point z i z i z i which is a simple pole Because we have cf the coefficients of Laurent series Res e i k z 1 z 2 i lim z i z i e i k z z 2 1 lim z i e i k z z i e k 2 i Res superscript e i k z 1 superscript z 2 i subscript normal z i z i superscript e i k z superscript z 2 1 subscript normal z i superscript e i k z z i superscript e k 2 i operatorname Res left frac e ikz 1 z 2 i right lim z to i z i frac e ikz z 2 1 lim z to i frac e ikz z i frac e k 2i the residue theorem yields J 2 π i e k 2 i π e k J normal 2 π i superscript e k 2 i π superscript e k J 2 pi i cdot frac e k 2i pi e k This does

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