Split (1)

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https://people.maths.bris.ac.uk/~matyd/GroupNames/217/D5xS4.html	Copied to clipboard ## G = D5×S4order 240 = 24·3·5 ### Direct product of D5 and S4 Aliases: D5×S4, A41D10, C5⋊S4⋊C2, (C5×S4)⋊C2, C51(C2×S4), (D5×A4)⋊C2, (C2×C10)⋊D6, C22⋊(S3×D5), (C5×A4)⋊C22, (C22×D5)⋊S3, SmallGroup(240,194) Series: Derived Chief Lower central Upper central Derived series C1 — C22 — C5×A4 — D5×S4 Chief series C1 — C22 — C2×C10 — C5×A4 — D5×A4 — D5×S4 Lower central C5×A4 — D5×S4 Upper central C1 Generators and relations for D5×S4 G = < a,b,c,d,e,f \| a5=b2=c2=d2=e3=f2=1, bab=a-1, ac=ca, ad=da, ae=ea, af=fa, bc=cb, bd=db, be=eb, bf=fb, ece-1=fcf=cd=dc, ede-1=c, df=fd, fef=e-1 > Subgroups: 468 in 66 conjugacy classes, 13 normal (all characteristic) C1, C2 [×5], C3, C4 [×2], C22, C22 [×6], C5, S3 [×2], C6, C2×C4, D4 [×4], C23 [×2], D5, D5 [×2], C10 [×2], A4, D6, C15, C2×D4, Dic5, C20, D10 [×5], C2×C10, C2×C10, S4, S4, C2×A4, C5×S3, C3×D5, D15, C4×D5, D20, C5⋊D4 [×2], C5×D4, C22×D5, C22×D5, C2×S4, S3×D5, C5×A4, D4×D5, C5×S4, C5⋊S4, D5×A4, D5×S4 Quotients: C1, C2 [×3], C22, S3, D5, D6, D10, S4, C2×S4, S3×D5, D5×S4 Character table of D5×S4 class 1 2A 2B 2C 2D 2E 3 4A 4B 5A 5B 6 10A 10B 10C 10D 15A 15B 20A 20B size 1 3 5 6 15 30 8 6 30 2 2 40 6 6 12 12 16 16 12 12 ρ1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 trivial ρ2 1 1 -1 1 -1 -1 1 1 -1 1 1 -1 1 1 1 1 1 1 1 1 linear of order 2 ρ3 1 1 -1 -1 -1 1 1 -1 1 1 1 -1 1 1 -1 -1 1 1 -1 -1 linear of order 2 ρ4 1 1 1 -1 1 -1 1 -1 -1 1 1 1 1 1 -1 -1 1 1 -1 -1 linear of order 2 ρ5 2 2 -2 0 -2 0 -1 0 0 2 2 1 2 2 0 0 -1 -1 0 0 orthogonal lifted from D6 ρ6 2 2 2 0 2 0 -1 0 0 2 2 -1 2 2 0 0 -1 -1 0 0 orthogonal lifted from S3 ρ7 2 2 0 -2 0 0 2 -2 0 -1+√5/2 -1-√5/2 0 -1+√5/2 -1-√5/2 1-√5/2 1+√5/2 -1+√5/2 -1-√5/2 1-√5/2 1+√5/2 orthogonal lifted from D10 ρ8 2 2 0 2 0 0 2 2 0 -1-√5/2 -1+√5/2 0 -1-√5/2 -1+√5/2 -1-√5/2 -1+√5/2 -1-√5/2 -1+√5/2 -1-√5/2 -1+√5/2 orthogonal lifted from D5 ρ9 2 2 0 2 0 0 2 2 0 -1+√5/2 -1-√5/2 0 -1+√5/2 -1-√5/2 -1+√5/2 -1-√5/2 -1+√5/2 -1-√5/2 -1+√5/2 -1-√5/2 orthogonal lifted from D5 ρ10 2 2 0 -2 0 0 2 -2 0 -1-√5/2 -1+√5/2 0 -1-√5/2 -1+√5/2 1+√5/2 1-√5/2 -1-√5/2 -1+√5/2 1+√5/2 1-√5/2 orthogonal lifted from D10 ρ11 3 -1 -3 1 1 -1 0 -1 1 3 3 0 -1 -1 1 1 0 0 -1 -1 orthogonal lifted from C2×S4 ρ12 3 -1 3 1 -1 1 0 -1 -1 3 3 0 -1 -1 1 1 0 0 -1 -1 orthogonal lifted from S4 ρ13 3 -1 3 -1 -1 -1 0 1 1 3 3 0 -1 -1 -1 -1 0 0 1 1 orthogonal lifted from S4 ρ14 3 -1 -3 -1 1 1 0 1 -1 3 3 0 -1 -1 -1 -1 0 0 1 1 orthogonal lifted from C2×S4 ρ15 4 4 0 0 0 0 -2 0 0 -1-√5 -1+√5 0 -1-√5 -1+√5 0 0 1+√5/2 1-√5/2 0 0 orthogonal lifted from S3×D5 ρ16 4 4 0 0 0 0 -2 0 0 -1+√5 -1-√5 0 -1+√5 -1-√5 0 0 1-√5/2 1+√5/2 0 0 orthogonal lifted from S3×D5 ρ17 6 -2 0 2 0 0 0 -2 0 -3-3√5/2 -3+3√5/2 0 1+√5/2 1-√5/2 -1-√5/2 -1+√5/2 0 0 1+√5/2 1-√5/2 orthogonal faithful ρ18 6 -2 0 -2 0 0 0 2 0 -3+3√5/2 -3-3√5/2 0 1-√5/2 1+√5/2 1-√5/2 1+√5/2 0 0 -1+√5/2 -1-√5/2 orthogonal faithful ρ19 6 -2 0 2 0 0 0 -2 0 -3+3√5/2 -3-3√5/2 0 1-√5/2 1+√5/2 -1+√5/2 -1-√5/2 0 0 1-√5/2 1+√5/2 orthogonal faithful ρ20 6 -2 0 -2 0 0 0 2 0 -3-3√5/2 -3+3√5/2 0 1+√5/2 1-√5/2 1+√5/2 1-√5/2 0 0 -1-√5/2 -1+√5/2 orthogonal faithful Permutation representations of D5×S4 On 20 points - transitive group 20T69 Generators in S20 (1 2 3 4 5)(6 7 8 9 10)(11 12 13 14 15)(16 17 18 19 20) (1 5)(2 4)(6 10)(7 9)(11 15)(12 14)(16 20)(17 19) (1 11)(2 12)(3 13)(4 14)(5 15)(6 16)(7 17)(8 18)(9 19)(10 20) (1 16)(2 17)(3 18)(4 19)(5 20)(6 11)(7 12)(8 13)(9 14)(10 15) (6 11 16)(7 12 17)(8 13 18)(9 14 19)(10 15 20) (6 11)(7 12)(8 13)(9 14)(10 15) G:=sub<Sym(20)\| (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20), (1,5)(2,4)(6,10)(7,9)(11,15)(12,14)(16,20)(17,19), (1,11)(2,12)(3,13)(4,14)(5,15)(6,16)(7,17)(8,18)(9,19)(10,20), (1,16)(2,17)(3,18)(4,19)(5,20)(6,11)(7,12)(8,13)(9,14)(10,15), (6,11,16)(7,12,17)(8,13,18)(9,14,19)(10,15,20), (6,11)(7,12)(8,13)(9,14)(10,15)>; G:=Group( (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20), (1,5)(2,4)(6,10)(7,9)(11,15)(12,14)(16,20)(17,19), (1,11)(2,12)(3,13)(4,14)(5,15)(6,16)(7,17)(8,18)(9,19)(10,20), (1,16)(2,17)(3,18)(4,19)(5,20)(6,11)(7,12)(8,13)(9,14)(10,15), (6,11,16)(7,12,17)(8,13,18)(9,14,19)(10,15,20), (6,11)(7,12)(8,13)(9,14)(10,15) ); G=PermutationGroup([(1,2,3,4,5),(6,7,8,9,10),(11,12,13,14,15),(16,17,18,19,20)], [(1,5),(2,4),(6,10),(7,9),(11,15),(12,14),(16,20),(17,19)], [(1,11),(2,12),(3,13),(4,14),(5,15),(6,16),(7,17),(8,18),(9,19),(10,20)], [(1,16),(2,17),(3,18),(4,19),(5,20),(6,11),(7,12),(8,13),(9,14),(10,15)], [(6,11,16),(7,12,17),(8,13,18),(9,14,19),(10,15,20)], [(6,11),(7,12),(8,13),(9,14),(10,15)]) G:=TransitiveGroup(20,69); On 30 points - transitive group 30T54 Generators in S30 (1 2 3 4 5)(6 7 8 9 10)(11 12 13 14 15)(16 17 18 19 20)(21 22 23 24 25)(26 27 28 29 30) (1 5)(2 4)(6 9)(7 8)(11 14)(12 13)(16 17)(18 20)(21 24)(22 23)(26 27)(28 30) (1 13)(2 14)(3 15)(4 11)(5 12)(16 22)(17 23)(18 24)(19 25)(20 21) (6 30)(7 26)(8 27)(9 28)(10 29)(16 22)(17 23)(18 24)(19 25)(20 21) (1 17 27)(2 18 28)(3 19 29)(4 20 30)(5 16 26)(6 11 21)(7 12 22)(8 13 23)(9 14 24)(10 15 25) (1 13)(2 14)(3 15)(4 11)(5 12)(6 20)(7 16)(8 17)(9 18)(10 19)(21 30)(22 26)(23 27)(24 28)(25 29) G:=sub<Sym(30)\| (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)(26,27,28,29,30), (1,5)(2,4)(6,9)(7,8)(11,14)(12,13)(16,17)(18,20)(21,24)(22,23)(26,27)(28,30), (1,13)(2,14)(3,15)(4,11)(5,12)(16,22)(17,23)(18,24)(19,25)(20,21), (6,30)(7,26)(8,27)(9,28)(10,29)(16,22)(17,23)(18,24)(19,25)(20,21), (1,17,27)(2,18,28)(3,19,29)(4,20,30)(5,16,26)(6,11,21)(7,12,22)(8,13,23)(9,14,24)(10,15,25), (1,13)(2,14)(3,15)(4,11)(5,12)(6,20)(7,16)(8,17)(9,18)(10,19)(21,30)(22,26)(23,27)(24,28)(25,29)>; G:=Group( (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)(26,27,28,29,30), (1,5)(2,4)(6,9)(7,8)(11,14)(12,13)(16,17)(18,20)(21,24)(22,23)(26,27)(28,30), (1,13)(2,14)(3,15)(4,11)(5,12)(16,22)(17,23)(18,24)(19,25)(20,21), (6,30)(7,26)(8,27)(9,28)(10,29)(16,22)(17,23)(18,24)(19,25)(20,21), (1,17,27)(2,18,28)(3,19,29)(4,20,30)(5,16,26)(6,11,21)(7,12,22)(8,13,23)(9,14,24)(10,15,25), (1,13)(2,14)(3,15)(4,11)(5,12)(6,20)(7,16)(8,17)(9,18)(10,19)(21,30)(22,26)(23,27)(24,28)(25,29) ); G=PermutationGroup([(1,2,3,4,5),(6,7,8,9,10),(11,12,13,14,15),(16,17,18,19,20),(21,22,23,24,25),(26,27,28,29,30)], [(1,5),(2,4),(6,9),(7,8),(11,14),(12,13),(16,17),(18,20),(21,24),(22,23),(26,27),(28,30)], [(1,13),(2,14),(3,15),(4,11),(5,12),(16,22),(17,23),(18,24),(19,25),(20,21)], [(6,30),(7,26),(8,27),(9,28),(10,29),(16,22),(17,23),(18,24),(19,25),(20,21)], [(1,17,27),(2,18,28),(3,19,29),(4,20,30),(5,16,26),(6,11,21),(7,12,22),(8,13,23),(9,14,24),(10,15,25)], [(1,13),(2,14),(3,15),(4,11),(5,12),(6,20),(7,16),(8,17),(9,18),(10,19),(21,30),(22,26),(23,27),(24,28),(25,29)]) G:=TransitiveGroup(30,54); On 30 points - transitive group 30T59 Generators in S30 (1 2 3 4 5)(6 7 8 9 10)(11 12 13 14 15)(16 17 18 19 20)(21 22 23 24 25)(26 27 28 29 30) (1 12)(2 11)(3 15)(4 14)(5 13)(6 28)(7 27)(8 26)(9 30)(10 29)(16 23)(17 22)(18 21)(19 25)(20 24) (1 13)(2 14)(3 15)(4 11)(5 12)(16 22)(17 23)(18 24)(19 25)(20 21) (6 30)(7 26)(8 27)(9 28)(10 29)(16 22)(17 23)(18 24)(19 25)(20 21) (1 17 27)(2 18 28)(3 19 29)(4 20 30)(5 16 26)(6 11 21)(7 12 22)(8 13 23)(9 14 24)(10 15 25) (6 21)(7 22)(8 23)(9 24)(10 25)(16 26)(17 27)(18 28)(19 29)(20 30) G:=sub<Sym(30)\| (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)(26,27,28,29,30), (1,12)(2,11)(3,15)(4,14)(5,13)(6,28)(7,27)(8,26)(9,30)(10,29)(16,23)(17,22)(18,21)(19,25)(20,24), (1,13)(2,14)(3,15)(4,11)(5,12)(16,22)(17,23)(18,24)(19,25)(20,21), (6,30)(7,26)(8,27)(9,28)(10,29)(16,22)(17,23)(18,24)(19,25)(20,21), (1,17,27)(2,18,28)(3,19,29)(4,20,30)(5,16,26)(6,11,21)(7,12,22)(8,13,23)(9,14,24)(10,15,25), (6,21)(7,22)(8,23)(9,24)(10,25)(16,26)(17,27)(18,28)(19,29)(20,30)>; G:=Group( (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)(26,27,28,29,30), (1,12)(2,11)(3,15)(4,14)(5,13)(6,28)(7,27)(8,26)(9,30)(10,29)(16,23)(17,22)(18,21)(19,25)(20,24), (1,13)(2,14)(3,15)(4,11)(5,12)(16,22)(17,23)(18,24)(19,25)(20,21), (6,30)(7,26)(8,27)(9,28)(10,29)(16,22)(17,23)(18,24)(19,25)(20,21), (1,17,27)(2,18,28)(3,19,29)(4,20,30)(5,16,26)(6,11,21)(7,12,22)(8,13,23)(9,14,24)(10,15,25), (6,21)(7,22)(8,23)(9,24)(10,25)(16,26)(17,27)(18,28)(19,29)(20,30) ); G=PermutationGroup([(1,2,3,4,5),(6,7,8,9,10),(11,12,13,14,15),(16,17,18,19,20),(21,22,23,24,25),(26,27,28,29,30)], [(1,12),(2,11),(3,15),(4,14),(5,13),(6,28),(7,27),(8,26),(9,30),(10,29),(16,23),(17,22),(18,21),(19,25),(20,24)], [(1,13),(2,14),(3,15),(4,11),(5,12),(16,22),(17,23),(18,24),(19,25),(20,21)], [(6,30),(7,26),(8,27),(9,28),(10,29),(16,22),(17,23),(18,24),(19,25),(20,21)], [(1,17,27),(2,18,28),(3,19,29),(4,20,30),(5,16,26),(6,11,21),(7,12,22),(8,13,23),(9,14,24),(10,15,25)], [(6,21),(7,22),(8,23),(9,24),(10,25),(16,26),(17,27),(18,28),(19,29),(20,30)]) G:=TransitiveGroup(30,59); On 30 points - transitive group 30T62 Generators in S30 (1 2 3 4 5)(6 7 8 9 10)(11 12 13 14 15)(16 17 18 19 20)(21 22 23 24 25)(26 27 28 29 30) (1 5)(2 4)(6 9)(7 8)(11 14)(12 13)(16 17)(18 20)(21 24)(22 23)(26 27)(28 30) (1 13)(2 14)(3 15)(4 11)(5 12)(16 22)(17 23)(18 24)(19 25)(20 21) (6 30)(7 26)(8 27)(9 28)(10 29)(16 22)(17 23)(18 24)(19 25)(20 21) (1 17 27)(2 18 28)(3 19 29)(4 20 30)(5 16 26)(6 11 21)(7 12 22)(8 13 23)(9 14 24)(10 15 25) (6 21)(7 22)(8 23)(9 24)(10 25)(16 26)(17 27)(18 28)(19 29)(20 30) G:=sub<Sym(30)\| (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)(26,27,28,29,30), (1,5)(2,4)(6,9)(7,8)(11,14)(12,13)(16,17)(18,20)(21,24)(22,23)(26,27)(28,30), (1,13)(2,14)(3,15)(4,11)(5,12)(16,22)(17,23)(18,24)(19,25)(20,21), (6,30)(7,26)(8,27)(9,28)(10,29)(16,22)(17,23)(18,24)(19,25)(20,21), (1,17,27)(2,18,28)(3,19,29)(4,20,30)(5,16,26)(6,11,21)(7,12,22)(8,13,23)(9,14,24)(10,15,25), (6,21)(7,22)(8,23)(9,24)(10,25)(16,26)(17,27)(18,28)(19,29)(20,30)>; G:=Group( (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)(26,27,28,29,30), (1,5)(2,4)(6,9)(7,8)(11,14)(12,13)(16,17)(18,20)(21,24)(22,23)(26,27)(28,30), (1,13)(2,14)(3,15)(4,11)(5,12)(16,22)(17,23)(18,24)(19,25)(20,21), (6,30)(7,26)(8,27)(9,28)(10,29)(16,22)(17,23)(18,24)(19,25)(20,21), (1,17,27)(2,18,28)(3,19,29)(4,20,30)(5,16,26)(6,11,21)(7,12,22)(8,13,23)(9,14,24)(10,15,25), (6,21)(7,22)(8,23)(9,24)(10,25)(16,26)(17,27)(18,28)(19,29)(20,30) ); G=PermutationGroup([(1,2,3,4,5),(6,7,8,9,10),(11,12,13,14,15),(16,17,18,19,20),(21,22,23,24,25),(26,27,28,29,30)], [(1,5),(2,4),(6,9),(7,8),(11,14),(12,13),(16,17),(18,20),(21,24),(22,23),(26,27),(28,30)], [(1,13),(2,14),(3,15),(4,11),(5,12),(16,22),(17,23),(18,24),(19,25),(20,21)], [(6,30),(7,26),(8,27),(9,28),(10,29),(16,22),(17,23),(18,24),(19,25),(20,21)], [(1,17,27),(2,18,28),(3,19,29),(4,20,30),(5,16,26),(6,11,21),(7,12,22),(8,13,23),(9,14,24),(10,15,25)], [(6,21),(7,22),(8,23),(9,24),(10,25),(16,26),(17,27),(18,28),(19,29),(20,30)]) G:=TransitiveGroup(30,62); On 30 points - transitive group 30T63 Generators in S30 (1 2 3 4 5)(6 7 8 9 10)(11 12 13 14 15)(16 17 18 19 20)(21 22 23 24 25)(26 27 28 29 30) (1 12)(2 11)(3 15)(4 14)(5 13)(6 28)(7 27)(8 26)(9 30)(10 29)(16 23)(17 22)(18 21)(19 25)(20 24) (1 13)(2 14)(3 15)(4 11)(5 12)(16 22)(17 23)(18 24)(19 25)(20 21) (6 30)(7 26)(8 27)(9 28)(10 29)(16 22)(17 23)(18 24)(19 25)(20 21) (1 17 27)(2 18 28)(3 19 29)(4 20 30)(5 16 26)(6 11 21)(7 12 22)(8 13 23)(9 14 24)(10 15 25) (1 13)(2 14)(3 15)(4 11)(5 12)(6 20)(7 16)(8 17)(9 18)(10 19)(21 30)(22 26)(23 27)(24 28)(25 29) G:=sub<Sym(30)\| (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)(26,27,28,29,30), (1,12)(2,11)(3,15)(4,14)(5,13)(6,28)(7,27)(8,26)(9,30)(10,29)(16,23)(17,22)(18,21)(19,25)(20,24), (1,13)(2,14)(3,15)(4,11)(5,12)(16,22)(17,23)(18,24)(19,25)(20,21), (6,30)(7,26)(8,27)(9,28)(10,29)(16,22)(17,23)(18,24)(19,25)(20,21), (1,17,27)(2,18,28)(3,19,29)(4,20,30)(5,16,26)(6,11,21)(7,12,22)(8,13,23)(9,14,24)(10,15,25), (1,13)(2,14)(3,15)(4,11)(5,12)(6,20)(7,16)(8,17)(9,18)(10,19)(21,30)(22,26)(23,27)(24,28)(25,29)>; G:=Group( (1,2,3,4,5)(6,7,8,9,10)(11,12,13,14,15)(16,17,18,19,20)(21,22,23,24,25)(26,27,28,29,30), (1,12)(2,11)(3,15)(4,14)(5,13)(6,28)(7,27)(8,26)(9,30)(10,29)(16,23)(17,22)(18,21)(19,25)(20,24), (1,13)(2,14)(3,15)(4,11)(5,12)(16,22)(17,23)(18,24)(19,25)(20,21), (6,30)(7,26)(8,27)(9,28)(10,29)(16,22)(17,23)(18,24)(19,25)(20,21), (1,17,27)(2,18,28)(3,19,29)(4,20,30)(5,16,26)(6,11,21)(7,12,22)(8,13,23)(9,14,24)(10,15,25), (1,13)(2,14)(3,15)(4,11)(5,12)(6,20)(7,16)(8,17)(9,18)(10,19)(21,30)(22,26)(23,27)(24,28)(25,29) ); G=PermutationGroup([(1,2,3,4,5),(6,7,8,9,10),(11,12,13,14,15),(16,17,18,19,20),(21,22,23,24,25),(26,27,28,29,30)], [(1,12),(2,11),(3,15),(4,14),(5,13),(6,28),(7,27),(8,26),(9,30),(10,29),(16,23),(17,22),(18,21),(19,25),(20,24)], [(1,13),(2,14),(3,15),(4,11),(5,12),(16,22),(17,23),(18,24),(19,25),(20,21)], [(6,30),(7,26),(8,27),(9,28),(10,29),(16,22),(17,23),(18,24),(19,25),(20,21)], [(1,17,27),(2,18,28),(3,19,29),(4,20,30),(5,16,26),(6,11,21),(7,12,22),(8,13,23),(9,14,24),(10,15,25)], [(1,13),(2,14),(3,15),(4,11),(5,12),(6,20),(7,16),(8,17),(9,18),(10,19),(21,30),(22,26),(23,27),(24,28),(25,29)]) G:=TransitiveGroup(30,63); D5×S4 is a maximal quotient of CSU2(𝔽3)⋊D5 Dic5.6S4 Dic5.7S4 GL2(𝔽3)⋊D5 D10.1S4 D10.2S4 A4⋊Dic10 Dic52S4 Dic5⋊S4 D10⋊S4 A4⋊D20 Matrix representation of D5×S4 in GL5(𝔽61) 0 1 0 0 0 60 17 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 1 , 0 60 0 0 0 60 0 0 0 0 0 0 60 0 0 0 0 0 60 0 0 0 0 0 60 , 1 0 0 0 0 0 1 0 0 0 0 0 0 60 1 0 0 0 60 0 0 0 1 60 0 , 1 0 0 0 0 0 1 0 0 0 0 0 0 1 60 0 0 1 0 60 0 0 0 0 60 , 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 , 60 0 0 0 0 0 60 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 G:=sub<GL(5,GF(61))\| [0,60,0,0,0,1,17,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,1],[0,60,0,0,0,60,0,0,0,0,0,0,60,0,0,0,0,0,60,0,0,0,0,0,60],[1,0,0,0,0,0,1,0,0,0,0,0,0,0,1,0,0,60,60,60,0,0,1,0,0],[1,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,60,60,60],[1,0,0,0,0,0,1,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,1,0,0],[60,0,0,0,0,0,60,0,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,1] >; D5×S4 in GAP, Magma, Sage, TeX D_5\times S_4 % in TeX G:=Group("D5xS4"); // GroupNames label G:=SmallGroup(240,194); // by ID G=gap.SmallGroup(240,194); # by ID G:=PCGroup([6,-2,-2,-3,-5,-2,2,80,1155,1810,916,1091,1637]); // Polycyclic G:=Group<a,b,c,d,e,f\|a^5=b^2=c^2=d^2=e^3=f^2=1,bab=a^-1,ac=ca,ad=da,ae=ea,af=fa,bc=cb,bd=db,be=eb,bf=fb,ece^-1=fcf=cd=dc,ede^-1=c,df=fd,fef=e^-1>; // generators/relations Export ׿ × 𝔽	2018-12-11 05:32:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9998825788497925, "perplexity": 5042.991155757338}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-51/segments/1544376823565.27/warc/CC-MAIN-20181211040413-20181211061913-00315.warc.gz"}
https://fykos.org/problems/start?tasktag=hydroMech&p=8	Search hydromechanics 2. Series 20. Year - 2. crushing impact Find a relation between the speed of meteorite (just before impact) and radius of created crater. Na problém narazil Honza Prachař při psaní textu Fyzikální olympiády. 2. Series 20. Year - E. waves on the water Using dimension (unit) analysis find the relation for the speed of waves on the water surface. Verify the theoretical relation and find unknown constants from measuring the speed of waves dependent on wavelength. Do not forget that there are two types of waves – one caused by gravitational field and second by surface tension. Úloha napadla Honzu Prachaře při čteni Feynmanoých přednášek z fyziky. 2. Series 20. Year - P. shaking the tea Explain why the tea in the box after it is shaken is separated such that larger pieces of tea leaves are on the surface. The solution can be enhanced using your observation. S úlohou přišel Petr Sýkora. 1. Series 20. Year - P. tree height Estimate the height of the trees on the planet. Think over all possible aspects influencing the height of the trees. Úlohu navrhla Zuzka Safernová. 5. Series 19. Year - 1. what a stability of a ship Scientists in NASA discovered special sediments of vegetable origin at Europa (Jupiter's moon). They are suitable for making very strong planks of oblong shape or triangular shape and are ideal for building a ship of height $h$, length $d$ and width 2$a$ (see figure 1) Help: captain to decide for which range of densities of ocean it is safe to sail such boat. Assume that planks have constant thickness and density $ρ_{m}$ and that the ship is hollow and has a deck. (discuss the case when the ship is not hollow and has constant density $ρ_{m}.)$ It is not necessary to show just one final equation, more important is practical recipe (including all required equations) how to make such calculations. Try to make it simple to understand and explain all approximations made. Vymyslel Pavel Brom při vzpomínce na historku o jedné nešťastně navržené lodi. 4. Series 19. Year - 1. competition of balloons This series of questions is dedicated to the research on the „planet of the balloons“. This year balloons are competing in the 'The higher, the better' competition. Each balloon has a piece of string attached to measure his height. All balloons have the same parameters and noone of them have won yet. The length density of string is 11 lufts per sprungl, density of atmosphere is 110101 lufts per cubic sprungl and the radius of each balloon is 10 sprungls and weight of balloon is 10 lufts. Every object in gravitation field of the planet increases its speed by 111 sprungls per temp. Calculate the maximum height which the referees will measure and how the balloon will move after reaching this height. Unlifted part of the string is laying freely on the ground. The competition happens at low altitudes, where density of atmosphere is approximately constant. Hint: Sprungl, luft and temp are units used on planet of the balloons. Each balloon has maximum of 1 string attached. Úlohu navrhl Petr Sýkora od Havránka. 4. Series 19. Year - 3. balloons on the merry go round Two balloons are fixed on a single string of length $l$. The string is threaded through a horizontal hollow tube of length $L$. Both balloons weight approximately the same, however one is slightly heavier. The tube starts spinning around vertical axis. What is the optimum position of this axis for the horizontal distance between balloons to be the biggest? Vymyslel Jirka a Kájínek špatně pochopil. 4. Series 19. Year - 4. wedding of two balloons At the wedding ceremony the new couple kisses. For balloons it means to connect it orifices to connect the inside gas into a single One. Describe what would happened after the orifices are connected. Do not forget that all balloons have the same parameters. Navrhl Petr Sýkora. 5. Series 18. Year - P. faster than water Is it possible for the boat to move faster than the water in the river? Justify you answer and assume laminar flow. Kapicova úloha 3. Series 18. Year - 4. with the glider over the channel One of well known glider pilots decided to cross the British Channel. In Calais he rented a plane to took him to the height h = 3 km and from there glided directly to England. As every pilot knows, glider's downward speed $v_{kles}$ depends on the forward speed $v_{dop}$ as in the image 2. What is the optimum speed to achieve the longest flight? When the pilot is 3/4 of its way to England strong wind starts to blow in direction from England to France at the speed 10 $ms^{-1}$. What is the optimum speed now? What is the maximum wind speed which allows him to come to England? And what is the sped of wind to allow for the safe return to France? Vymyslel Matouš. Partners Host Media partner Created with <love/> by ©FYKOS – webmaster@fykos.cz	2021-01-26 12:09:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5413687825202942, "perplexity": 2794.416863009334}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610704799741.85/warc/CC-MAIN-20210126104721-20210126134721-00287.warc.gz"}
https://ginger.readthedocs.io/en/latest/news/version_0.9.2.html	# Ginger Version 0.9.2¶ ## Ubuntu SNAP Packaging¶ Canonical have a new ‘universal’ package format they call Snap. Ginger has been rewritten to be Snap compliant. This is the fastest and simplest way to install, update and remove Ginger, at least on Ubuntu. (There are other competing formats, such as Flatpak, that would also be possible.) ## Command-Line Interface¶ The command-line interface has been substantially rewritten to make GNU readline part of the core functionality. This eliminates the dependency on rlwrap, although that remains a useful tool for developers e.g. for gvmtest. This simplifies the delivery and setup and the ginger executable is now a genuine executable and not a shebang script - the latter unfortunately being more limited in terms of scripting. The ginger executable now takes a ‘command’ argument that selects the actual program: % ginger <COMMAND> <OPTIONS> The program is selected by the very simple strategy of prefixing the command with “ginger-”. Hence, as an example, the following two command-line are identical: % ginger admin	2023-02-03 07:41:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.40863847732543945, "perplexity": 5337.023536218695}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764500044.16/warc/CC-MAIN-20230203055519-20230203085519-00352.warc.gz"}
https://www.lmfdb.org/Genus2Curve/Q/50000/a/200000/1	Properties Label 50000.a.200000.1 Conductor $50000$ Discriminant $200000$ Mordell-Weil group $$\Z \oplus \Z/{5}\Z$$ Sato-Tate group $F_{ac}$ $$\End(J_{\overline{\Q}}) \otimes \R$$ $$\C \times \C$$ $$\End(J_{\overline{\Q}}) \otimes \Q$$ $$\mathsf{CM}$$ $$\End(J) \otimes \Q$$ $$\Q$$ $$\overline{\Q}$$-simple yes $$\mathrm{GL}_2$$-type no Learn more Show commands: SageMath / Magma Simplified equation $y^2 + y = 2x^5$ (homogenize, simplify) $y^2 + z^3y = 2x^5z$ (dehomogenize, simplify) $y^2 = 8x^5 + 1$ (homogenize, minimize) sage: R.<x> = PolynomialRing(QQ); C = HyperellipticCurve(R([0, 0, 0, 0, 0, 2]), R([1])); magma: R<x> := PolynomialRing(Rationals()); C := HyperellipticCurve(R![0, 0, 0, 0, 0, 2], R![1]); sage: X = HyperellipticCurve(R([1, 0, 0, 0, 0, 8])) magma: X,pi:= SimplifiedModel(C); Invariants Conductor: $$N$$ $$=$$ $$50000$$ $$=$$ $$2^{4} \cdot 5^{5}$$ magma: Conductor(LSeries(C)); Factorization($1); Discriminant: $$\Delta$$ $$=$$ $$200000$$ $$=$$ $$2^{6} \cdot 5^{5}$$ magma: Discriminant(C); Factorization(Integers()!$1); G2 invariants $$I_2$$ $$=$$ $$0$$ $$=$$ $$0$$ $$I_4$$ $$=$$ $$0$$ $$=$$ $$0$$ $$I_6$$ $$=$$ $$0$$ $$=$$ $$0$$ $$I_{10}$$ $$=$$ $$8$$ $$=$$ $$2^{3}$$ $$J_2$$ $$=$$ $$0$$ $$=$$ $$0$$ $$J_4$$ $$=$$ $$0$$ $$=$$ $$0$$ $$J_6$$ $$=$$ $$0$$ $$=$$ $$0$$ $$J_8$$ $$=$$ $$0$$ $$=$$ $$0$$ $$J_{10}$$ $$=$$ $$200000$$ $$=$$ $$2^{6} \cdot 5^{5}$$ $$g_1$$ $$=$$ $$0$$ $$g_2$$ $$=$$ $$0$$ $$g_3$$ $$=$$ $$0$$ sage: C.igusa_clebsch_invariants(); [factor(a) for a in _] magma: IgusaClebschInvariants(C); IgusaInvariants(C); G2Invariants(C); Automorphism group $$\mathrm{Aut}(X)$$ $$\simeq$$ $C_2$ magma: AutomorphismGroup(C); IdentifyGroup($1); $$\mathrm{Aut}(X_{\overline{\Q}})$$ $$\simeq$$$C_{10}$magma: AutomorphismGroup(ChangeRing(C,AlgebraicClosure(Rationals()))); IdentifyGroup($1); Rational points All points: $$(1 : 0 : 0),\, (0 : 0 : 1),\, (0 : -1 : 1),\, (1 : 1 : 1),\, (1 : -2 : 1)$$ All points: $$(1 : 0 : 0),\, (0 : 0 : 1),\, (0 : -1 : 1),\, (1 : 1 : 1),\, (1 : -2 : 1)$$ All points: $$(1 : 0 : 0),\, (0 : -1 : 1),\, (0 : 1 : 1),\, (1 : -3 : 1),\, (1 : 3 : 1)$$ magma: [C![0,-1,1],C![0,0,1],C![1,-2,1],C![1,0,0],C![1,1,1]]; // minimal model magma: [C![0,-1,1],C![0,1,1],C![1,-3,1],C![1,0,0],C![1,3,1]]; // simplified model Number of rational Weierstrass points: $$1$$ magma: #Roots(HyperellipticPolynomials(SimplifiedModel(C))); This curve is locally solvable everywhere. magma: f,h:=HyperellipticPolynomials(C); g:=4*f+h^2; HasPointsEverywhereLocally(g,2) and (#Roots(ChangeRing(g,RealField())) gt 0 or LeadingCoefficient(g) gt 0); Mordell-Weil group of the Jacobian Group structure: $$\Z \oplus \Z/{5}\Z$$ magma: MordellWeilGroupGenus2(Jacobian(C)); Generator $D_0$ Height Order $$(1 : -2 : 1) - (1 : 0 : 0)$$ $$x - z$$ $$=$$ $$0,$$ $$y$$ $$=$$ $$-2z^3$$ $$0.821051$$ $$\infty$$ $$(0 : 0 : 1) - (1 : 0 : 0)$$ $$x$$ $$=$$ $$0,$$ $$y$$ $$=$$ $$0$$ $$0$$ $$5$$ Generator $D_0$ Height Order $$(1 : -2 : 1) - (1 : 0 : 0)$$ $$x - z$$ $$=$$ $$0,$$ $$y$$ $$=$$ $$-2z^3$$ $$0.821051$$ $$\infty$$ $$(0 : 0 : 1) - (1 : 0 : 0)$$ $$x$$ $$=$$ $$0,$$ $$y$$ $$=$$ $$0$$ $$0$$ $$5$$ Generator $D_0$ Height Order $$(1 : -3 : 1) - (1 : 0 : 0)$$ $$x - z$$ $$=$$ $$0,$$ $$y$$ $$=$$ $$-3z^3$$ $$0.821051$$ $$\infty$$ $$(0 : 1 : 1) - (1 : 0 : 0)$$ $$x$$ $$=$$ $$0,$$ $$y$$ $$=$$ $$z^3$$ $$0$$ $$5$$ BSD invariants Hasse-Weil conjecture: unverified Analytic rank: $$1$$ Mordell-Weil rank: $$1$$ 2-Selmer rank: $$1$$ Regulator: $$0.821051$$ Real period: $$11.84775$$ Tamagawa product: $$5$$ Torsion order: $$5$$ Leading coefficient: $$1.945523$$ Analytic order of Ш: $$1$$ (rounded) Order of Ш: square Local invariants Prime ord($$N$$) ord($$\Delta$$) Tamagawa L-factor Cluster picture $$2$$ $$4$$ $$6$$ $$5$$ $$1$$ $$5$$ $$5$$ $$5$$ $$1$$ $$1$$ Galois representations For primes $\ell \ge 5$ the Galois representation data has not been computed for this curve since it is not generic. For primes $\ell \le 3$, the image of the mod-$\ell$ Galois representation is listed in the table below, whenever it is not the maximal image $\GSp(4,\F_\ell)$. Prime $$\ell$$ mod-$$\ell$$ image Is torsion prime? $$2$$ 2.36.1 no $$3$$ 3.1296.1 no Sato-Tate group $$\mathrm{ST}$$ $$\simeq$$ $F_{ac}$ $$\mathrm{ST}^0$$ $$\simeq$$ $$\mathrm{U}(1)\times\mathrm{U}(1)$$ Decomposition of the Jacobian Simple over $$\overline{\Q}$$ Endomorphisms of the Jacobian Not of $$\GL_2$$-type over $$\Q$$ Endomorphism ring over $$\Q$$: $$\End (J_{})$$ $$\simeq$$ $$\Z$$ $$\End (J_{}) \otimes \Q$$ $$\simeq$$ $$\Q$$ $$\End (J_{}) \otimes \R$$ $$\simeq$$ $$\R$$ Smallest field over which all endomorphisms are defined: Galois number field $$K = \Q (a) \simeq$$ $$\Q(\zeta_{5})$$ with defining polynomial $$x^{4} - x^{3} + x^{2} - x + 1$$ Not of $$\GL_2$$-type over $$\overline{\Q}$$ Endomorphism ring over $$\overline{\Q}$$: $$\End (J_{\overline{\Q}})$$ $$\simeq$$ the maximal order of $$\End (J_{\overline{\Q}}) \otimes \Q$$ $$\End (J_{\overline{\Q}}) \otimes \Q$$ $$\simeq$$ $$\Q(\zeta_{5})$$ (CM) $$\End (J_{\overline{\Q}}) \otimes \R$$ $$\simeq$$ $$\C \times \C$$ Remainder of the endomorphism lattice by field Over subfield $$F \simeq$$ $$\Q(\sqrt{5})$$ with generator $$a^{3} - a^{2}$$ with minimal polynomial $$x^{2} - x - 1$$: $$\End (J_{F})$$ $$\simeq$$ $$\Z [\frac{1 + \sqrt{5}}{2}]$$ $$\End (J_{F}) \otimes \Q$$ $$\simeq$$ $$\Q(\sqrt{5})$$ $$\End (J_{F}) \otimes \R$$ $$\simeq$$ $$\R \times \R$$ Sato Tate group: F_{ab} Of $$\GL_2$$-type, simple	2023-01-28 13:50:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9599648714065552, "perplexity": 2003.819821702147}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-06/segments/1674764499634.11/warc/CC-MAIN-20230128121809-20230128151809-00711.warc.gz"}
https://math.stackexchange.com/questions/2813284/squaring-in-a-multivariable-equation/2814013	# Squaring in a multivariable equation I'm doing the following exercise: "If $x = \frac{y}{\sqrt{1 + y^2}}$, then express $y$ in terms of $x$." I know that we can solve it by squaring and doing the usual arithmetic operations until we get: $$y^2 = \frac{x^2}{1 - x^2}$$ Then taking the square root to get a final answer of: $$y = \pm \frac{x}{\sqrt{1 - x^2}}$$ This is the answer given in my textbook. However, I have some doubts because the first step, squaring both sides of the equation, is a non-reversible operation. There could be a solution to the squared equation where $x > 0$ and $y < 0$, which clearly wouldn't be a solution to the original equation. In equations with a single variable, we can sub our solutions into the original equation and see if they are correct, but what does it mean to square an equation when it contains two or more variables? • By the way; in this case, it is not possible that $x>0$ and $y<0$ because if $y<0$ then the LHS is $<0$. – Ovi Jun 10 '18 at 5:30 Your doubts are well-founded. The solutions of the squared equation are merely possible solutions of the original equations. Fortunately, you can check the answer by plugging it into the original equation. First, recall that the $\pm$ sign means there are two proposed solutions. Let’s consider the $+$ solution first: $$y = \frac{x}{\sqrt{1 - x^2}}. \tag1$$ Note that this solution implies $-1<x<1.$ Now we try substituting the expression on the right of Equation $(1)$ for $y$ in the original equation. The result is $$x = \frac{ \left( \frac{x}{\sqrt{1 - x^2}} \right) } {\sqrt{1+ \left( \frac{x}{\sqrt{1 - x^2}} \right)^2 }} = \frac{ \left( \frac{x}{\sqrt{1 - x^2}} \right) } {\sqrt{\frac{1}{1 - x^2} }} = x.$$ That is, the substitution results in an equation that is always correct, $x=x.$ Now let’s consider the $-$ solution: $$y = - \frac{x}{\sqrt{1 - x^2}}. \tag2$$ When we try substituting the expression on the right of Equation $(2)$ for $y$ in the original equation, the result is $$x = \frac{ \left( - \frac{x}{\sqrt{1 - x^2}} \right) } {\sqrt{1+ \left( - \frac{x}{\sqrt{1 - x^2}} \right)^2 }} = -\frac{ \left( \frac{x}{\sqrt{1 - x^2}} \right) } {\sqrt{\frac{1}{1 - x^2} }} = -x.$$ This is not correct unless $x=0.$ In fact, in the only case in which Equation $(2)$ can be true (the case $x=0,$ $y=0$), Equation $(1)$ is true. The conclusion is that Equation $(1)$ (the $+$ case) is the complete solution, and the $\pm$ in the book’s solution is incorrect. Another way to approach this question is to observe that $\sqrt{1 + y^2}$ is always positive in the original equation, and therefore $x$ and $y$ always have the same sign. In the proposed solution, $\sqrt{1 - x^2}$ must also be positive. The $+$ solution then ensures that $x$ and $y$ have the same sign, which is required. But the $-$ part of the $\pm$ sign, insofar as it provides an alternative “solution,” says that $x$ and $y$ have opposite signs, which is impossible, except when $x=0,$ in which case the $\pm$ sign has no effect. At this point you could still do the substitution to verify that the $+$ solution is correct, but there would be no need to check the $-$ alternative. • I'm still a bit confused, I thought that $\frac{\frac{x}{\sqrt {1 - x^2}}}{\sqrt{\frac{1}{1 - x^2}}}$ could evaluate to $\pm x$ (since the square roots can be either negative or positive). – Pig Jun 10 '18 at 11:36 • Sorry, didn't finish that comment. Second try: I'm still a bit confused, I thought that $\frac{\frac{x}{\sqrt {1 - x^2}}}{\sqrt{\frac{1}{1 - x^2}}}$ could evaluate to $\pm x$ (since the square roots can be either negative or positive). Same for the negative case. As for the second explanation, why must $\sqrt{1 + y^2}$ always be positive? Couldn't we have that $x > 0$, $y < 0$ and $\sqrt{1 + y^2} < 0$? – Pig Jun 10 '18 at 11:49 • $\sqrt v$ means the non-negative square root of $v$ for any non-negative $v.$ It’s that simple. If $\sqrt v$ could be negative then there would be no need to write $\pm \sqrt v$ in formulas. – David K Jun 10 '18 at 19:04 • Okay, thanks, I was confused when I read somewhere that the square root maps an input to two outputs. – Pig Jun 22 '18 at 14:14 • There's the operation of taking a square root of $v,$ which could be interpreted as solving for the value of $x$ in $x^2 = v$ (two solutions may exist), and then there's the symbol $\sqrt{\cdot},$ which according to the usual usage is a function (one output value). So when I say "$\sqrt v$ is the non-negative square root of $v$," implicitly I admit that when $v > 0$ you can say there is also a negative square root of $v$, which would be written $-\sqrt v.$ – David K Jun 22 '18 at 17:50 For the sets of solutions to be preserved you need equivalence relations ($\Leftrightarrow$) between the expressions. Whenever you square you get something called an implication which is not the same as an equivalence. If you only have implication ($\Leftarrow$ or $\Rightarrow$) then the set of solutions can grow or shrink depending on the direction of the implication. For example $x = 1 \Rightarrow x^2 = 1^2$ The solution set to the left hand side is $x\in\{1\}$ and to the right one $x\in\{-1,1\}$ and so we see that all elements of left hand solution set are contained in right hand solution set (but not the other way around). This means this algebraic manipulation risks introducing solutions which are not valid solutions to the first equation. You can compare to famous set theoretic examples. All men are mammals and Adam is a man, therefore Adam must be a mammal. But if we know Lassie is a mammal we can't for sure say he's a man. If for example you plug in $y= \pm \dfrac 34$ you get $x= \pm \dfrac 35$ and in total there are four correct cases possible as shown below; we can choose the sign in each case,i.e., from each quadrant depending on quadrant context. • But $y=-3/4$, $x=3/5$ doesn't satisfy the original equation. – Gerry Myerson Jun 9 '18 at 8:51 • Did I understand correctly? When evaluating $x$ for given $y$ the fraction has a plus or minus sign in the denominator due to sqrt sign in $\sqrt {1+y^2}$ from which positive sign only needs to be taken. – Narasimham Jun 9 '18 at 9:03 • The . $\sqrt{..}$ admits/includes by implication what appears to be an extraneous or spurious solution. – Narasimham Jun 13 '18 at 18:47	2019-08-18 07:39:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9249339699745178, "perplexity": 154.9369824049775}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027313715.51/warc/CC-MAIN-20190818062817-20190818084817-00433.warc.gz"}
https://www.ncertguess.com/unseen-passage-for-class-10-english-reading-skills/	# Unseen Passage for Class 10 English Reading Skills ## Unseen Passage for Class 10 English Reading Skills Syllabus Qs. 1-2. This section will have two unseen passages of a total length of 700-750 words. The arrangement within the reading section is as follows : Q. 1. A Factual passage 300-350 words with eight very short answer type questions. (8 marks) Q. 2. A Discursive passage of 350-400 words with four short answer type questions to test inference, evaluation and analysis and four MCQs to test vocabulary. (12 marks) ### Unseen Passage for Class 10 \| Reading Comprehension PDF Tips for solving Class 10 comprehension passages : 2. Focus on the relevant details and underline them with a pen or pencil. 3. Read the questions carefully and go back to the passage to find the answers. 4. The answers are generally in a logical sequence. 6. To find answers to the vocabulary based questions like synonyms, etc., replace the word with the meaning. If you find that it is the same in meaning, the answer is correct. 7. To find the correct option in Multiple Choice Questions, go through all the options. Re-read the passage and then tick the correct option. Objective : Local global comprehension of a text. To identify the main points of the text. Marking : 20 marks -1 mark for each correct answer. No penalty for spelling, punctuation or grammatical mistakes. ### Unseen Passage for Class 10 \| Factual Passages (8 marks each) Q. 1. Read the passage given below and answer the questions/complete the sentences that follow: (8 marks) Sniffer dog Tucker uses his nose to help researchers find out why a killer whale population off the northwest coast of the United States is on tKe decline. He searches for whale faeces floating on the surface of the water, which are then collected for examination. He is one of the elite team of detection dogs used by scientists studying a number of species including right whales and killer whales. Conservation canines are fast becoming indispensable tools for biologists according to Aimee Hurt, associate director and co-founder of Working Dogs for Conservation, based in Three Forks, Montana. Over the last few years, though, so many new conservation dog projects have sprung up that Hurt can no longer keep track of them all. Her organization’s dogs and their handlers are fully booked to assist field researchers into 2012. “Dogs have such a phenomenal sense of smell”, explained Sam Wasser, director of the Center for Conservation biology at the University of Washington in Seattle. He has worked with scat-detection dogs since 199(g). Scientists have been using Conservation Canines in their research since 199(g). These dogs have enabled them to non-invasively access vast amount of genetic and physiological information which is used to tackle conservation problems around the world. Such information has proved vital for determining the causes and consequences of human disturbances on wildlife as well as the actions needed to mitigate such impacts. The ideal detection dog is extremely energetic with an excessive play drive. These dogs will happily work all • day long, motivated by the expectation of a ball game as a reward for sample detection. The obsessive, high energy personalities of detection dogs also make them difficult to maintain as pets. As a result, they frequently find themselves abandoned to animal shelters, facing euthanasia. The programme rescues these dogs and offers them a satisfying career in conservation research. Unseen passage with questions and answers class 10 English: (1×8 = 8 marks) (Board 2014, Set 8L1922Q) (a) According to the text there are a few________ detection dogs like Tucker. (b) Tucker sniffs for whale________ (c) The dogs are special because they assist in research without ________ (d) The ideal detection dog ________ (e) The dogs expect________ as a reward of their hard work. (f) ________of these dogs make it difficult to keep them as pets. (g) These dogs find career in ________ (h) The word ‘euthanasia’ means ________ Ans. (a) elite team of. (b) farces floating on the surface of water. (c) invasion. (d) is extremely energetic with an excessive play drive. (e) a ball game. (f) The obsessive, high energy personalities. (g) conservation research. (h) painless killing. Q. 2, Read the following passage carefully: (8 marks) (NCT 2014) ### Unseen Poem for Class 10 \| Power Foods (1) Power foods are foods that provide rich levels of nutrients like fibre, potassium and minerals. With people becoming increasingly health conscious today, a lot of finess trainers encourage their clients to include these foods in their daily diet to increase muscle development. There are various ways of incorporating power foods in your daily diet. Of course, the key to enjoying power foods is proper preparation of these foods, the use of season-fresh foods, and indentifying your choice of flavour among power foods. (2) Some of the recommended power food combinations are those that are prepared in our kitchens on a regular basis. Take for instance, the combination of chickpeas and onions. This combination is a powerful source of iron which is required by the body to transport oxygen to its various parts. Iron deficiency can lead to anaemia, fatigue, brain fog and tiredness. A study by the Journal of Agricultural and Food Chemistry says that sulphur compounds in onion and garlic help in the absorption of iron and zinc from chickpeas. The combination is a hit with teenagers who need to be diligent about getting iron in their diet. A quick way to prepare this power food is to make a chickpea salad with chopped onions, chaat masala and cilantro. (3) Another favourite combination with power food takers is yoghurt and bananas. This makes for a perfect snack after a rough game of football. Exercising bums glucose and thus lowers blood sugar. Yoghurt is packed with proteins that help preserve muscle mass, and bananas are packed with carbohydrates that help in refuelling energy and preventing muscle soreness. A quick and easy recipe with bananas is a banana smoothie topped with cool yoghurt. (4) Among beverages, green tea is the best source of catechins that are effective in halting oxidative damage to cells. According to researchers at the Purdue University, adding a dash of lemon juice to green tea makes the catechins even more easily absorbable by the body. So, the next time you have instead of are friends serve them rounds of iced green tea with mint and lemon juice. Unseen passage with questions and answers class 10 English: . (1×8 = 8 marks) (a) What are power foods ? (b) What are the rules regarding the partaking of power foods ? (c) What is the advantage of including onions and garlic in our diet ? (d) Suggest a quick recipe with chickpea and onions. (e) Why is yoghurt and bananas, an enriching power food ? (f) Why is green tea a recommended power food ? (g) . What is the advantage of combining green tea with lemon juice ? (h) What is the key to enjoying power foods in a wholesome way ? Ans. (a) Power foods are foods which provide rich level of nutrients like fibre, potassium and minerals. (b) Power foods should be prepared properly using season-fresh foods and identifying one’s choice of flavour among power foods. (c) Onion and garlic help in the absorption of iron and zinc from the chickpeas. (d) A quick way to prepare chickpea and onions is to make a chickpea salad with chopped onions, chaat masala and cilantro. (e) Yoghurt is full of proteins that help preserve muscle mass and bananas are packed with carbohydrates that help in refuelling energy and preventing muscle soreness. (f) Green tea contains catechins which are effective in halting oxidative damage to cells. (g) Combining green tea with lemon juice helps the body to absorb catechins more easily. (h) Power foods can be enjoyed in a wholesome way by including them in our daily diet to increase muscle development. Q. 3. Read the passage given below and answer the questions/complete the sentences that follow: (8 marks) (CBSE Marking Scheme, 2014) A sparrow is a small bird which is found throughout the world. There are many different species of sparrows. Sparrows are only about four to six inches in length. Many people appreciate their beautiful song. Sparrows prefer to build their nests in low places-usually on the ground, clumps of grass, low trees and low bushes. In cities they build their nests in building nooks or holes. They rarely build their nests in high places. They build their nests out of twigs, grasses and plant fibres. Their nests are usually small and well-built structures. Female sparrows lay four to six eggs at a time. The eggs are white with reddish brown spots. They hatch between eleven to fourteen days. Both the male and female parents care for the young. Insects are fed to the young after hatching. The large feet of the sparrows are used for scratching seeds. Adult sparrows mainly eat seeds. Sparrows can be found almost everywhere, where there are humans. Many people throughout the world enjoy these delightful birds. The sparrows are some of the few birds that engage in dust bathing. Sparrows first scratch a hole in the ground with their feet, then lie in it and fling dirt or sand over their bodies with flicks of their winds. They also bathe in water, or in dry or melting snow. Water bathing is similar to dust bathing, with the sparrow standing in shallow water and flicking water over its back with its wings, also ducking its head under the water. Both activities are social, with up to a hundred birds participating at once, and is followed by preening and sometimes group singing. Unseen passage with questions and answers class 10 English: (1×8 = 8 marks) (Board 2014, Set PRE2N18) (a) The chief food for the adult sparrow is _________ (b) Sparrows live wherever _________ (c) The word, ‘species’ means_________ (d) Sparrows in high places._________ (e) _________ take care of the young sparrows. (f) Sparrows take bathe in _________ (g) Bathing for the sparrows is a _________ (h) Bathing is followed by_________ and_________ Ans. (a) seeds. (b) there are humans. (c) kinds. (d) rarely build their nests in high places. (e) Both parents. (f) dust, water Or snow. (g) social activity. , ‘ (h) preening and group singing. Q. 4. Read the passage given below and answer the questions/complete the sentences that follow: (8 marks) (Board 2014, Set QUD9VQW) A chimpanzee is one of the great apes and the nearest in intelligence to man. Scientists have examined its mental capacities and sent it into space in anticipation of man. Chimpanzees need little description. Being apes and not monkeys, they have no tails. Their arms are longer than their legs and they normally rim on all fours. They can also walk upright with toes turned outwards. When erect they stand 3-5 ft high. The hair is long and coarse, black except for a white patch near the rump. The face, ears, hands and feet are bare and except for the black face, the flesh is coloured. Chimpanzees exhibit great concern for each other. When chimpanzees meet after having been apart they greet each other in a very human way by touching each other or even clasping hands. Chimpanzees have amazing social discipline. When a dominant male arrives, the rest of the chimpanzees hurry to pay respect to it. The dominant male is not allowed to wrest food from his inferiors. The members of a party also spend considerable amount of time grooming each other and themselves. Mothers go through the fur of their babies for any foreign particles, dirt, and ticks and they aid each other when they are injured. Chimpanzees are the best tools users apart from man. Sticks 2-3 ft long are picked off the ground orbaalnfnjai branches and pushed into nests, then withdrawn and the honey or insects licked off. Stones are and ttecacfc nuts or as missiles to drive humans and baboons away from its food. Chimpanzees are not only toolunBUthut also toolmakers. They make their own rods by stripping the leaves off a twig or tear shreds off a chimpanzees learn all this by observing the older chimpanzees making and using them. So man is not the easily toolmaker, merely better at it than his relatives. Unseen passage with questions and answers class 10 English: (1 x 8 = 8 marks) (a) Chimpanzees are as_______ as men. (b) Chimpanzees greet each other by_______ each other. (c) Like man, chimpanzees are_______ (d) Chimpanzees_______ tails. (e) Baby chimpanzees learn, all by _______ (f) Chimpanzees have amazing _______ (g) The dominant male chimpanzees is not allowed _______ (h) The word ‘wrest’ means_______ Ans. (a) intelligent. (b) touching. (c) both tool users and toolmakers. (d) have no. (e) observing the older chimpanzees. (f) social discipline. (g) to take food from inferiors. (h) take away violently. Q.5. Read the following passage carefully: (8 marks) These days, it is not unusual to see people listening to music or using their electronic gadgets while crossing busy roads or travelling on public transports, regardless of the risks involved. I have often wondered why they take such risks : is it because they want to exude a sense of independence, or is it that they want to tell the world to stop bothering them ? Or is it that they just want to show how cool they are ? Whether it is a workman or an executive, earphones have become an inseparable part of our lives, sometimes even leading to tragicomic situations. The other day, an electrician had come to our house to fix something. We told him in detail what needed to be done. But after he left, I found that the man had done almost nothing. It later turned out that he could not hear our directions clearly because he had an earphone on. Hundreds of such earphones addicts commute by the Delhi Metro every day. While one should not begrudge anyone their moments of privacy or their love for music, the fact is ‘iPod oblivion’ can sometimes be very dangerous. Recently, I was travelling with my wife on the Delhi Metro. Since the train was approaching the last station, there weren’t too many passengers. In our compartment, other than us, there were only two women sitting cm the other side of the aisle. And then suddenly, I spotted a duffel bag. The bomb scare lasted for several minutes. Then suddenly, a youth emerged from nowhere and picked up the bag. When we tried to stop him, he looked at us, surprised. Then he took off his earpieces, lifted the bag, and told us that the bag belonged to him and that he was going to get off at the next station. We were stunned but recovered in time to ask him where he was all this while. His answer : he was in the compartment, leaning against the door totally immersed in the music. He had no clue about what was going on around him. When he got off, earplugs in his hand, we could hear strains of the song. Unseen passage with questions and answers class 10 English (A) Read the above passage and answer the questions that follow: (1 x 5 = 5 marks) (i) What reasons does the author offer for the people taking risks on the road ? (ii) Why didn’t the electrician carry out the work properly ? (iii) Why were the people in the Metro doubtful about the bag ? (iv) Why were the passengers stunned ? (v) Explain the term ‘earphone addicts’? (B) Find words from the paragraph indicated which are similar in meaning to the words given below: (1 x 3 = 3 marks) (Board Term-12012, Set EC2,046) (i) inspite of (para 1) (ii) absorbed (para 4) (iii) picked (para 3) Ans. (A) (i) Exude a sense of independence or to tell the world to stop bothering them. (ii) He did not hear the instructions carefully and so did not do the work properly. (iii) Nobody came forward to take the bag so, they doubted it to be a bomb scare. (iv) At the carelessness and behaviour of the young boy. (v) Persons who always wear earphones and keep listening to music. (B) (i) regardless (ii) immersed (iii) lifted Q. 6. Read the passage given below carefully: (8 marks) Title: Dreams to Reality (1) It was evening in the picturesque seaside town of Rameshwaram, on the southern edge of Tamil Nadu. A cool breeze was gently blowing in from the sea. Along with the sound of waves lapping against the shore could be heard the sweet sound of birds circling overhead. (2) Among the children playing on the beach was a boy with wavy hair and dreamy eyes. This youngster was Avul Pakir Jainulabdeen Abdul Kalam who later became the eleventh President of India. (3) While spending time with his friends, Abdul was attracted by the sound of the birds flying above. He carefully observed that a fledgling perched on a boat was trying to take off. It spread its wings, fluttered briefly and sprang up. The air seemed to give the needed thrust for its take off! The bird soared up into the sky. It steered its pace and course with great ease. How Abdul wished he could fly like those beauties in the air! (4) This passion for flying, aroused by the beautiful birds, later inspired Abdul Kalam to design India’s first rocket which successfully sent a satellite Rohini, into orbit on 18th July 1980. It was called the SLV-3 (Satellite Launch Vehicle). At the time when Abdul was growing up, no one had even dreamt of such a happening. (5) Rameshwaram, where Abdul was bom on 15th October 1931, was a small town with narrow streets lined with old houses made of limestone and brick. The town was famous for its Shiva Temple. Abdul stayed in the house with his father, mother, brothers and sister and led a secure and happy childhood. (6) Abdul’s father, Jainulabdeen was a pious man. He led an austere life without depriving his family of the basic comforts. (7) In this closely knit family, dinner was always a special meal. During dinner they exchanged views on a variety of topics ranging from family matters to spiritual subjects. (8) The main income for Abdul’s family came from ferrying pilgrims across the sea between Rameshwaram and Dhanushkodi. Pilgrims visiting Rameshwaram made it a point to visit Dhanushkodi, twenty kilometres away in the sea. Dhanushkodi has religious significance. (9) Ferrying pilgrims fetched good money and the family lived comfortably. However, a devastating cyclone lashed the shores of Rameshwaram and their boat was destroyed. The family lost their only source of livelihood in one swift, tragic stroke. (10) The enterprising young Abdul wanted to help the family through the crisis. He realized that there was demand for tamarind seeds. He decided he would collect them and sell them to a shop near his house. His family wanted him to concentrate on his studies. He said he would study as well as help his family. Reluctantly, everyone agreed. Even while studying or enjoying the evenings with his friends on the beach, he set aside some time to collect tamarind seeds and sell them to a nearby shop. For this he was paid a princely sum of one anna! (11) Besides selling tamarind seeds, he helped his cousin Samsuddin to sell the popular Tamil newspaper Dinamani to earn a little more money. At dawn, several bundles of the newspaper, printed in Madras (now Chennai), were thrown on to the platform of Rameshwaram railway station from passing trains. Trains did not stop at Rameshwaram statiqn during those days of the Second World War as almost all of them were commandeered to transport troops. (12) Abdul, after picking up the bundles marked for his area, rushed and handed them over to Samsuddin, who gave Abdul a small amount for the service he had rendered. There was a great demand for Dinamani because people wanted to know about India’s freedom struggle and the latest developments in the war. (13) This is how Abdul earned his first wages. However, it was the joy of being able to care for his family that Abdul cherished most. Even decades later, he recalls earning his own money for the first time, with immense pleasure and a sense of pride. Unseen passage with questions and answers class 10 English (A) On the basis of your reading of the above passage, answer the following questions: (1 x 6 = 6 marks) (i) Who does this passage tell us about ? (ii) What inspired Abdul to design a rocket ? (iii) . What was SLV-3 ? (iv) Where did Abdul spend his childhood ? (v) Why was dinner a special meal in the Kalam’s family ? (vi) What did Abdul cherish most about the memory of his first earning ? (B) Find the words from the passage which mean the same as the following: (1 x 2 = 2 marks) (i) disciplined/hard (para 8) (ii) hard working and bold (para 10) Ans. (A) (i) This passage tells us about Abdul Kalam Azad who later became the eleventh President of India. (ii) The flight of a fledgling inspired Abdul to design a rocket. (iii) SLV-3 or satellite launch vehicle was India’s first rocket which successfully sent a satellite Rohini into orbit on 18th July 1980. (iv) Abdul spent his childhood in a small town with narrow streets lined with old houses made of limestone and brick. (v) Dinner was a special meal in Kalam’s family because they all sat together and exchanged views on a variety of topics ranging from family matters to spiritual subjects. (vi) It was the joy of being able to care for his family that Abdul Kalam cherished the most. (B) (i) austere (ii) enterprising Q. 7. Read the following passage carefully: (8 marks) One day, I hopped in a taxi and we took off for the airport. We were driving in the right lane when suddenly a car jumped out of a parking space right in front of us. My taxi driver slammed on his brakes, skidded and missed the other car by just inches ! The driver of the other car whipped his head around and started yelling at us. My taxi driver just smiled and waved at the guy. And I mean he was really friendly. So I asked, ‘Why did you just do that ? This guy almost ruined your car and sent us to the hospital!’ This is when my taxi driver taught me what I now call the ‘The law of the Garbage Truck’. He explained that many people are like garbage trucks. They run around full of garbage, full of anger, and full of disappointment. As their garbage piles up, they need a place to dump it and sometimes they’ll dump it on you. Don’t take it personally. Just smile, wave, wish them well, and move on. Don’t take their garbage and spread it to other people at work, at home or on the streets. The bottom line is that successful people do not let garbage trucks take over their day. Life’s too short to wake up in the morning with regrets, so ‘Love the people who treat you right. Forgive the ones who don’t’. This attitude will help you sail through life. There will be fewer jerks and bumps. Learn to take the bad with the good for life can never be perfect. Acceptance of what is, is the solution. Don’t react, just accept and you will be a lot more happy. Life is ten percent what you make it and ninety percent how you take it! Unseen passage with questions and answers class 10 English (i) The narrator boarded a taxi (a________ when all of a sudden a car jumped (b)________ (2 marks) (ii) How did the taxi driver avert an accident ? (1 mark) (iii) To what did he compare the yelling driver ? (1 mark) (iv) The term ‘garbage’ in the context of the passage means________ (1 mark) (v) The narrator learnt an invaluable lesson from the taxi river. What was it ? (1 mark) (vi) Give words from the passage that means the same as : (1×2 = 2 marks) (a) to move briskly (para 1) (b) feeling of defeat (para 2) Ans. (i) (a) to go to the airport. , (b) out of a parking space right in front of them. (ii) The taxi driver slammed on his brakes, skidded and missed the other car by some inches. (iii) He compared the yelling driver to a garbage truck. – (iv) Full of frustration, full of anger and full of disappointment. (v) He learnt that one must love the people who treat us right and forgive the ones who don’t. (vi) (a) whipped (b) frustration Q. 8. Read the following passage carefully: (8 marks) ### Unseen Passage for Class 10 \| Voice of Love 1. I was a below average student. Both in schools and colleges, rarely my teachers knew me by name. I don’t think I was a dunce; just that I didn’t have an academic bent of mind. Since failures were more a habit than an exception, a below par performance never disturbed me. Neither were my teachers proud of me nor did I make my parents feel proud of me. 2. Of course, my parents being noble-hearted, never gave up on me. They never put me down in the presence of others. In fact, to shield me, they always projected a positive image of me to the world. 3. In 1984, while pursuing my graduation in Mathematics, I had failed in one of the subjects in my fourth semester. I already had three arrears. For the first time, hearing about my failure, I saw tears in my father’s eyes. This was my first experience of seeing my father cry. I couldn’t handle his crying. To withdraw myself I escaped to the terrace of my apartment. I suffered a fear like I have never known before. I was trembling. I never wanted my parents to ever cry again because of my failures. But, I was scared. ”Was it too late to begin in life,” I doubted. I was already over 18, just about 50% marks in my four semesters, 4 arrears to clear and 3 regular paper to face in my fifth semester, no talent, no special abilities, never won a prize in my life, not a single certificate I truly wanted to make it very big in life, if not for my sake, just to make up for all that I had put my parents through. My thoughts were haunting me, “Rajan, you don’t have any taste of success. You just don’t know what it is to succeed.” With tears flooding, I cried, “Can I still make it big in life or have I missed the bus ?” 4.My neighbour, Vijayaraghavan, who learnt about the sobbing of my soul, casually said, “Rajan, the harder you press the spring the faster it will bounce back. So what if your life has been pressed by failures for 18 years. Decide to bounce back and bounce back big in life. Even God will not stop you.” Bounce back I did ! I bounced back big to clear all the 7 papers in first class and ever since I smile at my failures. 5. After all what is resilience the number : To get up one more time when you have fallen. I do not remember of times when life has pushed me down, but every time I have bounced back big, for, I can hear the message lingering from within me even louder, “The harder you press the spring the faster it will bounce back. Bounce back and bounce back big.” 6.Ever since, I have given enough opportunities for my parents to cry cry they do, glad they have a son in me. Unseen passage with questions and answers class 10 English (A) On the basis of your reading of the passage complete the following: (1 x 6 = 6 marks) (i) The author was a below average student because he________ (ii) His failures and poor performance made his teachers and parents ________ (iii) The author couldn’t bear________ (iv) At the age of 18, the author felt a deep sense of fear and wondered whether ________ (v) After being inspired by his neighbour, he________ (vi) The message given by the author is ________ (B) Find words from the passage which are similar in meaning to the following: (1 x 2 = 2 marks) (i) ability to recover quickly (Para 5) (ii) difficult to forget (Para 3) Ans. (A) (i) didn’t have an academic bent of mind. (iii) his father crying/the sight of his father crying/his father in tears. (iv) he would be able to overcome his failures/he would be able to succeed. (v) he realized he could fight back/there is no age to learn to fight back/one’s will is important to learn to succeed. (vi) never give up/the harder you work the faster you bounce back/to get up one more time when you have fallen. Note : Other suitable responses to be accepted. (B) (i) resilience (ii) haunting ### Unseen Passage for Class 10\| Discursive Passages (12 marks each) Q. 1. Read the passage given below and answer the questions/complete the statements that follow: (12 marks) Long, long ago, in a big forest, there were many trees. Among the cluster of trees, there was a very tall pine tree. He was so tall that he could talk to the stars in the sky. He could easily look over the heads of the other trees. One day late in the evening, the pine tree saw a ragged, skinny girl approaching him. He could see her only because of his height. The little girl was in tears. The pine tree bent as much as he could and asked her : “What is the matter ? Why are you dying ?” The little girl, still sobbing, replied, “I was gathering flowers for a garland for goddess Durga, who I believe, would help my parents to overcome their poverty and I have lost my way”. The pine tree said to the little girl, “It is late in the evening. It will not be possible for you to return to your house, which is at the other end of the forest. Sleep for the night at this place.” The pine tree pointed out to an open cave-like place under him. The little girl was frightened of wild animals. The girl quickly crept into the cave-like place. The pine tree was happy and pleased with himself. He stood like a soldier guarding the place. The little girl woke up in the morning and was amazed to see the pine tree standing guard outside the cave. Then her gaze travelled to the heap of flowers that she had gathered the previous night. The flowers lay withering on the ground. The pine tree understood what was going on in the girl’s mind. He wrapped his branches around the nearby flower trees and shook them gently. The little girl’s eyes brightened. But a great surprise awaited her. The pine tree brought out a bag full of gold coins which had been lying for years in the hole in its trunk and gave it to the girl. With teary eyes she thanked her benefactor and went away. Unseen passage with questions and answers class 10 English (A) Answer the following questions : (2 x 4 = 8 marks) (a) Why was the girl crying ? (b) Where did the pine tree want the little girl to sleep for the night ? (c) Why was the little girl disappointed when she looked at the flowers and what did the tree do to make her happy ? (d) What lesson does this short story teach us ? (B) Do as directed : (4 marks) (Board 2014, Set 8L1922Q) (e) What is meant by the word ‘cluster’ ? (Para 1) (i) group (ii) team (iii) class (iv) party (f) What is meant by the word ‘approaching’ ? (Para 1) (i) calling (ii) touching (iii) coming close (iv) running towards (g) What is meant by the word ‘wild’ ? (Para 2) (i) cunning (ii) dirty (iii) unpolished (iv) dangerous (h) What is meant by the word ‘ withering ‘ ? (Para 2) (i) dead (ii) shrunk (iii) colourless (iv) unhappy Ans. (A) (a) The little girl was crying because she had lost her way while gathering flowers for a garland for goddess Durga. (b) The pine tree wanted the little girl to sleep in an open cave-like place under the tree itself. (c) The little girl was disappointed because the flowers that she had gathered the previous night lay withered on the ground. The tree wrapped his branches around the nearby flower trees and shook them gently so that the little girl may become happy getting so many flowers. (d) The short story teaches us to help others in their time of need. (B) (e) (i) group. (f) (iii) coming close. (g) (iii) unpolished. Q. 2. Read the following passage carefully: (12 marks) 1. After a long day out in the scorching sun at Nizamuddin railway station, having checked out every bit of garbage disposed off trains, a group of ragpickers gathered for a chat in a rain-swept shelter. This was no regular gathering for them. It was a Mother’s Day gathering. 2. Most children had never heard the word but grew emotional once they got to know what the day signified. Gifts for their mothers ranged from promises of not running away, to earning enough to assure their mothers of some comfort some day. 3. When it came to actual celebrations for the day, a group of boys at the centre run by Chetna, an NGO, near the station, surrounded Manjula Rai. Some even pulled her hair and the rowdy ones calmed down after a stem glance from her. For many boys and girls, this 47-year old social worker is a face recognized as a help at hand. For quite a few of them, Manjula is the ‘mother’ who influences them. 4. When they settled down to talk, the children at Chetna, related their aspirations with thoughtful intent. Vikas Kumar, who felt shy to pronounce the word Mother’s Day, said he was determined to secure some day, quality life for his mother, who was a domestic worker. His promise to his mother on Mother’s Day was to take out his mother from a rented house and to give her a home with a better life, some day. 5. Vikas left school when he was just 9, but is now a motivator for other boys to break free of ragpicking and study through die Open School System. Today, Vikas, who is a Class VII student, is the pride of the Centre and Manjula has been a part of his journey all through. She has been convincing the parents of these children to understand their evolution from being a child labourer to a teenager who can acquire skills for a better life. 6. Aslam for instance, who giggled as the boys tried to pronounce Mother’s Day, knew the words ‘Mom’ and ‘Dad’ having watched T.V. serials and Hindi films. According to him : “I keep running away from home and fight with my Mom. But on Mother’s Day, when I go back, I will not fight and stay with her.” The traumatic life stories related by these children become a self-explanation for their aggressive behaviour and their suppressed feelings. Unseen passage with questions and answers class 10 English (A) Answer the following questions in about 30-40 words : (2 x 4 = 8 marks) (a) What was special about Mother’s Day at the rain-swept shelter ? (b) What were the ragpickers’ reaction about Mother’s Day ? (c) How did the children celebrate the day at the Chetna premises ? (d) Why is Vikas’ case inspiring for the ragpickers ? (B) Do as directed: (4 marks) (NCT 2014) (e) When one is out in the scorching sun, it means the day is_________ (i) cool (ii) rainy (iii) very hot (iv) very dry (f) The ragpickers had understood about Mother’s Day when Manjula told them what it_________ (i) signified (ii) promised (iii) gifted (iv) created (g) When one hankers for something better, it becomes_________ (i) an emotional moment (ii) a dream moment (iii) an aspiration (iv) traumatic life story (h) A term in the passage which means the same as ‘compelling’ is_________ (i) convincing (ii) scorching (iii) understanding (iv) pronouncing Ans. (A) (a) At the rain swept shelter, a group of boys had gathered for a special occasion. It was a Mother’s Day gathering and not a regular gathering. (b) The ragpickers grew emotional as they had never heard the word before and thought of some gifts for their mothers in the form of promises. (c) The children at the Chetna premises surrounded Manjula Rai as for a few of them, she is the ‘mother’ who influences them. (d) Vikas’ case is inspiring for the ragpickers because he studied through the Open School System after breaking himself free of ragpicking. (B) (e) (iii) very hot. (f) (i) signified. (g) (iii) an aspiration. (h) (i) convincing. Q. 3. Read the passage given below and answer the questions/complete the statements that follow: (12 marks) “Something is coming uphill”, said Mother Wolf, twitching one ear. “Get ready”. The bushes rustled a little in the thicket and father Wolf dropped on his haunches under him, ready for his leap. Then, if you had been watching, you would have seen the most wonderful tiling in the world-the wolf checked in mid-spring. He made his leap before he saw what he was jumping at, and then he tried to stop himself. The result was that he had shot up straight into the air for four or five feet, landing almost where he had left the ground. “Man!” he snapped. “A man’s cub. Look!” Directly in front of him, holding on a low branch, stood a naked brown baby who could just walk-as soft and as dimpled a little atom as ever came to a wolf’s cave at night. He looked up into Father Wolf’s face and laughed. “Is that a man’s cub ?” said Mother Wolf. “I have never seen one. Bring it here. How little! How naked! How bold!” she said softly. The baby was pushing his way between the cubs to get close to the warm hide. “Aha! He is taking his meal with the others. And so this is a man’s cub. Now, was there ever a wolf that could boast of a man’s cub among her children ?” “I have heard now and again of such a thing, but never in our pack or in my time”, said Father Wolf. “He is altogether without hair, and I could kill him with a touch of my foot. But see, he looks up and is not afraid.” The moonlight was blocked out of the mouth of the cave, for Shere Khan’s great square head and shoulders were thrust into the entrance. Tabaqui, behind him, was squeaking: “My lord, my lord, it went in here!” “Shere Khan does us great honor”, said Father Wolf, but his eyes were very angry. “What does Shere Khan need ?” My quarry. A man’s cub went this way”, said Shere Khan. “Its parents have run off. Give it to me.” (A) Answer the following questions : (2 x 4 = 8 marks) (a) Why did Father Wolf drop on his haunches ? (b) Why did he stop in the middle of his jump ? (c) What did a man’s cub look like ? (d) What did it do to keep itself warm ? (B) Do as directed : (1 x 4 = 4 marks) (e) What is meant by the word ‘bold’ ? (para 3) (i) brave (ii) smart (iii) bright (iv) thrust (f) What is meant by the word ‘spring’ ? (para 1) (i) coil (ii) jump (iii) cushion (iv) bed (g) What is meant by the word ‘altogether’ ? (para 4) (i) combined (ii) cleverly (iii) completely (iv) quiet (h) What is the opposite of the word ‘uphill’ ? (i) bottom (ii) plain (iii) valley (iv) downhill (Board Term-12014, Set PRE2N18) Ans:(A) (a) To be ready for a leap. (b) Saw a man’s cub holding a low branch. (c) Naked brown baby, soft and dimpled. (d) Pushed his way between the wolf cubs to get close to their warm hide. (B) (e) brave. (f) jump. (g) completely. (h) downhill. (CBSE Marking Scheme, 2014) Q.4.Read the passage given below and answer the questions/complete the statements that follow: (12 marks) The ‘Little Tramp’, the unforgettable character Charlie Chaplin invented, was born purely by accident in 1915. While rushing to a film shoot in California, he grabbed clothes other people had left behind in the changing room. And when he emerged, he found he had created a personality everybody loved. A little guy in a bowler hat, a close-fitting jacket, a cane, outsize shoes and a brush-like moustache! Before long, Chaplin found himself a star. That puzzled him, for he saw himself essentially as a shy British Music hall comedian. The U.S. acknowledged him as its king of silent film comedy. Soon, so did crowds all over the world. But life wasn’t always a laugh for Charles Spencer Chaplin. Both his parents were Music Hall artists, who separated when Charlie was very young. His childhood was very sad, for his mother never earned enough to look after Charlie and his older brother, Sydney. Sometimes, Chaplin had to sleep on the streets and forage foe food in the garbage. Charlie took his first bow on stage when his mother made her last appearance. It happened when her voice broke during a song. Her son stepped on stage and sang a popular song. That’s when a star was bom. Through all these years of success, Charlie never forgot his troubled childhood. It made him recall a Christmas when he was denied two oranges and his bag of sweets for breaking a rule at the orphanage he went to after his mother’s death. It would have broken his heart, if the other children had not offered him a share of theirs. Spontaneously, the adult Chaplin gifted the orphanage with a motion picture machine and insisted that each child should have as many oranges and sweets as they pleased. (A) Answer the following questions: (2 x 4 = 8 ) (a) Which unforgettable character did Charlie Chaplin invent ? (b) Describe the personality created by Charlie, whom everybody loved. (c) What did Charlie see himself as ? (d) Give two reasons to show that his early life was very sad. (B) Do as directed: (1×4 = 4 marks) (Board 2014, Set QUD9VQW) (e) The word that means ‘to search for food’ is_________ (para 3) (i) stepped (ii) forage (iii) emerged (iv) orphanage (f) What is meant by the word ‘invented’ ? (para 1) (i) created (ii) wrote (iii) struggled (iv) laughed (g) Find the word in the passage opposite in meaning to ‘bold’ ? (para 2) (i) found (ii) shy (iii) silent (iv) puzzled (h) Find the word opposite in meaning to ‘failure’ ? (para 5) (i) troubled (ii) share (iii) motion (iv) success Ans.(A) (a) The’Little Tramp’. (b) A little guy in a bowler hat, a close-fitting jacket, a cane outsize shoes etc. (c) As a shy British Music hall comedian. (d) His parents separated when he was young-his mother could not earn enough-had to sleep the street- forage for food. (Any two) (B) (e) forage. (f) created. (g) shy. (h) success. Q.5.Read the passage given below: (12 marks) Cardamom, the Queen of all spices, has a history as old as the human race. It is the dried fruit of a herbaceous perennial plant. Warm humid climate, loamy soil rich in organic matter, distributed rainfall and special cultivation and processing methods all combine to make Indian cardamom truly unique in aroma, flavour, size and it has a parrot green colour. Two types of cardamom are produced in India. The first type is the large one, which has not much significance as it is not traded in the future market. It is cultivated in North-eastern area of the country. The second type is produced in the Southern states and these are traded in the future market. These are mainly cultivated in Kerala, Tamil Nadu and Karnataka. As per the future market rules, only 7 mm quality was previously traded in exchanges. But later, it relaxed its norms, and now 6 mm quality is also traded in the exchanges. The small variety, known for its exotic quality throughout the world, is now traded in India’s commodity future exchanges. Traditional auction markets also exist for trading in small cardamom in the country. Cardamom is an expensive spice, second only to saffron. It is often adulterated and there are many inferior substitutes from cardamom-related plants such as Siam cardamom, Nepal cardamom, Winged Java cardamom, etc. However, it is only Elettaria cardamom which is the true cardamom. Indian cardamom is known in two main varieties : Malabar cardamom and Mysore cardamom. The Mysore variety contains levels of cineol and limonene and hence is more aromatic. India was the world’s largest producer and exporter of cardamom till the 1980s. By 1990s Guatemala emerged as the leading producer and exporter of cardamom. The main harvest season of cardamom in India is between August-February. Cardamom reaches yielding stage two years after planting. The primary physical markets of cardamom are Kumily, Vandenmedu, Thekkady, Puliyarmala in Kerala and Bodynaikkannur and Cumbam in Tamil Nadu. Cardamom auctions also take place in Sakalespur and Sirsi in Karnataka. North India is the main market for cardamom produced in the country. Normally, domestic demand goes up during the major festivals such as Diwali, Dussehra and Eid. Colour, size and aroma are the major variables that shape cardamom prices in the Indian market. Cardamom price formation in India is also influenced by the output in Guatemala as that country controls the global markets. Kerala is the main producer of cardamom and contributes up to 60% in total production. Karnataka produces around 25% cardamom of the total production. Ooty is the main producer of cardamom in Tamil Nadu and contributes around 10-15% of the total production. Besides India, Guatemala also produces around 2,200 ton cardamom. On the other hand, India produces nearly 1,000-2,000 ton cardamom per year. Due to low quality of cardamom from Guatemala, it remains available at cheaper rates. Moreover, cardamom of Indian origin fetches $3-4 per kilogram higher rates than the ones from Guatemala. (A) On the basis of your reading of the above passage, answer the following questions. (2×4 = 8 marks) (a) Why is Indian cardamom unique ? (b) What is special about the Mysore quality of cardamom ? (c) What role does Guatemala play in the Indian market ? (d) Write two sentences on the harvesting of cardamom. (B) Find words from the options given below which mean the same as: (1×4 = 4 marks) (e) permanent: (i) unique (ii) perennial (iii) exotic (iv) old (f) fragrance: (i) herbaceous (ii) loamy (iii) aroma (iv) humid (g) earlier: (i) substitute (ii) exported (iii)main (iv) previously (h) make poor in quality by adding another substances : (i) adulterated (ii) emerged (iii) contribute (iv) remain Loading... Ans.(A) (a) Indian cardamom is unique because it requires warm humid climate, loamy soil rich in organic matter, distributed rainfall, special cultivation and processing methods. (b) The Mysore quality of cardamom is unique as it contains higher levels of cineol and limonene and is more aromatic. (c) Guatemala produces around 2200 ton cardamom and controls the global market. (d) The harvest season of the cardamom in India is between August and February. It reaches its yielding stage two years after planting. Ans.(B) (e) (ii) perennial (f) (iii) aroma (g) (iv) previously (h) (î) adulterated Q. 6. Read the following passage carefully: (1) Tom Sawyer found Monday morning miserable. He always found it so because it began another week’s slow suffering in school. Tom lay thinking. Presently it occurred to him that he wished he was sick; then he could stay home from school. Here was a vague possibility. But no ailment was found. Suddenly, he discovered something. One of his upper front teeth was loose. This was lucky ; he was about to begin to groan, when it occurred to him that his aunt would pull it out, and that would hurt. So, he thought he would hold the tooth in reserve for the present, and seek further. Then he remembered hearing the doctor tell about a certain things that threatened to make patients lose a finger. So, the boy eagerly drew his sore toe from under the sheet and held it up for inspection. It seemed worthwhile to chance it, so he fell to groaning with considerable spirit. (2) But Sid slept on. (3) Tom groaned louder. (4) No result from Sid! Sid snored on. (5) Tom said, “Sid, Sid” and shook him and began to groan again. Sid said : ” Tom ! Say, Tom !” (No response.) “Here, Tom ! Tom! What is the matter, Tom ?” And he shook him and looked in his face anxiously. (6) “Don’t groan so, Tom, it’s awful. How long have you been this way ?” (7) “Hours Ouch! Oh, don’t stir so, Sid, you’ll kill me”. (8) “Tom, why didn’t you wake me sooner ? Oh, Tom don’t! It makes my flesh crawl to hear you. Tom, What is the matter ?” (9) “I forgive you everything, Sid. (Groan.) Everything you’ve ever done to me. When I’m gone.” (10) “Oh, Tom, you am’t dying, are you? Don’t, Tom -oh, don’t. May be.” (11) “I forgive everybody, Sid. (Groan.) Tell’em so, Sid. (12) But Sid had snatched his clothes and gone. (13) Sid flew down – stairs and said : (14) “Oh, Aunt Polly, come ! Tom’s dying !” (15) “Dying!” (16) “Yes. Don’t wait – come quick !” (17) “Rubbish! I don’t believe it!” (18) But she fled up – stairs, nevertheless, with Sid and Mary at her heels. And her face grew white, too, and her lip trembled. When she reached the bedside she gasped out. (19) “You, Tom! Tom, what’s the matter with you ?” (20) “Oh, auntie, I’m.” (21) “What’s the matter with you – what is the matter with you, child ?” (22) “Oh, auntie, my sore toe’s mortified !” (23) The old lady sank down into a chair and laughed a little, then cried a little, then did both together. This restored her and she said : “Tom, what a turn you did give me ! Now you shut up that nonsense and climb out of this.” (24) The groans ceased .The boy felt a little foolish and he said : (25) “Aunt Polly, it seemed mortified, and it hurt so I never minded my tooth at all”. (26) “Your tooth, indeed ! What’s the matter with your tooth ?” (27) “One of them’s loose, and it aches perfectly awful.” (28) “There, there, now, don’t begin that groaning again. Open your mouth. Well – your tooth is loose, but you’re not going to die about that. Mary, get me a silk thread, and a chunk of fire out of the kitchen .” (29) Tom said : “Oh, please, auntie, don’t pull it out. It don’t hurt any more. Please don’t, auntie. I don’t want to stay home from school.” (30) “Oh, you don’t, don’t you ? So, all this row was because you thought you’d get to stay home from school and go for fishing ? Tom, Tom, I love you so, and you seem to try every way you can to break my old heart with your outrageousness.” By this time the dental instruments were ready. The old lady made one end of the silk thread fast to Tom’s tooth with a loop and tied the other to the bedpost. Then she seized, the chunk of fire and suddenly thrust it almost into the boy’s face. The tooth hung dangling by the bedpost, now. (From: The Adventures of Tom Sawyer by Mark Twain) Loading... (A) Answer the following questions: (a) (i) Why did Tom hate Monday mornings ? (ii) Why did he lay in bed thinking ? (b) Why did Tom decide to use his sore toe as an excuse ? (c) Why did Aunt Polly laugh and cry at the same time ? (d) How did Aunt Polly pull out Tom’s loose tooth ? (B) Find words from the passage which mean the same as the following: (e) unclear: (i) miserable (ii) hazy (iii) vague (iv) sore Q. 7. Read the following passage carefully: (12 marks) Title: Adoration by Man Other animals move about the world as nature made them. Why then, did man start to adorn himself by hanging things round the neck, arms, waist and legs or putting things on his head. (1) We can imagine many reasons. If an exceptionally strong or brave man succeded in killing an exceptionally large bear, might he not get the idea of boring a hole through one of its teeth with a sharp flint and lying the tooth round his neck in order to remind himself of his great achievement and to show his friends what a great man he was ? Gradually, it might have become the custom in that tribe for all strong and brave hunters to wear a bear’s tooth, and it might be regarded as a disgrance not to wear one and a sign that one was weak or very young. (2) Another man might make an ornament of a coloured shell or stone simply because he liked it or because its shape reminded him of something. Then if he happened to escape from some danger when he was wearing it he might think the ornament had something to do with it- that it had magic qualities. And his friends and relations would not be satisfied until they had an ornament of the same kind. (3) People who wear ornaments would soon learn to arrange them in different ways according to their size and colour in order to make them more decorative and impressive. A necklace found in Italy with the skeleton of young man of the Stone Age was quite elaborate. It consisted of stage’s teeth arranged at intervals with, between them, two upper rows made up of the vertebrae of a fish and row of shells. (4) Another reason why men might tie feathers, horns, skins and all kinds of other things td themselves would be in order to make themselves look fierce and more terrifying to animals or to the men of other tribes. (5) Objects such as sea-shells that came from a distance and were therefore, scarce for people living far inland— would come in time to have a special value and might be worn only by chiefs and their families in order to show that they were particularly important people. (6) Primitive tribes living today often associate themselves with some particular animal or bird, such as an eagle or lion, or with a particular place, such as a mountain or river. Man may have started doing this kind of thing very early in his history. Then, every member of a group of family may have worn something such as feathers, claws or even a stone or wooden object of a certain shape or colour, to represent the animal or mountain or whatever it might be that they believed themselves to be connected with. (7) So, as we have seen, clothing may have started as ornament or to distinguish one tribe from another or to show rank or because certain things were believed to have magical qualities. But in some places a time came when men and women began to wear clothes for other reasons. During the ice age, when the polar ice spread over far more of the world than it does today, some of the districts in which human beings were living became very cold and indeed. Man must have learnt that he would be more comfortable and more likely to survive, if he covered his body with the skins of animals. At first perphaps, he would simply tie a skin round his waist or over his shoulders but as time passed he learnt how to treat skin in order to make them softer and more supple and how to join them together in order to make better garments. (8) Flint tools have been found buried deep under the earth floors of caves in which prehistoric men sheltered When the weather became colder. Some of the tools were probably used to scrape the inner side of skins to make them soft. Stone Age people may also have softened skins in the same way that Eskimo women do today, by chewing them. The teeth of Eskimo women are often worn down to stumps by the constant chewing of seal skins. (9) Among the wonderful flint and bone tools and implements that later cave-men made, have been found some beautiful bone needles, some not bigger than those we use today: Although the people who made them had only flint tools to work with, some of the needles are finer and more beautifully shaped than those of Roman times. Loading... (A) On the basis of your reading of the above passage, complete the following statements briefly: (2×4 = 8 marks) (a) How did man start to adorn himself ? (b) What was the speciality about the necklace found in Italy ? (c) Why did man tie feather and skin to himself ? (d) What did man realise during the ice age ? (B) Find the words from the above passage which mean the same as the following: (1×4 = 4 marks) (g) Relating to the earliest times : (i) tribes (ii) particular (iii) primitive (iv) connected (h) rub against a hard surface : (i) scrape (ii) flint (iii) stumps (iv) implement Ans. (A) (a) Man started to adorn himself by hanging things around his neck, arms, waist and legs or by putting things on his head. (b) It consisted of stag’s teeth arranged at intervals with two upper rows made up of the vertebrae of a fish and one row of shells. (c) Man tied feather and skin to himself so that he could look fiercer and more terrifying to animals or to the men of other tribes. . (d) During the ice age, man realised that he would be more comfortable and more likely to survive if he covered his body with the skins of animals. Loading... Ans. (B) (e) (ii) ornament (f) (i) elaborate (g) (iii) primitive (h) (i) scrape Q. 8. Read the following passage carefully: (12 marks) (1) Last summer I boarded a flight from the IGI airport. The airplane waited at the runway in a queue to take off for one hour, with the engines running. A lot of aviation fuel was wasted. Carbon dioxide, nitrogen oxide and water vapour were released into the atmosphere. (2) With the entry of many players in the aviation industry, pollution has reached the skies as well. With the number of air line flights worldwide growing and expected to skyrocket over the coming decades, the problem of delayed arrival and departure will intensify. (3) The inefficiencies in the air and on the ground caused by the system also mean wastage of fuel and excessive of CO2. No doubt, the air travel industry is coming under scrutiny for its role in climate change. Though aviation industry contributes only 2% of the total C-emissions. But with the rapid economic growth and ever increasing affordability of air travel, this industry will only expand at a much faster pace than ever before, thus increasing the rate of carbon emissions. (4) Aircraft emission pollutes the air and threatens to become one of the largest contributors of global warming by 2050. At present, pollution from the aircrafts is less that 3% of the environmental pollution, but it is believed that aircraft emissions are currently one of the fastest growing contributors to global warming. (5) Aviable and a sustainable solution comes from the next generation jet bio fuels made from algae or coconuts. Another sustainable alternative would be to put an analog traffic-control system, which is installed in a few airports around the world. (6) Next Gen is the FAA’s (Federal Aviation Administration) long term plan to replace the current U.S. radar based air – traffic control system with one that operates using satellites and a global positioning system. Instead of a radar system, that updates the position of planes only as often as its dish rotates every 12 sec. or so, next Gen will use satellite data to locate planes in real time. Instead of relying on time consuming voice communication with a control tower, pilots will instantly know the location, speed and direction of the planes around them. Every minute saved from a flight plan means fuel saved and carbon emissions averted. And with jet fuel costing about$ 1.75 per gallon that save the airlines millions. (A) On the basis of your reading of the above passage, answer the following statements briefly: (2×4 = 8 marks) (a) What is the outcome of the entry of many players in the aviation industry ? (b) What did the inefficiencies in the air on the ground caused by the system mean ? (c) What does every minute saved from a flight plan means ________ (d) What is FAA’s long term plan ? (g) capable of working successfully: (i) sustainable (ii) viable . (iii) install (iv) update (h) move in a circle round a central position: (i) rotate (ii) analog (iii) satellite (iv) speed Ans. (A) (a) With the entry of many players in the aviation industry, pollution has reached the sky. Moreover, the problem of delayed arrival and departure will also intensify. (b) The inefficiencies in the air and on the ground caused by the system means wastage of fuel and excessive co2. (c) Every minute saved from a flight means fuel saved and carbon emission averted. (d) FAA’s long term plan is to replace the current US radar based on traffic control system with one that operates using satellites and a global positioning system. Ans. (B) (e) (iii) emission (f) (i) averted (g) (ii) viable (h) (i) rotate More Resources for CBSE Class 10 • NCERT Solutions • NCERT Solutions for Class 10 Science • NCERT Solutions for Class 10 Maths • NCERT Solutions for Class 10 Social • NCERT Solutions for Class 10 English • NCERT Solutions for Class 10 Hindi • NCERT Solutions for Class 10 Sanskrit • NCERT Solutions for Class 10 Foundation of IT • RD Sharma Class 10 Solutions <!– –>	2021-05-12 01:25:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.20758146047592163, "perplexity": 5941.427845187413}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-21/segments/1620243991693.14/warc/CC-MAIN-20210512004850-20210512034850-00290.warc.gz"}
http://mathhelpforum.com/statistics/196413-standard-deviation.html	1. ## Standard Deviation How would you explain... x is the standard deviation of the set of numbers (a,b,c,d,e). For each of the following sets, indicate which sets must have a standard deviation equal to x. (a+2, b+2, c+2, d+2, e+2) (a-2, b-2, c-2, d-2, e-2) (2a, 2b, 2c, 2d, 2e) (a/2, b/2, c/2, d/2, e/2) 2. ## Re: Standard Deviation Originally Posted by GIBETH How would you explain... x is the standard deviation of the set of numbers (a,b,c,d,e). For each of the following sets, indicate which sets must have a standard deviation equal to x. (a+2, b+2, c+2, d+2, e+2) (a-2, b-2, c-2, d-2, e-2) (2a, 2b, 2c, 2d, 2e) (a/2, b/2, c/2, d/2, e/2) Do you know how to evaluate standard deviations? You need to square root the average of all squared deviations from the mean... 3. ## Re: Standard Deviation Originally Posted by GIBETH How would you explain... x is the standard deviation of the set of numbers (a,b,c,d,e). For each of the following sets, indicate which sets must have a standard deviation equal to x. (a+2, b+2, c+2, d+2, e+2) (a-2, b-2, c-2, d-2, e-2) (2a, 2b, 2c, 2d, 2e) (a/2, b/2, c/2, d/2, e/2) The first 2 do. Think about what the standard deviation actually measures. 4. ## Re: Standard Deviation Standard deviation measures spread about the mean. Adding 2 to each value also increases the mean by 2 so difference between each value and the mean does'nt change. Hence standard deviation doesn't change. Similarly when subtracting 2.	2016-09-26 17:22:42	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9169750809669495, "perplexity": 3787.254645610223}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-40/segments/1474738660871.1/warc/CC-MAIN-20160924173740-00061-ip-10-143-35-109.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/644286/verification-and-help-to-simplify-an-argument-about-closure-of-some-sets	# Verification and help to simplify an argument about closure of some sets. Hi everyone I'd like to know if what I have so far is correct, I think is much work for something which is too simple I would appreciate any advice or whatever. Moreover, I have doubt in (3) and (4), hopefully are correct. Thanks in advance. Definition: (Closure) Let $X$ be a subset of $\mathbb{R}$. The closure of $X$, denote it $\overline{X}$, is define to be the set of all the adherent points of $X$. A point $x$ is said to be adherent to $X$, iff for each given $\varepsilon>0$ there is some $y\in X$ for which $\|x-y\|\le \varepsilon$. Lemma: The closure of $\mathbb{N}$ is $\mathbb{N}$. The closure of $\mathbb{Z}$ is $\mathbb{Z}$. The closure of of $\mathbb{Q}$ is $\mathbb{R}$. The closure of $\mathbb{R}$ is $\mathbb{R}$. The closure of $\varnothing$ is $\varnothing$. Proof: (1) First we have to show that $\overline{\mathbb{N}}=\mathbb{N}$. Clearly for each $n\in \mathbb{N}$, $n$ is $\varepsilon$-adherent to $\mathbb{N}$ for any $\varepsilon>0$. So, it lies in $\overline{\mathbb{N}}$, i.e., $\overline{\mathbb{N}}\supset \mathbb{N}$. Now we have to show that $\overline{\mathbb{N}}$ only contains element of $\mathbb{N}$. Suppose for the sake of contradiction that there exists an element $n'\in \overline{\mathbb{N}}$ such that $n'\notin \mathbb{N}$. Then $\|n-n'\|>0$ for all $n\in \mathbb{N}$. Let define the set of all $\|n-n'\|$, the set is non-empty and bounded below by zero. Then it has a greatest lower bound, let call it $b$. Clearly $b\ge 0$. We have to show that all the cases lead a contradiction. If $b>0$, setting $\varepsilon= b/2$, implies that $n'$ is not $\varepsilon$-adherent to any $n\in \mathbb{N}$, in particular $n'$ is not adherent to $\mathbb{N}$. If $b=0$, then either $b$ lies in the set or does not, if $b$ lies in the set, it is a least element and would imply $\|n-n'\|=0$ for some $n\in \mathbb{N}$ and so $n=n'$ which contradicts our hypothesis. If $0$ is not in the set, we shall show that we can reached a contradiction. We define recursively the sequence $n_0=0$ and $n_j=\text{min}(\{n>n_{j-1}: \|n-n'\|\le 1/j \})$, the sequence is well-define because the set $\{n>n_{j-1}: \|n-n'\|\le 1/j \}\not= \varnothing$ (if were empty it has a minimum element which is not zero, contradicting that $0$ is its greatest lower bound). Then for each $n_j$ we have $\|n_j-n'\|\le 1/j$, thus $(n_j)\rightarrow n'$, i.e., $(n_j)$ converges to $n$. But $(n_j)$ is a subsequence $\mathbb{N}$, so it has to be unbounded and therefore not convergent. In either case we have a contradiction. Thus $\overline{\mathbb{N}}\subset \mathbb{N}$ and hence $\overline{\mathbb{N}}= \mathbb{N}$ as desired. (2) For the second assertion notice that $\mathbb{Z}= \mathbb{N}\cup \mathbb{-N}$. The set $\mathbb{-N}$ define by $\mathbb{-N}=\{-n:n\in \mathbb{N}\}$. Thus we must have $\overline{\mathbb{Z}}=\overline{ \mathbb{N}\cup \mathbb{-N}}=\mathbb{N}\cup \overline{ \mathbb{-N}}$. But by the same argument as above we can show that $\overline{ \mathbb{-N}}=\mathbb{-N}$. Thus $\overline{\mathbb{Z}}=\mathbb{Z}$ as desired. The proof is based in the fact that $\overline{X} \cup \overline{Y}= \overline{X \cup Y}$, which will show below. (3) We have to show that $\overline{\mathbb{Q}}=\mathbb{R}$. Let $x\in \mathbb{R}$, and $\varepsilon>0$ be arbitrary. Then $\exists q\in \mathbb{Q}$ such that $x<q<x+\varepsilon$, i.e, $d(x,q)\le \varepsilon$ since $\varepsilon$ was arbitrary this shows that $x$ is an adherent point of $\mathbb{Q}$. For the other inclusion we have to show that $\overline{\mathbb{Q}}\subset \mathbb{R}$. Let $q'\in \overline{\mathbb{Q}}$ and let $f: \mathbb{N}\rightarrow \mathbb{Q}$ be a bijective map. Define a sequence of natural numbers recursively by setting $n_0=0$ and $$n_j= \text{min}\{n\in \mathbb{N}: d(q',q_f(n))\le 1/j \text{ and }\, n\not= n_i \text{ for all }\, i< j\}$$ Since $q'$ is an adherent point and $f$ is a bijective map the set is non-empty and each $n_j$ is well-defined. Thus the sequence $(q_{f(n_j)})$ converges to $q'$ and hence $q'$ is a real number as desired. (4) We shall show that $\overline{\mathbb{R}}=\mathbb{R}$. Let $x'\in \overline{\mathbb{R}}$ our task is prove that $x\in \mathbb{R}$. Let define $$X_n:=\{x\in \mathbb{R}: x'\le x\le x'+1/n \}$$ Clearly for each $n\ge 1$, the set $X_n$ is non-empty. Then using the AC$_\omega$ (Axiom of Countable Chocie) we can find a sequence $(x_n)_{n=1}^\infty$ such that $x_n \in X_n$. Thus $(x_n)\rightarrow x$ and by the completeness the result follows. (5) $\varnothing= \overline{\varnothing}$. Suppose for the sake of contradiction that $\overline{\varnothing} \not= \varnothing$, then it has an element $x'$. So $x'$ is adherent to the empty set, but this implies that there exist an element in $\varnothing$ for which $x'$ is $\varepsilon$-adherent for every $\varepsilon$, which is a contradiction. $\Box$ Now we shall show the two auxiliary claims which were used in the proof of the above lemma. Claim 1: If $X\subset Y \subset \overline{X}$. Then $\overline{Y}=\overline{X}$. Proof claim 1: Let $y'$ be an adherent point of $Y$, i.e., $y'\in \overline{Y}$ and let $\varepsilon>0$ be given. Thus, $\exists y \in Y\subset \overline{X}$ s.t. $d(y,y')\le \varepsilon$. So, $y$ is an adherent point of $X$. Thus $\exists x\in X$ s.t. $d(x,y)\le \varepsilon$ and therefore we must have $d(x,y')\le 2\varepsilon$ which shows that $y'$ is an adherent point of $X$, i.e., $\overline{Y}\subset \overline{X}$. Conversely if $x'$ is an adherent point of $X$. Then $\exists x\in X\subset Y$ s.t. $d(x,x')\le \varepsilon$, and so is adherent to $Y$. $\Box$ Claim 2: If $X,Y\subset \mathbb{R}$. Then $\overline{X} \cup \overline{Y}= \overline{X \cup Y}$. Proof claim 2: Let $x' \in \overline{X \cup Y}$, i.e., $x'$ is and adherent point of $X \cup Y$. Then $\exists x\in {X \cup Y}$ s.t. $d(x,x')\le \varepsilon, \forall\varepsilon>0$. Without loss of generality suppose that $x\in X$, so $x'\in \overline{X} \subset \overline{X} \cup \overline{Y}$. Thus $\overline{X \cup Y} \subset \overline{X} \cup \overline{Y}$. Now since $X \cup Y\subset \overline{X \cup Y} \subset \overline{X} \cup \overline{Y}$, we can conclude that $\overline{\overline{X \cup Y}} =\overline{X} \cup \overline{Y}$. But since $\overline{\overline{X \cup Y}}=\overline{X \cup Y}$ the result follows. $\Box$ Any suggestion? Do you think is correct? or Am I totally off track? Do exist for (4) a way to keep away the Axiom of Countable Choice I cannot find one? Please, any help would be great. • Is there anything wrong with my proof? – Jose Antonio Jan 20 '14 at 6:57 Let $x \in \overline{\mathbb{N}}$. Assume that $x \notin \mathbb{N}$, and pick maximal $m$, minimal $n$, both in $\mathbb{N}$, such that $m < x < n$. Then $\|x-m\| = x-m$ and $\|x-n\| = n-x$. Let $\epsilon = \min \{ (n-x)/2, (x-m)/2 \}$. Then there exists a natural number $k$ such that $\|x-k\| < \epsilon$. Contradiction. • OK, I understand good one :). For $\mathbb{Z}$ I think is better use a lemma to show $\overline{X\cup Y} = \overline{X} \cup \overline{Y}$. – Jose Antonio Jan 20 '14 at 1:08 • @Svinepels: I know this is an old topic, but I'm working on the same exercise and reading your answer I don't understand why after writing $\|x-k\|<\varepsilon$ you claim that this is a contradiction; could you explain this point in more detail? – lorenzo Dec 13 '16 at 23:44 • If $\|x-k\|<\epsilon$ then $k$ is closer to $x$ than both $m$ and $n$. This contradicts the fact that $m$ was chosen to be the biggest natural number less than $x$and $n$ the least natural number bigger than $x$. – Ulrik Dec 14 '16 at 21:02	2019-07-17 14:35:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9785075783729553, "perplexity": 53.88845032160982}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195525312.3/warc/CC-MAIN-20190717141631-20190717163631-00136.warc.gz"}
