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Funcion beta

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contiene ejercicios de la funciΓ³n beta para ingenieros y carreras a fines

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Funcion beta

1. 1. FunciΓ³n Beta 1π½ π₯, π¦ = π‘ π₯β1 (1 β π‘) π¦β1 ππ‘ ; π₯>0 π¦>0 0Si hacemos π‘ = π ππ2 π ππ‘ = 2 π ππ π cos π ππ πSi reemplazamos limites π‘ = 0 β π = 0 π‘=1 β π= 2Reemplazamos π 2π½ π₯, π¦ = 2 (π ππ2 π) π₯β1 1 β π ππ2 π¦β1 π ππ π cos π ππ 0 π 2π½ π₯, π¦ = 2 π ππ2π₯β1 π β cos2yβ1 π ππ 0 π1 2 π½ π₯, π¦ = π ππ2π₯β1 π β cos2yβ1 π ππ2 0Si hacemos 1 ππ’π‘= ππ‘ = 2 π π π‘ = 0 β π’ = β π¦ π π π‘=1β0=0 1+ π’ 1+ π’ 0 π₯β1 π¦β1 1 1 ππ’π½ π₯, π¦ = β 1β 2 β 1+ π’ 1+ π’ 1+ π’ β π₯β1 π¦β1 1 1+ π’β1 ππ’π½ π₯, π¦ = 2 0 1+ π’ 1+ π’ 1+ π’ β 1 π’ π¦ β1 ππ’π½ π₯, π¦ = π₯β1 β β 0 1+ π’ 1 + π’ π¦β1 1 + π’ 2 π’ π¦ β1 ππ’π½ π₯, π¦ = 1 + π’ π₯β1+π¦β1+2 π’ π¦β1 ππ’π½ π₯, π¦ = 1 + π’ π₯+π¦
2. 2. Teorema ΞxΞy π½ π₯, π¦ = ; π₯>0 π¦>0 Ξ x+yEjemplo π 2 tan π ππ 0 π/2 1/2 π ππ π ππ 0 cos π π/2 π ππ 1/2 π πππ  β1/2 π ππ 0Comparando π1 2 π½ π₯, π¦ = π ππ2π₯β1 π β cos2yβ1 π ππ2 0 1 1 3 π2π₯ β 1 = β 2π₯ = + 1 β 2π₯ = β π= 2 2 2 π 1 1 1 π2π¦ β 1 = β β 2π¦ = β + 1 β 2π¦ = β π= 2 2 2 πSi aplicamos el teorema Ξ 3 βΞ 1 1 4 4= β Ξ 3+1 2 4 4 Ξ 3 βΞ 1 1 4 4 4= β ; ππππ Ξ =Ξ 1 =1 2 Ξ 4 4 4
3. 3. 1 3 1= β Ξ β Ξ 2 4 4 1 1 1= β Ξ β Ξ 1β 2 4 4Aplicamos teorema de gamma 1 Ο= β Ο 2 sen 4 1 π= 2 2 2 π= 2 β π₯ π β1Resolver 0 1+π₯ ππ₯Por definiciΓ³n π’ π¦ β1 ππ’π½ π₯, π¦ = π₯ +π¦ 1+π’Comparandoy-1=p-1x+y=1y=px=1βpReemplazamos β π₯ πβ1 ππ₯ = π½ 1 β π, π 0 1+ π₯= π½ π, 1 β π
4. 4. Ξ p Ξ 1βp= Ξ p+1βp= Ξ p Ξ 1βpAplicamos teorema de gamma π= π ππ ππResolver β π 2π₯ ππ₯ ββ π 3π₯ + 1 2π’ = π 3π₯ β ln π’ = ln π 3π₯ β ln π’ = 3π₯ 1 1 ππ’π₯= ln π’ β ππ₯ = 3 3 ππ₯Evaluamos los lΓ­mitesCuando π₯ = β β π’=β π¦ π₯ = ββ β π’=0 1 β 2β ln π’ π 3 1 ππ’ 2 β 0 π’+1 3 π’ 2 1 β π 3 ln π’= 2 ππ’ 3 0 π’ π’+1Por propiedades de euler y logaritmos 2 β 1 π’3β π’β1= ππ’ 3 0 π’+1 2 β1 β 1 π’3= 2 ππ’ 3 0 π’+1
5. 5. π’ π¦ β1 ππ’Si comparamos con π½ π₯, π¦ = 1+π’ π₯ +π¦ 1 1 ππ¦β1= β β π¦= β +1β π= 3 3 π 2 ππ₯+ π¦ =2 β π₯ =2β β π= 3 π Reemplazamos1 4 2 π½ ,3 3 3 4 2 1 Ξ 3 Ξ 3= β 3 4 2 Ξ 3+3 1 1 2 1 Ξ 3 Ξ 3= β 3 3 6 Ξ 3 1 2 1 Ξ 3 Ξ 3= β 9 Ξ(2)Ξ 2 = 1! 1 1 2= βΞ Ξ 9 3 3 1 1 1= βΞ Ξ 1β 9 3 3Aplicamos teorema de gamma 1 Ο= β 9 sen Ο 3 1 Ο= β 9 3 2 2 Ο= β 9 3
