Power Series
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The document presents mathematical analysis of the Birch and Swinnerton-Dyer conjecture using power series. It derives formulas for the coefficients of the power series involving the sums and derivatives of the coefficients. It then applies these formulas and the Legendre polynomial to obtain an equation relating the coefficients to the value of the parameter ¦Ë.
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![(1-s2
)y??(s) - 2sy?(s) + ¦Ë y(s) = 0 (Legendre Polynomial)
[(1-12
)][c(2)] - 2(1)[c(2(1)-2)] + 6[(c)(12
- 2(1) + 1)] = 0
[(0)][c(2)] - 2(1)[c(0)] + 6[(c)(0)] = 0
¡Þ
y1 = c0 + ¡Æ c2ks2k
k=1
¦Ë = n(n+1) = 2(2+1) = 6
cn = (2n)!
2n
(n!)2
y1 = 1+ 3/2 (1)2
+ 4.375 (1)4
+¡+cnsn](https://image.slidesharecdn.com/solvingthebirchandswinnerton-150608211410-lva1-app6891/85/Power-Series-4-320.jpg)
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This document provides solutions to 40 problems involving techniques of integration. The problems cover a variety of integration techniques including substitution, integration by parts, and trigonometric substitutions. The solutions show the setup and evaluation of each integral, with many resulting in expressions involving common functions such as logarithms, inverse trigonometric functions, and hyperbolic functions.Calculo i
Calculo isalomon benito
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This document provides solutions to 40 problems involving techniques of integration. The problems cover a variety of integration techniques including substitution, integration by parts, and trigonometric substitutions. The solutions show the setup and evaluation of each integral, with many resulting in expressions involving common functions such as logarithms, inverse trigonometric functions, and hyperbolic functions.Recently uploaded (20)
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- 1. Power Series Analysis of the Birch and Swinnerton-Dyer Conjecture ¡Þ ¡Æ (n+1) cn+1 (s-1)(n+1)-1 n=0 =(1)c(1)(s-1)0 +(2)c(2)(s-1)(1) +(3)c(3)(s-1)(2) ¡ (n+1)cn+1 = cn, n= 0,1,2,3,¡, Or cn+1 = cn/(n+1) Hence c(1)=c(0), c(2)=c(1)/2, c(3)=c(2)/3 =(1)c(0)(s-1)0 +(2)/2 c(1)(s-1)1 +(3)/3 c(2)(s-1)2 ¡ ¡Þ = c(0) (s-1)0 + c(1)(s-1)1 + c(2) (s-1)2 ¡ = ¡Æ cn (s-1)n n=0
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- 4. (1-s2 )y??(s) - 2sy?(s) + ¦Ë y(s) = 0 (Legendre Polynomial) [(1-12 )][c(2)] - 2(1)[c(2(1)-2)] + 6[(c)(12 - 2(1) + 1)] = 0 [(0)][c(2)] - 2(1)[c(0)] + 6[(c)(0)] = 0 ¡Þ y1 = c0 + ¡Æ c2ks2k k=1 ¦Ë = n(n+1) = 2(2+1) = 6 cn = (2n)! 2n (n!)2 y1 = 1+ 3/2 (1)2 + 4.375 (1)4 +¡+cnsn