https://zbmath.org/?q=an%3A1166.47058	## On a hybrid method for a family of relatively nonexpansive mappings in a Banach space.(English)Zbl 1166.47058 Summary: We prove strong convergence theorems by the hybrid method given by W. Takahashi, Y. Takeuchi and R. Kubota [J. Math. Anal. Appl. 341, No. 1, 276–286 (2008; Zbl 1134.47052)] for a family of relatively nonexpansive mappings under weaker conditions. The method of the proof is different from the original one and it shows that the type of projection used in the iterative method is independent of the properties of the mappings. We also deal with the problem of finding a zero of a maximal monotone operator and obtain a strong convergence theorem using this method. ### MSC: 47J25 Iterative procedures involving nonlinear operators 47H09 Contraction-type mappings, nonexpansive mappings, $$A$$-proper mappings, etc. 47H05 Monotone operators and generalizations Zbl 1134.47052 Full Text: ### References: [1] Alber, Y.I., Metric and generalized projection operators in Banach spaces: properties and applications, (), 15-50 · Zbl 0883.47083 [2] Barbu, V.; Precupanu, T., Convexity and optimization in Banach spaces, Math. appl. (east eur. ser.), vol. 10, (1986), D. Reidel Publishing Co. Dordrecht · Zbl 0594.49001 [3] Beer, G., Topologies on closed and closed convex sets, (1993), Kluwer Academic Publishers Group Dordrecht · Zbl 0792.54008 [4] Butnariu, D.; Censor, Y.; Reich, S., Iterative averaging of entropic projections for solving stochastic convex feasibility problems, Comput. optim. appl., 8, 21-39, (1997) · Zbl 0880.90106 [5] Butnariu, D.; Iusem, A.N., Totally convex functions for fixed points computation and infinite dimensional optimization, Appl. optim., vol. 40, (2000), Kluwer Academic Publishers Dordrecht · Zbl 0960.90092 [6] Butnariu, D.; Reich, S.; Zaslavski, A.J., Asymptotic behavior of relatively nonexpansive operators in Banach spaces, J. appl. anal., 7, 151-174, (2001) · Zbl 1010.47032 [7] Butnariu, D.; Reich, S.; Zaslavski, A.J., Weak convergence of orbits of nonlinear operators in reflexive Banach spaces, Numer. funct. anal. optim., 24, 489-508, (2003) · Zbl 1071.47052 [8] Censor, Y.; Reich, S., Iterations of paracontractions and firmly nonexpansive operators with applications to feasibility and optimization, Optimization, 37, 323-339, (1996) · Zbl 0883.47063 [9] Goebel, K.; Reich, S., Uniform convexity, hyperbolic geometry, and nonexpansive mappings, Monogr. textbooks pure appl. math., vol. 83, (1984), Marcel Dekker Inc. New York · Zbl 0537.46001 [10] Halpern, B., Fixed points of nonexpanding maps, Bull. amer. math. soc., 73, 957-961, (1967) · Zbl 0177.19101 [11] Ibaraki, T.; Kimura, Y.; Takahashi, W., Convergence theorems for generalized projections and maximal monotone operators in Banach spaces, Abstr. appl. anal., 621-629, (2003) · Zbl 1045.47041 [12] Israel, M.M.; Reich, S., Extension and selection problems for nonlinear semigroups in Banach spaces, Math. japon., 28, 1-8, (1983) · Zbl 0531.47057 [13] Kimura, Y., On mosco convergence for a sequence of closed convex subsets of Banach spaces, (), 291-300 · Zbl 1095.49006 [14] Mann, W.R., Mean value methods in iteration, Proc. amer. math. soc., 4, 506-510, (1953) · Zbl 0050.11603 [15] Matsushita, S.; Takahashi, W., An iterative algorithm for relatively nonexpansive mappings by a hybrid method and applications, (), 305-313 · Zbl 1086.47055 [16] Matsushita, S.; Takahashi, W., Weak and strong convergence theorems for relatively nonexpansive mappings in Banach spaces, Fixed point theory appl., 37-47, (2004) · Zbl 1088.47054 [17] Matsushita, S.; Takahashi, W., A strong convergence theorem for relatively nonexpansive mappings in a Banach space, J. approx. theory, 134, 257-266, (2005) · Zbl 1071.47063 [18] Mosco, U., Convergence of convex sets and of solutions of variational inequalities, Adv. math., 3, 510-585, (1969) · Zbl 0192.49101 [19] Nakajo, K.; Takahashi, W., Approximation of a zero of maximal monotone operators in Hilbert spaces, (), 303-314 · Zbl 1264.47079 [20] Reich, S., Weak convergence theorems for nonexpansive mappings in Banach spaces, J. math. anal. appl., 67, 274-276, (1979) · Zbl 0423.47026 [21] Reich, S., Strong convergence theorems for resolvents of accretive operators in Banach spaces, J. math. anal. appl., 75, 287-292, (1980) · Zbl 0437.47047 [22] Reich, S., Approximating fixed points of nonexpansive mappings, Panamer. math. J., 4, 23-28, (1994) · Zbl 0856.47032 [23] Reich, S., A weak convergence theorem for the alternating method with Bregman distances, (), 313-318 · Zbl 0943.47040 [24] Shioji, N.; Takahashi, W., Strong convergence of approximated sequences for nonexpansive mappings in Banach spaces, Proc. amer. math. soc., 125, 3641-3645, (1997) · Zbl 0888.47034 [25] Takahashi, W., Convex analysis and approximation of fixed points, (2000), Yokohama Publ. Yokohama, (in Japanese) [26] Takahashi, W., Nonlinear functional analysis: fixed point theory and its applications, (2000), Yokohama Publ. Yokohama · Zbl 0997.47002 [27] Takahashi, W.; Kim, G.-E., Approximating fixed points of nonexpansive mappings in Banach spaces, Math. japon., 48, 1-9, (1998) · Zbl 0913.47056 [28] Takahashi, W.; Takeuchi, Y.; Kubota, R., Strong convergence theorems by hybrid methods for families of nonexpansive mappings in Hilbert spaces, J. math. anal. appl., 341, 276-286, (2008) · Zbl 1134.47052 [29] Tsukada, M., Convergence of best approximations in a smooth Banach space, J. approx. theory, 40, 301-309, (1984) · Zbl 0545.41042 [30] Wittmann, R., Approximation of fixed points of nonexpansive mappings, Arch. math. (basel), 58, 486-491, (1992) · Zbl 0797.47036 This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.	2022-06-26 12:55:25	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7481091618537903, "perplexity": 3581.3178589506906}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-27/segments/1656103205617.12/warc/CC-MAIN-20220626101442-20220626131442-00278.warc.gz"}
https://datascience.stackexchange.com/questions/8043/optimizing-parameters-for-a-closed-black-box-system	# Optimizing parameters for a closed (black-box) system I am working on a problem that involves finding optimal parameter values for a black-box system. This system consists of 6 inputs and produces a single output. I can compare the returned values to observed values to determine whether the system is well calibrated based on the parameter values I specified. The main problem that I am having is that I do not know how these parameters are used in the black-box system to produce the output. The parameters may be correlated so changing the value of one parameter may affect the way the others behave in the system. Since I do not know the function I am not sure how to optimize this problem efficiently. Question: what are some methods for optimization when the function is unknown? Edits: The variable types are each a vector of real numbers but we can make some assumptions if it is is helpful. The cost to run the black-box system is only time. Let's say that it takes 10 seconds to run the system but I happen to have five black boxes -- so if I can run my algorithm in parallel then I can cut down the run time. If the algorithm runs sequentially then I don't gain anything by having the extra boxes (expect perhaps choosing different starting positions). For this problem I am interested in learning about how to solve this type of problem -- not trying to brute force my way to a solution. That is to say I am less interested in obtaining an answer and more finding out how I would obtain the the answer. I think it is easy to say that given sufficient resources (time, processing power) that simply generating many many random numbers and choosing the combination that gives the best result is the simplest (and probably worst) approach. • You have a lot of options. Could you clarify what the variable types are here (I suspect just a vector of real numbers for input and single real number for output - but this is critical, if any values are discrete integers, it changes what may work)? Could you also clarify what it costs you to call the black box - i.e. how many calls per second it could cope with? Also, how much you value your own time understanding more sophisticated optimisation approaches compared with simpler more brute-force approaches. – Neil Slater Sep 8 '15 at 18:08 • I've amended the post to address your questions. If things are still ambiguous I can try to clarify them. – Ellis Valentiner Sep 8 '15 at 18:35 • Hmm 1 call per 2 seconds may not be fast enough for simple generic solvers. Your premise in the question that there is a single optimal way how to find an answer is not necessarily correct. Random guessing is surprisingly efficient with the amount of unknowns you present so far. Can we make some assumptions such as function is continuous and differentiable everywhere, and has some general global shape (so we worry about local minima close to global minimum, not trying to find deepest valley in the Alps)? – Neil Slater Sep 8 '15 at 18:39 • Sure. Let's assume that the function is continuous and differentiable. I suspect the global shape is Gaussian -- or at least unimodal. – Ellis Valentiner Sep 8 '15 at 18:52 ## 3 Answers There are lots of function optimising routines that could be applied, based on the description so far. Random search, grid search, hill-climbing, gradient descent, genetic algorithms, simulated annealing, particle swarm optimisation are all possible contenders that I have heard of, and I am probably missing a few. The trouble is, starting with next to zero knowledge of the black box, it is almost impossible to guess a good candidate from these search options. All of them have strengths and weaknesses. To start with, you seem to have no indication of scale - should you be trying input parameters in any particular ranges? So you might want to try very crude searches through a range of magnitudes (positive and negative values) to find the area worth searching. Such a grid search is expensive - if you have $k$ dimensions and want to search $n$ different magnitudes, then you need to call your black box $n^k$ times. This can be done in parallel though, and given you are confident that the function is roughly unimodal, you can start with a relatively low number of n (maybe check -10, -1, 0, +1, +10 for 15625 calls to your function taking roughly 8 hours 40 mins using 5 boxes). You may need to repeat with other params once you know whether you have found a bounding box for the mode or need to try yet more values, so this process could take a while longer - potentially days if the optimal value for param 6 is more like 20,000. You could also refine more closely, once you have a potential mode you might want to define another grid of values to search based around. This basic grid search might be my first point of attack on a black box system where I had no clue about parameter meaning, but some confidence that the black box output had a rough unimodal form. Given the speed of response you should be storing all input and output values in a database for faster lookup and better model building later. No point repeating a call taking 10 seconds when a cache could look it up in 1 millisecond. Once you have some range of values you think that a mode might be in, then it is time to pick a suitable optimiser. Given the information so far, I would be tempted to run either more grid search (with a separate linear scaling between values of each param) and/or a random search, constrained roughly to the boxes defined by the set of $2^6$ corner points around the best result found in initial order-of-magnitude search. At that point you could also consider graphing the data, to see if there is any intuition about which other algorithms could perform well. With the possibility of parallel calls, then gradient descent might be a reasonable guess, because you can get approximate gradients by adding a small offset to each param and dividing difference that causes in the output by it. In addition, gradient descent (or simple hill climbing) has some chance of optimising with less calls to evaluate the function than approaches that rely on many iterations (simulated annealing) or lots of work in parallel (particle swarm or genetic algorithms). Gradient descent optimisers as used in neural networks, with additions like Nesterov momentum or RMSProp, can cope with changes in function output "feature scale" such as different sizes and heights of peaks, ridges, saddle points. However, gradient descent and hill climbing algorithms are not robust against all function shapes. A graph or several of what your explorations are seeing may help you to decide on a different approach. So keep all the data and graph it in case you can get clues. Finally, don't rule out random brute-force search, and being able to just accept "best so far" under time constraints. With low knowledge of the internals of the black box, it is a reasonable strategy. Since no one has mentioned bayesian optimization yet: "Bayesian optimization is a sequential design strategy for global optimization of black-box functions." At the bottom of that wikipedia page are links to several bayesian optimization libraries like: Besides libraries using bayesian optimization there are also libraries using TPE (trees of parzen estimators), like hyperopt. Here's an image of fmfn's bayesian optimizer optimizing a non-linear 2d function. As you can see, a point close to the global optimum is found after relatively few iterations and the system tests more points in more promising (red) areas than in less promising (blue) areas - making it more efficient than random search: Here is a solution. A mathematical method is described in this paper • Link-only answers are discouraged; summarize what about these resources answers the question – Sean Owen May 18 '16 at 22:22	2020-01-19 12:47:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5433181524276733, "perplexity": 522.2832797027995}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250594603.8/warc/CC-MAIN-20200119122744-20200119150744-00351.warc.gz"}
http://mathhelpforum.com/statistics/179233-t-test.html	# Math Help - t - test 1. ## t - test ive calculated T=5.4 and the t(11,0.05)=2.2010 does this mean we accept or reject the null because T is greater than t? EDIT: I have found that the null will be rejected for this example. In another example, I have found a negative T value (T=-4.3, t(13,0.05) = 2.1604) for a two tailed test, am I correct in thinking that this should just be treated as a positive value and therefore the null hypothesis is again rejected? 2. Originally Posted by keelejody In another example, I have found a negative T value (T=-4.3, t(13,0.05) = 2.1604) for a two tailed test, am I correct in thinking that this should just be treated as a positive value and therefore the null hypothesis is again rejected? If your level of significance (that is , alpha) is 0.05, then for a two-tailed test, you need to calculate t(13,0.025). You have calculated the correct thing but mistakenly wrote it as "t(13,0.05)". Here \|T\| > t(13,.025). So you reject the null hypothesis.	2015-04-26 10:26:32	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8254446983337402, "perplexity": 550.0268301904616}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1429246654264.98/warc/CC-MAIN-20150417045734-00106-ip-10-235-10-82.ec2.internal.warc.gz"}
https://stacks.math.columbia.edu/tag/0EHR	Lemma 30.23.11. Let $X$ be a Noetherian scheme. Let $\mathcal{I}, \mathcal{J} \subset \mathcal{O}_ X$ be quasi-coherent sheaves of ideals. If $V(\mathcal{I}) = V(\mathcal{J})$ is the same closed subset of $X$, then $\textit{Coh}(X, \mathcal{I})$ and $\textit{Coh}(X, \mathcal{J})$ are equivalent. Proof. First, assume $X = \mathop{\mathrm{Spec}}(A)$ is affine. Let $I, J \subset A$ be the ideals corresponding to $\mathcal{I}, \mathcal{J}$. Then $V(I) = V(J)$ implies we have $I^ c \subset J$ and $J^ d \subset I$ for some $c, d \geq 1$ by elementary properties of the Zariski topology (see Algebra, Section 10.17 and Lemma 10.32.5). Hence the $I$-adic and $J$-adic completions of $A$ agree, see Algebra, Lemma 10.96.9. Thus the equivalence follows from Lemma 30.23.1 in this case. In general, using what we said above and the fact that $X$ is quasi-compact, to choose $c, d \geq 1$ such that $\mathcal{I}^ c \subset \mathcal{J}$ and $\mathcal{J}^ d \subset \mathcal{I}$. Then given an object $(\mathcal{F}_ n)$ in $\textit{Coh}(X, \mathcal{I})$ we claim that the inverse system $(\mathcal{F}_{cn}/\mathcal{J}^ n\mathcal{F}_{cn})$ is in $\textit{Coh}(X, \mathcal{J})$. This may be checked on the members of an affine covering; we omit the details. In the same manner we can construct an object of $\textit{Coh}(X, \mathcal{I})$ starting with an object of $\textit{Coh}(X, \mathcal{J})$. We omit the verification that these constructions define mutually quasi-inverse functors. $\square$ In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).	2022-05-17 20:22:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0, "math_score": 0.9824041128158569, "perplexity": 139.11433270286554}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662520817.27/warc/CC-MAIN-20220517194243-20220517224243-00646.warc.gz"}
https://leanprover-community.github.io/mathlib_docs/core/init/meta/well_founded_tactics.html	# mathlibdocumentation core.init.meta.well_founded_tactics theorem nat.lt_add_of_zero_lt_left (a b : ) : 0 < ba < a + b theorem nat.zero_lt_one_add (a : ) : 0 < 1 + a theorem nat.lt_add_right (a b c : ) : a < ba < b + c theorem nat.lt_add_left (a b c : ) : a < ba < c + b meta structure well_founded_tactics : Type • rel_tac : expr • dec_tac : Argument for using_well_founded The tactic rel_tac has to synthesize an element of type (has_well_founded A). The two arguments are: a local representing the function being defined by well founded recursion, and a list of recursive equations. The equations can be used to decide which well founded relation should be used. The tactic dec_tac has to synthesize decreasing proofs.	2020-10-30 07:30:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4621613621711731, "perplexity": 4419.032130860805}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-45/segments/1603107909746.93/warc/CC-MAIN-20201030063319-20201030093319-00464.warc.gz"}
https://stats.stackexchange.com/questions/359537/how-to-test-if-bandit-algorithm-is-converging	# How to test if bandit algorithm is converging? I have coded up the a multi-armed bandit algorithm based on algorithm 1 in the original LinUCB algorithm paper, but I am having trouble determining if it is working properly. My test setup is the following: • I created some random context data matrix, with D=np.random.random( (10000,3)), where each row is a training example (user) and each column is a feature for that user / training example. Each row is $x_t$ from the algorithm. • I then created 100 arms where each arm is represented by a Bernoulli distribution with varying $p$ (probability) values. In fact the logic of each arm's reward distribution is the following: if random.random() > self.p: return 0.0 else: return 1.0 • The different p-values of the arms are defined by np.random.random(100), which means I know what the best/worst arms are beforehand. The test then follows these steps: 1. Step through each row of $D$ from $x_1$ to $x_{10000}$ 2. At each $x_t$, go through every bandit arm, from $a_1$ to $a_{100}$ and estimate the payoff, $p_a$, using the formula from the paper. 3. Choose the arm with the highest estimated payoff, call this $a_t$. 4. With the recommended arm, $a_t$, sample its Bernoulli distribution using the logic above with its respective $p$-value. Call this $R_{t,a}$. At the same time, sample best arm's reward (arm with highest $p$ value), call this $R_t^$ 5. Calculate $Regret=R_t^-R_{t,a}$ and record this at each step $t$. Now, after running this, according to the paper, the cumulative regret over each time step should be logarithmic or $O(\sqrt{KdT} )$. But when I plot the cumulative sum of Regret over each timestep, it is simply linear. Where am I going wrong? What other methods could I use to tell if it is converging? Note: I haven't posted all my code as I wanted to make sure the broader approach made sense and was correct before going into the nitty-gritty details. All the way at the end of the right column on page 3, the LinUCB paper says: "we can adapt the analysis from [6] to show the following: if the arm set $\mathcal{A}_t$ is fixed and contains $K$ arms, then [...], and then prove the strong regret bound of $\tilde{O}(\sqrt{KdT})$, matching the state-of-the-art result [6] for bandits satisfying Eq. (2)." Now, that last part ("for bandits satisfying Eq. (2)") is slightly ambiguous in my opinion, it can be interpreted in two different ways: 1. That bound was the state-of-the-art for bandits satisfying Equation (2), OR 2. This particular paper also relies on Equation (2) I didn't go through the referenced paper (reference [6]) yet to see if the actual analysis also requires Equation (2) to hold, but I suspect it does. Right above Equation (2), the LinUCB paper does say: "[...] we assume the expected payoff of an arm $a$ is linear in its $d$-dimensional feature vector $\mathbf{x}_{t, a}$ with some unknown coefficient vector $\boldsymbol{\theta}_a^$; namely, for all $t$, $$\mathbf{E} \left[ r_{t, a} \vert \mathbf{x}_{t, a} \right] = \mathbf{x}^{\top}_{t, a} \boldsymbol{\theta}_a^$$" Generally, when a paper says "we assume" like that, that indicates that their later theoretical results rely on that assumption unless they later on specifically indicate for one of their theoretical results that it does not rely on that assumption. So, now it's time to see if that assumption actually holds in your case. Unless I'm misunderstanding a part of your description, it looks to me like the context vectors $\mathbf{x}_t$ are actually useless in some sense; they do not in reality have any relation whatsoever to the reward distribution. This kind of means that you have a standard Multi-Armed Bandit problem (not contextual), but it's "disguised" as a Contextual MAB problem. In your case, the rewards are completely independent from the context vectors, which means Equation 2 from the paper cannot hold; the expectation of the reward given a context vector cannot be a linear function of that context vector. The assumptions of the theoretical analysis leading to that regret bound do not hold, so the results do not hold either (at least, not necessarily; they still might by coincidence). Note that, in your case, under the assumption that you're learning a separate parameter vector $\boldsymbol{\theta}_a$ per arm $a$, there is a very easy way to make the assumption of Equation (2) hold; simply add a fourth feature to every feature vector $\mathbf{x}_t$ which always has a value of $1$ (very much like an "intercept" or "bias" term that's often used in things like Linear Regression, Logistic Regression, Neural Networks, etc.). Hopefully, the algorithm should then be able to learn that the three other features are completely useless for predictions, and learn to predict a consistent estimate of the expected rewards of all the arms independent of the remaining three features. The following procedure was described in the question for evaluating regret: 1. With the recommended arm, $a_t$, sample its Bernoulli distribution using the logic above with its respective $p$-value. Call this $R_{t,a}$. At the same time, sample best arm's reward (arm with highest $p$ value), call this $R_t^$ 2. Calculate $Regret=R_t^-R_{t,a}$ and record this at each step $t$. I was inclined to say that this was not 100% correct, but it actually does appear to be correct according to the formal definitions of regret. I didn't originally feel like it was 100% correct, because the best arm with respect to expected reward ($p$) is not necessarily the best arm in any single given round. This evaluation method can theoretically lead to negative regret (consider the case where a suboptimal arm was played, but that suboptimal arm randomly got "lucky" and produced a better reward in one particular timestep). Before looking up the formal definitions, I was inclined to say that a better evaluation method would be to generate the actual reward outcomes in a time step for all arms, and subtract the reward of the chosen arm from the best possible reward in that timestep to compute that timestep's regret. In comparison to your evaluation method, this approach; • can never result in a negative regret • is more "strict" / punishing for suboptimal strategies • results in a regret of precisely $0$ for a "cheating" or "oracle" strategy that always knows which arm is the best and picks the best When comparing the performance of multiple different algorithms, both variants of computing regret would always lead to the same ordering of algorithms' performance levels. • Thanks so much. This makes sense and explains why even if though the estimated $\theta_a$ are not converging, the algorithm still manages to find the best arms. It is acting like a traditional multi-armed bandit problem, but without any use of the contexts as you described. I will try adding a bias as you mention to see if the code is correct. Is my method of calculating regret correct? – guy Jul 29, 2018 at 18:29 • @guy That's an... interesting question which I kinda mistakenly glossed over initially. Edited my thoughts on that into my answer, but it's not quite a conclusive answer. Jul 29, 2018 at 18:53 • That's just what I expect though, I might be wrong. The best thing to do is to try to evaluate what happens in such cases empirically. Jul 30, 2018 at 16:52 • @guy Hmm if I look at papers such as this one, it looks like cumulative regret plots that at least "look" linear aren't that uncommon in practice. It may simply take more time to learn adequately for a problem with 100 arms. You could try lowering to e.g. 10 arms, see what happens. It looks like proper tuning of the $\alpha$ parameter can also have a significant influence. Jul 31, 2018 at 8:16 • @guy great, thanks for the update! Enjoyed this discourse, Dennis and guy, thanks! Dec 13, 2018 at 0:20	2022-08-16 01:35:22	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8264482021331787, "perplexity": 570.4632060390567}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-33/segments/1659882572215.27/warc/CC-MAIN-20220815235954-20220816025954-00771.warc.gz"}
https://plosjournal.deepdyve.com/lp/springer-journals/on-discrete-versions-of-two-accola-s-theorems-about-automorphism-1Ln7RaJnXR	# On discrete versions of two Accola’s theorems about automorphism groups of Riemann surfaces On discrete versions of two Accola’s theorems about automorphism groups of Riemann surfaces In this paper we give a few discrete versions of Robert Accola’s results on Riemann surfaces with automorphism groups admitting partitions. As a consequence, we establish a condition for $$\gamma$$ γ -hyperelliptic involution on a graph to be unique. Also we construct an infinite family of graphs with more than one $$\gamma$$ γ -hyperelliptic involution. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Analysis and Mathematical Physics Springer Journals # On discrete versions of two Accola’s theorems about automorphism groups of Riemann surfaces , Volume 7 (3) – Jul 25, 2016 11 pages Loading next page... /lp/springer-journals/on-discrete-versions-of-two-accola-s-theorems-about-automorphism-1Ln7RaJnXR Publisher Springer Journals Copyright Copyright © 2016 by Springer International Publishing Subject Mathematics; Analysis; Mathematical Methods in Physics ISSN 1664-2368 eISSN 1664-235X DOI 10.1007/s13324-016-0138-4 Publisher site See Article on Publisher Site ### Abstract In this paper we give a few discrete versions of Robert Accola’s results on Riemann surfaces with automorphism groups admitting partitions. As a consequence, we establish a condition for $$\gamma$$ γ -hyperelliptic involution on a graph to be unique. Also we construct an infinite family of graphs with more than one $$\gamma$$ γ -hyperelliptic involution. ### Journal Analysis and Mathematical PhysicsSpringer Journals Published: Jul 25, 2016	2022-01-20 13:59:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4805108904838562, "perplexity": 1200.5004302094922}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320301863.7/warc/CC-MAIN-20220120130236-20220120160236-00385.warc.gz"}
http://physics.stackexchange.com/tags/orbital-motion/new	# Tag Info ## New answers tagged orbital-motion 6 The orbit of a satellite is determined by 6 parameters, the orbital elements: The eccentricity $e$ The semimajor axis ($a$) The inclination ($i$) The longitude of the ascending node ($\Omega$) The argument of periapsis ($\omega$) The mean anomaly at a specific time ($M_0$) The plane of reference is the equator, and the reference direction is the Vernal ... 5 Actually, both are quite possible. In the general case, for an arbitrary elliptical orbit, what you'll tend to find is that B is true (granted there will be some precession, but not usually in line with the Sun). However, it is possible to set up an orbit (such as a Sun-synchronous one) in which A is true. But orbits such as this require planning and precise ... 4 The velocity of the orbiting space junk is a vector, with both a radial and a tangential component. $$\vec{v}_f = \dot{r}_f\hat{r} + r_f\dot{\theta}_f\hat{\theta}$$ (my $r_f$ is your $r$) The equation for conservation of angular momentum involves only the tangential component of velocity, because it comes from the cross product of the radius vector and the ... 7 $\hat r, \hat \theta, \hat z$ form an orthogonal, right-handed triad of basis vectors in 3d. Taking the cross-product with them is no more exotic or unusual than doing so for Cartesian basis vectors. $\hat r \times \hat \theta = \hat z$, and so on. 0 For each body, you keep its velocity, position, and any other changing properties in variables. Whether you organize them as having all properties of each object in a struct (or object) associated with that body, as in Brandon Enright's comment, or keep an array of all x coordinates, an array of all y coordinates, etc. each indexed by body, doesn't matter. ... 2 No, the right ascension of the mean Sun is NOT zero at the vernal equinox. It is in fact nearly identical to the ecliptic longitude of the mean Sun (the difference is due to UT vs ephemeris time), and this is defined such that it coincides with the ecliptic longitude of the apparent Sun when the Earth is at perihelion. So that should be the starting time to ... 1 The trajectories of the two bodies each trace out separate ellipses. The barycenter is located at the left focus of one trajectory, and the right focus of the other. The orbit of the two bodies is the ellipse generated by the polar graph of the re distance between the bodies as a function of the azimuthal angle. This ellipse is larger than either of the ... 0 There are many different types of orbits according to General Relativity - more that that according to Newtonian mechanics. For example: http://arxiv.org/abs/1207.7041 Characterization of all possible orbits in the Schwarzschild metric revisited (skip right to the figures in the end of the article). Elliptical path is one particular case of these. 0 The Oberth effect takes advantage of when you add something to a quantity that is squared then the effect will be larger, the larger the quantity is initially. It works when a body is under the gravitational influence of another, usually larger e.g. a spacecraft around the Earth. In the central body approximation, the energy equation of a (keplerian) orbit ... 1 This link answers the question well: http://en.wikipedia.org/wiki/Bertrand%27s_theorem#Radial_harmonic_oscillator It sidesteps some of the difficulties you would otherwise have with this class of problem by looking at the potential, which is a metric of the square of the radius, which eliminates the square root, which can then be isolated with each ... 3 What you're looking for is Bertrand's theorem which states that for central force field in $1/r^p$, only $p=2$ (newtonian field) and $p=-1$ (harmonic field) give closed orbit whatever the initial conditions are (and assuming you are bounded to the field). For sure you can find some initial conditions that give closed orbits (for example, all $p\in[-1;2]$ ... 0 Look at it this way. An electron is a charged particle. A moving, electrically charged particle creates a magnetic field, and the particle itself already has an electric field. If the particle is accelerating, then you're going to have a ripple effect from the electric and magnetic field, or an electromagnetic ripple, ie an electromagnetic wave. So an ... 2 On a uniform circular orbit, even if the speed does not change in norm, it does change in direction so that the speed vector change over time and $\frac{d\vec{v}}{dt}\neq\vec{0}$. In fact, in polar coordinates, you have $$\vec{a} = \frac{d\vec{v}}{dt} = -\frac{v^2}{R}\,\vec{e_r}$$ Imagine a car taking a turn at constant speed: if the turn is left, you feel ... 5 In uniform circular motion: $a=\frac{v^2}{r}$ 2 Let's generalize your idea and see if it can be more propellant-efficient, at least in principle. Call your two orbits $\mho_1$ and $\mho_2$. Both have semi-major axis $a_2$ and $a_2$ and inclinations $i_1$ and $i_2$ respectively. As per the problem, $$a_1<a_2$$ $$i_1=i_2$$ and any phasing issues may be ignored. Also, $$r_{p1} = r_{a1}$$ r_{p2} = ... Top 50 recent answers are included	2013-05-23 18:33:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8381229639053345, "perplexity": 332.52208165970524}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2013-20/segments/1368703682988/warc/CC-MAIN-20130516112802-00068-ip-10-60-113-184.ec2.internal.warc.gz"}
http://math.stackexchange.com/questions/148017/odd-binomial-coefficients?answertab=active	# Odd Binomial Coefficients? By Newton's Formula: $$(a+b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k}b^k$$ Proof that every $\dbinom{n}{k}$ is odd if and only if $n=2^r-1$. I have already shown that if $n$ is of the form $2^r-1$, having used the property $$\binom{n-1}{k} = \binom{n}{k}-\binom{n}{k-1}+ \binom{n}{k-2} - \cdots \pm \binom{n}{0}.$$ But I have not been able to demonstrate "$\Rightarrow$". - @Mario: You are using $k$ twice. I think it would be better if you change it to $n = 2^m - 1$. – user17762 May 22 '12 at 3:59 Consider polynomial $(x+1)^n$ on $\mathbb{Z}_{2}[x]$. – y zhao May 22 '12 at 4:20 See this answer for a follow up to the hint by y zhao. A (somewhat compressed) proof of Lucas' Theorem cited by Arturo is in there. – Jyrki Lahtonen May 22 '12 at 4:57 We will prove that $p$ does not divide $\dbinom{n}{k}$ for any $k \in \{0,1,2,\ldots,n\}$ iff $n = p^m-1$. Write $n$ in base $p$ as $$n = \sum_{i=0}^{l} n_i p^i$$ The highest power of $p$ that divides $n!$ is $$\left \lfloor \frac{n}{p} \right \rfloor + \left \lfloor \frac{n}{p^2} \right \rfloor + \cdots + \left \lfloor \frac{n}{p^l} \right \rfloor = \sum_{i=1}^{l} n_i p^{i-1} + \sum_{i=2}^{l} n_i p^{i-2} + \cdots + \sum_{i=l}^{l} n_i p^{i-l}\\ = \sum_{i=1}^{l} n_i \left( p^{i-1} + p^{i-2} + \cdots + 1\right) = \sum_{i=0}^{l} n_i \left( p^i - 1 \right) = n - \sum_{i=0}^{l} n_i$$ The power of $p$ that divides the binomial coefficient $\dbinom{n}{k}$ is nothing but $$(n - \sum_{i=0}^{l} n_i) - (k - \sum_{i=0}^{l} k_i) - (n -k - \sum_{i=0}^{l} (n-k)_i) = \sum_{i=0}^{l} k_i + \sum_{i=0}^{l} (n-k)_i - \sum_{i=0}^{l} n_i$$ Hence, $p \not \vert \dbinom{n}{k}$ if and only if $$\sum_{i=0}^{l} k_i + \sum_{i=0}^{l} (n-k)_i - \sum_{i=0}^{l} n_i = 0$$ i.e. $$\sum_{i=0}^{l} n_i = \sum_{i=0}^{l} k_i + \sum_{i=0}^{l} (n-k)_i$$ This means that $n = p^m - 1$ for some $m$. - @ArturoMagidin aah, yes. Will update it. – user17762 May 22 '12 at 4:23 The equation $$\sum_{i=0}^{l} n_i = \sum_{i=0}^{l} k_i + \sum_{i=0}^{l} (n-k)_i$$ is what Arturo says as no carry. – user17762 May 22 '12 at 4:27 @ArturoMagidin You do not need to delete the comments in future. – user17762 May 22 '12 at 4:29 Marvis: I have a question, $$\sum_{i=1}^l \left \lfloor{\frac{n}{p^i}} \right \rfloor.$$ The right side, with subindexation with the $\sum$. – Mario De León Urbina May 23 '12 at 1:20 @Mario De León Urbina I don't understand your comment. Can you elaborate it? – user17762 May 23 '12 at 1:50 This follows easily from Kummer's Theorem, that the highest power of a prime $p$ that divides $\binom{n}{m}$ is equal to the number of "carries" when adding $n-m$ and $m$ in base $p$. In particular, $\binom{n}{m}$ is odd if and only if there are no carries when adding in base $2$. If the binary expression for $n$ has any $0$s, then selecting $m$ to have a $1$ at precisely the first $0$ and $0$s elsewhere gives a value with $\binom{n}{m}$ even. So the expression for $n$ must consist entirely of $1$s, i.e., $n$ must be of the form $n^r-1$ for some $r$. (Note that this argument shows that the same conclusion holds if "odd" and $2$ are replaced by "prime to $p$" and $p$".) The result also follows from the Lucas' Theorem, which describes the remainder of $\binom{n}{m}$ when divided by a prime $p$.	2014-09-23 14:37:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9121706485748291, "perplexity": 279.66268208700296}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-41/segments/1410657138980.37/warc/CC-MAIN-20140914011218-00148-ip-10-234-18-248.ec2.internal.warc.gz"}
https://mathoverflow.net/questions/202178/reference-request-for-division-algebras-over-mathbbq-pt	Reference request for division algebras, over $\mathbb{Q}_{p}((t))$ I was looking for a possible reference that would answer the following question, Let $\mathbb{Q}_{p}$ be the $p$-adic numbers and $\mathbb{Q}_{p}((t))$ be the field of Laurent polynomials over $\mathbb{Q}_{p}$. Does anyone know of a reference that addresses the following question, "Are all division algebras over $\mathbb{Q}_{p}((t))$ cyclic?" • By "division algebra" I assume you mean a finite dimensional central algebra over $\mathbb{Q}_p((t))$, which is a divsion ring. Is that correct? – Pace Nielsen Apr 7 '15 at 1:28 Here is the closest result I know. If the degree of the division algebra is a prime $q \ne p$, an affirmative answer has been given over finite extensions of $\mathbb Q_p(t)$ by Saltman's paper Cyclic algebras over $p$-adic curves. • Saltman's paper is working with the arithmetically trickier $\mathbb{Q}_{p}(t)$. But it is a good place to start. Thanks. – TheNumber23 Apr 7 '15 at 15:01	2020-05-28 18:53:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8300092220306396, "perplexity": 189.31898651481313}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-24/segments/1590347399830.24/warc/CC-MAIN-20200528170840-20200528200840-00282.warc.gz"}
https://hal.archives-ouvertes.fr/L2C/hal-01949587	# Effective approach to lepton observables: The seesaw case * Corresponding author Abstract : In the absence of direct evidence of new physics, any ultraviolet theory can be reduced to its specific set of low-energy effective operators. As a case study, we derive the effective field theory for the seesaw extension of the Standard Model, with sterile neutrinos of mass $M>m_W$. We systematically compute all Wilson coefficients generated at one loop. Hence, it becomes straightforward to (i) identify the seesaw parameters compatible with the smallness of neutrino masses; (ii) compute precision lepton observables, which may be sensitive to scales as large as $M\sim 10^3$ TeV; and (iii) establish sharp correlations among those observables. We find that the flavour-conserving Wilson coefficients set an upper bound on the flavour-violating ones. The low-energy limits on $\mu\to e$ and $\tau\to e,\mu$ transitions suppress flavour violation in $Z$ and Higgs decays, as well as electric dipole moments, far beyond the experimental reach. The bounds from the universality of $G_F$ and the invisible $Z$ width are more stringent than present and future limits on $\tau\to e,\mu$ transitions. We also present a general spurion analysis, to compare the seesaw with different models, thus assessing the discriminating potential of the effective approach. Document type : Journal articles https://hal.archives-ouvertes.fr/hal-01949587 Contributor : L2c Aigle <> Submitted on : Monday, December 10, 2018 - 11:27:24 AM Last modification on : Monday, July 1, 2019 - 11:52:44 AM ### Citation Rupert Coy, Michele Frigerio. Effective approach to lepton observables: The seesaw case. Physical Review D, American Physical Society, 2019, 99 (9), pp.095040. ⟨10.1103/PhysRevD.99.095040⟩. ⟨hal-01949587⟩ Record views	2019-07-23 15:33:44	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7812110781669617, "perplexity": 1866.8549055200438}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195529480.89/warc/CC-MAIN-20190723151547-20190723173547-00174.warc.gz"}