6. 6. Resolver 3 ππ₯ 1 π₯β1 3β π₯ 3 1 1 β β π₯β1 2 3β π₯ 2 ππ₯ 1Sea x β 1 = 2y ο  x = 2y+1 ο  dx = 2dyCuando x = 1 y = 0 cuando x=3 y=1 1 1 1 β β= 2π¦ 2 3 β 2π¦ + 1 2 2ππ¦ 0 1 1 1 1 β β=2 2 2 (π¦)β2 3 β 2π¦ + 1 2 ππ¦ 0 1 2 β 1 β 1= π¦ 2 3 β 2π¦ β 1 2 ππ¦ 2 0 1 2 1 1= π¦ β2 2 β 2π¦ β 2 ππ¦ 2 0 1 1 2 β 1 β 2= π¦ 2 2 1β π¦ ππ¦ 2 0 1 2 1 1 1= π¦ β2 2β2 1β π¦ β 2 ππ¦ 2 0 1 2= π¦ β 1/2 (1 β π¦)β 1/2 ππ¦ 2 2 0Sea x - 1 = - Β½ ο x=Β½ y β 1 = - Β½ ο  y= Β½Luego 1 1 1 1 1 1 Ξ 2 Ξ 2 Ξ 2 Ξ 2 1 1 π½ , = 1 1 ο  ο  Ξ Ξ 2 2 Ξ 2 +2 Ξ(1) 2 2= πβ π= π Rta
7. 7. Ejercicio especialResolver 1 π β1 π₯ 1 β π₯ πβ1 ππ₯ 0 π₯ + π π +π π+1 π₯Sugerencia π¦= π+π₯π¦ π+ π₯ = π+1 π₯ Derivada de un cocienteπ¦π + π¦π₯ = π + 1 π₯ π(π + 1 β π¦) β π¦π(β1)π¦π = π + 1 π₯ β π¦π₯ π π + 1 β π¦ + π¦ππ¦π = π + 1 β π¦ π₯ π 2 + π β π¦π + π¦π π¦π π2 + ππ₯= π+1β π¦ π(π + 1) π π + 1 ππ¦ππ₯ = π+1β π¦ 2 π π π₯ = 0 β π¦ = 0Reemplazamos π π π₯ = 1 π β1 πβ1 π¦π π¦π π¦π 1= 1 1β π π+1 π+1β π¦ π+1β π¦ π+1β π¦ π +π 2 ππ¦0 π¦π π+1β π¦ π+1β π¦+ π π + 1 β π¦ = π¦π π¦π π β1 π + 1 β π¦ β π¦π πβ1 π + 1 = π¦π + π¦ 1 π + 1 β π¦ π β1 π+1β π¦ π π+1 π +π 2 ππ¦ π+1= π¦ π+10 π¦π + π π + 1 β π¦ π+1β π¦ π+1β π¦ π+1 = π¦ π+1 π¦π π β1 π + 1 β π¦ β π¦π πβ1 1 π+1β π¦ π β1 π + 1 β π¦ πβ1 π(π + 1) 1= π¦ 2 + π β π¦π π +π ππ¦0 π¦π + π (π + 1 β π¦)2 π + 1 β π¦ π +π π β1 π¦π π + 1 β π¦ β π¦π πβ1 π 2 + π 1 π + 1 β π¦ π β1+πβ1+2 ππ¦0 π 2 + π π +π π + 1 β π¦ π +π
8. 8. π β1 π¦π π + 1 β π¦ β π¦π πβ1 π2 + π 1 π + 1 β π¦ π +π ππ¦0 π 2 + π π +π π + 1 β π¦ π +π 1 π β1 π¦π π + 1 β π¦ β π¦π πβ1 π 2 + π π+1β π¦ π +π ππ¦0 π + 1 β π¦ π +π π 2 + π π +π 1 π β1 πβ1 π¦π π + 1 β π¦ β π¦π π2 + π ππ¦0 π 2 + π π +π 1 π β1 πβ1 π¦π π + 1 β π¦ β π¦π 2 + π π +πβ1 ππ¦0 π 1 π β1 πβ1 πβ1 π¦ π π + 1 β π¦ β π¦π ππ¦0 π π + 1 π +πβ1 1 π β1 πβ1 πβ1 π¦ π π + 1 β π¦ β π¦π π +πβ1 ππ¦0 π π + 1 π +πβ1 π +πβ1 βπ +1 π +πβ1βπ +1 π π β π = π = π 1 π β1 πβ1 π¦ π + 1 β π¦ β π¦π π ππ¦0 π π + 1 π +πβ1Como m, n, r son constantes son sacadas de la integral 1 1 π β1 πβ1 π π +πβ1 π¦ π + 1 β π¦ β π¦π ππ¦ π π+1 0 π + 1 β π¦ β π¦π = π + 1 β π¦(1 + π)= π + 1 (1 β π¦)Nos queda entonces 1 1 π β1 πβ1 π π +πβ1 π¦ π+1 (1 β π¦) πβ1 ππ¦ π π+1 0 π + 1 πβ1 1 π β1 πβ1 π π¦ 1β π¦ ππ¦ π π + 1 π +πβ1 0 1 1 π β1 π π π¦ (1 β π¦) πβ1 ππ¦ π π+1 0
9. 9. Si comparamos con 1 π½ π₯, π¦ = π‘ π₯β1 1 β π‘ π¦ β1 ππ‘ ; π₯>0 π¦>0 0π₯β1= πβ1 β π₯ = ππ¦β1= πβ1 β π¦ = πReemplazamos los nuevos valores 1= π π½(π, π) β¦ Rta π π+1 π