https://www.physicsforums.com/threads/most-earthlike-exoplanet-yet-eso-team-23-april.167340/	# Most earthlike exoplanet yet (ESO team 23 April) 1. Apr 24, 2007 ### marcus 2. Apr 24, 2007 ### neutrino Why hasn't this news appeared in more reliable sources of astronomy-related news? And more importantly, where is the press release from ESO? http://www.eso.org/outreach/press-rel/pr-2007/ 3. Apr 24, 2007 ### neutrino 4. Apr 24, 2007 ### baywax That's so cool. Its kind of errie thinking about some red dwarf system, 20 light years away, with a planet whizzing around the sun in 13 days and the possibility that it has oceans and everything oceans come with. Any chance that SETI is directing some of its efforts toward the Gliese 581 system? I'd think that would be a good candidate for eaves dropping. 5. Apr 24, 2007 ### OSalcido http://news.yahoo.com/s/ap/20070424/ap_on_sc/habitable_planet where the heck did he pull this from? how can he say it is "probably full of water"? I mean, Venus and Mars are in the solar system's goldilocks zone.... and there's no appreciable amount of water on either one of them. that's a 66% chance of not having water, right? am i missing something? 6. Apr 25, 2007 ### baywax I just want to thank Marcus for posting this info a day or two ago. Today its all over the front page of the papers. I feel pretty ahead of the times having seen it first here on PF Actually the national news had some good animations showing the probabilities involved with the Gliese 581 system a day after I saw this thread. Good working it Marcus! Last edited: Apr 25, 2007 7. Apr 25, 2007 ### cesiumfrog The front cover of The Australian (newspaper) today includes "WE ARE NOT ALONE Earth-like planet found" which seems a sensationalist leap from the actual (p.3) story content "It's too soon to be certain that GL581c is a rocky planet like Earth or a gas planet like Neptune." Accompanied by the highly detailed (photograph-like) "illustration".. 8. Apr 26, 2007 ### OSalcido Furthermore..... its not even around a sun-like yellow star. It's orbiting a tiny Red Dwarf and it's much closer to it's star. I know some reddwarfs are prone to violent solar flares that would blast this planet into a barren moonscape if it ever got hit by one. 9. Apr 26, 2007 ### Chronos Actually, red dwarf stars are probably the best place to look for earthlike exoplanets. They are extremely abundant and extremely long lived. And planets orbiting them are much easier to detect due to their relatively small mass [which magnifies the gravitational perturbations] and low intrinsic brightness. The part of the article about abundant water is purely speculative, albeit not unreasonable. Water appears to be fairly abundant in our solar system and the universe in general. A nice, temperate planet with decent gravity is the logical place to look for it in liquid form. It may already be detected in the atmosphere of at least one exosolar planet: http://www.astrobio.net/news/article2298.html 10. Apr 26, 2007 ### Staff: Mentor Well, it's PopSci - science for appeal to the general public, so the statements are designed to capture the imagination. I don't think Venus is in the Goldilocks zone since the atmospheric and surface temperatures is rather high, and the atmosphere is not too hospitable to most life with we are familiar. I think Mars is too far out, so the night time temps would be too low, but perhaps under the surface life might have a chance. Whatever water Venus has is mostly tied up as sulfuric acid, so little actual free water might exist. Apparently the atmosphere (which is mostly CO2 (~96%) and N2 (~4%)) has traces of water vapor. Good points, and it is about 0.073 AU or 11 million km from the red dward, yet the astronomers are apparently projecting average surface temperature of 0–40 °C (32–104 °F). The atmosphere would certainly be denser than earth's. http://en.wikipedia.org/wiki/Gliese_581_c I wonder about the magnetic field. 11. Apr 26, 2007 ### dimensionless I would say that it is worthy of reporting. I would also point out that water is very plentiful in this solar system. 12. Apr 26, 2007 ### matt.o Actually, water is one of the most abundant molecules in the Universe. It is found in star formation regions and cold molecular clouds. The 22GHz water maser line is the strongest spectral line observed in the radio Universe. 13. Apr 26, 2007 ### Wallace Slightly tangential question here, but is anyone aware of the state of spectral observations of extra solar planets? I know it's never been done but I think there are efforts to try and do it, but I'm not much of an observer so I'm not sure of the state of the art. It seems logical to me that the presence of $$O_2$$ in the atmosphere is a far clearer indicator of life than the existence of liquid water on the surface, but much harder to infer of course! 14. Apr 27, 2007 ### Garth In terms of relative abundance in the 'cosmic mix', the most abundant element is hydrogen (~75%), then helium {~23%) and the next most abundant element is oxygen (~1%), after which comes carbon (~0.5%). Everything else, including the silicon and iron of our own Earth planet, is to be found in the remaining ~ 0.5%! (% by mass). It is therefore to be expected that water is actually the most abundant molecule in the universe! The trick is finding it in liquid form. Garth Last edited: Apr 27, 2007 15. Apr 27, 2007 ### Garth I agree that the signature of free oxygen is life's 'smoking gun' to look out for. This paper on today's ArXiv may be pertinient. Detailed Models of super-Earths: How well can we infer bulk properties?. Fancy a swim anybody? Garth Last edited: Apr 27, 2007 16. Apr 27, 2007 ### baywax Hello Wallace... I dug this up for you. Its just an abstract but goes into some detail with regard to your question. From: http://www.cosis.net/abstracts/COSPAR04/02478/COSPAR04-A-02478.pdf You might have to email the M. Ollivier to get the main body of the text. Also a thought to consider is that life may be starting in the form of anaerobic bacteria on the planet in question and would therefore not require O2. Last edited: Apr 27, 2007 17. Apr 27, 2007 ### baywax Some what related....... from Science Actualitiés, Friday, April, 27 (French) http://www.cite-sciences.fr/francai...u/question_actu.php?id_article=3303&langue=an Sorry, just another quote from the same page. Last edited: Apr 27, 2007 18. Apr 30, 2007 ### marcus The Udry et al technical journal article for this just became (finally!) available on the arxiv. http://arxiv.org/abs/0704.3841 The HARPS search for southern extra-solar planets XI. Super-Earths (5 & 8 M_Earth) in a 3-planet system S. Udry (1), X. Bonfils (2), X. Delfosse (3), T. Forveille (3), M. Mayor (1), C. Perrier (3), F. Bouchy (4), C. Lovis (1), F. Pepe (1), D. Queloz (1), J.-L. Bertaux (5) (Submitted on 29 Apr 2007) Revised version resubmitted to A&A Letters, 5 pages, 4 figures "This Letter reports on the detection of two super-Earth planets in the Gl581 system, already known to harbour a hot Neptune. One of the planets has a mass of 5 M_Earth and resides at the 'warm' edge of the habitable zone of the star. It is thus the known exoplanet which most resembles our own Earth. The other planet has a 7.7 M_Earth mass and orbits at 0.25 AU from the star, close to the 'cold' edge of the habitable zone. These two new light planets around an M3 dwarf further confirm the formerly tentative statistical trend for i) many more very low-mass planets being found around M dwarfs than around solar-type stars and ii) low-mass planets outnumbering Jovian planets around M dwarfs." Authors' institutions: ((1) Observatoire de Geneve, Université de Geneve, Switzerland, (2) Centro de Astronomia e Astrofisica da Universidade de Lisboa, Portugal, (3) Laboratoire d'Astrophysique, Observatoire de Grenoble, Universite J. Fourier, France, (4) Institut d'Astrophysique de Paris, France, (5) Service d'Aéronomie du CNRS/IPSL, Verrières-le-Buisson, France) To me, the farther out one at the "cold" edge of hab zone sounds like more fun than the one we were discussing earlier. 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https://www.physicsforums.com/threads/crazy-resume-descriptions.317315/	Crazy resume descriptions! 1. May 31, 2009 Pengwuino I googled some examples for resumes a few days ago and I noticed something quite silly. A lot of them had in their descriptions or summarys or whatever, phrases such as "hard-working" "highly competent", "very diligent", "goal oriented", "driven" and all these sort of catch phrases. Now we all know a resume is where you "sell" yourself to your employer but my god, do these stupid little phases actually have any effect on employers? DISCUSS!!! 2. May 31, 2009 rootX I learned that it's usually: "hard-working" demonstrated with ... (maybe past work experience) And under, Work experience: - a point for proving why you are hard worker. It doesn't look silly to me because you are just trying to highlight the skills you ve learned from work, volunteer etc. 3. May 31, 2009 Cyrus I'm working on a project. We are hiring some undergrads, and I must say the few that actually gave a resume were full of padding. They included such things as: Camp Counselor (I dont care), Current memeber of xyz techincal institudes (again, I dont care), Edit for school paper (I don't care). Since such little information of actual importance was given, the only thing I looked at was their GPAs. I understand its hard for the younger students to fill a resume since the are so 'green' when it comes to doing actual things, but sheesh. I'm not looking for a camp counselor or someone that writes for the school post. Don't pad your resume folks, its easy to spot a mile away. Keep it short, keep it simple. Tell me the skills you have that are relevant. 4. May 31, 2009 Pengwuino :rofl: :rofl: That's so awesome. I love this guy. ps. just kidding pps. not about the awesomeness though 5. May 31, 2009 Cyrus I'm dead serious. Joining the society of xyz engineers requires filling up an application, paying a yearly subscription and getting magazines in the mail each month. Why do I need to know that information? 6. May 31, 2009 Pengwuino I wonder hwy it's all emphasized in college admissions? I think that's most people's first serious attempt to impress an "employer". 7. May 31, 2009 Ivan Seeking Staff Emeritus Resume writing is an art. I took a resume writing class when I finished college and found it to be remarkably helpful, but there is no one best approach. IIRC, the rule of thumb is that you have about 10-15 seconds to make an impression. So it is important to be specific about what you have to offer and get it out there fast in a highly compact [efficient] format. Using generalized language may detract from more important details to be included. For example, rather than saying "hard-working", try to demonstrate that by listing one's accomplishments. Rather than saying "goal oriented", I would state the goals one actually has and how it applies to the indicated career path. Driven? To do what? I don't remember the exact details anymore, but I recall that before taking the class, my resume took one page. After taking the class, it took four pages or so that had to be condensed into one [the 10 second rule]. It was the difference between saying what I think someone wants to hear, and really thinking about what I had to offer. Last edited: May 31, 2009 8. May 31, 2009 Pengwuino I like that thinking. Reading some of these really padded resume examples was kind of pathetic. With all the fluff in the introduction (which took a good 20 seconds to read!), once you're at the meat of the resume, you're already convinced this person's all show. I suppose the logic follows from how people say interviews work, that is you're basically hired within the first minute or so of the interview most of the time (in a job where a LOT of people are interviewed). 9. May 31, 2009 swat4life You want some insider's advice? This is coming from a former classmate in HR who recruited for McKinsey and Company + my own experience going through dozens of resumes looking to fill positions with my company. First - MAKE YOUR ELECTRONIC RESUMES VERY, VERY KEYWORD-FRIENDLY. Most days, recruiters type in specific keywords when they are looking for resumes; you could be the brightest bunny in the basket, but if those words aren't on your CV/Resume, no one will see it. Find out what the buzz words are for your industry and USE THOSE WORDS. To not do so, is professional suicide. SECOND -I usually give a resume an immediate 30 second once over. If I don't immediately see something that strikes me, well I move on. The biggest mistake people make is using a resume to describe "actions" vs. "results". What do I mean? Let's say I am looking for a business analyst. He/she will run macros in excel, research business trends, do powerpoint presentations (or whatever). Most people submit a resume that looks something like: -ran macros excel models for xyz company -completed powerpoint presentation to display results OK. What's the problem here? There will be probably 50 other resumes that said the same thing. How does one distinguish? Also, I didn't get you to do all those things just because clients pay me every time I log a completed excel spreadsheet or powerpoint presentation. I did it to achieve....RESULTS! The difference between the "just fine" resume above and one that jumps out at me, might look something like: -ran macros excel models for xyz company. Created models which reduced project completion time by 10% and saved company $10,000. -research business trends. Created business trend templates for this kind of research which increased project efficiency by 5% over the course of 6 months (or whatever). -completed powerpoint presentation to display results As a result of work, project manager exceeded clients' expectation and secured 5 more clients which resulted in additional sales revenues of$500,000 for the company (or whatever). Can you see the difference?! The second person/resume has shown me how he/she has had a direct impact on THE BOTTOM LINE, on the M-O-N-E-Y, COMPANY EFFICIENCY, His/her employers major in goals. The problem is that most employees aren't aware enough (or bosses for that matter)....they don't spend as much time as they should getting to understand how their actions have directly impacted the company. Almost every employee has made some impact -however big or small - on the bottom line of the company and they should do their best to demonstrate that. Once you do that, you become less of a financial risk, you make it easier for the person making the hiring decision to go with you and in general you really stand out..... Edit: If you have a physics/science degree and are hunting for non science jobs, you should definitely sell your quant. skills. It's easy for an employer to envision a physics student being good at quantitative stuff. However, one must SPELL IT OUT FOR PEOPLE. Also, it pays (and perhaps literally) to say something about how you wanted to study physics to learn the practical application to addressing REAL WORLD problems. For instance, in a cover letter you might say that, you learned that 1/3rd of the usa's economy consists of products directly derived from the practical application of quantum physics - MRIs, Transistors, etc. You wanted to learn how to apply scientific theory to the successful solving of real world problems (or something like that) Last edited: May 31, 2009 10. May 31, 2009 Ivan Seeking Staff Emeritus A note to our graduating physics students: I learned something rather remarkable after college: In order to "sell" my physics degree, I had to explain what physics students learn. That made ALL the difference in the world. As it turns out, even many engineers don't know what a physics grad brings to the table. So I tried to sprinkle specifics into the language that helped to connect my education to the jobs being sought. One day I had lunch with a former physics professor and was talking about what I had learned. It happened that she was in charge of the graduating students and was so surprised by the story that I was asked to come back to college and give a talk to the graduating class. 11. May 31, 2009 TheStatutoryApe When I took my resume into a temp agency the woman I spoke with thought I had it professionally done. I hadn't even thought it was that good. My work isn't easily quantifiable so I had to come up with creative ways of describing it in the context of a resume. I figured if I can describe my work in an elegant and succinct manner than they may at least think I am intelligent enough to do it. 12. May 31, 2009 Pengwuino Easy! I've been having to explain what the hell I've been doing the last 5 years for... the last 5 years! 13. May 31, 2009 Cyrus This is a very good post, and brought things to mind that I didn't think about until you mentioned it. Thank you. Last edited: May 31, 2009 14. May 31, 2009 TheStatutoryApe So what do you do if you have no way of quantifying the results of your work? 15. May 31, 2009 swat4life Again, speaking as an employer, NEVER underestimate the power of good personal brand management. Look at television - fluff works old friend.... To elucidate, I am not saying substance doesn't matter. What I am saying is that a pretty package is more likely to get someone to open it to see the real value inside. Confidence, ambition and passion jumps out on the page (even when I look at it for 30 seconds). And let's face it, employers want workers who really, really, really want the job and are willing to do what is necessary to convince you why they are the right person. Focus on results - as stated previously, results sell. Esp. numbers (percentages, dollar signs, etc). 16. May 31, 2009 swat4life There's always a way. It doesn't have to be down to the 3rd decimal place bar (i.e. in terms of quantifying it). For instance, let's say you were tasked with created excel spreadsheets. This job normally took about 1 hour. Well, you decided to save time and created a template that reduced the time to say 30 minutes. Result? You created a 100% efficiency increase for that task.... These kind of examples abound. This is why **smart companies codify and formalize any task that is repeated more than 3 times. Because once you have a standard process for doing something on paper, it goes from being employee-specific knowledge (which leaves when they leave) to intellectual capital for the company - that can be valued by the say as an asset and even put on the balance sheet. Once you came up with that simple way of doing that job more efficiently, and the company wrote it down. If there are 100 other people across 10 offices doing the same thing, I can send that puppy to each employee via an email attachment and after a little bit of training...voila....I've multiplied efficiency by 100x... hope this helps Last edited: May 31, 2009 17. Jun 1, 2009 TheStatutoryApe Actually I seriously mean no way to quantify it. I work as a security guard. The only number I could come up with (I had already been told numbers are important) for my resume was the number of pages I write in my report. I could try the number of employees I have trained but I doubt I could remember. Also many of them were fired (most of our guards get fired) so I can't really show any great success in my training abilities. No records are kept of crime statistics so I can't really show results in regard to prevention. I very rarely come into contact with the sort of people we take as clients so have never actually gotten us any. I've proposed a few ideas in regards to things that we can do to improve our operations but my bosses are stuborn and much prefer to have things done their way. Essentially all I have is that I know my job well and do it. 18. Jun 1, 2009 Cyrus You saved x dollars for your client because you prevented a theft from happening. 19. Jun 1, 2009 Pengwuino and X lives since they would have all been packing... 20. Jun 1, 2009 Jimmy Snyder When I was starting out, I could do anything. Now I can do something. That's what goes on my resume.	2018-03-21 08:09:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.30506429076194763, "perplexity": 1633.3032428930712}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-13/segments/1521257647584.56/warc/CC-MAIN-20180321063114-20180321083114-00206.warc.gz"}
http://jurnalkanun.dbp.my/apset-question-ybrg/quintic-polynomial-example-6db5f1	We now consider the question of solv-ing for the roots of … Conservapedia - Recent changes [en] Parksc onjectured the correct formula for the number of degree d rational curves in a Calabi-Yau quintic. for the polynomial shown below, find f(-1). Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. (The "-nomial" part might come from the Latin for "named", but this isn't certain.) Anatomy of a polynomial function In physics and chemistry particularly, special sets of named polynomial functions like Legendre , Laguerre and Hermite polynomials (thank goodness for the French!) ... example of a quintic with one real root.) In other words, it must be possible to write the expression without division. A practical method to determine such positivity and nonnegativity is presented in Section 6. lois theory is important because it associates to each polynomial a group (called its Galois group) that encodes this arithmetic structure. I think after quintic it becomes cumbersome to name them (since the prefixes become increasingly more complex). The interpretation of this theorem is that, based only on the sum of the three real roots Finding the constant . Below is a list of quintic polynomial words - that is, words related to quintic polynomial. Why is the Quintic Unsolvable? For example, from string-theoretic considerations, Candelas, de la Ossa, Green, and Parkes conjectured the correct formula for the number of degree d rational curves in a Calabi-Yau quintic. 153 restrictions on degree). Third-degree polynomial functions with three variables, for example, produce smooth but twisty surfaces embedded in three dimensions. The example shown below is: Galois Theory and the Insolvability of the Quintic Equation Daniel Franz 1. Write a polynomial of the lowest degree with real coefficients and with zeros 6-3i (multiplicity 1) and 0 ( multiplicity 5) algebra2. process: calculate its coefficients and T minimizing. Files are available under licenses specified on their description page. A polynomial function of degree 5 (a quintic) has the general form: y = px 5 + qx 4 + rx 3 + sx 2 + tx + u. quintic polynomial. Fred Akalin September 26, 2016 (This was discussed on r/math and Hacker News.). Quintic Polynomial-Type A. 18,799 results, page 17 math. Human translations with examples: 展開, 多項式, 多項式時間, 五次方程式, 多項回帰式. The highest power of the variables in a polynomial is termed as its degree. For example, one of the solutions for the polynomial x5 5 = 0 is 5 p 5. Overview. Meaning of Quintic with illustrations and photos. state the number of positive real zeros, negative real zeros, and imaginary zeros for g(x) = 9x^3 - 7x^2 +10x - 4 . Quintic polynomial with only the 5 th degree and constant terms. From the graph we see that when x = 0, y = −1. And again, by knowing where to look on these figures, mathematicians can learn more about their underlying polynomial structure. 26 Nov 2015 (10) Thus, I feel like "degree seven" or "seventh degree" polynomial is more appropriate. Solution: In general form, we can write it as $$1{x^{ - 1}} + 0$$. The general form of a quintic function is given below: .... quintic equation A polynomial equation in which the highest power of the variable is five. close all. f(x)=3x^4-x^3+4x-2 . This page was last edited on 17 September 2020, at 16:15. Pronunciation of Quintic and it's etymology. All structured data from the file and property namespaces is available under the Creative Commons CC0 License; all unstructured text is available under the Creative Commons Attribution-ShareAlike License; additional terms may apply. Quintic Function Polynomials do cover a lot of portion in maths. Polynomial Names. Generation of lateral movement High speed trajectories at high speed, d(t) and s(t) can be chosen independently. The statement means that we cannot provide a single radical formula for every general quintic polynomial as we can for the quadratic polynomials. We will show that the Galois group is S 5 and thus by our theorem from class (Thm. For example, the polynomial + −, which can also be written as + −, has three terms. Solution . In mathematical language, by quintic function, one means to refer a polynomial of degree 5. Theorem 1 provides an easy criterion for the solvability of a general quintic polynomial (see the examples below). It only takes a minute to sign up. check it against collision. 10.20 in the notes) the polynomial fis not solvable by radicals! An icosahedral iterative scheme for solving the quintic. No symmetry. It turns out that when we draw the graph corresponding to a linear polynomial, we will get a straight line – hence the name linear. SOLVING THE QUINTIC BY ITERATION Fig. Three points of inflection. Therefore, to show the statement is true, it is Math. 1. 1. Contextual translation of "quintic polynomial" into Japanese. Later, Schoof and Washington [SW] showed that these units were fundamental units. General form of a quintic. Examples? These algorithms exhibit much of the flexibility of smooth dynamical systems (in fact they are discrete approximations to the Newton vector field). Clearly, the degree of this polynomial is not one, it is not a linear polynomial. Introduction Polynomial equations and their solutions have long fascinated math-ematicians. For example, when the starting point and the … Four extrema. Find the local maximum and minimum values and saddle point(s) of the function. quintic polynomials, we can nd a radical solution. 1. Substituting these values in our quintic gives u = −1. Note:- Let F be a field. This type of quintic has the following characteristics: One, two, three, four or five roots. quintic polynomial: A 5th degree polynomial.The lowest order polynomial, for which there is no general formula for finding the roots.. How to find the equation of a quintic polynomial from its graph 26 Mar 2016 (7) Squaring the circle - a reader's approach 12 Feb 2016 (1) Butterfly map of the world 08 Dec 2015 Reuleaux triangles 08 Dec 2015 Is a 1x1 matrix a scalar? hw31.m . fprintf(‘Enter coefficients for a quintic polynomial of the form:\n\ty = ax^5 + bx^4 + cx^3 + dx^2 + ex + f\n’); quintic polynomial for positive reals in Section 5. denotes real monic polynomials (polynomials with real coefficients in which quintic the leading coefficient is 1). clc. The "poly-" prefix in "polynomial" means "many", from the Greek language. are the solutions to … A quintic polynomial Consider the polynomial f(x) = x5 6x+ 3 2Q[x]. Related words - Quintic synonyms, antonyms, hypernyms and hyponyms. Solving Quintic EquationsOverviewBy the nineteenth century, mathematicians had long been interested in solving equations called polynomials. Example sentences containing Quintic We'll find the easiest value first, the constant u. Let KˆC denote the splitting eld and Gthe Galois group. Solved Example: Example 1: Is $$\frac{1}{x}$$ a linear polynomial? However, Paolo Ruffini (1765-1822) and Niels Abel (1802-1829) proved that some polynomials could not be solved by previously known methods. How is an exponential function different from a polynomial function? Hi there! Definition of Quintic in the Fine Dictionary. So I guess a hypothetical equation that involves both variables ##x## and ##y##, like ##y^2+x^3+2x^2+x+5=0## should not to be called a cubic equation just because the highest monomial degree is 3 and because the name cubic equation is reserved for … clear. Partly in response, Evariste Galois (1811-1832) developed a new way of analyzing and working with these … ... For example, we write e3 = ¯e3 25d4 2, where ¯e3 is a polynomial function of the coeﬃcients a,b,c,d and e, and we use ¯e3 instead of e3 in our theorems. A quintic polynomial through the same points and the same time interval will always lead to a smaller cost. cost function: g(T)=T, h(d1)=d1^2. Quintic Polynomial. Note on a polynomial of Emma Lehmer Henri Darmon September 9, 2007 1 Abstract In [Leh], Emma Lehmer constructed a parametric family of units in real quintic ﬁelds of prime conductor p = t4 +5t3 +15t2 +25t+25, as translates of Gaussian periods. 10.20 in the notes) the polynomial fis ... Give an example of an irreducible cubic polynomial in Q[x] that has Galois group A 3. Inflection points and extrema are all distinct. If you are really interested in the prefixes look here under "ordinal". A nonzero, nonunit element of D[x] that is not irreducible over D is called reducible over D. Example 1:- The polynomial f ( x) 2 x 2 4 is irreducible over Q but reducible over Z and is irreducible over R but reducible over C. Example 2:- The polynomial x 2 1 is irreducible over Z 3 but reducible over Z 5 . A Polynomial can be expressed in terms that only have positive integer exponents and the operations of addition, subtraction, and multiplication. It's easiest to understand what makes something a polynomial equation by looking at examples and non examples as shown below. What is an example of a quintic polynomial function that has exactly four zeros? Calculus. The first term has a degree of 5 (the sum of the powers 2 and 3), the second term has … B. Quintic Polynomial Interpolation In the case where the trajectory is more stringent and the constraint condition is increased, the cubic polynomial interpolation can’t satisfy the requirement, and the high order polynomial is used for interpolation. We will show that the Galois group is S 5 and thus by our theorem from class (Thm. For example, ##y=ax^3+bx^2+cx+d## is called the cubic function and when ##y=0##, it becomes the cubic equation ##ax^3+bx^2+cx+d=0##. A quintic polynomial Consider the polynomial f(x) = x5 6x+ 3 2Q[x]. 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Three terms = x5 6x+ 3 2Q [ x ] ) Contextual translation of quintic polynomial we..., it is not a linear polynomial easiest to understand what makes something a polynomial is a...	2021-10-25 13:19:21	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7462418079376221, "perplexity": 1026.7570749535457}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323587711.69/warc/CC-MAIN-20211025123123-20211025153123-00693.warc.gz"}
https://math.stackexchange.com/questions/2653777/extending-a-one-to-one-correspondence-between-sets	# Extending a one to one correspondence between sets In proving the theorem: Every infinite set is equivalent to one of its proper subsets, I am confused about the following: We consider an infinite set $M$, which always contains a countable subset, which is denoted $A := \{a_1, a_2, \dots \}$. We may partition $A$ into two countable subsets: $$A_1 := \{a_1, a_3, a_5 \dots \}, \qquad A_2 := \{a_2, a_4, a_6 \dots \}$$ and we have a one-to-one correspondence between $A$ and $A_1$ given by $a_n \to a_{2n-1}$. We can then extend this correspondence to a one-to-one correspondence between the two sets: $$A \cup (M - A) = M, \qquad A_1 \cup (M-A) = M-A_2$$ by simply assigning $x$ itself to each $x \in M-A$. I don't quite understand this extension, how are we allowed to do this? This is a proof taken from Introductory Real Analysis by Kolmogorov and Fomin. • What do you mean by "being allowed"? It is not a matter of permissions. – Andrés E. Caicedo Feb 17 '18 at 3:02 • @AndrésE.Caicedo as in what is the property that permits us to perform this extension – dimebucker Feb 17 '18 at 3:04 You have a bijection between the elements of $A$ and $A_{1}$. That is, you have a map $f\colon A\longrightarrow A_{1}$ which is bijective. You define another map, $g\colon A\cup (M-A) \longrightarrow A_{1}\cup (M-A)$, such that $$g(x)= f(x) \quad \text{if} \quad x\in A,\quad \text{and} \quad g(x)=x \quad \text{if}\quad x\in M-A.$$ Note that $A\cap (M-A)=\varnothing$, so $g$ is well-defined and $g$ is a bijection.	2019-09-22 14:03:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9441020488739014, "perplexity": 123.71811479995655}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514575515.93/warc/CC-MAIN-20190922135356-20190922161356-00363.warc.gz"}
http://www.physicsforums.com/showthread.php?p=4263254	# What is {ℝ}? by kokolovehuh Tags: P: 23 Hi, Someone I know tried to convey me the meaning of {$ℝ$}, stating it represents a set of real numbers. But using notation, {$ℝ$}, is implying that the real space is (improperly) contained in a set, and I don't think this makes any logical sense. On the other hand, we can say {$x \in S \| \forall S \in ℝ$}, etc....or simply $x \in ℝ$. Another way of thinking about this is instead of putting your foot in a sock, you are putting the sock into your foot and it's disturbing. Am I right or wrong? Thanks Sci Advisor HW Helper PF Gold P: 3,172 It is perfectly valid to put sets inside of other sets. For example, we may consider the set of all subsets of the complex numbers ##\mathbb{C}##. This is called the power set of ##\mathbb{C}## and is sometimes given the notation ##\mathcal{P}(\mathbb{C})## or ##2^\mathbb{C}##. Any subset of ##\mathbb{C}## is an element of ##\mathcal{P}(\mathbb{C})##. For example, we have ##\mathbb{R} \subset \mathbb{C}## and ##\mathbb{R} \in \mathcal{P}(\mathbb{C})##. We can form subsets of ##\mathcal{P}(\mathbb{C})## in the usual way, by putting elements of ##\mathcal{P}(\mathbb{C})## into a set. Thus ##\{\mathbb{Z}, \mathbb{Q}, \mathbb{R}\} \subset \mathcal{P}(\mathbb{C})##, and a special case is a subset containing only one set, such as your example: ##\{\mathbb{R}\} \subset \mathcal{P}(\mathbb{C})##. P: 23 Thanks bjunniii, you have a good point. However, let me rephrase my doubt. We want to use a notation to represent a set of all real number, say $X$. It is immediately apparent that $x \in X \subseteq ℝ$ for some real number $x$. In this case, we are not not considering any stronger set, for instance, $P(ℂ)$ as you mentioned. Now having limited ourselves to real space, it is rather redundant to say the set is represented as {$ℝ$} because since $ℝ$ is not a proper subspace in this case. This is the reason why I said "this does not make any logical sense"; I am ridiculed by those curly brackets! Instead, we could simply write $X \in ℝ$ that give a much more direct and sensible idea of what space we are talking about. Do you agree? Mentor P: 18,040 What is {ℝ}? The set $\{\mathbb{R}\}$ is a set which contains only one element. Its element is $\mathbb{R}$. There is no reason why such a construction would not be allowed. It is true, however, that sets like $\{\mathbb{R}\}$ don't play a big role in mathematics. P: 820 Quote by kokolovehuh But using notation, {$ℝ$}, is implying that the real space is (improperly) contained in a set, and I don't think this makes any logical sense. Unless you state otherwise, all set theory is done in ZFC, and one of its axioms is the axiom of pairing. It asserts: given any two sets A and B, there exists set C with exactly those two elements, i.e. C = {A, B}. So it does logically exist (in this case A = B = ℝ). P: 23 @micromass, @pwsnafu, I see what you are saying. Overall, you have convinced me {$ℝ$} is possible. But, The notation with curly bracket is directly defining the single element in this set as the real space which is essentially another set. I was simply saying there is no necessity to put real space as a subset of a set in the first place; there are no other disjoint elements. Quote by kokolovehuh Hi, Someone I know tried to convey me the meaning of {$ℝ$}, stating it represents a set of real numbers	2014-07-30 07:06:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8740031719207764, "perplexity": 410.40159445886667}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510268734.38/warc/CC-MAIN-20140728011748-00437-ip-10-146-231-18.ec2.internal.warc.gz"}
https://www.baeldung.com/cs/prime-number-algorithms	## 1. Overview Prime numbers have always been an interesting topic to dive into. However, no one has been able to find a clean and finite formula to generate them. Therefore, mathematicians have relied on algorithms and computational power to do that. Some of these algorithms can be time-consuming while others can be faster. In this tutorial, we’ll go over some of the well-known algorithms to find prime numbers. We’ll start with the most ancient one and end with the most recent one. Most algorithms for finding prime numbers use a method called prime sieves. Generating prime numbers is different from determining if a given number is a prime or not. For that, we can use a primality test such as Fermat primality test or Miller-Rabin method. Here, we only focus on algorithms that find or enumerate prime numbers. ## 2. Sieve of Eratosthenes Sieve of Eratosthenes is one of the oldest and easiest methods for finding prime numbers up to a given number. It is based on marking as composite all the multiples of a prime. To do so, it starts with as the first prime number and marks all of its multiples (). Then, it marks the next unmarked number () as prime and crosses out all its multiples (). It does the same for all the other numbers up to : However, as we can see, some numbers get crossed several times. In order to avoid it, for each prime , we can start from to mark off its multiples. The reason is that once we get to a prime in the process, all its multiples smaller than have already been crossed out. For example, let’s imagine that we get to . Then, we can see that and have already been marked off by and . As a result, we can begin with . We can write the algorithm in the form of pseudocode as follows: In order to calculate the complexity of this algorithm, we should consider the outer loop and the inner loop. It’s easy to see that the former’s complexity is . However, the latter is a little tricky. Since we enter the loop when is prime, we’ll repeat the inner operation number of times, with being the current prime number. As a result, we’ll have: In their book (theory of numbers), Hardy and Wright show that . Therefore, the time complexity of the sieve of Eratosthenes will be . ## 3. Sieve of Sundaram This method follows the same operation of crossing out the composite numbers as the sieve of Eratosthenes. However, it does that with a different formula. Given and less than , first we cross out all the numbers of the form less than . After that, we double the remaining numbers and add . This will give us all the prime numbers less than . However, it won’t produce the only even prime number (). Here’s the pseudocode for this algorithm: We should keep in mind that with as input, the output is the primes up to . So, we divide the input by half, in the beginning, to get the primes up to . We can calculate the complexity of this algorithm by considering the outer loop, which runs for times, and the inner loop, which runs for less than times. Therefore, we’ll have: This looks like a lot similar to the complexity we had for the sieve of Eratosthenes. However, there’s a difference in the values can take compared to the values of in the sieve of Eratosthenes. While could take only the prime numbers, can take all the numbers between and . As a result, we’ll have a larger sum. Using the direct comparison test for this harmonic series, we can conclude that: As a result, the time complexity for this algorithm will be . ## 4. Sieve of Atkin Sieve of Atkin speeds up (asymptotically) the process of generating prime numbers. However, it is more complicated than the others. First, the algorithm creates a sieve of prime numbers smaller than 60 except for . Then, it divides the sieve into separate subsets. After that, using each subset, it marks off the numbers that are solutions to some particular quadratic equation and that have the same modulo-sixty remainder as that particular subset. In the end, it eliminates the multiples of square numbers and returns along with the remaining ones. The result is the set of prime numbers smaller than . We can express the process of the sieve of Atkin using pseudocode: It is easy to see that the first three loops in the sieve of Atkin require operations. To conclude that the last loop also runs in time, we should pay attention to the condition that will end the loop. Since when and , which is a multiple of a square number, are greater than , we get out of the loops, they both run in time. As a result, the asymptotic running time for this algorithm is . Comparing this running time with the previous ones shows that the sieve of Atkin is the fastest algorithm for generating prime numbers. However, as we mentioned, it requires a more complicated implementation. In addition, due to its complexity, this might not even be the fastest algorithm when our input number is small but if we consider the asymptotic running time, it is faster than others. ## 5. Conclusion In this article, we reviewed some of the fast algorithms that we can use to generate prime numbers up to a given number.	2023-03-21 13:43:33	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8750276565551758, "perplexity": 248.87121246937858}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296943698.79/warc/CC-MAIN-20230321131205-20230321161205-00352.warc.gz"}
http://www.fabiocozzolino.eu/change-ios-status-bar-color-xamarin-forms/	Xamarin.Forms is a great way to abstract the local implementation. But, you know, sometimes you need to perform some small adjustments based on the underlying OS. Is this the case of the iOS Status Bar style. Xamarin.Forms implements, in the NavigationRenderer, a way to automatically set the style (dark/default or light) by checking the Luminosity value of the BarTextColor property. Check the following code: if (statusBarColorMode == StatusBarTextColorMode.DoNotAdjust \|\| barTextColor.Luminosity <= 0.5) { // Use dark text color for status bar UIApplication.SharedApplication.StatusBarStyle = UIStatusBarStyle.Default; } else { // Use light text color for status bar UIApplication.SharedApplication.StatusBarStyle = UIStatusBarStyle.LightContent; } For full code, check the NavigationRenderer source on github Xamarin.Forms project. Now, if in your code you set the BarTextColor to black, then the status bar style will be the Default (dark). Otherwise, the status bar style will be light (white). So, if we want to set the BarTextColor, in a classic Xamarin.Forms App class constructor, we’ll have something like this: public App() { InitializeComponent(); { BarBackgroundColor = Color.FromHex("#043596"), BarTextColor = Color.White }; } But, this is not enough. In iOS, we have two ways to set the StatusBarStyle: • at application level • at view controller level The default behavior is at view controller level. Since Xamarin.Forms sets the StatusBarStyle at application level, we need to set the related property in the Info.plist. So, open the info.plist file, go in the Source tab and add a new entry with this values: Name = UIViewControllerBasedStatusBarAppearance Type – boolean Value – No And we have done! Now start the app and check the result: Nice and simple!	2019-05-19 21:19:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22470447421073914, "perplexity": 5957.078464123487}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232255165.2/warc/CC-MAIN-20190519201521-20190519223521-00290.warc.gz"}
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For example: I prepare my files, open the Magit status buffer, stage the file via s, commit the file ... 6 views ### sql-interactive-mode get product name for the buffer I have 2 connections to different servers through sql interactive mode: (setq sql-connection-alist '((mysql_server (sql-product 'mysql) (sql-server "server1") (sql-user &... 44 views ### Move the fringe indicators close to the window text When I use olivetti-mode I'd like to have grey margins and the fringe indicators close to the window text, like in this picture: The problem is that when the fringes are wide the indicators end up ... 46 views ### emacsclient closes immediately on "ERROR: Beginning of buffer" I have an app where I call emacsclient to do some editing on an org file. When it calls it, the frame flashes for an instant, then closes immediately. On stderr, I get the following message: ERROR:... 35 views ### How to control the number of characters for the SHA1 displayed in Magit's log view? 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This is in Linux Emacs built from master of 2022-... 30 views ### sort: How to specify the order in which multiple sorts are performed In the following example, there are two criteria for sorting -- orders with an inactive status should appear last in the last, and all orders should be organized by order number from smallest to ... 85 views ### How to disable Emacs bidi? I want to disable Emacs from ever displaying text right-to-left (i.e. ignore Unicode control sequences to change text orientation). How to do it? 36 views ### WARNING: the following functions are not known to be defined: x-display-pixel-width, x-display-pixel-height, x-show-tip. emacs-nox I'm using Emacs in terminal mode. I am not getting any error, what I am getting is just a warning, the following functions are not known to be defined: x-display-pixel-width, x-display-pixel-height, x-... 16 views ### Buffer display issue with unicode characters on WSL2 I have a file with a single Unicode character in it: 🕯 (Unicode character CANDLE, codepoint 128367, #o372557, #x1f56f). When I open this file via emacs -Q and hit C-l, the display buffer has issues. ... 23 views 19 views ### custom-set-faces common color variables [duplicate] I want to fine-tune my syntax highlighting, and have created a separate colors.el file that gets loaded around the beginning of init.el. I often use the same color for multiple entries, and it gets ... 10 views ### Can an agenda deadline warning be temporarily dismissed w/o editing org-deadline-warning-days? I have many repeating tasks with deadline warnings. Sometimes I just want to see the reminder and dismiss it until the next day, i.e. remove it from that day's agenda. Currently, I manually subtract 1 ... 30 views ### No colored png icons in treemacs (Doom Emacs) After reinstalling Doom Emacs several times I can’t find any way to make treemacs use the colored png icons of it’s own default theme. I always get monochrome icons. How do I solve this? 27 views ### How to configure Emacs to list the most recently used (MRU) completion options first? I find autocomplete features in Emacs very helpful. Every time I type a partial word, it infers what the word I have in mind based on some history it has seen so far. I am not exactly sure how. I've ... 14 views ### How to setup GTK emacs to open files in its current instance instead of opening a popup window? Spent a lot of time on this and tried various solutions on the internet. Please help. Using the following emacs version. Want to run it in GUI mode on an Ubuntu 20.04 server, same issue happens on ... 30 views ### Is there a way to get the name of the current file being opened? I'm trying to conditionally load configuration in my init.el. How can I get the filename (preferable the extension) of the current file being opened? For example, if I run emacs foo.rs, I'd like to ... 38 views ### How to avoid this question: "Non-existent agenda file. [R]emove from list or [A]bort?" When I create an .org file (C-x C-f, find-file), the following appears in the mini-buffer: "Non-existent agenda file. [R]emove from list or [A]bort?". The question is not asked when creating ... 23 views ### Is time-stamp broken? I use GNU Emacs 27.1 (build 1, x86_64-pc-linux-gnu, GTK+ Version 3.24.23) of 2020-11-09 (bundled with Ubuntu 21.10) with this init file: % cat ~/.emacs.d/init-1.el (add-hook 'before-save-hook 'time-... 33 views ### Is there a way to add display annotation to the buffer text so that it's searchable? Suppose I have a text data file with various data values which the computer can parse easily, but it's not easy to read for a human. In order to make the file easier to read I'd add display only ... 8 views ### How do I make org-indent-mode put me at the right indent when I make a new line? This is the current situation: Heading Sub-heading * Sub-headings content1<return> \| I entered that text and now want to add a second line to the content by pressing return. My ... 20 views ### org-agenda not displaying agendaorg-agenda-skip: Wrong number of arguments: (0 . 1), 2 I am having trouble using org-agenda with org-journal. Essentially invoking org-agenda and trying to list all todos does not work as intended. I have included the debug trace and associated ... 30 views ### Org-roam raises "Selecting deleted buffer" error after formatting laptop I recently formatted my laptop's hard drive and reinstalled my OS (Manjaro in dual boot with Windows 11). I restored my entire emacs configuration on Manjaro and everything seems to work properly, ... 15 views ### Terminal layout is broken when I turn on lsp I use lsp-mode when I write code for Rust, C++ with this init.el settings: (use-package lsp-mode ;; reformat code and add missing (or remove old) imports :hook ((rust-mode c++-mode) . lsp) :bind ... 40 views ### set variable only if its name is already interned Is there a built-in function/macro of Emacs that sets a variable only if it is already defined? I mean something like that: (cl-defmacro set-if-interned (symbol-name value &optional (use-obarray ... 9 views ### (set-eshell-alias!) for mv causes error: missing destination file or directory In doom emacs, I'm setting some aliases in my config as follows: (set-eshell-alias! "ls" "exa --icons \$*" "doomsync" "~/.emacs.d/bin/doom sync" ... 15 30 50 per page	2022-01-29 01:24:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7779607772827148, "perplexity": 8028.206373439347}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-05/segments/1642320299894.32/warc/CC-MAIN-20220129002459-20220129032459-00390.warc.gz"}
http://www.varsitytutors.com/gre_subject_test_math-help/domain-range	# GRE Subject Test: Math : Domain & Range ## Example Questions ← Previous 1 ### Example Question #1 : Calculus What is the domain of ? Explanation: Step 1: We need to determine what kind of function we have here. We have a rational function. Step 2: Since we have a rational function, the denominator cannot be equal to . We will equal the denominator to and find the values of that make the bottom zero. Square both sides: Add to both sides of the equal sign: Simplify: Take the square root of both sides However, we said earlier that these two solutions cannot be values of x, so we must change the sign: Step 3: We need to write the solution in interval notation form. The smallest number we can plug in for is , and the biggest is cannot be and , but it can be anything else. So, we should have three intervals: 1) Between and , which is written as 2) Between and , which is written as 3) Between and , which is written as The full solution in interval notation is ### Example Question #2 : Calculus What is the domain of: Explanation: Step 1: We have a square root here, where the inside of the radical must be greater or equal than , so we get a real answer. Step 2: Set the inside equation greater or equal than zero... Step 3: Solve for x... ... Write this in interval notation. is the leftmost boundary, infinity is the right boundary. has a bracket with it, infinity does not.. ### Example Question #3 : Calculus What is the domain and the range for this set of numbers? domain is and the range is domain is and the range is domain is and the range is domain is and the range is domain is and the range is Explanation: The domain is the set of all first elements of ordered pairs (x-coordinates). The range is the set of all second elements of ordered pairs (y-coordinates). Given this set of ordered pairs , the domain is and the range is Note: If a value of a coordinate is repeated, it only gets listed once in the set. ### Example Question #4 : Calculus The surface area of a cube is the total outside area of that cube. A cube has a side length of at least The surface area is a function of side length. What is the domain (side length) and the range (surface area)? ; ; ; ; ; Explanation: The surface area of a cube is the total outside area of that cube. The formula to find the Area of a cube is . A cube has a side length of at least inches. The surface area is a function of side length. What is the domain (side length) and the range (surface area)? The independent variable is the length of the side. The domain consists of numbers that represent the length of sides. The problem states that the length of the sides is at least inches. So, the domain is all numbers greater than or equal to four. The range consists of the numbers that corresponds with the chosen values in a function. To find the range, find the Area of the cube with the length of the side measuring inches; this is the smallest value that (or the side length measurement), can be equal to. ### Example Question #1 : Domain & Range What is the domain of the function ? the domain is all the domain is all the domain is all the domain is all the domain is all Explanation: For a function defined by an expression with variable the domain of is the set of all real numbers that the variable can take such that the expression defining the function is real. The expression defining function contains a square root. . The domain is the set of all first elements of ordered pairs or Because there cannot be a negative inside the radical sign, set the inside of the radical and solve. The expression under the radical has to satisfy the condition for the function to take real values. Solve: Subtract from both sides of the equation. Divide both sides of the equation by . Because you are dividing both sides of the equation by a negative integer this will change the sign from to . The domain or d is all ### Example Question #6 : Calculus What is the range of the function if the domain is ? Explanation: The domain is the set of all first elements of ordered pairs (x-coordinates). The range is the set of all second elements of ordered pairs (y-coordinates). The domain, or all values of . In order to get the range or all values of insert the given values of (the domain) into the equation and solve for (the range). The range for the function with the domain of is . ### Example Question #7 : Calculus What is the domain and range for this graph? The domain is ; the range is . The domain is ; the range is . The domain is ; the range is . The domain is ; the range is . The domain is ; the range is . Explanation: The ordered pairs represented on the graph are: The domain is the set of all first elements of ordered pairs (x-coordinates). The range is the set of all second elements of ordered pairs (y-coordinates). The domain is ; the range is . ### Example Question #8 : Calculus Find the domain for the function ? the domain is all the domain is all the domain is all the domain is all the domain is all Explanation: For a function defined by an expression with variable the domain of is the set of all real numbers that the variable can take such that the expression defining the function is real. The expression defining function contains a square root. The expression under the radical has to satisfy the condition for the function to take real values. Solve Add to both sides of the equation. Divide both sides by . The domain is all ### Example Question #9 : Calculus Find the domain and the range of the inverse of a relation with this set of coordinates: The domain is and the range is . The domain is and the range is . The domain is and the range is . The domain is and the range is . The domain is and the range is . Explanation: To find the domain and the range of the inverse of a relation with this set of coordinates, first find the domain and range of the set of coordinates given. The domain is all of the -coordinates and the range is all of the -coordinates. Remember, if a coordinate is the same, it is only listed once. The domain of is and the range is . However, the question is asking for the domain and the range of the inverse. So, the values of (the domain) will now become the values of , (the range) and the values of (the range) will become the values of , or the domain. The domain for the inverse is and the range is . ### Example Question #10 : Calculus What is the range of the function , when the domain is ? Explanation: The domain is the set of all first elements of ordered pairs (x-coordinates). The range is the set of all second elements of ordered pairs (y-coordinates). domain or values of Plug all of the values of that have been given into the equation Even though some of the values of are repeated, they are only listed once in the set.. The range, or all values of for the function are . ← Previous 1	2016-12-04 18:31:46	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8058035969734192, "perplexity": 920.4432724161795}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-50/segments/1480698541361.65/warc/CC-MAIN-20161202170901-00237-ip-10-31-129-80.ec2.internal.warc.gz"}
https://www.learnpick.in/questions/details/25827/p-a-thin-walled-spherical-vessel-1-m-inner-diameter-and-10	## MyPrivateTutor is now LearnPick Post a Question Question: A thin-walled spherical vessel (1 m inner diameter and 10 mm wall thickness) is made of a material with \|σY\| = 500MPa in both tension and compression. The internal pressure P$_y$ at yield, based on the von Mises yield criterion, if the vessel is floating in space, is approximately............ Please explain the answer in brief • A 500MPa • B 250MPa • C 100MPa • D 20MPa Posted by: Gaurav M. 7 months ago From Our Qualified and Experienced Tutors and Trainers Post a Question • #### Post Answer & Earn Points Get 5 Credits for each answer. The best one gets 25.	2018-04-20 22:12:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4486541152000427, "perplexity": 8918.783481647064}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125944742.25/warc/CC-MAIN-20180420213743-20180420233743-00210.warc.gz"}
http://educ.jmu.edu/~waltondb/ModelCalculus/appendix-numbers-arithmetic.html	# SectionA.1Numbers, Sets and Arithmetic¶ permalink Numbers started as a conceptual way to quantify count objects. Later, numbers were used to measure quantities that were extensive, such as the geometric ideas of length, area and volume. Arithmetic was developed to provide a numeric representation of physical operations. Combining two quantities or collections corresponds to addition. Repeated addition corresponds to multiplication. Repeated multiplication corresponds to powers or exponents. As these ideas developed, inverse operations were invented to help solve problems, including subtraction, division and roots. However, the introduction of each inverse operation required an extension of the idea of number. # SubsectionA.1.1Integers, Rational Numbers and Real Numbers Numbers conceptually begin with the natural numbers, which are the numbers $1, 2, 3, \ldots$. The set of all natural numbers is represented by the symbol $\mathbb{N}$: \begin{equation} \mathbb{N} = \{ 1, 2, 3, \ldots \}. \end{equation} If we include the number 0, we get all possible counting numbers, represented by the symbol $\mathbb{N}_0$: \begin{equation} \mathbb{N}_0 = \{ 0, 1, 2, 3, \ldots \}.\end{equation} Both of these sets have an infinite number of elements because there is no upper bound (i.e., for any number you find, there is always a number greater). The natural and counting numbers are used most basically for ordering and counting elements in sets or collections of objects. For example, consider a set consisting of the basic suits of a standard deck of playing cards, namely hearts, diamonds, spades, and clubs, $\{\heartsuit, \diamondsuit, \spadesuit, \clubsuit\}$. We can count the number of elements by associating each element in the set with one of the natural numbers in order: \begin{gather} 1 \mapsto \heartsuit,\\ 2 \mapsto \diamondsuit,\\ 3 \mapsto \spadesuit,\\ 4 \mapsto \clubsuit. \end{gather} This ordering (which is admittedly arbitrary) allows us to refer to the first, second, third or fourth element in our set. When numbers are used to order elements of a set in this way, we are thinking of numbers as ordinal numbers. Because the greatest number used in our ordering was 4, we can say that the number of elements in our set is 4. When numbers are used to count the number of elements in a set, we are thinking of numbers as cardinal numbers. Both of these ideas can be extended to sets with infinities of elements. It is not in the scope of this text to deal with these issues, but they are typically addressed in more general discussions of set theory. However, the elementary ideas of the arithmetic operations of addition and multiplication are often introduced using the ideas of counting. Addition is first defined by joining sets of known size and asking how many elements are in the combined set. For example, $3+5$ is interpreted in this context as joining a set of three elements, say $\{a,b,c\}$, to another set of five elements, say $\{z,y,x,w,v\}$, to get a combined set $\{a,b,c,v,w,x,y,z\}$. The size of this new set is 8. The equation \begin{equation}3+5=8\end{equation} is interpreted in this context, namely that “the number of elements in a set formed by joining a set with 3 elements and a set with 5 elements” (3+5) is the same as “the number of elements as a set formed from 8 elements” (8). Multiplication is first defined as adding a certain number of groups of the same size. For example, $3 \times 5$ is interpreted as creating a set consisting of 3 groups of 5 elements, which is a new set with 15 elements. This leads to the equation \begin{equation}3 \times 5=15.\end{equation} Once the arithmetic operations of addition and multiplication are introduced, inverse operations of subtraction and division soon follow. Subtraction corresponds to taking away from a set with $5-3$ being interpreted as starting with a set of 5 elements and removing a subset of 3 elements, so that \begin{equation}5-3=2.\end{equation} Division corresponds to determining how many groups of a certain size are in a particular set, with $12 \div 4$ counting the number of groups of size 4 that can be formed from a set of size 12, so that \begin{equation}12 \div 4 = 3.\end{equation} Inverse operations have the property that when performed consecutively, the original value remains unchanged. That is, $a+b-b=a$ for any values of $a$ and $b$ because you join and then remove a set of size $b$ to a set of size $a$, resulting in a set of the original size $a$. Similarly, $a \times b \div b = a$ because the set $a \times b$ has $a$ groups of sets with $b$ elements. The problem that arises is that using only natural or counting numbers, there are expressions that have no valid interpretation. For example, the inverse property suggests that $3-5+5=3$, but the intermediate calculation $3-5$ does not make sense using counting numbers because there is no interpretation for how to remove 5 elements from a set with only 3 elements. Similarly, although the inverse property suggests $5 \div 3 \times 3 = 5$, the intermediate calculation $5 \div 3$ has no whole number interpretation because grouping 5 into sets of size 3 results in one group of 3 and another group of 2 (the remainder). In order to resolve this complication, the idea of number itself is extended. Negative numbers resolve the challenge for subtraction. The set of all integers, also called the whole numbers, introduces both positive and negative counting numbers and is represented by the symbol $\mathbb{Z}$: \begin{equation} \mathbb{Z} = \{ \ldots, -3, -2, -1, 0, 1, 2, 3, \ldots \}. \end{equation} Although we originally thought of subtraction as the inverse operation of addition, the introduction of negative numbers motivated the idea of a number itself having and inverse with respect to addition, or an additive inverse. Every positive and negative number of the same size are inverses because the add to zero, \begin{equation}a+-a=0.\end{equation} With additive inverses, the concept of subtraction is equivalent to addition by an additive inverse, \begin{equation}a-b = a+-b.\end{equation} The advantage of this perspective is that it makes clear how to subtract negative numbers — subtracting a negative number is defined as adding the inverse of the negative number, or adding the corresponding positive number. Just as subtraction led to the development of negative numbers, division motivates the need to extend numbers from just the integers to rational numbers. As soon as we leave the world of whole numbers, our sense of arithmetic actually changes from counting elements in a set to measuring divisible quantities (like length or volume). The standard representation of numbers on a number line illustrates this directly by thinking of numbers as measuring a directed length from an origin (the number 0). There always must be a unit length which corresponds to the distance between 0 and 1. Positive numbers are to the right and negative numbers are to the left. A new interpretation of number requires a new interpretation of arithmetic. Addition of numbers corresponds to combining lengths, with 3+5 meaning we find the number which is found by starting at 0 (the origin), moving three units to the right (to find 3) and then moving five more units to the right (to add 5). Since this the same as the number 8 (starting at 0 and moving 8 units to the right), we know 3+5=8. Multiplication (by integers) will still mean repeated addition, just repeating the displacement interpretation of addition instead of groups. If we consider the property of consecutive inverse operations, we know that we should get $1 \div 5 \times 5 = 1$. So if we think about the intermediate value $a=1 \div 5$ (which is not an integer), we can see that it is a value such that $a \times 5 = 1$. In the geometric interpretation, $a$ is a length such that when it is repeated five times, we recover the unit length. This is the unit fraction $\frac{1}{5}$, which is also called the multiplicative inverse (or reciprocal) of the integer 5. Other fractions have a similar interpretation, such as $3 \div 4 = \frac{3}{4}$ (dividing three into fourths) being the length such that when it is repeated four times is equivalent to three units. Just as subtraction was found (or defined) to be equivalent to addition by an additive inverse, division is also equivalent to multiplication by a multiplicative inverse. Given any non-zero number $a \ne 0$, the multiplicative inverse $\div a$ is that number so that \begin{equation}a \cdot \div a = 1.\end{equation} Then division $a \div b$ is defined by \begin{equation}a \div b = a \cdot \div b.\end{equation} This process allows us to define the rational numbers $\mathbb{Q}$. The rational numbers are formed by considering all of the integers, their multiplicative inverses, and all sums and products of those values. It is most commonly defined by \begin{equation}\mathbb{Q} = \{ p \div q : p \in \mathbb{Z}, q \in \mathbb{N} \}. \end{equation} That is, it consists of all fractions defined by integers. My goal is not to provide an exhaustive explanation of arithmetic and these representations. That would require, for example, an explanation of what it means to multiply and divide negative numbers. However, let it suffice to say that multiplying two negative numbers together will be a positive number while multiplying a positive number and a negative number will result in a negative number. Other mathematical operations introduce the need for even more extensions to the idea of number. For example, the mathematical operation of squaring a number has an inverse operation of the square root. The square of a rational number is still a rational number, but the square root of a rational number is not necessarily rational. The most famous historical example is $\sqrt{2}$, which was proved not to be a rational number (according to legend) by the Greek philosopher Pythagoras. The existence of irrational numbers was a closely guarded secret by his followers, the Pythagoreans. The set of real numbers is the set of both rational and irrational numbers and is represented by the symbol $\mathbb{R}$. Complex numbers extend the real numbers to include square roots of negative numbers by introducing $i=\sqrt{-1}$ and is defined as \begin{equation} \mathbb{C} = \{ a+bi : a, b \in \mathbb{R} \}. \end{equation} Every real number $a \in \mathbb{R}$ is also complex with $b=0$. We will be working almost exclusively with the real numbers. So we very often think in terms of the real number line, which is a continuous and connected curve. Every point on the number line corresponds to a particular real number. Locations correspond to rational numbers if they can be exactly represented using fractional units. An irrational number can never be exactly represented using fractional units. > A set is a mathematical collection of objects. The objects that are in the set are called elements of the set. Set notation uses curly braces $\{$ and $\}$ with a description of the elements that belong to the set. When the set has a finite number of elements, we can just list them between the braces. Like other mathematical objects, we can use symbols to represent the set in the same way that variables can be represented by symbols. ##### ExampleA.1.3 The set that contains the odd digits could be written \begin{equation} O = \{ 1, 3, 5, 7, 9 \}. \end{equation} The set that contains the even digits could be written \begin{equation} E = \{ 0, 2, 4, 6, 8 \}. \end{equation} The set that contains the prime digits could be written \begin{equation} P = \{ 2, 3, 5, 7 \}. \end{equation} Note: The symbols (names) for these sets, $O,E,P$, are just used as examples. We could have used any other symbols that might have been convenient. The symbol $\in$ is a logical operator used to say that an element is in a set. Using the example sets above, we would say $3 \in O$ (read as “ 3 is in $O$”) since 3 is an odd digit. But $3 \not \in E$ (read as “3 is not in $E$”). Most useful sets can not be described by listing all of the elements. Instead, we define sets according to a logical rule that describes when an element is in the set. Such sets start with what is called a universal set that classifies what type of elements are being considered. For example, a set containing numbers might have a universal set $\mathbb{Z}$ (only integers) or $\mathbb{R}$ (all real numbers). A typical set would be defined with a statement like the following, \begin{equation} A = \{ x \in U : \hbox{logical statement about $x$} \}, \end{equation} where $A$ is the symbol for the set being defined, $U$ is the universal set, $x$ is a symbol being used to represent an arbitrary element of $U$, and the statement is how you decide if $x \in A$. ##### ExampleA.1.4 To define the set of all real numbers between -1 and 1, we would write \begin{equation}A = \{ x \in \mathbb{R} : -1 \lt x \lt 1 \}. \end{equation} To define the set of positive real numbers, we would write \begin{equation}B = \{ x \in \mathbb{R} : x \gt 0 \}. \end{equation} We can combine sets to create new sets using unions and intersections. Suppose that $A$ and $B$ represent any two sets. The union of the sets, written $A \cup B$, is the combination of sets that includes elements that are in at least one of the sets: \begin{equation} A \cup B = \{ x : x \in A \hbox{ or } x \in B \}. \end{equation} The intersection of the sets, written $A \cap B$, is the combination of sets that includes only elements that are in both of the sets, also thought of as the overlap of the sets: \begin{equation} A \cap B = \{ x : x \in A \hbox{ and } x \in B \}. \end{equation} ##### ExampleA.1.5 Using the sets defined in the examples above, we have the following statements. The union of $O$ (odd digits) and $P$ (prime digits) gives \begin{equation} O \cup P = \{ 1, 2, 3, 5, 7, 9 \}, \end{equation} which is the same as $O \cup \{2\}$. The intersection of $A$ (real numbers between -1 and 1) and $B$ (positive real numbers) gives \begin{equation} A \cap B = \{ x \in \mathbb{R} : 0 \lt x \lt 1 \}.\end{equation} Because most sets being studied in calculus come from the real numbers, the universal set is often not explicitly stated. So the following represent the same set: \begin{equation} \{ x \in \mathbb{R} : 1 \lt x \lt 3 \} = \{ x : 1 \lt x \lt 3 \}. \end{equation} One of the most common type of sets in calculus is the interval. An interval is a set of real numbers representing a connected segment of the real number line. Intervals are defined by their end points. An open interval does not include the end points while a closed interval does include the end points. An open interval with end points $a$ and $b$ with $a \lt b$ is represented by the notation using round parentheses \begin{equation} (a,b) = \{ x : a \lt x \lt b \}. \end{equation} A closed interval with the same end points is represented by similar notation using square brackets \begin{equation} [a,b] = \{ x : a \le x \le b \}. \end{equation} If only one end point is included, then the notation uses both parentheses and brackets: \begin{gather} [a,b) = \{ x : a \le x \lt b \}, \\ (a,b] = \{ x : a \lt x \le b \}. \end{gather} A set that consists of disjoint intervals can be represented with interval notation using unions. Consider, for example, the interval $[1,5]$ and remove the two values 2 and 4. This is no longer a single interval but consists of three disjoint intervals, namely $[1,2)$, $(2,4)$, and $(4,5]$. We can write this set as a union of the three intervals. \begin{equation} \{ x \in [1,5] : x \ne 2 \hbox{ and } x \ne 4 \} = [1,2) \cup (2,4) \cup (4,5]. \end{equation} Notice how the set defined on the left uses curly brackets because the set is defined using a rule, but the interval notation on the right does not include curly brackets because the interval notation defines everything about the sets. # SubsectionA.1.3Algebra Properties One of the guiding principles in interpreting arithmetic in different representations is that we expect the fundamental properties of arithmetic to be satisfied. These include the commutative and associative properties of addition and multiplication and the distributive properties of multiplication over addition. That is, for every system of numbers, we expect the following properties to hold for any numbers $a,b,c$. Property Description $a+b = b+a$ Addition is Commutative $(a+b)+c = a+(b+c)$ Addition is Associative $a+0 = a$ Zero is Additive Identity $a+-a = 0$ Additive Inverse Property $-a = -1 \cdot a$ Finding Additive Inverse $a \cdot b = b \cdot a$ Multiplication is Commutative $(a \cdot b)\cdot c = a \cdot (b \cdot c)$ Multiplication is Associative $a \cdot 0 = 0$ Multiplication by Zero $a \cdot 1 = a$ One is Multiplicative Identity $a \cdot \div a = 1$ Multiplicative Inverse Property $\div a = \frac{1}{a}, \: a \ne 0$ Finding Multiplicative Inverse $a \cdot (b+c) = a \cdot b + a \cdot c$ Left Distributive Property $(a+b) \cdot c = a \cdot c + b \cdot c$ Right Distributive Property These basic rules establish the basic properties of arithmetic (and algebra) over the real numbers. (Advanced mathematics considers other structures that have some but not all of the same properties in a subject called abstract algebra.) Other consequences of these properties are often used in algebra. We list some of these below for reference.	2017-03-27 16:30:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9497532248497009, "perplexity": 196.01142653895127}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-13/segments/1490218189490.1/warc/CC-MAIN-20170322212949-00303-ip-10-233-31-227.ec2.internal.warc.gz"}
https://www.byteflying.com/archives/3967	# C#LeetCode刷题之#788-旋转数字（Rotated Digits） X is a good number if after rotating each digit individually by 180 degrees, we get a valid number that is different from X. Each digit must be rotated – we cannot choose to leave it alone. A number is valid if each digit remains a digit after rotation. 0, 1, and 8 rotate to themselves; 2 and 5 rotate to each other; 6 and 9 rotate to each other, and the rest of the numbers do not rotate to any other number and become invalid. Now given a positive number N, how many numbers X from 1 to N are good? Input: 10 Output: 4 Explanation: There are four good numbers in the range [1, 10] : 2, 5, 6, 9.Note that 1 and 10 are not good numbers, since they remain unchanged after rotating. Note:N will be in range [1, 10000]. ```public class Program { public static void Main(string[] args) { var N = 10; var res = RotatedDigits(N); Console.WriteLine(res); N = 68; res = RotatedDigits2(N); Console.WriteLine(res); } private static int RotatedDigits(int N) { var res = 0; for(int i = 0; i < N; i++) { if(IsGoodDigit(i + 1)) res++; } return res; } private static bool IsGoodDigit(int n) { //先给出映射列表 var dic = new Dictionary<char, char>() { {'0', '0'}, {'1', '1'}, {'2', '5'}, {'5', '2'}, {'6', '9'}, {'8', '8'}, {'9', '6'} }; //创建 StringBuilder 加速字符串运算 var sb = new StringBuilder(n.ToString()); //循环计算所有字符串 for(var i = 0; i < sb.Length; i++) { //不包含时，根据题意直接返回 false if(!dic.ContainsKey(sb[i])) return false; else { //进行“旋转” sb[i] = dic[sb[i]]; } } //跟原串不一样时返回 true return n.ToString() != sb.ToString(); } private static int RotatedDigits2(int N) { var res = 0; for(int i = 0; i < N; i++) { res += IsGoodDigit2(i + 1) ? 1 : 0; } return res; } private static bool IsGoodDigit2(int n) { //转换成字符串 var bit = n.ToString(); //包含3，4，7时，直接判定 false //因为旋转后无效了 if(bit.Contains('3') \|\| bit.Contains('4') \|\| bit.Contains('7')) return false; //包含2，5，6，9时，直接判定 true //因为旋转后值肯定变了，并且代码执行到此处 //说明原串中不包含3、4、7，不可能会无效 if(bit.Contains('2') \|\| bit.Contains('5') \|\| bit.Contains('6') \|\| bit.Contains('9')) { return true; } //其它所有情况直接判定 false 即可 //包含 0，1，8 却不能被上述代码命中 //肯定不是好数 return false; } }``` ```4 28``` (1)	2022-09-28 10:33:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 2, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3457620441913605, "perplexity": 602.1449774989942}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335190.45/warc/CC-MAIN-20220928082743-20220928112743-00359.warc.gz"}
https://cs.ericyy.me/laff-linear-algebra/week-4.html	# Week 4 - Matrix-Vector to Matrix-Matrix Multiplication ## 1. Preparation ### 1.1. Partitioned Matrix-vector Multiplication • $\displaystyle \left( \begin{array}{ c \| c \| c \| c } A_{0,0} & A_{0,1} & \cdots & A_{0,N-1} \\ A_{1,0} & A_{1,1} & \cdots & A_{1,N-1} \\ \vdots & \vdots & \ddots & \vdots \\ A_{M-1,0} & A_{M-1,1} & \cdots & A_{M-1,N-1} \end{array} \right) \left( \begin{array}{ c } x_0 \\ x_1 \\ \vdots \\ x_{N-1} \end{array} \right) = \left( \begin{array}{ c } A_{0,0} x_{0} + A_{0,1} x_{1} + \cdots + A_{0,N-1} x_{N-1} \\ A_{1,0} x_{0} + A_{1,1} x_{1} + \cdots + A_{1,N-1} x_{N-1} \\ \vdots \\ A_{M-1,0} x_{0} + A_{M-1,1} x_{1} + \cdots + A_{M-1,N-1} x_{N-1} \end{array} \right)$ • Two different algorithms to calculate Matrix-vector Multiplication • First one, is calculating by rows. • $\psi_1 := a_{10}^T x_0 + \alpha_{11} x_1 + a_{12}^T x_2 + \psi_1$ • Second one is by columns. • $y_0 := \chi_1 a_{01} + y_0$ • $\psi_1 := \chi_1 \alpha_{11} + \psi_1$ • $y_2 := \chi_1 a_{21} + y_2$ ### 1.2. Transposing a Partitioned Matrix • $\left( \begin{array}{c \| c \| c \| c} A_{0,0} & A_{0,1} & \cdots & A_{0,N-1} \\ A_{1,0} & A_{1,1} & \cdots & A_{1,N-1} \\ \vdots & \vdots & & \vdots \\ A_{M-1,0} & A_{M-1,1} & \cdots & A_{M-1,N-1} \end{array} \right)^ T = \left( \begin{array}{c \| c \| c \| c} A_{0,0}^ T & A_{1,0}^ T & \cdots & A_{M-1,0}^ T \\ A_{0,1}^ T & A_{1,1}^ T & \cdots & A_{M-1,1}^ T \\ \vdots & \vdots & & \vdots \\ A_{0,N-1}^ T & A_{1,N-1}^ T & \cdots & A_{M-1,N-1}^ T \end{array} \right).$ • Example ### 1.3. Matrix-Vector Multiplication with Special Matrices • When doing calculation, we always concern flops and memops. And later we will prove that, memops is much slower that flops. • Let's take an example. If we want calculate $y := A^T x + y$. Normally, we will do, set $B = A^T$, then $y := B x +y$ • Which means, we are going to spend the most time to transpose matrix A to B, and little time computing. • What we want is, compute with $A^T$ without actually transposing. • The solution is, we can simply use columns of A for the dot products in the dot product based algorithm for $y := Ax + y$. #### Triangular Matrix-Vector Multiplication • Let $U \in \mathbb{R}^{n \times n}$ be an upper triangular matrix and $x \in \mathbb{R}^n$ be a vector. Consider • We notice that $u^T = 0$ (a vector of two zeroes) and hence we need not compute with it. • Let's calculate the flops: • The calculate step: $\psi_1 := u_{11} x_1 + u_{12}^T x_2 + \psi_1$, (without $u_{10}^T x_0$) • if we set the size of $U_{00}$ to $k$,then the size of $U_{02}$ and $U_{20}$ should be $n - k - 1$. • flops($u_{11} x_1$) = 1, flops($u_{12}^T x_2$) = $2(n-k-1)$, flops($u_{11} x_1 + u_{12}^T x_2$) = 1 • So the flops = $\sum_{i=0}^k (2 + 2(n - k - 1)) = n*(n+1)$ #### Symmetric Matrix-Vector Multiplication • We purposely chose the matrix on the right to be symmetric. We notice that $a_{10}^T = a_{01}$ , $A_{20}^T = A_{02}$ , and $a_{12}^T = a_{21}$. • So we just need to change the step of calculation • By rows: • from $\psi_1 := a_{10}^T x_0 + \alpha_{11} x_1 + a_{12}^T x_2 + \psi_1$ • to $\psi_1 := a_{01} x_0 + \alpha_{11} x_1 + a_{12}^T x_2 + \psi_1$ • By columns: • from • $y_0 := \chi_1 a_{01} + y_0$ • $\psi_1 := \chi_1 \alpha_{11} + \psi_1$ • $y_2 := \chi_1 a_{21} + y_2$ • to • $y_0 := \chi_1 a_{01} + y_0$ • $\psi_1 := \chi_1 \alpha_{11} + \psi_1$ • $y_2 := \chi_1 (a_{11}^T)^T + y_2$ ## 2. Composing linear transformations • Let $L_ A: \mathbb {R}^ k \rightarrow \mathbb {R}^ m$ and $L_ B: \mathbb {R}^ n \rightarrow \mathbb {R}^ k$ both be linear transformations and, for all $x \in \mathbb{R}^n$, define the function $L_ C: \mathbb {R}^ n \rightarrow \mathbb {R}^ m$ by $L_ C( x ) = L_ A( L_ B( x ) )$. Then $L_ C( x )$ is a linear transformations. ## 3. Matrix-matrix multiplication • $A B = A \left( \begin{array}{c \| c \| c \| c } b_0 & b_1 & \cdots & b_{n-1} \end{array} \right) = \left( \begin{array}{c \| c \| c \| c } A b_0 & A b_1 & \cdots & A b_{n-1} \end{array} \right).$ • If $\begin{array}{c} C = \left( \begin{array}{c c c c } \gamma _{0,0} & \gamma _{0,1} & \cdots & \gamma _{0,n-1} \\ \gamma _{1,0} & \gamma _{1,1} & \cdots & \gamma _{1,n-1} \\ \vdots & \vdots & \vdots & \vdots \\ \gamma _{m-1,0} & \gamma _{m-1,1} & \cdots & \gamma _{m-1,n-1} \\ \end{array} \right) , \quad A = \left( \begin{array}{c c c c } \alpha _{0,0} & \alpha _{0,1} & \cdots & \alpha _{0,k-1} \\ \alpha _{1,0} & \alpha _{1,1} & \cdots & \alpha _{1,k-1} \\ \vdots & \vdots & \vdots & \vdots \\ \alpha _{m-1,0} & \alpha _{m-1,1} & \cdots & \alpha _{m-1,k-1} \\ \end{array} \right), \\ \text{and} \quad B = \left( \begin{array}{c c c c } \beta _{0,0} & \beta _{0,1} & \cdots & \beta _{0,n-1} \\ \beta _{1,0} & \beta _{1,1} & \cdots & \beta _{1,n-1} \\ \vdots & \vdots & \vdots & \vdots \\ \beta _{k-1,0} & \beta _{k-1,1} & \cdots & \beta _{k-1,n-1} \\ \end{array} \right). \end{array}$ • Then C = AB means that $\gamma _{i,j} = \sum _{p=0}^{k-1} \alpha _{i,p} \beta _{p,j}$ • A table of matrix-matrix multiplications with matrices of special shape is given at the end of this unit. ### 3.1. Outer product • Let $x \in \mathbb{R}^m$ and $y \in \mathbb{R}^n$. Then the outer product of x and y is given by $xy^T$. Notice that this yields an $m \times n$ matrix: \begin{aligned} xy^T &= \left( \begin{array}{c} \chi _0 \\ \chi _1 \\ \vdots \\ \chi _{m-1} \end{array} \right) \left( \begin{array}{c} \psi _0 \\ \psi _1 \\ \vdots \\ \psi _{n-1} \end{array} \right)^ T = \left( \begin{array}{c} \chi _0 \\ \chi _1 \\ \vdots \\ \chi _{m-1} \end{array} \right) \left( \begin{array}{c c c c} \psi _0 & \psi _1 & \cdots & \psi _{n-1} \end{array} \right) \\ &= \left( \begin{array}{c c c c} \chi _0 \psi _0 & \chi _0 \psi _1 & \cdots & \chi _0 \psi _{n-1} \\ \chi _1 \psi _0 & \chi _1 \psi _1 & \cdots & \chi _1 \psi _{n-1} \\ \vdots & \vdots & & \vdots \\ \chi _{m-1} \psi _0 & \chi _{m-1} \psi _1 & \cdots & \chi _{m-1} \psi _{n-1} \end{array} \right). \end{aligned} • The cost of memops of matrix-matrix multiplication is $2kmn$.	2019-02-16 01:03:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 52, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000076293945312, "perplexity": 1186.1479636085708}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550247479729.27/warc/CC-MAIN-20190216004609-20190216030609-00169.warc.gz"}
http://math.stackexchange.com/questions/359528/create-a-generating-function	# Create a Generating function Let $P$ be the set of permutations all of whose cycles are of even length. Prove that the exponential generating function for $P$ is $\dfrac{1}{\sqrt{1-x^2}}$. - What have you done? Where did it go wrong? What other generating functions have you seen? What have you tried? – mixedmath Apr 12 '13 at 15:47 A bit of notation to make the answer clearer. I shall write $P_n$ to be the set of all partitions of the set $\{1,\ldots,n\}$ and $\sigma=\{S_1,\ldots,S_k\} \in P_n$ to mean that $\sigma$ is a partition of the set $\{1,\ldots,n\}$ into "parts" $S_1,\ldots,S_k$ Define $$a_n= \begin{cases} (n-1)! & \text{if n is even and n\geq2} \\ 0 & \text{otherwise} \end{cases}$$ and $b_n=1$ for all $n$. Let $A(x)=\sum\limits_{i=0}^\infty\frac{a_nx^n}{n!}$ and $B(x)=\sum\limits_{n=0}^\infty\frac{a_nx^n}{n!}$. Then the exponential generating series for $P$ is $B(A(x))=\sum\limits_{n=0}^\infty\frac{c_nx^n}{n!}$, where $$c_n=\sum\limits_{\sigma=\{S_1,\ldots,S_k\} \in P_n}b_k a_{\|S_1\|}a_{\|S_2\|}\cdots a_{\|S_k\|}=\sum\limits_{\sigma=\{S_1,\ldots,S_k\} \in P_n,\ \|S_i\|\text{ even}}(\|S_1\|-1)! \cdots (\|S_k\|-1)!,$$ which is exactly the number of permutations of the set $\{1, \ldots,n\}$ with all cycles even! Note that $B(x)=\exp(x)$ and $$A(x)=\sum\limits_{n \geq 2,\text{ even}}^\infty \frac{(n-1)!x^n}{n!}=\sum\limits_{n \geq 2,\text{ even}}^\infty\frac{x^n}{n}=\frac{1}{2}(\ln(1+x) - \ln(1-x))= \ln \left(\frac{1}{\sqrt{1-x^2}}\right),$$ Hence the generating function for $P$ is $B(A(x))=\exp\left(\ln \left(\frac{1}{\sqrt{1-x^2}}\right)\right)=\dfrac{1}{\sqrt{1-x^2}}$ - This is definitely straightforward using symbolic combinatorics as formalized by Flajolet and Sedgewick. Let $\mathcal{Q}$ be the combinatorial class of permutations consisting only of even cycles. We have the combinatorial specification $$\mathcal{Q} = \mathfrak{P}(\mathfrak{C}_2(\mathcal{Z}) + \mathfrak{C}_4(\mathcal{Z}) + \mathfrak{C}_6(\mathcal{Z}) + \cdots)$$ where $\mathcal{Z}$ is the singleton class. Translating this into generating functions we obtain for the required EGF $G(z)$ that $$G(z) = \exp\left(\frac{z^2}{2} + \frac{z^4}{4} + \frac{z^6}{6}+\cdots\right) = \exp \left(\frac{1}{2}\left(\frac{(z^2)}{1} + \frac{(z^2)^2}{2} + \frac{(z^2)^3}{3} +\cdots\right)\right) \\= \exp\left(\frac{1}{2} \log \frac{1}{1-z^2}\right) = \left(\frac{1}{1-z^2}\right)^{1/2} = \frac{1}{\sqrt{1-z^2}}.$$ -	2015-11-30 15:27:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9264412522315979, "perplexity": 144.8745082005964}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-48/segments/1448398462686.42/warc/CC-MAIN-20151124205422-00326-ip-10-71-132-137.ec2.internal.warc.gz"}
http://mathoverflow.net/questions/136188/when-does-order-matter-when-decomposing-a-boundedly-generated-group	# When does order matter when decomposing a boundedly generated group A group $G$ is said to be boundedly generated if (it is finitely generated and) there exists a finite family of cyclic subgroups (not necessarily normal or distinct) $\lbrace C_i \rbrace_{i =1, \ldots, k}$ such that $G = C_1 \cdots C_k$. Simplest examples of boundedly generated groups are finite groups and polycyclic groups. Free groups are not boundedly generated, and $SL_2(\mathbb{Z})$ also isn't (since passing to finite index subgroups does not change this property). A priori, the order in the product matters. Assume $G = C_1 \cdots C_k$ is a boundedly generated group so that $G = C_{\sigma(1)} \cdots C_{\sigma(k)}$ for any permutation $\sigma \in Sym_k$. In particular, one may then assume the $C_i$ are distinct. Except virtually polycyclic groups, which group has this property? - "boundedly generated group where the order of the product does not matter": I don't understand what you mean by this. – Yves Cornulier Jul 9 '13 at 18:30 oh sorry, that was pretty unclear indeed... – Antoine Jul 9 '13 at 19:07 ok, it's clearly defined now. – Yves Cornulier Jul 9 '13 at 19:29 Actually, are there constructions of groups with bgp, which are not lattices in Lie/alberaic groups? (Polycyclic groups and groups of integer points in higher rank algebraic groups are the only examples I am aware of.) – Misha Jul 10 '13 at 15:30 @Misha: there are Muranov's construction of boundedly generated infinite f.g. groups that are simple. – Yves Cornulier Jul 10 '13 at 15:42 show 1 more comment	2014-04-16 14:12:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8530830144882202, "perplexity": 951.5320296430955}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609523429.20/warc/CC-MAIN-20140416005203-00065-ip-10-147-4-33.ec2.internal.warc.gz"}
https://ask.sagemath.org/questions/34994/revisions/	I want to find Resultant of trinomials $x^i-x^j-1$ of this form. Is it possible to do symbolic calculation for this in sage ? I want to find Resultant of trinomials $x^i-x^j-1$ of this form. form $x^i-x^j-1$. Is it possible to do symbolic calculation for this in sage ? ?	2020-04-03 00:31:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.35065504908561707, "perplexity": 201.77894670947953}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370509103.51/warc/CC-MAIN-20200402235814-20200403025814-00133.warc.gz"}
http://adsense.blogspot.com/2009/09/touch-ups-for-your-ad-units.html	You may remember that last year, we added arrows to the bottom of cost-per-click (CPC) ad units. These arrows allow users to browse through additional relevant ads, helping them find exactly they're looking for. We've been testing slight updates to the look of these arrows, and our experiments have shown improvements in the user experience. As a result, we'll soon be making a few minor aesthetic changes, including darkening the arrows to make them more visible and orienting all arrows to point left and right. In addition, to help users understand what the arrows do, hovering over the arrows will soon show the labels 'previous ads' and 'next ads'. You'll see these changes appear in CPC ad units in all languages gradually over the next few days.	2015-05-04 16:03:25	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8877856731414795, "perplexity": 1816.7757394678927}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1430454626420.21/warc/CC-MAIN-20150501043026-00067-ip-10-235-10-82.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/2239696/a-limit-with-arctan-function-and-euler-varphi-function	# A limit with arctan function and Euler $\varphi$ function The following limit has stumbed upon a while. I don't have the experience to work with these kind of limit as well as these special functions like $\varphi$. Prove that $$\lim_{n \rightarrow +\infty} \sum_{k=1}^{n} \frac{\varphi(k) \arctan \left(\frac{k}{n} \right)}{k (n+k)} = \frac{3 \log 2}{4\pi}$$ where $\varphi$ is the Euler's totient function. How would one go to prove it? • In what context did you encounter this limit? – Greg Martin Apr 18 '17 at 7:19 • Actually it was passed to me.. !! Question solved by the way by finding an article by Omran Kouba! No way I would have thought what he writes in the article by myself! – Tolaso Apr 18 '17 at 7:23 We can use the fact that $$\frac{\phi(k)}k = \sum_{d\mid k} \frac{\mu(d)}d,$$ where $\mu$ is the Möbius function and the sum is over all positive divisors $d$ of $k$. Then \begin{align} \sum_{k=1}^{n} \frac{\varphi(k) \arctan \left(\frac{k}{n} \right)}{k (n+k)} &= \sum_{k=1}^{n} \frac{\arctan \left(\frac{k}{n} \right)}{n+k} \sum_{d\mid n} \frac{\mu(d)}d \\ &= \sum_{d\le n} \frac{\mu(d)}d \sum_{\substack{k\le n \\ d\mid k}} \frac{\arctan \left(\frac{k}{n} \right)}{n+k} \\ &= \sum_{d\le n} \frac{\mu(d)}{d^2} \sum_{\substack{j\le n/d}} \frac1{n/d} \frac{\arctan \left(\frac{j}{n/d} \right)}{1+\frac j{n/d}}, \end{align} where we used the change of variables $k=jd$ in the last equality. For each fixed $d$, the inner sum is a Riemann sum for $\int_0^1 \frac{\arctan x}{1+x}\,dx$; since this integrand is bounded and $\sum_{d=1}^\infty \frac{\mu(d)}{d^2}$ converges absolutely to $\frac6{\pi^2}$, the dominated convergence theorem implies that \begin{align} \lim_{n\to\infty} \sum_{d\le n} \frac{\mu(d)}{d^2} \sum_{\substack{j\le n/d}} \frac1{n/d} \frac{\arctan \left(\frac{j}{n/d} \right)}{1+\frac j{n/d}} &= \sum_{d=1}^\infty \frac{\mu(d)}{d^2} \lim_{n\to\infty} \sum_{\substack{j\le n/d}} \frac1{n/d} \frac{\arctan \left(\frac{j}{n/d} \right)}{1+\frac j{n/d}} \\ &= \sum_{d=1}^\infty \frac{\mu(d)}{d^2} \int_0^1 \frac{\arctan x}{1+x}\,dx \\ &= \frac6{\pi^2} \int_0^1 \frac{\arctan x}{1+x}\,dx, \end{align} This reduces the problem to the evaluation $\int_0^1 \frac{\arctan x}{1+x}\,dx = \frac{\pi\log 2}8$, which is itself nontrivial but which has been answered here before (modulo an integration by parts which is natural to try). • The integral you mention boils down to this question math.stackexchange.com/questions/172111/… . You seem to have forgotten a summation sign at the line: $$\lim_{n\to\infty} \sum_{d\le n} \frac{\mu(d)}{d^2} \sum_{\substack{j\le n/d}} \frac1{n/d} \frac{\arctan \left(\frac{j}{n/d} \right)}{1+\frac j{n/d}} = \sum_{d=1}^\infty \frac{\mu(d)}{d^2} \lim_{n\to\infty} \frac1{n/d} \frac{\arctan \left(\frac{j}{n/d} \right)}{1+\frac j{n/d}}$$ – Tolaso Apr 18 '17 at 7:33	2019-08-21 12:28:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 1.0000075101852417, "perplexity": 137.10231748332754}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315936.22/warc/CC-MAIN-20190821110541-20190821132541-00204.warc.gz"}
http://the-write-words.com/viewtopic/archive.php?1233d1=coordinate-geometry-questions-pdf	The questions can be answered using A alone but not using B alone. Notably, in geometry, learning the application of concepts is very important to confidently arrive at the answers. Which of the following is true? The area of ABCD (in cm2) is, AB = BC = 2CH = 2CD = EH = FK = 2HK = 4KL = 2LM = MN. Also Read: Right approach to learn Geometry. a) x – 6y […] Coordinate Geometry, coordinate geometry problems, Coordinate plane, Slope Formula, Equation of a Line, Slopes of parallel lines, Slope of perpendicular lines, Midpoint Formula, Distance Formula, examples and step by step solutions, questions and answers . The distance between AB and the tangent at E is 5 cm. It is known that u2:v2:w2 is equal to Area A: Area B: Area C, where Area A, Area B and Area C are the areas of triangles A1A2A3, B1B2B3, and C1C2C3 respectively. Questions on Geometry for CAT exam is a crucial topic. Occasionally, questions from polygons, coordinate geometry and mensuration have also appeared. A. Sprinter A traverses distances A1A2, A2A3, and A3A1 at an average speed of 20, 30 and 15 respectively. Coordinate Geometry. Rate Us. The wording, diagrams and figures used in these questions have been changed from the originals so that students can have fresh, relevant problem solving practice even if they have previously worked through the related exam paper. B. C. 4. a y = 4x − 1 b y = 1 To help you with this, we have written various articles to help you understand not only the concepts but also some new tricks and shortcut formulas. Related: HOME . Q.1. SSC CGL Coordinate Geometry Questions with Solutions PDF: Download SSC CGL Coordinate Geometry questions with answers PDF based on previous papers very useful for SSC CGL exams. Where would A and C be when B reaches point B3? Find the area of the triangle formed by the vertices (4, 5), (10, 12) and (-3, 2) A. Each chapter builds on the skills reviewed in the chapters that precede it. Slope 5. The proportion of the sheet area that remains after punching is: ABC Corporation is required to maintain at least 400 Kilolitres of water at all times in its factory, in order to meet safety and regulatory requirements. (c) Find the equation of line CD. Impossible to find from the given information.. $$\frac{{\left( {2\sqrt 2 - 1} \right)}}{2}$$, $$\frac{{\left( {3\sqrt 2 - 1} \right)}}{2}$$, $$\frac{{\left( {2\sqrt 2 - 1} \right)}}{3}$$, $$\frac{{\left( {6 - \pi } \right)}}{8}$$, $$\frac{{\left( {4 - \pi } \right)}}{4}$$, $$\frac{{\left( {14 - 3\pi } \right)}}{6}$$. Amit wants to reach N2 from S1. Coordinate Geometry. a (3, 1) and (5, 5) b (4, 7) and (10, 9) c (6, 1) and (2, 5) d (−2, 2) and (2, 8) e (1, 3) and (7, −1) f (4, 5) and (−5, −7) g (−2, 0) and (0, −8) h (8, 6) and (−7, −2) 2 Write down the gradient and y-intercept of each line. The same string, when wound on the exterior four walls of a cube of side n cm, starting at point C and ending at point D, can give exactly one turn (see figure, not drawn to scale). The (x,y) Coordinate Plane 3. What is the ratio of the perimeter of $$\Delta ADC$$ to that of the $$\Delta BDC$$? Chapter 16: Coordinate Geometry Plot and ﬁnd points on the coordinate plane; ﬁnd the slope, midpoint, and distance of line segments. How to Use This Book Start at the beginning. At least 20% of CAT questions each year are from Geometry alone. 25 Very important Coordinate Geometry objective questions for SSC exams. The ratio of the sum of the lengths of all chord roads to the length of the outer ring road is. Pythagorean Triplets: Concepts and Tricks. B: The tank weights 30,000 kg when empty, and is made of a material with density of 3 gm/cc. The world questions can be answered by using either statement alone information given in the chapters that it. 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Includes nine questions on Geometry for important topics for all chapters in Class 9 Coordinate Geometry based on simple., we have listed around 85 Geometry questions with answers which have appeared in year. *, [ PDF ] Geometry questions of Pythagoras ’ theorem for the same inner diameter of the of!	2021-03-06 23:28:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5174564719200134, "perplexity": 1169.7389983420092}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178375529.62/warc/CC-MAIN-20210306223236-20210307013236-00163.warc.gz"}
https://stats.stackexchange.com/questions/385818/given-sample-variance-and-sample-size-test-two-variance-which-one-holds	# Given sample variance and sample size, test two variance which one holds If I am given the sample size and the sample variances of a Normal population, how can I test the hypothesis: H0: $$\sigma^2=\sigma_0^2$$, H1: $$\sigma^2=\sigma_1^2$$ • Perhaps an $F$-test? – user2974951 Jan 7 at 10:35	2019-06-16 19:14:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 2, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9178587794303894, "perplexity": 1014.8176389713135}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-26/segments/1560627998291.9/warc/CC-MAIN-20190616182800-20190616204800-00199.warc.gz"}
http://bradleyarms.co.uk/journal/blog/proc-lifetest-plots-729f49	# proc lifetest plots The covariate effect of $$x$$, then is the ratio between these two hazard rates, or a hazard ratio(HR): $HR = \frac{h(t\|x_2)}{h(t\|x_1)} = \frac{h_0(t)exp(x_2\beta_x)}{h_0(t)exp(x_1\beta_x)}$. In such a case, PROC LIFETEST creates a plot of the cumulative incidence function (CIF) estimates. The estimate of survival beyond 3 days based off this Nelson-Aalen estimate of the cumulative hazard would then be $$\hat S(3) = exp(-0.0385) = 0.9623$$. One way to fix this would with the usage of annotation. Perform search. If the observed pattern differs significantly from the simulated patterns, we reject the null hypothesis that the model is correctly specified, and conclude that the model should be modified. run; proc corr data = whas500 plots(maxpoints=none)=matrix(histogram); Finally, we see that the hazard ratio describing a 5-unit increase in bmi, $$\frac{HR(bmi+5)}{HR(bmi)}$$, increases with bmi. Most of the variables are at least slightly correlated with the other variables. Note: A number of sub-sections are titled Background. However, one cannot test whether the stratifying variable itself affects the hazard rate significantly. The censoring information above the X axis is moved outside the graph. Chapter 22, Any serious endeavor into data analysis should begin with data exploration, in which the researcher becomes familiar with the distributions and typical values of each variable individually, as well as relationships between pairs or sets of variables. refline 0 / axis=x; However, some customization can be achieved without any annotation by using some additional statements and options. The Subjects at Risk table is shown closer to the survival curves using the LOCATION=INSIDE option. It is not at all necessary that the hazard function stay constant for the above interpretation of the cumulative hazard function to hold, but for illustrative purposes it is easier to calculate the expected number of failures since integration is not needed. During the next interval, spanning from 1 day to just before 2 days, 8 people died, indicated by 8 rows of “LENFOL”=1.00 and by “Observed Events”=8 in the last row where “LENFOL”=1.00. run; The Kaplan_Meier survival function estimator is calculated as: $\hat S(t)=\prod_{t_i\leq t}\frac{n_i – d_i}{n_i},$. Thanks for your suggestion. xaxistable atrisk / x=tatrisk class=stratum colorgroup=stratum valueattrs=(weight=bold); Additionally, a few heavily influential points may be causing nonproportional hazards to be detected, so it is important to use graphical methods to ensure this is not the case. The survival function is undefined past this final interval at 2358 days. Particular emphasis is given to proc lifetest for nonparametric estimation, and proc phreg for Cox regression and model evaluation. If no options are requested, PROC LIFETEST computes and displays the product-limit estimate of the survivor function; and if an ods graphics on statement is specified, a plot of the estimated survivor function is also displayed. scatter x = hr y=dfhr / markerchar=id; run; proc phreg data = whas500; class gender; These are indeed censored observations, further indicated by the “” appearing in the unlabeled second column. PROC LIFETEST data=test1dts outsurv=_surv alphaqt=0.05 ties=EFRON plot=(s); time pfstmpfscen(1); strata treatment; Run; See Figure 1.Sample Kaplan-Meier Curve (Note – Adding the option plot=(s) with create a simple Kaplan-Meier Curve.) SGPLOT code for Survival Graph: SG_Survival_Plot. Provided the reader has some background in survival analysis, these sections are not necessary to understand how to run survival analysis in SAS. Customizing Survival Plots in In such cases, the correct form may be inferred from the plot of the observed pattern. While a user may get different mileage from the results, this exercise will show you how you can use such statements and options creatively to get different results. The same procedure could be repeated to check all covariates. Trending. We generally expect the hazard rate to change smoothly (if it changes) over time, rather than jump around haphazardly. To do so: It appears that being in the hospital increases the hazard rate, but this is probably due to the fact that all patients were in the hospital immediately after heart attack, when they presumbly are most vulnerable. Data that are structured in the first, single-row way can be modified to be structured like the second, multi-row way, but the reverse is typically not true. We could test for different age effects with an interaction term between gender and age. It is possible that the relationship with time is not linear, so we should check other functional forms of time, such as log(time) and rank(time). where $$d_i$$ is the number who failed out of $$n_i$$ at risk in interval $$t_i$$. It appears that for males the log hazard rate increases with each year of age by 0.07086, and this AGE effect is significant, AGEGENDER term is negative, which means for females, the change in the log hazard rate per year of age is 0.07086-0.02925=0.04161. keylegend 's' / linelength=20; model lenfolfstat(0) = gender age;; run; ; See the section Modifying the Layout and Adding a New Inset Table in Here we use proc lifetest to graph $$S(t)$$. Unless the seed option is specified, these sets will be different each time proc phreg is run. run; proc lifetest data=whas500 atrisk outs=outwhas500; Let’s interpret our model. Enabling ODS Graphics and the Default Kaplan-Meier Plot Tree level 6. However, nonparametric methods do not model the hazard rate directly nor do they estimate the magnitude of the effects of covariates. Re: How to change X axis in CIF plot (either in proc lifetest or proc phreg)? Posted 07-19-2018 04:55 PM (2127 views) \| In reply to Reeza Thanks - I was playing with your old code before I knew what I was doing and I think I changed it already so I do not have the original template. Plots of the covariate versus martingale residuals can help us get an idea of what the functional from might be. The changed syntax is highlighted in the code. ODS Graphics Template Modification. Graphs of the Kaplan-Meier estimate of the survival function allow us to see how the survival function changes over time and are fortunately very easy to generate in SAS: The step function form of the survival function is apparent in the graph of the Kaplan-Meier estimate. Thus far in this seminar we have only dealt with covariates with values fixed across follow up time. We can similarly calculate the joint probability of observing each of the $$n$$ subject’s failure times, or the likelihood of the failure times, as a function of the regression parameters, $$\beta$$, given the subject’s covariates values $$x_j$$: $L(\beta) = \prod_{j=1}^{n} \Bigg\lbrace\frac{exp(x_j\beta)}{\sum_{iin R_j}exp(x_i\beta)}\Bigg\rbrace$. I want to remove the frame of the survival curve in the topright corner. Notice in the Analysis of Maximum Likelihood Estimates table above that the Hazard Ratio entries for terms involved in interactions are left empty. These two observations, id=89 and id=112, have very low but not unreasonable bmi scores, 15.9 and 14.8. For example, if $$\beta_x$$ is 0.5, each unit increase in $$x$$ will cause a ~65% increase in the hazard rate, whether X is increasing from 0 to 1 or from 99 to 100, as $$HR = exp(0.5(1)) = 1.6487$$. Once outliers are identified, we then decide whether to keep the observation or throw it out, because perhaps the data may have been entered in error or the observation is not particularly representative of the population of interest. On the right panel, “Residuals at Specified Smooths for martingale”, are the smoothed residual plots, all of which appear to have no structure. In the code below, we show how to obtain a table and graph of the Kaplan-Meier estimator of the survival function from proc lifetest: Above we see the table of Kaplan-Meier estimates of the survival function produced by proc lifetest. proc loess data = residuals plots=ResidualsBySmooth(smooth); Because of this parameterization, covariate effects are multiplicative rather than additive and are expressed as hazard ratios, rather than hazard differences. Currently loaded videos are … if lenfol > los then in_hosp = 0; where $$R_j$$ is the set of subjects still at risk at time $$t_j$$. run; proc phreg data=whas500 plots=survival; The corresponding tests are known as the log-rank test and the Wilcoxon test, respectively. title2 h=0.8 'With Number of Subjects at Risk'; Without this offset, long values in the table may get clipped. The template that PROC LIFETEST is using is the Graph template that you see when you run ODS TRACE. 1469-82. See the section Displaying Survival Summary Statistics in The WHAS500 data are stuctured this way. Note: The terms event and failure are used interchangeably in this seminar, as are time to event and failure time. In the code below, we model the effects of hospitalization on the hazard rate. This confidence band is calculated for the entire survival function, and at any given interval must be wider than the pointwise confidence interval (the confidence interval around a single interval) to ensure that 95% of all pointwise confidence intervals are contained within this band. However, we have decided that there covariate scores are reasonable so we retain them in the model. This analysis proceeds in much the same was as dfbeta analysis, in that we will: We see the same 2 outliers we identifed before, id=89 and id=112, as having the largest influence on the model overall, probably primarily through their effects on the bmi coefficient. If ODS Graphics is enabled, PROC LIFETEST also displays a plot of the estimated survivor function. Hosmer, DW, Lemeshow, S, May S. (2008). Only as many residuals are output as names are supplied on the, We should check for non-linear relationships with time, so we include a, As before with checking functional forms, we list all the variables for which we would like to assess the proportional hazards assumption after the. scatter x=time y=censored / markerattrs=(symbol=plus) name='c'; In this model, this reference curve is for males at age 69.845947 Usually, we are interested in comparing survival functions between groups, so we will need to provide SAS with some additional instructions to get these graphs.	2021-02-26 16:36:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6607893109321594, "perplexity": 1262.29078589056}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-10/segments/1614178357929.4/warc/CC-MAIN-20210226145416-20210226175416-00179.warc.gz"}
https://leetcode.ca/2016-04-27-149-Max-Points-on-a-Line/	# Question Formatted question description: https://leetcode.ca/all/149.html 149 Max Points on a Line Given n points on a 2D plane, find the maximum number of points that lie on the same straight line. Example 1: Input: [[1,1],[2,2],[3,3]] Output: 3 Explanation: ^ \| \| o \| o \| o +-------------> 0 1 2 3 4 Example 2: Input: [[1,1],[3,2],[5,3],[4,1],[2,3],[1,4]] Output: 4 Explanation: ^ \| \| o \| o o \| o \| o o +-------------------> 0 1 2 3 4 5 6 @tag-hashtable @tog-top100 # Algorithm A slope can be calculated between two given points. Each slope represents a straight line. For each straight line, bring in all the points to see if they are collinear and calculate the number. Two special cases need to be considered. • One is that when two points coincide, a straight line cannot be determined, but this is also a collinear case and requires special treatment. • The second is the case where the slope does not exist. Since the slope k of the two points (x1, y1) and (x2, y2) is expressed as (y2-y1) / (x2-x1), then the slope does not exist when x1 = x2, This collinear situation requires special treatment TreeMap is used to record the mapping between the slope and the number of collinear points. The first case of coincident points assumes its slope is INT_MIN, and the second case assumes its slope is INT_MAX, so TreeMap can be used for mapping. A variable duplicate is also needed to record the number of coincident points, and finally only needs to be added to the number in the TreeMap to get the total number of collinear points. # Code Java • import java.util.HashMap; import java.util.Map; public class Max_Points_on_a_Line { public static void main(String[] args) { double a = 0.0 / (-1); System.out.println("a:" + a); double b = 0.0 / (1); System.out.println("b:" + b); System.out.println("a == b ? " + (a == b)); // output: true HashMap<Double, Integer> hm = new HashMap<Double, Integer>(); hm.put(a, 1); hm.put(b, 1); // @note:@memorize: so, different keys System.out.println("hashmap keys: " + (hm.keySet())); // output: [-0.0, 0.0] } /** * Definition for a point. * class Point { * int x; * int y; * Point() { x = 0; y = 0; } * Point(int a, int b) { x = a; y = b; } * } / public class Solution { public int maxPoints(Point[] points) { if (points == null \|\| points.length == 0) return 0; // line slope => number of points HashMap<Double, Integer> hm = new HashMap<Double, Integer>(); int max = 0; // have to touch every possible line, so O(logN) for (int i = 0; i < points.length; i++) { int duplicate = 1; int vertical = 0; for (int j = i + 1; j < points.length; j++) { // @note:@memorize: handle duplicates and vertical if (points[i].x == points[j].x) { if (points[i].y == points[j].y) { duplicate++; // itself } else { vertical++; // vertical line at the same x position } } else { double slope = points[j].y == points[i].y ? 0.0 : (1.0 (points[j].y - points[i].y)) / (points[j].x - points[i].x); hm.merge(slope, 1, (a, b) -> a + b); } } // record max number of points for one line for (Integer count : hm.values()) { if (count + duplicate > max) { max = count + duplicate; } } // @note:@memorize: special case for vertical line points max = Math.max(vertical + duplicate, max); hm.clear(); // @note: api, better than create a new hashmap? } return max; } } } • // OJ: https://leetcode.com/problems/max-points-on-a-line/ // Time: O(N^3) // In worse case, when visiting ith point, there will be O(i^2) // lines, but all the lines are of constant size 2. So in // sum it's O(N^3), not O(N^4). // Space: O(N^2) namespace std { template <> struct hash<Point> { std::size_t operator () (const Point &p) const { return hash<int>()(p.x) ^ hash<int>()(p.y); } }; } bool operator==(const Point& a, const Point& b) { return a.x == b.x && a.y == b.y; } class Solution { private: bool onSameLine(Point &a, Point &b, Point &c) { return (long long)(a.x - b.x) * (a.y - c.y) == (long long)(a.x - c.x) * (a.y - b.y); } public: int maxPoints(vector<Point>& points) { if (points.size() <= 2) return points.size(); unordered_map<Point, int> m; for (auto p : points) m[p]++; vector<pair<Point, int>> ps(m.begin(), m.end()); vector<vector<int>> lines(1, vector<int>{ 0, 1 }); int N = ps.size(); for (int i = 2; i < N; ++i) { auto &p = ps[i].first; for (auto &line : lines) { auto &p1 = ps[line[0]].first, &p2 = ps[line[1]].first; if (!onSameLine(p1, p2, p)) continue; for (int neighbor : line) bad[neighbor] = 0; line.push_back(i); } for (int j = 0; j < i; ++j) { if (bad[j]) lines.emplace_back(vector<int>{ j, i }); } } int ans = 2; for (auto line : lines) { int cnt = 0; for (auto i : line) cnt += ps[i].second; ans = max(ans, cnt); } return ans; } }; • # Definition for a point. # class Point(object): # def __init__(self, a=0, b=0): # self.x = a # self.y = b class Solution(object): def maxPoints(self, points): """ :type points: List[Point] :rtype: int """ def gcd(a, b): while b: a, b = b, a % b return a ans = 1 d = {} points.sort(key=lambda p: (p.x, p.y)) for i in range(0, len(points)): if i > 0 and (points[i].x, points[i].y) == (points[i - 1].x, points[i - 1].y): continue overlap = 1 for j in range(i + 1, len(points)): x1, y1 = points[i].x, points[i].y x2, y2 = points[j].x, points[j].y ku, kd = y2 - y1, x2 - x1 if (x1, y1) != (x2, y2): kg = gcd(ku, kd) ku /= kg kd /= kg d[(ku, kd, x1, y1)] = d.get((ku, kd, x1, y1), 0) + 1 else: overlap += 1 ans = max(ans, overlap) ans = max(ans, d.get((ku, kd, x1, y1), 0) + overlap) return min(ans, len(points))	2022-12-09 20:35:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.281162828207016, "perplexity": 14570.893810910758}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446711475.44/warc/CC-MAIN-20221209181231-20221209211231-00762.warc.gz"}
http://popflock.com/learn?s=Vector_meson_dominance	Vector Meson Dominance Get Vector Meson Dominance essential facts below. View Videos or join the Vector Meson Dominance discussion. Add Vector Meson Dominance to your PopFlock.com topic list for future reference or share this resource on social media. Vector Meson Dominance In physics, vector meson dominance (VMD) was a model developed by J. J. Sakurai[1] in the 1960s before the introduction of quantum chromodynamics to describe interactions between energetic photons and hadronic matter. In particular, the hadronic components of the physical photon consist of the lightest vector mesons, ${\displaystyle \rho }$, ${\displaystyle \omega }$ and ${\displaystyle \phi }$. Therefore, interactions between photons and hadronic matter occur by the exchange of a hadron between the dressed photon and the hadronic target. ## Background Hadronic contribution to the photon propagator in the VMD model Measurements of the interaction between energetic photons and hadrons show that the interaction is much more intense than expected by the interaction of merely photons with the hadron's electric charge. Furthermore, the interaction of energetic photons with protons is similar to the interaction of photons with neutrons[2] in spite of the fact that the electric charge structures of protons and neutrons are substantially different. According to VMD, the photon is a superposition of the pure electromagnetic photon (which interacts only with electric charges) and vector meson. Just after 1970, when more accurate data on the above processes became available, some discrepancies with the VMD predictions appeared and new extensions of the model were published.[3] These theories are known as Generalized Vector Meson Dominance theories (GVMD). ## VMD and Hidden Local Symmetry Whilst the ultraviolet description of the standard model is based on QCD, work over many decades has involved writing a low energy effective description of QCD, and further, positing a possible "dual" description. One such popular description is that of the hidden local symmetry.[4] The dual description is based on the idea of emergence of gauge symmetries in the infrared of strongly coupled theories. Gauge symmetries are not really physical symmetries (only the global elements of the local gauge group are physical). This emergent property of gauge symmetries was demonstrated in Seiberg duality[5] and later in the development of the AdS/CFT correspondence.[6] In its generalised form, Vector Meson Dominance appears in AdS/CFT, AdS/QCD, AdS/condensed matter and some Seiberg dual constructions. It is therefore a common place idea within the theoretical physics community. ## Criticism Measurements of the photon-hadron interactions in higher energy levels show that VMD cannot predict the interaction in such levels. In his Nobel lecture[7]J.I. Friedman summarizes the situation of VMD as follows: "...this eliminated the model [VMD] as a possible description of deep inelastic scattering... calculations of the generalized vector-dominance failed in general to describe the data over the full kinematic range..." The Vector Meson Dominance model still sometimes makes significantly more accurate predictions of hadronic decays of excited light mesons involving photons than subsequent models such as the relativistic quark model for the meson wave function and the covariant oscillator quark model.[8] Similarly, the Vector Meson Dominance model has outperformed perturbative QCD in making predictions of transitional form factors of the neutral pion meson, the eta meson, and the eta prime meson, that are "hard to explain within QCD."[9] And, the model accurately reproduces recent experimental data for rho meson decays.[10] Generalizations of the Vector Meson Dominance model to higher energies, or to consider additional factors present in cases where VMD fails, have been proposed to address the shortcomings identified by Friedman and others.[11][12] ## Notes 1. ^ Sakurai, J.J (1960). "Theory of strong interactions". Annals of Physics. Elsevier BV. 11 (1): 1-48. doi:10.1016/0003-4916(60)90126-3. ISSN 0003-4916. 2. ^ Bauer, T. H.; Spital, R. D.; Yennie, D. R.; Pipkin, F. M. (1978-04-01). "The hadronic properties of the photon in high-energy interactions". Reviews of Modern Physics. American Physical Society (APS). 50 (2): 261-436. doi:10.1103/revmodphys.50.261. ISSN 0034-6861. 3. ^ Sakurai, J.J.; Schildknecht, D. (1972). "Generalized vector dominance and inelastic electron-proton scattering--the small ${\displaystyle \omega }$? region". Physics Letters B. Elsevier BV. 40 (1): 121-126. doi:10.1016/0370-2693(72)90300-0. ISSN 0370-2693. 4. ^ Bando, Masako; Kugo, Taichiro; Yamawaki, Koichi (1988). "Nonlinear realization and hidden local symmetries". Physics Reports. Elsevier BV. 164 (4-5): 217-314. doi:10.1016/0370-1573(88)90019-1. ISSN 0370-1573. 5. ^ Seiberg, N. (1995). "Electric-magnetic duality in supersymmetric non-Abelian gauge theories". Nuclear Physics B. Elsevier BV. 435 (1-2): 129-146. arXiv:hep-th/9411149. doi:10.1016/0550-3213(94)00023-8. ISSN 0550-3213. 6. ^ Maldacena, Juan (1999). "The Large N limit of superconformal field theories and supergravity". International Journal of Theoretical Physics. Springer Science and Business Media LLC. 38 (4): 1113-1133. arXiv:hep-th/9711200. doi:10.1023/a:1026654312961. ISSN 0020-7748. 7. ^ Friedman, Jerome I. (1991-07-01). "Deep inelastic scattering: Comparisons with the quark model". Reviews of Modern Physics. American Physical Society (APS). 63 (3): 615-627. doi:10.1103/revmodphys.63.615. ISSN 0034-6861. 8. ^ See, e.g., The COMPASS Collaboration, "Measurement of radiative widths of a2(1320) and ?2(1670)" (March 11, 2014) https://arxiv.org/pdf/1403.2644v1.pdf 9. ^ Yaroslav Klopot, Armen Oganesian and Oleg Teryaev, "Axial anomaly and vector meson dominance model" (4 December 2013) https://arxiv.org/abs/1312.1226 10. ^ D. García Gudiño, G. Toledo Sánchez, "Determination of the magnetic dipole moment of the rho meson" (27 May 2013) https://arxiv.org/abs/1305.6345 11. ^ V. A. Petrov, "On Vector Dominance" (20 Dec. 2013) https://arxiv.org/pdf/1312.5500v1.pdf 12. ^ Stefan Leupold and Carla Terschlusen, "Towards an effective field theory for vector mesons" (11 Jun 2012) (also analyzing the circumstances where VMD is and is not successful in predicting experimental outcomes) https://arxiv.org/abs/1206.2253	2020-04-03 19:27:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8240300416946411, "perplexity": 3998.14675876226}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370518622.65/warc/CC-MAIN-20200403190006-20200403220006-00293.warc.gz"}
https://www.cableizer.com/documentation/h_soil/	# Heat transfer coefficient wall-soil Assuming uniform temperatures at mudline and the outer surface of the pipe (i.e. Dirichlet boundary conditions), the heat transfer coefficient equivalent to the thermal resistance of the surrounding soil is commonly calculated from the following expression (Carslaw1959). This expression is appropriate for deeply buried pipes. When the top of line is close the soil surface (i.e. when the pipe is just barely buried), the heat transfer coefficient increases to infinity. Symbol $h_{soil}$ Unit W/(K.m$^2$) Formulae $\frac{2 k_{4}}{D_{ext} \alpha_{0}}$ $D_{ext}$ Used in $D_{soil}$ $U_{buried}$	2020-08-06 19:18:38	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9038137197494507, "perplexity": 961.9760638758264}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439737019.4/warc/CC-MAIN-20200806180859-20200806210859-00588.warc.gz"}
https://proofwiki.org/wiki/Definition:Riemann_Zeta_Function	# Definition:Riemann Zeta Function The analytic continuation of $\zeta$ ## Definition The Riemann Zeta Function $\zeta$ is the complex function defined on the half-plane $\Re \paren s > 1$ as the series: $\displaystyle \map \zeta s = \sum_{n \mathop = 1}^\infty \frac 1 {n^s}$ ## Analytic Continuation By Analytic Continuations of Riemann Zeta Function, $\zeta$ has a unique analytic continuation to $\C \setminus \set 1$. This analytic continuation is still called the Riemann zeta function and still denoted $\zeta$. ## Also see • Results about the Riemann $\zeta$ function can be found here. ## Source of Name This entry was named for Georg Friedrich Bernhard Riemann. ## Historical Note The Riemann zeta function was discussed by Bernhard Riemann in his $1859$ article Ueber die Anzahl der Primzahlen under einer gegebenen Grösse. Studying it led Jacques Salomon Hadamard to his proof of the Prime Number Theorem. All of the statements made in that paper have now been proved except for one. The last remaining statement which has not been resolved is the Riemann Hypothesis. When Riemann first investigated this function, he used $s$ instead of the more typical complex variable $z$. To this day, $s = \sigma + i t$ is still typically used as the complex variable in investigations of the zeta function.	2019-03-23 19:34:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9750133752822876, "perplexity": 769.333437221695}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-13/segments/1552912202924.93/warc/CC-MAIN-20190323181713-20190323203713-00533.warc.gz"}
https://www.love2d.org/forums/viewtopic.php?f=14&t=86466	## Get mouse coordinates on a rotated "grid" Show off your games, demos and other (playable) creations. Jasoco Inner party member Posts: 3651 Joined: Mon Jun 22, 2009 9:35 am Location: Pennsylvania, USA Contact: ### Get mouse coordinates on a rotated "grid" So I'm working on this thing... (Everyone's doing it these days) And I got to thinking, it would be neat if I could figure out what tile the mouse is actually over. But I don't know the math to do that. The "ground" image (The checkerboard grid image) is a simple Löve transformation where it's rotated at the angle I want first, then scaled to give it a fake perspective. Here's another simpler example with the Y scaling and fake perspective scaling turned off: So how would I go about figuring out where the mouse is given that the corner shown above is 0,0 and the image's x and y offset is the position of the player? (The purple square) Surely it's pretty simple. But Google isn't helping and only seems to have answers for dealing with plain old isometric grids. My grid has free-rotation. Edit: How in the world did I end up putting this in Games and Creations? Sorry I meant to put it in Support. Last edited by Jasoco on Thu Mar 14, 2019 11:34 am, edited 1 time in total. monolifed Party member Posts: 127 Joined: Sat Feb 06, 2016 9:42 pm ### Re: Get mouse coordinates on a rotated "grid" I think it is screen space/coordinates/position to world space/coordinates/position transformation/projection problem So maybe this helps? https://stackoverflow.com/questions/769 ... ace-coords https://stackoverflow.com/questions/467 ... oordinates Nelvin Party member Posts: 114 Joined: Mon Sep 12, 2016 7:52 am Location: Germany ### Re: Get mouse coordinates on a rotated "grid" pgimeno Party member Posts: 2211 Joined: Sun Oct 18, 2015 2:58 pm ### Re: Get mouse coordinates on a rotated "grid" Nelvin wrote: Thu Mar 14, 2019 8:04 am Have you tried https://love2d.org/wiki/love.graphics.i ... sformPoint ? Yeah but this may be off by 1 frame if used within love.update anywhere but in love.draw. Edit: And that's the best case. Actually the problems may be worse. It won't work unless that's the last transformation in love.draw. Instead, create a Transform object with love.math.newTransform. Instead of performing the deformations at draw time with love.graphics.scale etc, update the Transform object as needed. At draw time, use love.graphics.replaceTransform to set the LÖVE transformation to the Transform object. When you need the mouse position, use Transform:inverseTransformPoint. dusoft Party member Posts: 106 Joined: Fri Nov 08, 2013 12:07 am ### Re: Get mouse coordinates on a rotated "grid" Could this help? https://github.com/kikito/gamera I think it contains all the helper functions you need. You don't have to even use, just check the code. Jasoco Inner party member Posts: 3651 Joined: Mon Jun 22, 2009 9:35 am Location: Pennsylvania, USA Contact: ### Re: Get mouse coordinates on a rotated "grid" Nelvin wrote: Thu Mar 14, 2019 8:04 am Have you tried https://love2d.org/wiki/love.graphics.i ... sformPoint ? I can't seem to get this to work. Maybe I'm using it wrong because there's no sample code. It just returns the normal screen coordinates non-rotated or distorted. dusoft wrote: Thu Mar 14, 2019 10:12 pm Could this help? https://github.com/kikito/gamera I think it contains all the helper functions you need. You don't have to even use, just check the code. I can't find anything in the code that would help. Edit: In the meantime I am using a really roundabout and rudimentary (and expensive) method to get the currently clicked tile. Basically I first create a canvas the same size as the world map and draw a grid of uniquely colored squares for each tile using the following formula: Code: Select all for x = 0, self.worldWidth+1 do for y = 0, self.worldHeight+1 do lgr.setColor((x4) / 255, 0, (y4) / 255) lgr.rectangle("fill", x * self.tileSize, y * self.tileSize, self.tileSize, self.tileSize) end end It also creates a screen sized canvas that won't be displayed and clears it as green then using the same transformation data (rotation, scaling, offset) it draws the color map to that canvas. Then when I press and hold the mouse down it uses Löve's functions for getting the color of a pixel in an image data which in 11.x requires me to first convert the canvas to imagedata then getPixel() of that image data. It causes a slight frame rate hiccup but it's not enough to ruin anything. Once it receives the RGB value of the pixel at the current coordinate of the "grid map canvas" it first checks to see if the Green value is set to 1. If it is, the pixel is outside the map (I clear the canvas to green before drawing the transformed gradient map to it. The gradient map only uses the red and blue channels.) If not then it takes its Red and Blue values and multiplies them by 255 then divides by 4 to get the X and Y tile coordinate. (The 4 is because the map is 64x64 and 64 is one fourth 256 and I just wanted the gradient to look pretty. If I left out the multiplying by 4 part it would have looked too dark, but would still work the same. Also if I made the map 128x128 instead I'd just use 2 instead. It's all just for show really.) And amazingly this method works. It's overkill of course and expensive. Sadly it means I cant do live updating of the selected tile without losing frame rate. (creating image data and getting a pixel color is expensive as I said) But it works perfectly for at least being able to click a tile. I did all that work just to get around having to know math. lol However I'll gladly keep trying to figure out if there's a legit way to do it using math. Here's what it looks like behind the scenes: Screen Shot 2019-03-15 at 1.49.00 AM.png (516.92 KiB) Viewed 7578 times Nelvin Party member Posts: 114 Joined: Mon Sep 12, 2016 7:52 am Location: Germany ### Re: Get mouse coordinates on a rotated "grid" Hm so far haven't used it (my game is built using 0.10.2) but I've just tried a small hacky version and it seems to work as expected. See the .love file attached. It works both, using the transform object and the helper function of love.graphics. The one in love.graphics obviously only as long as it represents the desired state, i.e. if you push/pop in your love.draw at the end and try to transform the coordinates somewhere else, it won't work. Attachments transform.love pgimeno Party member Posts: 2211 Joined: Sun Oct 18, 2015 2:58 pm ### Re: Get mouse coordinates on a rotated "grid" Jasoco wrote: Thu Mar 14, 2019 4:23 am So how would I go about figuring out where the mouse is given that the corner shown above is 0,0 and the image's x and y offset is the position of the player? (The purple square) You need to compensate for that too, by adding the player's position to the result. dusoft Party member Posts: 106 Joined: Fri Nov 08, 2013 12:07 am ### Re: Get mouse coordinates on a rotated "grid" Jasoco wrote: Fri Mar 15, 2019 4:54 am dusoft wrote: Thu Mar 14, 2019 10:12 pm Could this help? https://github.com/kikito/gamera I think it contains all the helper functions you need. You don't have to even use, just check the code. I can't find anything in the code that would help. Maybe just use that library as it offers all the basic functions you are having difficulties to code yourself? But maybe I am misunderstanding what you need. dusoft Party member Posts: 106 Joined: Fri Nov 08, 2013 12:07 am ### Who is online Users browsing this forum: No registered users and 11 guests	2020-08-04 19:28:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.42424359917640686, "perplexity": 2357.5208846079736}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735882.86/warc/CC-MAIN-20200804191142-20200804221142-00451.warc.gz"}
https://www.winmoney4u.info/sweepstakesqvc/qui-a-gagne-5000-pch-une-semaine-pour-la-vie-tirages-au-sort-kernersville.html	There is a little-known phenomenon for binomial GLMs that was pointed out by Hauck & Donner (1977: JASA 72:851-3). The standard errors and t values derive from the Wald approximation to the log-likelihood, obtained by expanding the log-likelihood in a second-order Taylor expansion at the maximum likelihood estimates. If there are some \hat\beta_i which are large, the curvature of the log-likelihood at \hat{\vec{\beta}} can be much less than near \beta_i = 0, and so the Wald approximation underestimates the change in log-likelihood on setting \beta_i = 0. This happens in such a way that as \|\hat\beta_i\| \to \infty, the t statistic tends to zero. Thus highly significant coefficients according to the likelihood ratio test may have non-significant t ratios. Les amateurs de tirages au sort Whoever you are, whatever you're looking for, we have the perfect place for you. Our 28,974,234 listings include 6,226,750 listings of homes, apartments, and other unique places to stay, and are located in 153,519 destinations in 227 countries and territories. Booking.com B.V. is based in Amsterdam, the Netherlands and is supported internationally by 198 offices in 70 countries. knnTree Construct or predict with k-nearest-neighbor classifiers, using cross-validation to select k, choose variables (by forward or backwards selection), and choose scaling (from among no scaling, scaling each column by its SD, or scaling each column by its MAD). The finished classifier will consist of a classification tree with one such k-nn classifier in each leaf.	2019-11-18 20:13:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7588131427764893, "perplexity": 1890.2382897498142}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496669813.71/warc/CC-MAIN-20191118182116-20191118210116-00233.warc.gz"}
http://prob140.org/textbook/Chapter_03/03_Equality.html	## Equality We know what it means for two numbers to be equal: they are at the same spot on the number line. Equality of random variables, however, can be of more than one kind. ### Equal Two random variables $X$ and $Y$ defined on the same outcome space are equal if their values are the same for every outcome in the space. The notation is $X = Y$ and it means that $$X(\omega) = Y(\omega) \text{ for all } \omega \in \Omega$$ Informally, this says that no matter what the outcome, if $X$ is 10 then $Y$ must be 10 too; if $X$ is 11, $Y$ must be 11, and so on. An example will make this clear. Let $N_H$ be the number of heads in three tosses of a coin, and let $N_T$ be the number of tails in the same three tosses. Now consider the new random variable $M = 3 - N_T$. The two random variables $N_H$ and $M$ are equal. For every possible outcome of the three tosses, the value of $N_H$ is equal to the value of $M$. We write this simply as $N_H = M$. Equivalently, $N_H = 3 - N_T$. ### Equal in Distribution $N_H$ and $N_T$, as defined above, are not equal. For example, $$N_H(\text{TTT}) = 0 ~~~ \text{but} ~~~ N_T(\text{TTT}) = 3$$ However, there is a sense in which the number of heads "behaves in the same way" as the number of tails. The two random variables have the same probability distribution. The outcome space is three_tosses: coin = make_array('H', 'T') three_tosses = list(product(coin, repeat=3)) three_tosses [('H', 'H', 'H'), ('H', 'H', 'T'), ('H', 'T', 'H'), ('H', 'T', 'T'), ('T', 'H', 'H'), ('T', 'H', 'T'), ('T', 'T', 'H'), ('T', 'T', 'T')] There are only eight outcomes, so it is easy to inspect the table and write the distributions of $N_H$ and $N_T$. Both take the values 0, 1, 2, and 3, with probabilities 1/8, 3/8, 3/8, and 1/8 respectively. This distribution is shown in the table below. dist = Table().values(np.arange(4)).probabilities(make_array(1, 3, 3, 1)/8) dist Value Probability 0 0.125 1 0.375 2 0.375 3 0.125 We say that $N_H$ and $N_T$ are equal in distribution. In general, two random variables $X$ and $Y$ are equal in distribution if they have the same probability distribution. That is, they have the same set of possible values and the same probabilities for all those values. Equality in distribution is denoted as $$X \stackrel{d}{=} Y$$ ### Relation between the Equalities Equality is stronger than equality in distribution. If two random variables are the same, outcome by outcome, then they must have the same distribution because they are the same function on the outcome space. That is, for any two random variables $X$ and $Y$, $$X = Y \implies X \stackrel{d}{=} Y$$ But as the example of heads and tails in three tosses shows, the converse need not be true. ### Example: Two Cards Dealt from a Small Deck A deck contains 10 cards, labeled 1, 2, 2, 3, 3, 3, 4, 4, 4, 4. Two cards are dealt at random without replacement. Let $X_1$ be the label on the first card and $X_2$ be the label on the second card. Question 1. Are $X_1$ and $X_2$ equal? Answer 1. No, because the outcome could be 31 in which case $X_1 = 3$ and $X_2 = 1$. Question 2. Are $X_1$ and $X_2$ equal in distribution? Answer 2. Let's find the two distributions and compare. Clearly the possible values are 1, 2, 3, and 4 in each case. The distribution of $X_1$ is easy: $$P(X_1 = i ) = \frac{i}{10} , ~~ i = 1, 2, 3, 4$$ When a distribution is defined by a formula like this, you can define a function that does what the formula says: def prob1(i): return i/10 You can create a probability distribution object for $X_1$ using values as before but now with the probability_function method. The argument to probability_function is the name of the function that takes $i$ as its argument and returns $P(X_1 = i)$. possible_i = np.arange(1, 5, 1) dist_X1 = Table().values(possible_i).probability_function(prob1) dist_X1 Value Probability 1 0.1 2 0.2 3 0.3 4 0.4 Convince yourself that the function prob2 below returns $P(X_2 = i)$ for each $i$. The event has been partitioned according to the value of $X_1$. def prob2(i): if i == 1: return (9/10)(1/9) else: return (i/10)((i-1)/9) + ((10-i)/10)*(i/9) dist_X2 = Table().values(possible_i).probability_function(prob2) dist_X2 Value Probability 1 0.1 2 0.2 3 0.3 4 0.4 The two distributions are the same! Here is yet another example of symmetry in sampling without replacement. The conclusion is $$X_1 \stackrel{d}{=} X_2$$	2019-11-17 02:10:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8638134598731995, "perplexity": 269.8878860593804}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496668782.15/warc/CC-MAIN-20191117014405-20191117042405-00166.warc.gz"}
https://b-ok.org/book/2134124/503c78	Main Multicriteria Scheduling: Theory, Models and Algorithms # Multicriteria Scheduling: Theory, Models and Algorithms , Scheduling and multicriteria optimisation theory have been subject, separately, to numerous studies. Since the last twenty years, multicriteria scheduling problems have been subject to a growing interest. However, a gap between multicriteria scheduling approaches and multicriteria optimisation field exits. This book is an attempt to collect the elementary of multicriteria optimisation theory and the basic models and algorithms of multicriteria scheduling. It is composed of numerous illustrations, algorithms and examples which may help the reader in understanding the presented concepts. This book covers general concepts such as Pareto optimality, complexity theory, and general method for multicriteria optimisation, as well as dedicated scheduling problems and algorithms: just-in-time scheduling, flexibility and robustness, single machine problems, parallel machine problems, shop problems, etc. The second edition contains revisions and new material. Year: 2006 Edition: 2 Publisher: Springer-Verlag Berlin Heidelberg Language: english Pages: 360 / 368 ISBN 13: 978-3-540-24789-0 File: PDF, 6.76 MB ## You may be interested in ### Handbook on Scheduling - From Theory to Applications Year: 2007 Language: english File: PDF, 25.07 MB ### Integration der Unsicherheitsaspekte in die schedule-Optimierung : empirische Modellierung unter Anwendung der Fuzzy-Theorie am Beispiel des Luftverkehrs Year: 2006 Language: german File: PDF, 2.79 MB ### Schedule-Based Dynamic Transit Modeling: theory and applications Year: 2004 Language: english File: PDF, 9.47 MB ### Scheduling Theory. Multi-Stage Systems Year: 1994 Language: english File: PDF, 11.87 MB ### Automated Scheduling and Planning: From Theory to Practice Year: 2013 Language: english File: PDF, 5.90 MB You can write a book review and share your experiences. Other readers will always be interested in your opinion of the books you've read. Whether you've loved the book or not, if you give your honest and detailed thoughts then people will find new books that are right for them. 1 ### Principles of Law Relating to International Trade Year: 2006 Language: english File: PDF, 7.26 MB 2 ### Enabling Technologies for Wireless E-Business Year: 2006 Language: english File: PDF, 10.33 MB Multicriteria Scheduling Second Edition Vincent T’kindt Jean-Charles Billaut Multicriteria Scheduling Theory, Models and Algorithms Translated from French by Henry Scott Second Edition with 138 Figures and 15 Tables 123 Associate Professor Vincent T’kindt, Professor Jean-Charles Billaut Université François-Rabelais de Tours Laboratoire d’Informatique 64 avenue Jean Portalis 37200 Tours France Translator Henry Scott www.hgs-scientific-translations.co.uk Library of Congress Control Number: 2005937590 ISBN-10 3-540-28230-0 2nd ed. Springer Berlin Heidelberg New York ISBN-13 978-3-540-28230-3 2nd ed. Springer Berlin Heidelberg New York ISBN 3-540-43617-0 1st ed. Springer Berlin Heidelberg New York This work is subject to copyright.All rights are reserved, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microﬁlm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer-Verlag.Violations are liable for prosecution under the German Springer is a part of Springer Science+Business Media springeronline.com © Springer-Verlag Berlin Heidelberg 2002, 2006 Printed in Germany The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a speciﬁc statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Cover design: Erich Kirchner Production: Helmut Petri Printing: Strauss Offsetdruck SPIN 11538080 Printed on acid-free paper – 42/3153 – 5 4 3 2 1 0 Preface to the second edition It is a real pleasure for us to present the second edition of this book on multicriteria scheduling. In this preface we would like to introduce the reader with the improvements made over the first edition. During the writing of the first edition of this book we were focused on putting in it all the results, algorithms and models necessary for the reader to tackle correctly the field of multicriteria scheduling, which is at the crossroad of several research domains: from multicriteria optimisation to scheduling. Writing a second edition is a totally different exercise since we concentrate more on refining, augmenting and, in a sense, making growing the existing manuscript. some chapters as Chapters 5 and 7. Besides, new significant research results published since the first edition have been included into existing chapters of that second edition. We review hereafter the most important changes. Chapters 2 and 4 now include a survey on the complexity of counting and enumeration optimisation problems with application to multicriteria scheduling. These two chapters provide theoretical tools for evaluating the complexity of the enumeration of the set of strict Pareto optima. Chapter 4 also includes new real-life applications of multicriteria scheduling. Chapter 5 has been drastically revised and now provides a general unified framework for Just-in-Time scheduling problems. Besides, classic optimal timing algorithms, which calculate optimal start times of operations when the jobs order is fixed, are now presented. At last, chapter 6 is a new chapter dealing with robustness in multicriteria scheduling. This research area has been subject to a growing interest in the literature since the last ten years, notably when considering a criterion of flexibility or robustness in addition to a classic scheduling criterion. Henceforth, the aim of some scheduling problems become to increase the robustness of the calculated solution for its pratical use. Providing flexibility is a way to ensure a certain robustness when unexpected events occur in the shop. We hope that this new edition will become an important tool and a practical guide for novice an senior researchers that work on multicriteria scheduling. V. T'KINDT and J.-C. BILLAUT Tours (Prance), October 15th 2005 Preface to the first edition Prom Theory to Practice, there is a world, and scheduUng does not escape this immutable rule. For more than fifty years, theoretical researches on scheduling and complexity theory have improved our knowledge on both a typology of academic problems, mainly involving a single criterion, and on their solving. Though this work is far from being completed, a few famous books have been a major breakthrough. The typology will be all the more useful as it takes more and more realistic constraints into account. This is just a matter of time. The relevance of some single criteria, their equivalence and their conflict have been studied... Yet, numerous genuine problems, even outside the realm of scheduling, do not square with these single criterion approaches. For example, in a production shop, minimising the completion time of a set of jobs may be as interesting as retaining a maximum fragmentation of idle times on an easily damaged machine and minimising the storage of in-process orders. Moreover, even though the optimal solutions to the F2\\Cmax yielded by S.M. Johnson's famous algorithm are numerous, they are far from appearing equivalent to the decision maker when their structure is analysed. A genuine scheduling problem, in essence, involves multiple criteria. Besides, more general books on Decision Aid in a multicriteria environment have been published and a pool of researchers have long tackled the problem. Undoubtedly, a synthesis book offering a state-of-the-art on the intersection of both the fields of Scheduling and Multicriteria Decision Aid and providing a framework for tackling multicriteria scheduling problems is a must. I am most happy to present this book. It is divided in four parts: - the first one deals with research on scheduling, now an important branch of operational research. - the second one presents theories on Decision Aid and Multicriteria Optimisation as well as a framework for the resolution of multicriteria scheduling problems. VIII Preface - the third and fourth parts involve a tremendous work since they contain state-of-the-arts on multicriteria scheduhng problems. Numerous works and resolution algorithms are detailed. In my opinion, this book will become a reference book for researchers working on scheduling. Moreover, I am convinced it will help PhD students suitably and quickly embark on a fascinating adventure in this branch of Operational Research. May they be numerous in joining us... I very warmly thank MM. Vincent T'kindt and Jean-Charles Billaut for their tenacity in writing this significant book, and Springer-Verlag publishing for entrusting them. Professor C. PROUST Tours (Prance), february 22th 2002 The authors are very grateful to all the people who have directly or indirectly contributed to the birth of this book. Professor Christian Proust is at the root of this research and is undoubtedly the grandfather of this book. We would also like to thank the members of the research team "Scheduling and Control" of the Laboratory of Computer Science of the University of Tours for creating a friendly environment and thus for having promoted the emergence of this book. In this vein, all the technical and administrative persons of the E3i school have also to be thanked. At last, we would like to thank Professor Jacques Teghem of the "Faculte Polytechnique de Mons" for having provided excellent ideas and remarks which have helped in improving this book. Contents 1. Introduction to scheduling 1.1 Definition 1.2 Some areas of application 1.2.1 Problems related to production 1.2.2 Other problems 1.3 Shop environments 1.3.1 Scheduling problems without assignment 1.3.2 Scheduling and assignment problems with stages 1.3.3 General scheduling and assignment problems 1.4 Constraints 1.5 Optimality criteria 1.5.1 Minimisation of a maximum function: "minimax" criteria 1.5.2 Minimisation of a sum function: "minisum" criteria . . . 1.6 Typologies and notation of problems 1.6.1 Typologies of problems 1.6.2 Notation of problems 1.7 Project scheduling problems 1.8 Some fundamental notions 1.9 Basic scheduling algorithms 1.9.1 Scheduling rules 1.9.2 Some classical scheduling algorithms 5 5 6 6 7 7 8 8 9 9 12 13 13 14 14 16 17 18 21 21 22 2. Complexity of problems and algorithms 2.1 Complexity of algorithms 2.2 Complexity of problems 2.2.1 The complexity of decision problems 2.2.2 The complexity of optimisation problems 2.2.3 The complexity of counting and enumeration problems 2.3 Application to scheduling 29 29 32 33 38 40 48 3. Multicriteria optimisation theory 3.1 MCDA and MCDM: the context 3.1.1 MultiCriteria Decision Making 53 53 54 X 4. Contents 3.1.2 MultiCriteria Decision Aid 3.2 Presentation of multicriteria optimisation theory 3.3 Definition of optimality 3.4 Geometric interpretation using dominance cones 3.5 Classes of resolution methods 3.6 Determination of Pareto optima 3.6.1 Determination by convex combination of criteria 3.6.2 Determination by parametric analysis 3.6.3 Determination by means of the e-constraint approach . 3.6.4 Use of the Tchebycheff metric 3.6.5 Use of the weighted Tchebycheff metric 3.6.6 Use of the augmented weighted Tchebycheff metric . . . 3.6.7 Determination by the goal-attainment approach 3.6.8 Other methods for determining Pareto optima 3.7 Multicriteria Linear Programming (MLP) 3.7.1 Initial results 3.7.2 AppHcation of the previous results 3.8 Multicriteria Mixed Integer Programming (MMIP) 3.8.1 Initial results 3.8.2 Application of the previous results 3.8.3 Some classical algorithms 3.9 The complexity of multicriteria problems 3.9.1 Complexity results related to the solutions 3.9.2 Complexity results related to objective functions 3.9.3 Summary 3.10 Interactive methods 3.11 Goal programming 3.11.1 Archimedian goal programming 3.11.2 Lexicographical goal programming 3.11.3 Interactive goal programming 3.11.4 Reference goal programming 3.11.5 Multicriteria goal programming 54 55 57 60 62 64 64 70 72 76 79 81 86 91 92 93 93 94 94 95 97 100 100 101 106 107 108 Ill Ill Ill 112 112 A n approach to multicriteria scheduling problems 4.1 Justification of the study 4.1.1 Motivations 4.1.2 Some examples 4.2 Presentation of the approach 4.2.1 Definitions 4.2.2 Notation of multicriteria scheduling problems 4.3 Classes of resolution methods 4.4 Application of the process - an example 4.5 Some complexity results for multicriteria scheduling problems 113 113 113 114 118 118 121 122 123 124 Contents 5. 6. XI Just-in-Time scheduling problems 5.1 Presentation of Just-in-Time (JiT) scheduling problems 5.2 Typology of JiT scheduling problems 5.2.1 Definition of the due dates 5.2.2 Definition of the JiT criteria 5.3 A new approach for JiT scheduling 5.3.1 Modelling of production costs in JiT scheduling for shop problems 5.3.2 Links with objective functions of classic JiT scheduling 5.4 Optimal timing problems 5.4.1 The l\di,seq\Fe{f'',E^) problem 5.4.2 The Poo\prec, fi convex\ ^ ^ fi problem 5.4.3 The l\fi piecewise linear\Fi{Y^^ fi^ ^ . 7^) problem . . . 5.5 Polynomially solvable problems 5.5.1 The l\di = d> Y.Vi\F(>{E,f) problem 5.5.2 The l\di = d unknown^nmit\F£{E^T^d) problem 5.5.3 The l\pi C [pijpj HN, di = d non restrictive\Fe(E,T, CC"^) problem ^ ._^ 5.5.4 The P\di = d non restrictive^nmit\F£{E^T) problem . 5.5.5 The P\di = d unknown^ nmit\Fe{E^T) problem 5.5.6 The P\di = d unknown,pi = p,nmit\F(>{E, T^d) problem 5.5.7 The R\pi^j € [Pi,j;Pij],cfi = d unknown\Fi{T,E, CC"^) problem 5.5.8 Other problems 5.6 TVP-hard problems 5.6.1 The l\di, nmit\Fe(E'',T^)_pioblem 5.6.2 The F\prmu,di,nmit\Fe{E'^,T^) problem 5.6.3 The P\di = d non restrictive, nmit\fmax{E , T ) problem 5.6.4 Other problems 5.7 Open problems 5.7.1 The Q\di = d unknown, nmit\Fi{E,T) problem 5.7.2 Other problems 135 135 136 136 137 139 Robustness considerations 6.1 Introduction to flexibility and robustness in scheduling 6.2 Approaches that introduce sequential flexibility 6.2.1 Groups of permutable operations 6.2.2 Partial order between operations 6.2.3 Interval structures 6.3 Single machine problems 6.3.1 Stability vs makespan 6.3.2 Robust evaluation vs distance to a baseline solution... 193 193 195 195 197 199 201 201 202 141 145 147 147 149 153 153 153 155 157 157 159 165 169 170 173 173 176 178 182 188 188 189 XII Contents 6.4 6.5 Flowshop and jobshop problems 203 6.4.1 Average makespan of a neighbourhood 203 6.4.2 Sensitivity of operations vs makespan 203 Resource Constrained Project ScheduUng Problems (RCPSP) 204 6.5.1 Quality in project scheduling vs makespan 204 6.5.2 Stability vs makespan 205 7. Single machine problems 7.1 Polynomially solvable problems 7.1.1 Some l\di\C, /max problems 7.1.2 The l\si,pmtn,nmit\Fe{C^Pmax) problem 7.1.3 The l\pi € [pi;Piidi\Fe{Tmax.'CC^) problem 7.1.4 The l\pi e [pi',Piidi\Fe(C,CC'^) problem 7.1.5 Other problems 7.2 J\fV-hdiid problems_. 7.2.1 The l\di\T, C problem 7.2.2 The l\rupi £ [pi;pj H N\Fe{Cmax,CC^) problem 7.2.3 The l \| n , p i G [pi'.Pi] n N\Fe(Ü'^.CC'") problem 7.2.4 Other problems 7.3 Open problems .^ 7.3.1 The l\di\Ü,Tmax problem 7.3.2 Other problems 207 207 207 215 216 219 219 222 222 223 225 226 230 230 234 8. Shop problems 235 8.1 Two-machine flowshop problems 235 8.1.1 The F2\prmu\Lex{CmaxjC) problem 235 8.1.2 The F2\prmu\Fi{Cmax^ C)problem 250 8.1.3 The F2\prmu,ri\Fe{Cmax,C) problem 256 8.1.4 The F2\prmu\e{C/Cmax) problem 256 8.1.5 The F2\prmu,di\#{Cmax,Trnax) problem 262 8.1.6 The F2\prmu, di\#{Cmax,U) problem 265 8.1.7 The F2\prmu,di\#{Cmax^T) problem 267 8.2 m-machine flowshop problems 270 8.2.1 The F\prmu\Lex{Cmax2_C) problem 270 8.2.2 The F\prmu\#{Cmax,C) problem 272 8.2.3 The F\prmu,di\e{Cmax/Tmax) problem 277 8.2.4 The F\pij € [pij;Pi^j],prmu\Fe{Cmax, CC"^) problem. 280 8.2.5 The F\pi^j =Pie [2uPi],prmu\#{Cmax,'CC^) problem281 8.3 Jobshop and Openshop problems 284 8.3.1 Jobshop problems 284 8.3.2 The 02\\Lex{Cmax,C) problem 284 8.3.3 The 03\\Lex{Cmax,C) problem 286 Contents 9. Parallel machines problems 9.1 Problems with identical parallel machines 9.1.1 The P2\pmtn,di\e{Lmax/Cmax) problem 9.1.2 The P3\pmtn,di\e{Lma^/Cmax) problem 9.1.3 The P2\di\Lex{Trriax, U) problem 9.1.4 The P\|di\|#(C,[/)^roblem 9.1.5 The P\pmtn\Lex{C, Cmax) problem 9.2 Problems with uniform parallel machines 9.2.1 The Q\pi = p\e{fmaxl9max) problem 9.2.2 The Q\pi = v\ei^/fmax) problem 9.2.3 The Q\vmtn\e{C/Cmax) problem 9.3 Problems with unrelated parallel machines 9.3.1 The R\pi^j € [^i^j,PißFt(C,'CC^) problem 9.3.2 The R\pmtn\e{Fi{Imax,'M)/Cmax) problem XIII 287 287 287 290 293 295 296 297 297 302 303 310 310 311 10. Shop problems with assignment 315 10.1 A hybrid flowshop problem with three stages 315 10.2 Hybrid flowshop problems with k stages 316 10.2.1 The HFk, (PM(^))f^i\|\|F^_(C^ax,C) problem 316 10.2.2 The HFk, lPM^^"^)\^^\\e{C/Crnax) problem 318 10.2.3 The HFk, {PM^^^(t))t^i \| r f ^, d f ^ \e{Cmax/Tmax) problem 318 A. Notations A.l Notation of data and variables A.2 Usual notation of single criterion scheduling problems 323 323 323 B. 329 329 330 333 333 334 Synthesis on multicriteria scheduling problems B.l Single machine Just-in-Time scheduhng problems B.2 Single machine problems B.3 Shop problems B.4 Parallel machines scheduling problems B.5 Shop scheduling problems with assignment References 335 Index 357 List of algorithms and mathematical formulations The The The The The The The The The The The The The The The The The The The The The The The The The The The The The The The The The The algorithm EELl of [Lawler, 1973] algorithm EJMl of [Moore, 1968] algorithm ESJl of [Johnson, 1954] algorithm HCDSl of [Campbell et al., 1970] algorithm HNEHl of [Nawaz et a l , 1983] algorithm ESSl of [Sahni, 1979] algorithm EGTWl of [Garey et al., 1988] algorithm ECSl of [Chretienne and Sourd, 2003] algorithm EJKl of [Kanet, 1981a] algorithm EPSSl of [Panwalker et a l , 1982] mathematical formulation ECLTl of [Chen et a l , 1997] algorithm ESAl of [Sundararaghavan and Ahmed, 1984] algorithm EEMl of [Emmons, 1987] algorithm EEM2 of [Emmons, 1987] algorithm ECCl of [Cheng and Chen, 1994] algorithm HOMl of [Ow and Morton, 1988] algorithm HZIEl of [Zegordi et al., 1995] algorithm HLCl of [Li and Cheng, 1994] algorithm HLC2 of [Li and Cheng, 1994] mathematical formulation EFLRl of [Pry et al., 1987b] algorithm HEM3 of [Emmons, 1987] algorithm EWGl of [VanWassenhove and Gelders, 1980] algorithm EHVl of [Hoogeveen and van de Velde, 2001] algorithm ERVl of [Vickson, 1980b] mathematical formulation ECLT2 of [Chen et al., 1997] algorithm HGHPl of [Gupta et al., 1999a] algorithm HGHP2 of [Gupta et al., 1999a] algorithm HCRl of [Rajendran, 1992] algorithm ECRl of [Rajendran, 1992] algorithm EGNWl of [Gupta et al., 2001] algorithm HGNWl of [Gupta et al., 2001] algorithm HTGBl of [T'kindt et al., 2003] algorithm HTMTLl of [T'kindt et al., 2002] algorithm HNHHl of [Nagar et a l , 1995b] 23 24 24 25 26 27 148 152 154 156 158 160 162 164 169 175 178 180 181 188 190 208 216 218 220 232 233 236 238 242 243 246 249 252 XVI The The The The The The The The The The The The The The The The The The The The The The The The The The The The The The List of algorithms and mathematical formulations algorithm HSUl of [Sivrikaya-Serifoglu and Ulusoy, 1998] algorithm ESUl of [Sivrikaya-Serifoglu and Ulusoy, 1998] algorithm HCLl of [Chou and Lee, 1999] algorithm ESKl of [Sayin and Karabati, 1999] algorithm ESK2 of [Sayin and Karabati, 1999] algorithm EDCl of [Daniels and Chambers, 1990] algorithm HDC3 of [Daniels and Chambers, 1990] algorithm ELYJl of [Liao et al., 1997] algorithm HLYJl of [Liao et al., 1997] algorithm ELYJ2 of [Liao et a l , 1997] mathematical formulation ESHl of [Selen and Hott, 1986] mathematical formulation E J W l of [Wilson, 1989] algorithm HGRl of [Gangadharan and Rajendran, 1994] algorithm HGR2 of [Gangadharan and Rajendran, 1994] algorithm HCR3 of [Rajendran, 1995] algorithm HCR5 of [Rajendran, 1994] algorithm HDC4 of [Daniels and Chambers, 1990] algorithm ECSl of [Cheng and Shakhlevich, 1999] algorithm HSHl of [Sarin and Hariharan, 2000] algorithm ELYl of [Leung and Young, 1989] algorithm ETMMl of [Tuzikov et al., 1998] algorithm ETMM2 of [Tuzikov et al., 1998] algorithm ETMM3 of [Tuzikov et al., 1998] algorithm EMPl of [Mc Cormick and Pinedo, 1995] algorithm EMP2 of [Mc Cormick and Pinedo, 1995] mathematical formulation ETBPl of [T'kindt et al., 2001] algorithm ETBP2 of [T'kindt et a l , 2001] mathematical formulation ERMAl of [Riane et al., 1997] mathematical formulation ERMA2 of [Riane et a l , 1997] algorithm EVBPl of [Vignier et al., 1996] 253 255 257 258 260 264 265 268 269 270 272 273 274 274 276 278 279 283 294 298 300 301 303 308 309 312 313 317 319 321 Introduction Scheduling theory first appears in the mid 1950s. Since then the problems addressed become closer to industrial applications, thus increasing in complexity. The layout of the shops taken into account are closer and closer to those met in practice: we encounter shops where the machines are found in different multiple copies, shops where an operation may require several resources simultaneously, or with multipurpose machines, etc. At the same time the embedded constraints are more and more concrete: many authors take into account release dates, the preemption of the jobs, the resource availabilities, etc. Paradoxically, the literature shows that in the majority of the problems addressed, schedules are only evaluated by a single criterion. During the diflFerent phases of planning different criteria can be considered. At a strategic level, at the long term planning phase with several years in view, the objectives concern minimising the costs related to the investment plans for materials, finance, or personel, related to the choice of new directions, or the launching of publicity campaigns. For tactical planning at the medium term phase with several months in view, the objectives always focus on minimising the costs: stock costs (supply or interruption of stocks), costs of getting supplies, costs of modifying production capacity, launching costs, costs of modifying production systems and certain commercial costs ([Merce, 1987], [Giard, 1988]). At the short term planning phase (with the order of a week in view), or scheduling phase, several objectives require the attention of the production executive: above all he must consider the delays that satisfy the customer, next, he must minimise the work-in-process costs in the shop, and finally he must minimise the manufacturing costs related to the time spent to set up the machines or idle periods of the machines. Therefore, a scheduling problem involves multiple criteria. Bernard Roy emphasises ([Roy, 1985]), that taking account of several criteria enables us to propose to the decision maker a more realistic solution. This still holds when solving scheduling problems in an applied context. Literature is dedicated in abundance to the study of multicriteria problems, whatever their field of application. Numerous theoretical works have been developed on multicriteria decision making. The purpose of this book is to provide a 2 Introduction survey, based on a proposed methodology, of the existing methods for solving multicriteria scheduling problems, considering both methods of multicriteria optimisation and scheduling fields. This book is divided into five major parts each devoted to particular themes. The first two chapters are devoted to the rudiments. Chapter 1 sets out the scheduling problems as encountered in the literature. It presents the shop layouts and the classic constraints and criteria. The notation used throughout this book, as well as the notation of scheduling problems, based on that of Graham, Lawler, Lenstra and Rinnooy Kan ([Graham et al., 1979] [Blazewicz et al., 1996]) are provided. We present a new typology, as well as several classifications. Chapter 2 reviews the basic concepts of the complexity of algorithms and the complexity classes of problems. The following two chapters are devoted to multicriteria decision making and multicriteria optimisation, and introduce multicriteria scheduling problems. It opens up a new approach to the resolution of multicriteria scheduling problems. Chapter 3 presents some important concepts related to methodologies of multicriteria decision aids. A large part of the difficulty in solving a multicriteria problem is linked to the way in which the criteria are taken into account. Optimisation techniques, which help in taking account of the criteria are also presented. Chapter 4 presents an approach to the tackling of multicriteria scheduling problems. This approach is divided into three phases. In the first phase the decision maker indicates what constraints define his problem as well as the criteria to be taken into account. The second phase, of taking account of criteria, consists in choosing a resolution approach, i.e. the method which is going to be called upon to calculate a solution. The decision maker also indicates the type of algorithm which he wants to implement: a priori^ interactive or a posteriori algorithm. This phase enables an objective function to be defined for the scheduling problem. The last phase consists of solving the identified scheduling problem. Its resolution leads to the best The next two chapters are chapters devoted to a particular thematic, whatever the configuration of the shop. In Chapter 5 we are concerned with "Just-in-Time" scheduling problems. Both general considerations and technical issues are investigated in this chapter. Chapter 6 focuses on robustness considerations in scheduling when multiple criteria are involved. The next two chapters of this book are devoted to the presentation of multicriteria scheduling problems depending on the shop configuration. Chapter 7 is devoted to single machine problems, which category of problems is undoubtly the most addressed in the literature on multicriteria scheduling. Introduction 3 Chapter 8 is devoted to shop problems, i.e., flowshop, jobshop and openshop problems. The last two chapters are dedicated to the presentation of multicriteria scheduling and assignment problems. Chapter 9 is devoted to multicriteria parallel machines scheduling problems, whilst Chapter 10 is devoted to multicriteria hybrid flowshop scheduling problems. 1. Introduction to scheduling 1.1 Definition Scheduling problems are encountered in all types of systems, since it is necessary to organise and/or distribute the work between many entities. We find in every book in the literature a definition of a scheduling problem as well as its principal components. Among these definitions we can quote the following one [Carlier and Chretienne, 1988]: "Scheduling is to forecast the processing of a work by assigning resources to tasks and fixing their start times. [...] The different components of a scheduling problem are the tasks, the potential constraints, the resources and the objective function. [...] The tasks must be programmed to optimise a specific objective [...] Of course, often it will be more realistic in practice to consider several criteria." Another definition has been put forward by [Pinedo, 1995]: "Scheduling concerns the allocation of limited resources to tasks over time. It is a decision-making process that has as a goal the optimization of one or more objectives." A statement of scheduling problems can be found in [Gotha, 1993]. This article sets out the resolution approaches and the traditional scheduling problems. We can find in [Lee et al., 1997] a presentation of the current problems as well as more recent resolution methods. In the above definitions, the task (or operation) is the entity to schedule. In this book we deal with jobs to schedule, each job is broken down into a series of operations. When all the jobs contain only a single operation we speak of a mono-operation problem. By contrast, we speak of a multi-operation problem. The operations of a job may be connected by precedence constraints. In this case the set of operations of a job and their precedence constraints define the routing of this job. We are also dealing with the resource or machine (this latter term is more often used in the context of shop scheduling). We consider generally that the resources are of two types: renewable or consumable. Renewable resources become available again after use (machine, file, proces- 6 1. Introduction to scheduling sor, personel, etc.), whereas non renewable resources disappear after use (money, raw materials, etc.). Among the renewable resources we can distinguish between the disjunctive resources, which can only perform one operation at a time and the cumulative resources which can process a limited number of operations simultaneously. The case of cumulative resources is being studied more and more as for example in shop scheduling problems [Carlier and Latapie, 1991], in project scheduUng problems and in batch scheduling problems ([Potts and Kovalyov, 2000]). Frequently, to solve a scheduling problem, we are also caused to solve an assignment problem, where it concerns in addition specifying the resources to process the operations. We can separate the criteria to optimise into two types: those relating to completion time and those relating to costs. In the category of completion time related criteria we find for example those which measure the completion time of the whole schedule and those which measure tardiness of jobs in relation to their due date. In the category of cost related criteria we may cite those which represent cost of machine use and those which represent cost allied to waiting time of operations before and/or after they are processed. 1.2 Some areas of application Scheduling problems are encountered at all levels and in all sectors of activity. Generally, we can distinguish between those of manufacturing production and those in computer systems or project management. 1.2.1 Problems related to production We encounter scheduling problems in Flexible Manufacturing Systems (FMS). Numerous definitions of an PMS are found in the literature. For [Liu and MacCarthy, 1996]: "i4n FMS comprises three principal elements: computer controlled machine tools; an automated transport system and a computer control system.'''' These problems are broadly covered in the literature and most often in a well defined application class. Besides, this very broad problem encompasses other problems related to Robotic Cell Scheduling and Scheduling of Automated Guided Vehicles (AGV). Equally, electroplating and chemical shops have their peculiarities in scheduling problems. The latter are also called Hoist Scheduling Problems. These shops are characterised by the presence of one or more travelling cranes sharing the same physical area and which are ordered to transport the products for treatment in tanks. In general, the soaking time in a tank is bounded by a minimum and a maximum {the interval processing time)., transport time 1.3 Shop environments 7 is not negligible and the operations must be carried out without waiting time. These problems are very common in industry and the "simple" cases (monorobot, single batch tanks, etc.) have been well solved by now. Scheduling problems in car production lines, so called Car Sequencing Problems, are encountered in assembly shops where certain equipment (or options) must be assembled in the different models of vehicles. These problems have constraints and peculiarities of their own. Knowing a sequence of vehicles undergoing treatment, the problem is to determine the type of the next vehicle programmed. We have to take account of a group of constraints connected principally to the assembly options for these vehicles and to the limited movement of the tools along the production Une. 1.2.2 Other problems We encounter scheduling problems in computer systems. These problems are studied in different forms by considering mono or multi processor systems, with the constraints of synchronisation of operations and resource sharing. In these problems, certain operations are periodic others are not, some are subject to due dates, others to deadlines. The objective is often to find a feasible solution, i.e. a solution which satisfies the constraints. Literature abounds on these problems. In fact, in spite of appearances they are very close to those encountered in manufacturing systems ([Blazewicz et al., 1996]). Timetable scheduling problems concern all educational establishments or universities, since they involve timetabling of courses assuring the availability of teachers, students and classrooms. These problems are just as much the object of studies. Project scheduling problems comprise a vast literature. We are interested more generally in problems of scheduling operations which use several resources simultaneously (money, personel, equipment, raw materials, etc.), these resources being available in known amounts. In other words we deal with the multi-resource scheduling problem with cumulative and non-renewable resources ([Brucker, 2004],[Herroelen et al., 1998b],[Herroelen et al., 2001]). 1.3 Shop environments When confronted with a scheduling problem, one has to identify it before tackling it. Acknowledging that the problem is complicated and to know if it is already solved in the literature, we must use a recognised notation. For that purpose, shop "models" have been set up, which differ from each other by composition and organisation of their resources. We denote by n the number 8 1. Introduction to scheduling of jobs to schedule, by Ji the job number i, by n^ the number of operations of job Ji, by Oi^j the operation j of job J^, by m the number of machines and by Mk the machine number k. A complete synthesis of the notations is given in appendix A. 1.3.1 Scheduling problems without assignment The problem is to find a processing start time for each operation. Several types of arrangement are traditionally encountered: • single machine: Only a single machine is available for the processing of jobs. It concerns a basic shop or one in which a single machine poses a real scheduling problem. Besides, resolution of more complex problems is often achieved by the study of single machine problems. We can find an area of direct application in computing, if we think of the machine as the single processor of the computer. The jobs to be processed are necessarily mono-operation. • flowshop (F): several machines are available in the shop. The characteristic of this type of shop is that the jobs processed in it use machines in the same order: they all have the same processing routing. In a permutation flowshop we find in addition that each machine has the same sequence of jobs: they cannot overtake each other. • jobshop (J): several machines are available in the shop. Each job has a route of its own, i.e. it uses the resources in its own order. • openshop (O): several machines are available in the shop. The jobs do not have fixed routings. They can, therefore, use the machines in any order. • mixed shop (X): several machines are available in the shop. Some jobs have their own routing and others do not. 1.3.2 Scheduling and assignment problems with stages The machines are grouped in well defined stages and a machine belongs to one stage only. In all cases the machines of a stage are capable of performing the same operations. To carry out one operation it is necessary to choose one among the available machines and, therefore, the problem is twofold, assigning one machine to each operation and sequencing the operations on the machines. At each stage we can differentiate between the following configurations: • the machines are identical (P): an operation has the same processing time on all the machines. • the machines are uniform (Q): the processing time of an operation Oi,j on the machine Mk is equal to Pi,j^k = Qi.jhk where qi^j is for example a number of components in the operation Oi^j to be processed, and Vk is the number of components which the machine Mk can process per unit of time. 1.4 Constraints 9 • the machines are unrelated or even independent (R): the processing time of the operation Oij on the machine Mk is equal to Pi,j,/c, and is a data of the problem. Of course, just as the assignment of Oij is unknown, so is its processing time. Globally, "traditional" scheduling and assignment problems correspond to the following configuration: • parallel machines (P/Q/R): there is only one stage and the jobs are mono-operation. • hybrid flowshop (HF): all the jobs have the same production routing, and therefore use the stages in the same order. • general jobshop (GJ): each job has a route of its own. • general openshop (GO): the jobs do not have a fixed routing. It is easily possible to generalise these problems by supposing that each operation can only use its own subset of the resources of the performing stage. 1.3.3 General scheduling and assignment problems This is the most general case where we suppose that each operation has its own set of machines on which it can be processed. No assumption is made on these sets of resources. We can differentiate several cases: • the jobs are mono-operations, and we are confronted by a problem of parallel machines with general assignment. We find these problems in the literature ([Brucker, 2004]) under the name "Multi Purpose Machines Scheduling Problems" ( P / Q / R MPM SP). • the jobs follow a processing order. It is difiicult in this case to distinguish between flowshop and jobshop since the groups of machines used by these jobs are not comparable. This is what is called shops with general assignment problems (" General Shop MPM SP"). • the jobs do not follow a fixed routing. This is the case in openshop with general assignment problems ("Openshop MPM SP"). 1.4 Constraints A solution of a scheduling problem must always satisfy a certain number of constraints, be they explicit or implicit. For example, in a flowshop problem it is implicit that the jobs are processed according to the routing and therefore an operation cannot start while its precedent remains uncompleted. On the other hand, the occurrence of different release dates constitutes a constraint which must be stated precisely. In this section we describe the explicit con- 10 1. Introduction to scheduling straints encountered most frequently in scheduling. A summary is given in appendix A. There are several types of constraints related to the due dates. There are those due dates which we do not wish to pass by, even if we tolerate to completing afterwards. They correspond to an agreed commitment which is negotiable. There are those due dates which are imperatives, also called deadlines, and which cannot be passed. Typically, they correspond to an unveiling date when the manufacturer must present his product, or even the departure date of the delivery lorry. These dates cannot be passed by: therefore, no tardiness can be permitted. Problems where we encounter these constraints are usually decision problems: these dates can or cannot be respected. When we can, either we are satisfied with a feasible solution or in addition we try to minimise a criterion. Constraints relating to start times are equally various. Of course, there is the release date of the product. Sometimes, it corresponds to the date of the order. Equally, we can find a release date associated with a specific operation of a job. This date can correspond to the arrival of supplies for the operation. Finally, it can happen that the start time of a particular operation is imposed. In this case it is a matter of meeting with the customer for him to witness the implementation of the operation which he regards as critical in the manufacturing process. These two latter definitions correspond to problems rarely dealt with in the literature. We list below some constraints met frequently in the literature. • pmtn indicates that preemption is authorised. Here it is possible to forsee interuption of an operation so that, possibly it can be taken up next by another resource. • split indicates that splitting is authorised. Here it is possible to forsee splitting of the operation into lots, which can be performed on one or several machines, possibly simultaneously. • prec indicates that the operations are connected by precedence constraints. This heading gives different particular cases according to the nature of the constraints: prec to describe the most general case, tree, in-tree, outtree, chains and sp-graph (for series-parallel graph ; see [Pinedo, 1995] or [Brucker, 2004]) to denote particular cases. • batch indicates that the operations are grouped in batches. Two types of batch constraints are differentiated in the literature: the first called sometimes s-batch concerns serial batches where the operations constituting a batch are processed in sequence and the second of type p-batch concerns parallel batches where the operations constituting a batch are processed in parallel on a cumulative resource. In both cases, the completion time of an operation is equal to the completion time of the batch. In the first case. 1.4 Constraints • • • • • • • • • • • 11 the duration of the batch is equal to the sum of the processing times of the operations which constitute it, whereas in the second case its duration is equal to the longest processing time of the operations in the batch. no-wait indicates that the operations which constitute each job follow each other without any waiting. p r m u (permutation) indicates that the operations occur on the machines in the same order. In other words, they cannot overtake themselves (this is true solely for flowshop problems). di = d indicates that all the due dates are identical. Likewise di — d for Pi = P indicates that the processing times are all identical. We often encounter this constraint with p = 1. Snsd and Rnsd indicate that the setup and removal times on the resources before and after each processing, respectively, must be taken into account. These preparation times are independent of the sequence of operations. Ssd and Rsd indicate that the setup and removal times on the resources before and after each processing, respectively, must be taken into account. These preparation times are dependent of the sequence of operations. ai^,i2 indicates that a minimum time lag must be respected between the jobs Jii and 7^2, if the jobs are mono-operation. Otherwise, we use Ciii,ji,i2j2 ^o indicate a minimum time lag which must be respected between the operation 0^^ j ^ and the operation Oi^j^- If ^^^^ value is positive, we model, for example, a drying time between two successive operations, or else a transport time. In the latter case the resource is available to process the following operation during the transport. If this value is negative, it indicates that it is possible to carry out an overlap, i.e. to start an operation before its precedent in the routing is completely finished. Of course, this is possible when a job is composed of lots of items and it is not necessary to wait for the end of a lot on a machine to start the operation on the following machine. blcg (balancing) is a constraint peculiar to parallel machine shops, translating the fact that the machines must complete processing of jobs which are assigned to them at the same time. This constraint may be imposed when it is necessary to change the type of manufacture on all the machines simultaneously. block (blocking) is a constraint indicating that the shop has a limited stock area between the machines. Then we note bk the stock capacity between machine Mk and machine Mk-\.i* recre (recirculation) is a constraint which indicates that a job may be processed several times on the same machine. unavailj translates the case where all the resources are not available all the time, but only during well defined periods. It is a matter of timetables translating periods of opening/closing of the factory, periods of planned maintenance, of hohdays, etc. Two types of operations can be associated 12 1. Introduction to scheduling with this problem: interruption and resumption of an operation as soon as possible (unavailj-resumable) or else the operation is not started if it is going to be interrupted (unavailj-nonresumable). In the latter case we can have problems of unfeasibility. We now state the difference between routing and precedence constraints: • a routing is a document which precisely describes the set of operations necessary for ending up with a final product: machine, processing time, tools, particular conditions, etc. This routing contains, of course, the order in which the operations must be processed, possibly with the help of a precedence graph (in the case of non identical routings). Two successive operations in a routing indicate a flow of material between machines or a set of machines. • precedence relations between operations indicate simply that the start of an operation is conditioned by the end of all the preceding operations. No notion of flow is attached, a priori, to this constraint and it may simply be a matter of severe technological constraints. Two operations linked by a precedence relation may correspond to two distinct jobs. In computer systems this distinction does not really exist since the routing of a job does not have a strong sense as in production systems. We have only a set of operations to schedule knowing that these ones can be connected by precedence constraints. Often it is assumed that these constraints are associated to communication times between the operations. We can also consider the existence of a communication media, or server, thus inducing disjunctive constraints. 1.5 Optimality criteria In order to evaluate schedules we can use a certain number of criteria. Occasionally we want a criterion to be close to a certain reference value. Here we are at the frontier between the notions of criteria and constraints. If a constraint represents a fact which definitely must be respected, optimising a criterion allows rather a certain degree of freedom. For example, stating that no job should be late regarding its due date leaves no margin in the schedule calculation. We may even find a situation where no feasible schedule exists. On the other hand, minimising the number of late jobs allows us to guarantee that there will always be a solution even though to achieve this certain operations might be late. Prom a practical point of view the difference between a criterion and a constraint is only apparent to the decision maker who initiates a schedule calculated by an algorithm. Certain criteria are equivalent and that is why they are presented jointly: minimising one or the other leads to the same optimal solution even if the criterion value is not the same in the two cases. Some scheduling problems 1.5 Optimality criteria 13 have no criterion to be minimised. In this case we are deaUng with a feasibiUty problem, also called a decision problem: does a solution which satisfies the constraints exist ? We can classify criteria into two large families: '^minimaj^^ criteria, which represent the maximum value of a set of functions to be minimised, and ''minisum!^ criteria, which represent a sum of functions to be minimised. A summary of the criteria presented below is given in appendix A. 1.5.1 Minimisation of a maximum function: "minimax" criteria We present "minimax" criteria which are most frequently met in the literature. The most traditional is without doubt the criterion measuring the completion time of the whole jobs. This criterion is denoted by Cmax and is called "makespan^\ We define Cmax = max (C^), with Ci being the complei=l,...,n tion time of the job J^. To simplify the notation, we write "max" for " max " i=l,...,n when there is no possible ambiguity. Cmax is the total length or duration of a schedule, i.e. it is the completion time of the last scheduled job. We also encounter other criteria based solely on the completion times of jobs, such as criteria: • -^max = max(Fi) with Fi = Ci — rf. the maximum time spent in the shop, or even yet, the duration of resting, with r^ the release date of the job Ji, • -^max = max(//c): with Ik the sum of idle times on resource M/j. Equally, we encounter in the literature criteria which are based on the due dates d^, Vz = 1, ...,n, of jobs. Notably, we find criteria: • -^max = max(Li) with Li = d — di'. the maximum lateness, • ^max = max(Ti) with Tj = max(0; Ci — di): the maximum tardiness, • -E'max = max(£^j) with Ei = max(0; di — Ci): the maximum earliness. Generally, fmax refers to an ordinary "minimax" criterion, which is a non decreasing function of the completion times of jobs. This is not the case for the criterion Emax1.5.2 Minimisation of a s u m function: " m i n i s u m " criteria "Minisum" criteria are usually more difficult to optimise than "minimax" criteria. This is confirmed from a theoretical point of view for certain spen cial problems ([Ehrgott, 1997]). We write "X]" for " ^ " when there is no ambiguity. Among the minisum criteria, we meet criteria: 14 1. Introduction to scheduling • C to designate ^J2^i ^^^^iT^^^^ criterion represents the average completion time or total completion time of jobs. • C to designate ^Y^WiCi^ ^ ^ J ] WiCi or else "^WiCi. This criterion represents the average weighted completion time or total weighted completion time of jobs. • F to designate ^ S ^ i ^^ J2^i- Optimising this criterion is equivalent to optimising the criterion C It is the same for the criterion F regarding to the criterion C . • T to designate ^J2'^i ^^ Z^^i- This criterion designates the average tardiness or total tardiness of jobs. • T to designate ^^WiTi, vlJ^jX^^i^i ^^ S'^i^i- This criterion designates the average weighted tardiness or total weighted tardiness of jobs. • i7 to designate ^ Ui which is the number of late jobs with Ui = 1 ii the job Ji is late and 0 otherwise. • U to designate ^^WiUi, ^ ^ ^WiUiOi Y^WiUi which is the weighted number of late jobs. • £^ is the average earliness of jobs. • E the average weighted earliness of jobs. In a general way, / designates an ordinary "minisum" criterion which is usually a non decreasing function of the completion times of jobs. This is not the case for criterion E. 1.6 Typologies and notation of problems Concerning scheduling problems, we distinguish between their typology and their notation. A typology is a classification of the problems according to their nature. In scheduling it is usually based on the machines environment and on the jobs particularities. A notation enables us to refer quickly to a problem. Thus it is possible to construct a database of the set of problems treated in the literature. The traditional notations in scheduling are clearly based on existing typologies. 1.6.1 Typologies of problems Different typologies of scheduling problems exist in the literature. We present in figure 1.1a typology which generalises that of [Mac Carthy and Liu, 1993] and which brings together the problems introduced in section 1.3. Concerning scheduling problems, the objective is to determine a sequence on each machine and a start time for each operation. In scheduling and assignment problems with stages we can define, independently of each operation, stages of machines. A machine belongs to only one stage. Then, we combine 1.6 Typologies a n d n o t a t i o n of problems Single Machine -4- Non duplicated machines All the jobs are monooperation Non duplicated Flowshop <- • All the jobs have the same routing Non duplicated Common sets of machines and all the jobs have the e routing Hybri^ Flowshop' All the jobs have the same routing All the jobs are monooperation Shop Problems with Common sets of machines General Assignment and all the jobs have General Jobshop All the jobs have routings machines All the jobs have routings All the jobs have routings Openshop <• Parallel Machines with General Assignment Common sets of - Parallel Machines <- All the jobs are monooperation Jobshop General Scheduling and Assignment Scheduling and Assignment with stages Scheduling 15 Non duplicated machines -, . ^ . General Openshop ^ Common sets of < :;i;^:dd^ JDpenshop with General Assignment F i g . 1.1. Typology of scheduling problems (1) each operation with a stage, and an operation can be processed by any machine of its stage. Therefore, we add an assignment problem to t h e initial scheduling problem. We must then not only find a start time for t h e operations b u t also an assignment of t h e operations on t h e machines. T h e same is true for general scheduling and assignment problems where a set or pool of machines is detailed for each operation. Of course, a machine may participate in several pools. An operation may be processed by any machine in its pool. T h e foregoing typology uses t h e machines environment and operations t o differentiate between problems. Other typologies exist ([Blazewicz et al., 1986]). Notably, we can consider problems according to different characteristics (figure 1.2): 1. deterministic vs. stochastic. In t h e case where all t h e characteristics of the problem (processing time of each operation, release dates, etc.) are well known, we speak of a deterministic problem. Conversely, some of these characteristics may b e random variables of known probability law. In this case we speak of a stochastic problem (see [Pinedo, 1995]). 2. unitary vs. repetitive. If t h e operations appear to be cyclical, we are dealing with a repetitive problem. Conversely, if each operation corresponds to a unique product t h e problem is said to b e unitary. 3. static vs. dynamic. If all t h e d a t a of t h e problem are known a t t h e same time we speak of a static problem. For some problems, a schedule m a y have been calculated and being processed when new operations arrive in 16 1. Introduction to scheduling the system. Then the foregoing schedule has to be re-established in "real time". These problems are said to be dynamic. Unitary Repetitive Deterministic- Stochastic t Static Dynamic Fig. 1.2. Typology of scheduling problems (2) These two typologies are complementary since it is possible to handle, for example, a deterministic flowshop problem, whether it be unitary or static. As we shall see in the following section, traditional notation of scheduling problems are the mirror image of these typologies. 1.6.2 N o t a t i o n of p r o b l e m s Two notations exist for referencing scheduling problems. Despite the fact that the oldest was proposed by [Conway et al., 1967], the notation most used in the literature was introduced by [Graham et al., 1979] (see a detailed description in [Blazewicz et al., 1996]). This notation is divided into three fields: a\|/3\|7. Field a refers to the typology presented in figure 1.1 and describes the structure of the problem (see section 1.3). It breaks down into two fields: a = a i a 2 . The values of a i and a2 refer to the machines environment of the problem and possibly to the number of available machines. Field ß contains the explicit constraints of the problem. See section 1.4 for some possible such constraints. Field 7 contains the criterion/criteria to be optimised (see section 1.5). Concerning a more complete presentation of the different possible criteria, the interested reader may refer to [Rinnooy Kan, 1976]. This field is detailed in 1.7 Project scheduling problems 17 chapter 4 for the multicriteria case. Appendix A presents a summary of the most current values which can take the fields a, /?, and 7. [Vignier et al., 1999] propose an extension of the notation for hybrid flowshop problems. For these ones the field a breaks down as follows: a = (a3af^)^ii. The values a^OTA ^ represent the configuration of each stage. Other extensions of the notation exist. We can quote works, notably the one of [Baptiste et al., 2001] who broaden the notation to hoist scheduling problems. When we address problems where machines are of the type ^^batch", [Jolai Ghazvini, 1998] and [Oulamara, 2001] similarly propose an extension of the notation. 1.7 Project scheduling problems Project scheduling problems have been extensively studied in the literature. They are usually separated from problems occuring in shop environments, since they have their own particularities. Several papers review the literature on project scheduling (see [Herroelen et al., 1998a], [Herroelen et al., 1998b], [Brucker et al., 1999], [Kolisch and Padman, 2001], [Tavares, 2002]). In project scheduling problems we consider the scheduling of a set of operations which are also called activities. Each operation has a processing time and the operations are connected by precedence constraints. These ones are usually represented by an "activity-on-the-node" network, where an edge represents a finish-start precedence relationship between two operations. To process the operations we distinguish between two situations. When the operations can be performed without any resource, we meet two classical problems in the literature. In the first one we have to compute a schedule of the operations which minimises the completion time of the whole project, also called makespan. In the second problem we associate to each operation a cash flow value and we compute a schedule which maximises the net present value of the project. When operations require resources, we deal with a Resource-Constrained Project Scheduling Problem (RCPSP). We can distinguish between the renewable resources, the non-renewable resources, the partially renewable resources (these are renewable ones during a known time period) and the doubly constrained resources (these are non-renewable resources with the added limitation of consumption for known time periods). Besides, we associate to each resource, whatever its type, a limited capacity per time unit and each operation requires one or several resources in known amounts. In the basic RCPSP we have only renewable resources and the problem is to minimise the makespan. This problem is a generalisation of the jobshop scheduling problem with makespan minimisation. We can consider extensions of this basic problem by allowing the preemption of operations, or by imposing time 18 1. Introduction to scheduling lags between the processing of two consecutive operations. Another classical extension of the basic RCPSP consists in defining for each operation a minimum and a maximum processing time. This is related to the presence of at least one non-renewable resource. The more we use of this resource to process an operation, the lower is its processing time. Therefore, the exact processing time of each operation has to be calculated. The aim is to minimise the total requirement of the non-renewable resources, or if this total requirement is limited, to minimise the makespan of the project. A presentation of other classical models can be found in [Herroelen et al., 1998b]. Various extensions to the three-field notation presented in section 1.6.2 exist. The two major are due to [Herroelen et al., 1998b] and [Herroelen et al., 2001] for the first extension and [Brucker et al., 1999] for the second one. 1.8 Some fundamental notions The notions which are presented in this section refer to the characterisation of dominant sets for certain scheduling problems. We say that a subset of schedules is dominant for a problem if and only if, whatever the data of the problem, an optimal solution is contained in this subset. Definition 1 introduces the notion of regular criterion in the case of a minimisation problem. Definition 1 Let S be the set of solutions. A criterion Z is a regular criterion if and only if Z is an increasing function of the completion times of jobs, i.e. if and only if: Vx,j/ € S, Ci{x) < Ci{y), Vi = 1, ...,n, ^Z{Cx{x),...,Cn{x))<Z{Ci{y),...,Cn{y)) For the criteria presented in section 1.5, we deduce the following result, which is not difficult to prove. Corollary 1 The criteria Cmax, C, C , Lmax, Tmax, T, T , U and U criteria Imaxj ^maxj E and E are not regular. are regular. The We distinguish four classes of schedules (figure 1.3). The schedules with insertion of machine idle times constitute an interesting class for the minimisation of certain non regular criteria. This is the case for many Just-in-Time scheduling problems. Definition 2 A schedule belongs to the class of schedules with insertion of machine idle times if and only if before each scheduled operation, the machines are voluntarily left idle during a positive or null period. 1.8 Some fundamental notions 19 Fig. 1.3. Inclusion of classes of schedules Some examples of schedules with insertion of machine idle times are presented in figures 1.4a, 1.4b and 1.4c. Definition 3 Let X G S be a schedule and Sx be the set of schedules having the same sequences of operations on the machines as x. A schedule x belongs to the class of semi-active schedules if and only if ßy G Sx such that Ci{y) < Ci{x), Vi = 1, ...,n, with at least one strict inequality. We note that the class of semi-active schedules is a subclass of the class of schedules with insertion of machine idle times. We say that the semi-active schedules are "left shifted". Figures 1.4b and 1.4c present some semi-active schedules. Definition 4 A schedule x e S belongs to the class of active schedules if and only if ßy £ S such that Ci{y) < Ci{x), Vi = l,...,n, with at least one strict inequality. Active schedules are equally semi-active. Figure 1.4c presents an active schedule. We say also that a schedule is active if it is impossible to start earlier the processing of an operation without delaying another. We can note that the definition of active schedules may be interpreted from a multicriteria point of view: the class of active schedules is the set of solutions which are not dominated for the n completion times Ci. Definition 5 A schedule x E S belongs to the class of non delayed schedules if and only if no operation is kept waiting while a machine is available to process it Non delayed schedules are equally active schedules. Figure 1.4c presents a non delayed schedule. One important result, presented in lemma 1, relates regular criteria to the class of active schedules. 20 1. Introduction to scheduling A P2\|prec\|C Ji Pi J. 3 h h h h 6 problem 1 5 4 - data M, M2 J, 3£ AW: h h J4 m M, M M2 ' j^ C„ -12 (a) A schedule with inserted machine idle times (b) A semi-active schedule M, h Ma h \h\ Vn M ^^^Ä^B h ^M (c) A both active and non delayed schedule F i g . 1.4. Illustration of different classes of schedules 1.9 Basic scheduling algorithms 21 L e m m a 1 [Baker, 1974] For optimisation problems of a regular criterion, the set of active schedules is dominant Lemma 1 implies that the search for an optimal solution of the optimisation problem of a regular criterion, may be limited to the set of active schedules. For multicriteria problems this result remains equally true if we optimise several regular criteria. However, it becomes invalid if at least one criterion is not regular, since this is the case for some problems where criteria E and T are minimised. This is the case in Just-in-Time scheduling. 1.9 Basic scheduling algorithms This section is intended to present some of the basic scheduling algorithms for single criterion problems. These algorithms are referred to throughout the book. More complex algorithms can be found in books dedicated to scheduling (see for instance [Tanaev et al., 1994a], [Tanaev et al., 1994b], [Pinedo, 1995], [Blazewicz et al., 1996] and [Brucker, 2004]). 1.9.1 Scheduling rules Several scheduling rules, optimal or heuristic, have been proposed in the literature. They are very often used in heuristic applications to industrial problems, given their simplicity and the little calculation time which they require ([Morton and Pentico, 1993]). Among the most traditional rules, we find the rule SPT which enables us to compute an optimal active schedule for the 1\\C problem. Rule SPT: {Shortest Processing Time first) sequences the jobs in increasing order of their processing time. The converse rule is the rule LPT {Longest Processing Time first). The 1\\C problem is solved optimally with the rule WSPT. Rule WSPT: (Weighted Shortest Processing Time first) sequences the jobs in increasing order of their ratio pt/wi. When we consider the due dates and the minimisation of criterion Lmax ? the corresponding single machine problem denoted by l\di\Lmaxj can be solved optimally by calculating an active schedule using the rule EDD. Rule EDD: {Earliest Due Date first) sequences the jobs in increasing order of their due date di. We notice that this rule also solves the l\|rfj\|Tmaa; problem optimally. Addition of the release dates r«, i = l,...,n, of the jobs no longer enables us to solve these problems optimally by simply considering an adaptation of 22 1. Introduction to scheduling these rules. For example, the rule EST {Earliest Start Time first) sequences the jobs in increasing order of their earliest start time, and breaks ties in favour of the job with the smallest processing time. This rule does not solve optimally the l\|ri\|C problem. Likewise, no simple sort based on the weights, the processing times or the due dates, can solve optimally the l\|ri\|C and l\ri^di\Lmax problems, since these problems aie AfV-haid. Generalisation of these rules to parallel machines problems necessitates the addition of a job assignment rule to the machines. In the case of identical machines, we generally use the assignment rule FAM {First Available Machine first), which assigns a job to the first available machine. The rule SPT-FAM solves optimally the P\\C problem by considering the jobs in the increasing order of their processing time, and assigning them in turn to the earliest available machine. The rules WSPT-FAM and EDD-FAM give way to the respective heuristic algorithms for the P\\C and P\di\Lmax problems. In the case of proportional machines, we often consider the assignment rule FM {Fastest Machine first) which consists of assigning a job to the fastest machine among those available. The rule SPT-FM solves optimally the Q\\C problem. The rules WSPT-FM and EDD-FM give heuristic algorithms for the Q\\C and Q\di\Lmax problems, respectively. When preemption of jobs is authorised, the rule SPT-FM becomes SRPT-FM {Shortest Remaining Processing Time on Fastest Machine first) and solves optimally the Q\pmtn\C problem. It consists of scheduling the job with the smallest remaining processing time, on the fastest machine among those available, preempting when necessary. The rule LRPT-FM optimally solves the Q\pmtn\Cmax problem. 1.9.2 Some classical scheduling algorithms Lawler's algorithm for the l\prec\fmax problem Consider the problem where n jobs have to be scheduled on a single machine. A set of precedence constraints between jobs is defined and no preemption is allowed. Let fi be an increasing function of the completion time of Ji, Vz = 1,..., n. The objective is to minimise the maximum cost function defined by fmax = inax {fi{Ci)). [Lawler, 1973] proposes an optimal polynomial i=l,...,n time algorithm which iteratively schedules a job by starting from the last position. At the first iteration the jobs that have no successor are candidates and can therefore be scheduled in the last position. Notice that the completion time of the last position is equal to P = pi +p2 + ••• +Pn- Thus, we schedule in the last position the candidate job Ji which has the lowest cost Ci {P) among the candidate jobs. By setting P = P — piwe can iterate this process for the 1.9 Basic scheduling algorithms 23 previous position. The complete algorithm, denoted by EELl, is presented in figure 1.5. ALGORITHM EELl /* T is the set of jobs to schedule / 5 = 0; n i=i For i = n down to 1 Do F = {Ji eT/Ji has no successor in T}; If (F = 0) Then I The problem is not feasible; End If; _ _ Let Jee Fhe such that fe{P) = min {fk{P)); / Break ties by choosing the job with the greatest processing time / S={Je}//S; T = T-{Jeh Ce=_P; P = P-pe', End For; Print 5; [Lawler, IQTäT Fig. 1.5. An optimal algorithm for the l\prec\fmax problem Moore's algorithm for the l\|di\|f/ problem Consider the problem where n jobs have to be scheduled on a single machine and each job Ji has a due date di. No preemption is allowed. The objective is to minimise the number of late jobs, denoted by U. [Moore, 1968] provides an optimal polynomial time algorithm to solve this problem. It starts with the schedule obtained by the rule EDD. Let Jk be the first tardy job in this schedule, i.e. all jobs scheduled before are early or on time. Moore's algorithm puts Jk on time by removing the preceding job with the greatest processing time. The latter is scheduled late and is not considered anymore. This process is iterated until we have no late jobs in the schedule, except those which have been previously removed and voluntarily put late. The number of late jobs is equal to the number of removed jobs. The algorithm, denoted by EJMl, is presented in figure 1.6. 24 1. Introduction to scheduling ALGORITHM EJMl / T is the set of jobs to schedule / / We assume that c?i < (^2 < ... < c^n / S = (Ji, J2,..., Jn); Tardy = 0; While {3Je € S such that d > de) Do Let k be such that Cs[k] > ds[k] and Vi < A;, Cs[i] < ds[i]': Let j be such that j < k and P5[j] = max {ps[i])'i i=l,...,k S = S-{Jj}; Tardy = Tardy//{Jj}; jEnd While; U=\Tardy\\ _ Print S//Tardy and [/; [Moore, IQGST Fig. 1.6. An optimal algorithm for the l\|c?i\|[7 problem Johnson's algorithm for the F2\prmu\Cmax problem Consider a two-machine flowshop problem where n jobs have to be scheduled. They first have to be processed on machine Mi and next on machine M2. As the makespan criterion is a regular criterion we are restricted to the set of permutation schedules which is a dominant set. [Johnson, 1954] proposes a sufiicient condition of optimality and derives an 0(nlog(n)) time optimal algorithm. It proceeds by scheduling first the jobs such that pi^i < pi^2 according to the increasing order of the Pi,i's. The remaining jobs are scheduled last according to the decreasing order of the Pi,2's. This algorithm, denoted by ESJl, is presented in figure 1.7. It is often also referred to as algorithm J. ALGORITHM ESJl 7 T is the set of jobs to schedule / Let U = {Jie T/pi,i < pi,2}; Let V = {Jie T/pi,i > pi,2}; Sort U by increasing values of the values pi,i; Sort V by decreasing values of the values pi,2; S = U//V; ^max ^^ y^max\^)i Print S and C:i:riax'', [Johnson, 1954] Fig. 1.7. An optimal algorithm for the F2\prmu\Cmax problem 1.9 Basic scheduling algorithms 25 Campbell, Dudek and Smith's heuristic for the F\prmu\Cmax problem Consider a flowshop problem where n jobs have to be scheduled on m machines. The jobs have the same routing, and we assume that they are first processed on machine Mi, next on machine M2, etc. Even if the set of permutation schedules is not dominant for this problem, [Campbell et al., 1970] restrict to this set and propose an heuristic to minimise the makespan criterion. This algorithm proceeds by building (m—1) fictitious two-machine problems and by solving each one using Johnson's algorithm ([Johnson, 1954]). Therefore at most (m — 1) distinct permutation schedules are built. The best one regarding the m-machine problem is retained. This detailed heuristic, denoted by HCDSl, is presented in figure 1.8. ALGORITHM HCDSl / T is the set of jobs to schedule / / J refers to Johnson's algorithm / For 7 =: 1 to (m - 1) Do / Building of a fictitious two-machine problem / j Pi,i = ^Phk and p-,2 = fc=l m X I ^'^' k=m—j-\-l Let 5^ be the sequence obtained by algorithm ESJl on the fictitious problem; End For: Let S^ be the schedule such that Cmax{S^) = min (Omax(»^ ))5 i=l,...,(m-l) /* Notice that the makespan is calculated by considering the m machines / Print S^ and Cmax(S^); [Campbell et al., 1970] Fig. 1.8. An heuristic algorithm for the F\prmu\Cmax problem Nawaz, Enscore and Ham's heuristic for the F\prmu\Cmax problem [Nawaz et al., 1983] consider the same permutation flowshop problem as the one of Campbell, Dudek and Smith. They propose an heuristic based on a jobinsertion scheme. Initially, the jobs are sorted by decreasing sums of process771 ing times on the machines, i.e. by decreasing order of the values /]pi,j' The heuristic considers only the two first jobs and retains the permutation schedule, among the two possible ones, which has a minimal value of criterion Cmax' This is the starting partial schedule. It next inserts the third job of 26 1. Introduction to scheduling the initial sorting, by trying all the possible positions in the partial schedule. The one which has a minimal makespan value is retained. This process is iterated until all the jobs are scheduled. The heuristic is presented in figure 1.9. Other classical algorithms for flowshop scheduling problems can be found in [Proust, 1992]. ALGORITHM HNEHl / T is the set of jobs to schedule / m m / We assume that y j p i , i ^ ••• > Y^Pn.j / Let S be the best permutation schedule among (Ji, J2) and (J2, Ji); For i = 3ton Do / Insertion of job Ji / Vfc = 1,..., \|5\|, 5^^ is the partial schedule with job Ji inserted in position k in S; V£ = 1,..., \Sl let S^ be such that Cmax{S^) = _min {Cmax{S^))] End For; Print S and CmaxjS); [Nawaz et aL, 1983] Fig. 1.9. An heuristic algorithm for the F\prmu\Cmax problem Sahni's algorithm for the P\pmtn/di\— problem Consider a scheduling problem where n independent jobs have to be scheduled on m parallel identical machines. Preemption of jobs is authorised but no job can simultaneously be processed b^ more than one machine. Each job Ji has associated with it a deadline di and the aim is to compute a feasible schedule, if it exists. [Sahni, 1979] proposes an optimal algorithm to solve this problem. This algorithm, denoted by ESSl, is presented in figure 1.10. Notice that this problem can also be solved by reducing it to a network flow problem ([Horn, 1974]). 1.9 Basic scheduling algorithms ALGORITHM ESSl / We assume that rfi > ... > dn / / Cj^: the completion time of the last job on Mj Vj: C f = 0,Vj = l,...,m; For i = 1 to n Do / * We schedule job Ji / Let L = {j/Cf < di}', If ((L = 0) or (di - m i n ( C f ) < pi)) Then .,m / Print "No feasible schedule exists"; END End if; Let A; G L be such that Cjf = max-,eL(CJ^); If (di - C f > Pi) Then / * Job Ji is entirely scheduled on Mj / c^ cf + Pi\ Else / Job Ji is processed by more than one machine / Let Li = {j e L/di - Cf < Pi}; Let La = {j € L/di - Cf > pi}; Let a € 1/2 be such that C^ = maxj^LiiCi^); Let /? 6 Li be such that Cp' = min^gii {Cj); C^ = C^ + Cß {pi-di+C^); = di] End If; End For; Print the calculated schedule; [Sahni, igTOf Fig. 1.10. An optimal algorithm for the P\p7ntn,di\— problem 27 2. Complexity of problems and algorithms This chapter presents an introduction to the theory of complexity. Decision problems, optimisation problems, couting problems and enumeration problems are defined, and complexity classes associated to these problems are introduced. These classes aim at qualifying the diSiculty of solving problems. We first start with some considerations on the complexity of solution algorithms. 2.1 Complexity of algorithms The complexity of an algorithm lies in estimating its processing cost in time (time complexity) or in the required space memory (spatial complexity). Set apart for certain particular algorithms, as for example dynamic programming algorithms which usually take up a lot of memory space, spatial complexity has been less considered than time complexity. In both cases it is possible to propose a theoretical complexity and a practical complexity. Theoretical complexity reflects an independent estimate on the machine which processes the algorithm. It is less accurate than the practical complexity which enables us to calculate the cost of the algorithm for a given computer. For the latter case, time complexity is obtained using an estimation of the calculation time for each instruction of the program. The advantage of theoretical complexity is that it provides an estimation independent of the calculation time for the machine. In the remainder of this section we use the term complexity to refer to the time complexity of an algorithm. This complexity is established by calculating the number of iterations done by the algorithm during its processing. The number of iterations depends on the size of the data, noted Length, and possibly the magnitude of the largest element, noted Max, belonging to these data. If the number of iterations is bounded by a polynomial function of Length then the algorithm is of polynomial complexity. If this function is limited by a polynomial of Max and Length, then we say that the algorithm is of pseudo-polynomial complexity. In other cases the algorithm is said to be of exponential complexity. 30 2. Complexity of problems and algorithms More precisely, we can distinguish minimal, average, and maximal complexities in order to translate complexity in the best case, the average case or in the worst case respectively. These latter two actually are interesting and the easiest to calculate is maximal complexity. On the other hand, average complexity requires a statistical analysis of the processing of the algorithm by function of the input data. Example. We can illustrate these notions by the example presented in figure 2.1. The maximum number of iterations is equal to n and the minimum number to 1. The average complexity itself depends on the probability p that the element is found in a given position. We suppose that this probability follows a uniform law, i.e. p = ^ . Thus, the average complexity is equal to p(l + 2 + 3 + ... + n) = \^ = ^ ^ . We notice that the calculation is only valid if we are sure that the element belongs to the list. In the opposite case, the calculation of the average complexity is even more complicated as it causes the law of generation of the elements of the list to intervene. Search for an element belonging to a list / elt the searched element / / n is the list size / / list is the list of elements / / We assume that elt belongs to the list */ i = 0; While {elt ^ list\i]) Do z = i-h 1; End While; Fig. 2 . 1 . Search for an element in a non sorted list To calculate the complexity of an algorithm, it is possible either to count the number of iterations, as we have done in the above example, or to break up the algorithm into sub-algorithms of known complexity. In this latter case, we can multiply or add the complexities according to the structure of the program. By using the above example, we can propose an algorithm which searches k elements in a list. Its average and maximal complexities are then of the order of fe x n. In the case of spatial complexity, the calculation cannot be performed on t h e algorithm itself -we cannot count the number of iterationsb u t rather on the d a t a it uses. T h e theoretical complexity of an algorithm is usually a function of Max, of Length and of addition and multiplying constants. Very often, we resume this complexity by the expression of the t e r m which gives its asymptotic value. For 2.1 Complexity of algorithms 31 example, if the maximal complexity of an algorithm is Max^ + a x Length + c, we say that it is in 0{Max^ + Length). More precisely, the notation 0(6) means that the complexity has an upper limit set by a linear function in b whereas the notation 0(6) enables to specify that the complexity is equivalent to 6. The simplification by the notation 0(6) may lead to paradoxical situations. In effect, an algorithm A in 0{Max'^) may be slower than an algorithm B in 0(2^^^) for certain problems. Let us take, for example a scheduling problem where Max is the number of jobs n and let us suppose that the complexity of the algorithm A is 2^^°^n^ and that of the algorithm B is 2'^. It is then obvious that for n < 1000, the algorithm A causes more iterations than the algorithm B. This remark does not imply that the theoretical complexity of an algorithm is lacking in interest. Indeed, algorithm A remains more sensitive to the calculating machines than algorithm B since between two machines of different power, algorithm B attaches little difference regarding the size of the largest problem it can solve, which is not the case with algorithm A, We give in table 2.1 the complexities of the best algorithms available to solve some classical problems. Notice that for those examples the average complexity is equal to the maximal complexity. Table 2.1. Some types of algorithms and their complexity Algorithm t o . . . Search an element belonging to a list of n elements Add an element in a non sorted list of n elements Add an element in a sorted list of n elements Perform a dichotomic search in an interval [min; max] of integer values Sort a list of n elements (fu1 sion sort) Maximal complexity 0(n) 0(1) 0{n) 0(log(max — min)) 0(nlog(n)) The complexity of a well written algorithm may sometimes be improved to the detriment of the spatial complexity: it is possible to reduce the computational time of an algorithm by increasing the size of the data. However, such a step often leads to adding new functions uniquely dedicated to the management of these data. So, the equilibrium to find is between the size of the used data and the complexity of the algorithm. It is clear that this complexity cannot be indefinitely 32 2. Complexity of problems and algorithms broken down in order to get at the end an algorithm which complexity is null. Solving a problem implies a minimum algorithmic complexity. But what is the minimum algorithmic complexity required to solve a given problem ? The algorithm provided in figure 2.1 solves the problem of searching an element in a non sorted list in 0{n) time, which is equivalent to say that it is solvable in a polynomial time of the size n. Can we guarantee that this is the case for any problem, i. e. there always exists a polynomial time algorithm to solve it ? If the answer to this vaste question is yes then all problems are easy to solve: we just have to find the correct algorithm and solve it. Otherwise it means there are some problems which are intractable: do not think about solving in polynomial time these problems. In this case we are only able to solve small-size problems; for the bigger one we should let the computer running for hundred years ! Unfortunately we do not know the answer to the above question. This is quite confusing: given a problem on which we try unsuccessfully to design a polynomial time algorithm, must we continue in this way or should we give up because such an algorithm does not exist? Complexity theory provides useful elements to establish the complexity of problems. This theory assumes that there are some problems which can be solved in polynomial time whilst others cannot. Given this, complexity classes exist which help in deciding if we must look for a polynomial time algorithm or not. This is the matter of the next section. 2.2 Complexity of problems Complexity theory proposes a set of results and methods to evaluate the intrinsic complexity of the problems. A problem belongs to a class of complexity, which informs us of the complexity of the "best algorithm" able to solve it. Hence, if a given problem is shown to belong to the class of "easy" problems then it means that we are able to exhibit a polynomial time algorithm to solve it. Usually this is a good news but unfortunately this does not often happen for complex problems. Accordingly, if a problem belongs to the class of hard problems, it cannot be solved in polynomial time which, said differently, implies that for some instances the required CPU time to solve it becomes "exponential". Along the years, numerous complexity classes have been defined and can be separated depending on the type of problems they address to. Basically we distinguish between decision problems, search and optimisation problems, and counting and enumeration problems. In this section we present these kinds of problems and provide the existing complexity classes. Notice that counting and enumeration problems are not often considered in the literature. 2.2 Complexity of problems 33 2.2.1 The complexity of decision problems Complexity theory brings our attention to decision problems. Basically, the complexity classes presented in this section have been firstly dedicated to such problems. Complexity theory is based, at the roots, on language theory and Turing machines but can be presented less formally in terms of algorithms. The reader interested in a very detailed presentation of complexity theory is referred to [Garey and Johnson, 1979] or [Papadimitriou, 1995]. Let us first define a decision problem . Decision problem 77: • Input data , or instances, noted I. The set of all the instances is noted • Question such that for each instance I G D77, the answer R € {yes\ no). The set of instances / for the problem 77, for which the answer to the question is yes^ is noted Yn- It is possible to propose a grammar G equivalent to the problem 77, by encoding all the instances by an encoding scheme e (see figure 2.2). For example, with the binary coding scheme all the terminal and non terminal elements will be binary numbers. With set Y/j of the decision problem, we can associate a set of chains produced by the grammar: the language 7/(77, e). Thus, we have transformed the decision problem into a grammar. If the answer to the question is yes for an instance, then the chain corresponding to this instance belongs to the language L{n,e). In order to know this, we propose a Turing machine. Taking the input chain, this machine exits in an accepting state if the chain belongs to L(77, e). We see that for the decision problem this Turing machine is equivalent to a "^o/t'e" procedure which returns true or false. The complexity of a decision problem thus depends on the "best" Turing machine which we can propose. The encoding scheme used influences equally the efiiciency of the proposable Turing machine. In general, we consider a reasonable encoding, i.e. which does not pointlessly complicate the obtained grammar. For all reasonable encoding schemes, the proposable Turing machines are judged equivalent ([Garey and Johnson, 1979]). Before introducing the classes of problems, we must define for the decision problem 77 two functions: Length[I] and Max[I], with 7 € Dn- The function Length represents the size of the instance 7, i.e. the length of chains produced by the grammar G. The function Max enables us to know the magnitude of the instance 7. In general, we consider that this function returns the magnitude of the largest integer, if it exists, occurring in 7. For example, if 7 = {3; 4; 6; 8; 14}, we have Length[I] = 5 and Max[I] = 14. The functions Length and Max are supposed to be calculable in polynomial time. It is then possible to define several classes of problems according to the difläculty in finding an answer to the decision problem. The definition 34 2. Complexity of problems and algorithms Decision problem n Using an encoding scheme e we can provide a; which corresponds to the instances of Grammar G > We refer to UJ\,e) as the language associated to the chains produced by G and which correspond to the «ofYn Yas No We search a Turing machine able to decide if a chain produced by G belongs to UJl,e) V Solve (I) I I I I I M II II IM II The existence of a Turing machine is < equivalent to the existence of a decision algorithm for problem n J I I I I I \| T I I I I I II A Turing machine Fig. 2.2. Decision/grammar duality problem of these classes requires the notion of deterministic and non deterministic Turing machines ([Hopcroft and Ullman, 1979]). Definition 6 A decision problem U belongs to the class V if the following holds: an encoding scheme e exists, such that for all instances I of 11, we can construct for the corresponding grammar a one-tape deterministic Turing machine, capable of checking if the chain corresponding to I belongs to the language. This is equivalent to a ^^yes" answer to the decision problem U. In this case the resolution time is a polynomial function of Length[I], This definition does not prevent the number of possible solutions from being exponential. We simply have the equivalence: a decision problem iJ is in 7^ if and only if an algorithm exists which enables us to calculate in polynomial time a solution which has the answer yes. Not all decision problems belong to the class V. A more general class exists, which is introduced in the following definition. Definition 7 A decision problem 77 belongs to the class AfV if a non deterministic polynomial one-tape Turing machine exists which reaches an accepting state in a finite number of iterations when it takes upon entry a chain of language L{n, e). The number of iterations is upper bounded by a polynomial function of Length. 2.2 Complexity of problems 35 A non deterministic polynomial one-tape Turing machine can be seen as a Turing machine having two modules: a divination module allowing us to construct a solution and an evaluation module capable of calculating if the answer R is yes in a time which is a polynomial function of Length. Obviously, we have V C J\fV. A conjecture in complexity theory which has never been demonstrated to this day, rests on the non inclusion ofAfV in V. We suppose in the following that V ^ ÄfV, which implies that decision problems which do not belong to V exist. This leads to the definition of the class AfVC of AfVcomplete problems, which is a sub-class of ÄfV. For this we must introduce the notion of polynomial reduction (or transformation), denoted by oc. Definition 8 A polynomial reduction oc of a decision problem 11' towards a decision problem n is a function such that: • V/' instance of U', a {!') is an instance of 11 and is calculable in polynomial time. • MI' instance of 11', the answer for the instance I' is yes for the problem n' if and only if the answer for the instance oc (/') is yes for the problem n. This means that a deterministic one-tape Turing machine exists, which is capable of calculating oc starting with a chain x generated by the grammar G. Moreover, x € L{n',e') ^ a {x) € L{n,e). If we consider two decision problems 77 and IT', 11' oc 11 means that the problem IJ' reduces polynomially towards the problem 77, which implies that n is at least as difficult to solve as 11'. A polynomial reduction, or transformation, of a problem U' towards a problem 77 can be seen as a function (in the algorithmic sense) which: 1. solves 77', i.e. which is able to verify if an instance I' of 77' leads to an 2. to solve 77', transforms in polynomial time the instance 7' in an instance 7 of 77, and calls a resolution function of 77. The answer returned by this last function is the answer to the problem 77'. Definition 9 A problem U is MV-complete if and only if 11 £ MV and \in' G MV, 3 oc such that n' oc n. This definition implies that the class MVC of ATP-complete problems contains the most difficult problems to solve. Indeed, if an A/^P-complete problem 77 is solvable in polynomial time, then all the problems of MV are so since all reduce to 77, and therefore V = MV. We say that an ATP-complete problem is solvable in polynomial time if and only if V=J\fV. The first problem to have been demonstrated A/'P-complete is due to [Cook, 1971]. In practice, to demonstrate that a problem 77 is A/'T^-complete, it is sufficient to demonstrate either that: 36 2. Complexity of problems and algorithms 1. i7 G NV and that a polynomial reduction oc and an ATT^-complete problem n' exist such that i J ' a 77, or that 2. there exists an ATP-complete sub-problem 11' of 77. 77' is a sub-problem of 77 if and only if: • 77 and 77' have the same question. • the set of instances of 77' is included in the set of instances of 77. • the set of instances of 77' for which the answer is yes is included in the set of instances of 77 for which the answer is yes. In the class of A/^T^-complete problems we can distinguish two types of problems: problems ATP-complete in the weak sense (or ordinary sense) and problems A/'T'-complete in the strong sense. Definition 10 A problem 11 is weakly MV-complete if it is MV-complete and if it is possible to find an algorithm to solve it such that its complexity is a polynomial function of Max[I] and of Length[I], V7 instance of 11. We say then that 77 is solvable in pseudo-polynomial time. If an A/''P-complete problem 77 is such that p a polynomial function of Length exists, for which V7 instance of 77, Max[I] < p{Length[I]), then 77 cannot be solved by a pseudo-polynomial time algorithm. Otherwise, this pseudopolynomial algorithm would be a polynomial algorithm, which contradicts the fact that 77 is jVT^-complete. If such a polynomial p exists, then we say that 77 is A/^'P-complete in the strong sense. If p does not exist, then 77 is called a number problem. More precisely, the definition of strong ATP-completeness can be introduced, extending this result. Definition 11 Let n be a decision problem and p a polynomial defined over a set of integer values. We define lip the sub-problem of 11 which is such that: 1. The set of instances of lip, denoted by D^, is included in Dn2. yi eD^jj, Max[I] < p{Length\i]). The problem 11 is AfV-complete in the strong sense if: L n eNV. 2. A polynomial p exists such that Lip is MV-complete. A decision problem 77 is weakly ATP-complete if we can show either that: 1. 77 is A/'P-complete and there exists an algorithm that solves it requiring a computational time upper bounded by a polynomial of Max[I] and Length[I] for every instance 7, or that 2. 77 is A/'P-complete and a polynomial reduction ex of 77 exists towards a weakly ATT^-complete problem 77', or that 2.2 Complexity of problems 37 3. 77 is A/'T^-complete and iJ is a sub-problem of a weakly jVP-complete problem. It is diflFerent to show that a problem is strongly ATP-hard. As for proving A/'P-completeness in the weak sense, suppose that we have a known strongly AfV-couiplete problem U' and a polynomial time reduction oc such that 77' oc 77. It is not possible to conclude that 77 is strongly A/^P-complete for the following reasons. We note D^ = { a {1')^!' G Dn'}- Two cases can occur: 1. A polynomial p exists such that V7 € DJj^ Max[I] < p{Length[I]). 2. A polynomial p does not exist such that V7 € 7>^, Max[I] < p{Length[I]), In the first case, the corresponding problem 77p is necessarily strongly J\fVcomplete, otherwise we have found a polynomial time algorithm to solve problem 77'. In the second case, 77p is a number problem. Thus, we cannot decide if 77 is strongly A/'P-complete. Therefore, to establish strong ATT^-completeness we have to consider special reductions. A pseudo-polynomial reduction is one such special reduction and is defined below. Definition 12 A pseudo-polynomial reduction from a decision problem 77' towards a decision problem 11 is a function ocs such that: 1. V7' G Dw, r G Yw if and only if ocs (7') G Yn2. (Xs can be calculated in polynomial time of Max[r] and 3. It exists a polynomial qi such that MI' G Dn', qi{Length[(Xs (7')]) > length'[r]. 4' It exists a polynomial q2 such that V7' G Dn', Max[(Xs (7')] < q2{Max'[riLength'[r]), Length[r]. Conditions 3 and 4 of the above definition ensure that with the instances built by (X5, 77p does not correspond to a number problem. Thus, it enables us to prove that if problem 77' is strongly A/'P-complete, then problem 77 is also. To demonstrate that a decision problem 77 is strongly A/''P-complete, it is thus sufficient to show that either: 1. 77 G Afp and that a strongly ATT^-complete problem 77' and that a pseudo-polynomial transformation oCs exist such that 77' oCs 77, or that 2. 77 is ATP-complete and that a polynomial p exists such that V7 instance of 77, Max[I] < p{Length[I]), or that 3. 77 G AfV possesses a strongly A/^P-complete sub-problem. 38 2. Complexity of problems and algorithms 2.2.2 The complexity of optimisation problems We now turn to a more general class of problems for which the aim is not to decide on the feasibility of an instance but to calculate a solution to that one. These problems are generally referred to as search problems. A search problem SP is more formally defined as follows. Search problem SP: • Input data, or instances, noted I. The set of instances is noted Dsp^ • A set of solutions 5 / for each instance / G Dsp^ defined by means of the question. An algorithm is said to solve a search problem if, given an instance / G Dsp^ it returns the answer "no" if Si is empty and otherwise returns a solution s E Sj. A decision problem U can be considered as a particular search problem for which Sj = {yes} if / € l / j and 5 / = 0 otherwise. However, we do not usually search for an arbitrary solution in 5 / but for a solution which optimises a given objective function. In this case, the search problem turns to an optimisation problem and the aim becomes to calculate any optimal solution. Formally, an optimisation problem O is defined as follows. Optimisation problem O: • Input data, or instances, noted / . The set of instances is noted Do^ • For each instance / € Do, a set of optimal solutions 5 / , i.e. solutions which optimise a given objective function (also called criterion). An algorithm is said to solve an optimisation problem if, given an instance / G Do, it returns the answer ^^no^^ if 5 / is empty and otherwise returns an optimal solution 5 G 5 / . It is not difficult to associate a decision problem with an optimisation problem by searching for a solution which has a better value than a given bound K. Henceforth, the question of the decision problem is '^Does a solution exist with a criterion value lower than K ?', with K an input of the problem. If this decision problem is A/'T^-complete, then at least so is the optimisation problem. Starting from the results presented in section 2.2.1 it is possible to derive straight complexity classes for optimisation problems. This is achieved by using a generalisation of polynomial reductions: the polynomial Turing reductions. A polynomial Turing reduction OCT of a problem O towards a problem O', from the algorithmic point of view, is an algorithm A which verifies the following three properties: 2.2 Complexity of problems 39 1. A solves O, i.e. calculates for an instance / a solution of Sj if it exists. 2. A uses a procedure S which solves the problem 0\ 3. If 5 solves the problem O' in polynomial time, then A solves O in polynomial time. The complexity of the procedure S is not important in defining the polynomial Turing reduction. What matters is that if S is polynomial then A is also. We notice that the notion of polynomial Turing reduction generalises the notion of polynomial reduction in the sense that procedure S can be used iteratively. In the remainder of this chapter we denote by OCT any polynomial reduction or polynomial Turing reduction, if there is no ambiguity. Definition 13 An optimisation problem O is J\fV-hard if another optimisation problem O' AfV-hard and a polynomial Turing reduction ofO' towards O exist. The problem O is at least as difficult to solve as the problem O'. This definition is also true if O' is an A^T^-complete decision problem, and the reduction used a polynomial reduction. We say that an ATP-hard problem cannot be solved in polynomial time unless V=NV. An optimisation problem O is ATP-hard if we can show that: 1. A polynomial Turing reduction OCT and an ATT^-hard optimisation problem O' exist such that O' OCT O. 2. A polynomial Turing reduction OCT (which is not a simple polynomial reduction) and an ATT^-complete decision problem 11' exist such that n' OCT O, 3. O contains an ATP-hard sub-problem. Similarly we can demonstrate that a problem is weakly MV-hoxd. Besides, we can deduce the following property. Property 1 Let us consider two optimisation problems O and O'. If 1. yr instance of Do', 3 / an instance of Do such that Sj C Sp. 2. I can be constructed in polynomial time starting with V. then a polynomial Turing reduction exists such that O' OCT O. Considering strong A/'T^-hardness, similar results to those for proving strong A^'P-completeness have to be stated. Thus polynomial Turing reductions are not sufficient to prove strong A/'P-hardness, and similarly the notion of pseudo-polynomial Turing reduction must be considered. A/'T^-hardness is definitely a general complexity state, an AfV-hdud problem being at least as hard as any problem in class ÄfV. The class of AfV-hdiid 40 2. Complexity of problems and algorithms problems is not a class specifically dedicated to optimisation problems and even decision problems can belong to it. For instance, the Kth largest subset problem (see [Garey and Johnson, 1979] for a formal definition) is a decision problem which cannot be proved to belong to class ÄfV but which can be reduced from Partition problem. Consequently, the Kth largest subset problem is not A/'P-complete but AfV-haid. The central question is whether an optimisation problem can belong to J\fV or not. If so, it is straightforward that AfV-hard optimisation problems are AT'P-complete and the complexity classes introduced for decision problems are relevant for optimisation problems. But, the definition of class AfV implies that given a solution, we are capable of checking in polynomial time if it is optimal or not. Henceforth, except for polynomialy solvable problems, there are few chance that an optimisation problem belongs to AfV since often this checking step is as hard as solving the optimisation problem itself. This is the main reason why, in the literature, complexity of "hard" optimisation problems is often established in term of AfV-handness instead of TVP-completeness. It is remarkable that for optimisation problems, notably, the class MVO has been introduced to overcome that unsuitability of class MV. A problem belongs to class MVO if any solution can be evaluated, according to the criteria, in polynomial time. And we still have V C MVO, Henceforth, it can be easily seen that a ATT^-hard optimisation problem is ATPO-complete. More formally, the completeness inside class MVO is, at the lower level, defined by mean of an straigth extension of the polynomial reduction introduced for class MV. Henceforth, a polynomial reduction between an optimisation problem O' towards an optimisation problem O can be seen as an algorithm capable of changing, in a time bounded by a polynomial of Lengthy every instance of V of O' into an instance / of O. Besides, this reduction can also change in polynomial time any solution for instance / into a solution for instance / ' . A particular polynomial reduction, namely the AP-reduction, already enables to define the completeness of class MVO (see [Ausiello et al., 1999]). This reduction has been originally introduced in the context of approximation algorithms. 2.2.3 The complexity of counting and enumeration problems In terms of complexity, when we introduced the complexity classes dedicated to decision or optimisation problems, we were only interested in deciding of the feasibility or finding one solution. Not to count the number of solutions to the problem, nor to enumerate them. In the literature we commonly distinguish between the problem of counting the number of solutions and the problem of enumerating them. The latter is also referred to as a generation problem. In this section we consider the counting and enumeration problems associated to optimisation problems, even if w	2020-01-22 02:08:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6241855025291443, "perplexity": 3087.5583015117836}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-05/segments/1579250606269.37/warc/CC-MAIN-20200122012204-20200122041204-00252.warc.gz"}
https://stats.stackexchange.com/questions/174829/whats-the-name-for-a-time-series-with-constant-mean?noredirect=1	# What's the name for a time series with constant mean? Consider a random process $\{X_t\}$ for which the mean $\mathbb{E}(X_t)$ exists, and is constant, for all times $t$, i.e. $\mathbb{E}(X_t)=\mathbb{E}(X_{t+\tau})$ for all times $t$ and time shifts (or "lags") $\tau$. I impose no further conditions on higher moments nor on the distribution function. How can I describe such a process? It is only stationary in a weaker sense than "weakly stationary" (i.e. second-order stationarity). Other forms of stationarity have many names — I could also add "wide-sense stationary" or "covariance stationary" for the weak case, for example. So I'd expect several possible terms to be applicable, but all ones I can think of have drawbacks. • First-order stationary, or stationary to order one, is analogous to "second-order stationary" and the "stationary to order $n$" formulation often used for higher moments. But while I have seen "first-order stationary" used for processes with constant mean (e.g. here) it is commonly used with a different meaning in signal processing, the field that provides the majority of search engine hits. Every signal processing book I checked defined a process to be first-order stationary iff the first-order distribution function is invariant over time, i.e. $F_{X(t)}(x)=F_{X(t+\tau)}(x)$ for all times $t$, shifts $\tau$ and values $x$. This is quite a different condition to requiring an invariant mean — so long as the mean exists, it is a far stricter condition. They also defined "second-order stationarity" to refer to the second-order distribution function satisfying $$F_{X(t_1),X(t_2)}(x_1, x_2) = F_{X(t_1 + \tau),X(t_2 + \tau)}(x_1, x_2)$$ for all times $t_1$, $t_2$, shifts $\tau$, and values $x_1$ and $x_2$; this is (assuming the appropriate moments exist) a stronger condition than requiring means and covariance at any given lag to be independent of time, for which they reserved the term "wide-sense stationarity". Clearly one must be eagle-eyed whether "$n^\text{th}$-order" refers to distributions or moments, with great potential for confusion. As far I can see "first-order stationary", in particular, is mostly used in the distributional sense. Perhaps we can disambiguate, but I found no search engine hits for e.g. "first-order moment stationary" and just one relevant hit for "first-moment stationary". • Mean stationary may work by analogy to "covariance stationary", but I found it hard to establish prior usage. Search results were swamped by "zero-mean stationary process", which is quite different. I did find about a dozen relevant results for mean-value stationary being used in the sense I desire, too low to be the conventional terminology. • Constant level seems at first sight quite unambiguous, since "level" is widely understood to refer to "mean response" (e.g. in a regression context). However, take a random walk (without drift) $X_t = \sum_{i=0}^{t}\varepsilon_i$ where $\{\varepsilon_i\} \sim \text{WN}(0, \sigma^2)$. We know that in the population $\mathbb{E}(X_t)=0$ for all $t$, yet, in any particular realisation of $\{X_t\}$, the persistence of shocks produces a "drunkard's walk" which can stray far from the mean. When we can see multiple realisations, as illustrated, the fact the true mean remains zero is clearer; if we saw only one particular sample then, for most of the series below, "constant level" would not be the description that immediately springs to mind! Moreover, the search term "constant level time series" in Google scholar found only two papers, so it doesn't seem to be used in an adjectival way. How might I fill out the sentences "$X_t$ is a [...] process" or "$X_t$ is [...]" in a clear and unambiguous manner? Is there another term I have missed, or will one of the above — perhaps after suitable clarification — work well enough? I thought "first-moment stationary" had admirable clarity but its usage is clearly in the minority; I liked "mean stationary" for similar reasons, but found it hard to establish evidence of prior use. • If anyone wants the R code for the plot, try k <- 8; n <- 30; x <- apply(matrix(rnorm(kn), nrow=n), 2, cumsum) ; matplot(x, col=1:k, type="o", lty=1, pch="x", xlab=expression("Time index, "t), ylab=expression("Random walk, "*x[t])) – Silverfish Sep 30 '15 at 13:05 • The Wikipedia article on stationary process provides useful language for this. It may help address your concerns about whether terminology is conventional. – whuber Sep 30 '15 at 13:12 • I am somewhat used to calling such processes mean stationary but this is only personal experience. – Richard Hardy Sep 30 '15 at 13:12 • @whuber Indeed. Wikipedia does not make the distinction between "second-order" and "covariance" stationary that the signal processing books do, and if its summary of Priestley is accurate then "up to order one" ought to work okay - it appears pretty clear that the Wikipedia authors weren't coming from a signal processing background, though. Something I didn't note in my Q was that when I looked at books on geospatial statistics, they used "first-order stationary" (over space rather than time indices, but same idea) for the kind of "mean stationary" that I want. – Silverfish Sep 30 '15 at 13:21 • If you want to be as clear as possible, "weakly first-order stationary" would likely be correctly understood. If you want to be brief, you can always define your own terms. For instance, at the outset you can say "in this paper/book/monograph, a time series process $X_t$ will be called misty when the expectation $\mathbb{E}(X_t)$ exists, is finite at all times $t$, and is constant with respect to $t$; this constant is called the mean of $X_t$." Then you can safely (and gleefully) write "Let $X_t$ be a misty process with mean $\mu$..." without any fear of misunderstanding. – whuber Sep 30 '15 at 13:29	2020-08-14 04:29:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8355026841163635, "perplexity": 925.4701774325443}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439739177.25/warc/CC-MAIN-20200814040920-20200814070920-00591.warc.gz"}
https://tex.stackexchange.com/questions/490666/customize-list-of-symbols/494236	# Customize List of Symbols I would like to create such a list of symbols: So I have to have three columns and on top two sections for greek and for latin symbols... Is \listofsymbols even a good choice? Does anyone have an idea how to accomplish it? This is my code so far: % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = % Preamble % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = \documentclass[12pt,a4paper,twoside]{report} %Seitenlayout \usepackage[top=2.5cm, bottom=2cm, left=2.5cm, right=2.5cm,paper=a4paper]{geometry} %Deutsche Umlaute \usepackage{ngerman} %Mathematische Symbole und Zeichen \usepackage{amsmath} \usepackage{amssymb} \usepackage{mathrsfs} %Symbolverzeichnis \usepackage[final]{listofsymbols} %draft = Entwurfsmodus, no = Symbol wurde nicht verwendet %final = nur die verwendeten Symbole werden angezeigt % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = % Anfang von Dokumententext % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = \begin{document} \pagenumbering{Roman} % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = % Verzeichnisse % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = \tableofcontents \newpage \opensymdef \newsym[Dichte der Luft]{rhoL}{\mathbb{\rho}_{L}} \newsym[Globalstrahlung]{G}{\text{G}} \closesymdef \listofsymbols \newpage % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = % Beginn des Inhalts % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = \pagenumbering{arabic} \chapter{Einleitung} \rhoL \G \end{document} I ended up with this, which was fine for me, but doesn't have a sort function: % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = % Preamble % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = \documentclass[12pt,a4paper,twoside]{report} %Seitenlayout \usepackage[top=2.5cm, bottom=2cm, left=2.5cm, right=2.5cm,paper=a4paper]{geometry} %Deutsche Umlaute \usepackage{ngerman} %Mathematische Symbole und Zeichen \usepackage{amsmath} \usepackage{amssymb} \usepackage{mathrsfs} % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = % Symbolverzeichnis % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = \usepackage{longtable} \newcommand\nomenclature[3]{#1 & #2 & #3\\} \usepackage{array} % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = % Anfang von Dokumententext % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = \begin{document} \pagenumbering{Roman} % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = % Verzeichnisse % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = \tableofcontents % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = % Einbettung des Symbolverzeichnisses % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = \newpage \chapter{Symbolverzeichnis} \chaptermark{Symbolverzeichnis} \begingroup \renewcommand{\arraystretch}{1.275} \textbf{Griechische Symbole} \begin{longtable}[l]{@{}p{114pt}@{}p{114pt}@{}p{225pt}@{}} \nomenclature{\textbf{Symbol}}{\textbf{Einheit}}{\textbf{Bezeichnung}} \endhead % all the lines above this will be repeated on every page \nomenclature{ρ\textsubscript{L}}{[kg/m\textsuperscript{3}]}{Dichte der Luft} \nomenclature{N}{kg}{DThe number of angels per needle point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .} \nomenclature{A}{kg}{The area of the needle point} \nomenclature{sigma}{kg}{The total mass of angels per unit area} \nomenclature{m}{kg}{The mass of one angel} \end{longtable} \textbf{Lateinische Symbole} \begin{longtable}[l]{@{}p{114pt}@{}p{114pt}@{}p{225pt}@{}} \nomenclature{\textbf{Symbol}}{\textbf{Einheit}}{\textbf{Bezeichnung}} \endhead % all the lines above this will be repeated on every page \nomenclature{G}{[W/m\textsuperscript{2}]}{Globalstrahlung} \nomenclature{N}{kg}{The number of angels per needle point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .} \nomenclature{A}{kg}{The area of the needle point} \nomenclature{sigma}{kg}{The total mass of angels per unit area} \nomenclature{m}{kg}{The mass of one angel} \end{longtable} \endgroup \newpage % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = % Beginn des Inhalts % = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = = \pagenumbering{arabic} \chapter{Einleitung} \end{document}	2019-11-12 08:27:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3531211018562317, "perplexity": 21.7789853332196}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496664808.68/warc/CC-MAIN-20191112074214-20191112102214-00389.warc.gz"}
https://math.stackexchange.com/questions/3492541/dimension-of-solution-space-of-system-of-linear-equations	# Dimension of solution space of system of linear equations I have the system of linear equations \begin{align} x_1+x_3=0 \\ -4x_1+2x_2=0 \\ -13x_1+4x_2-5x_3=0 \end{align} and I have converted it to an augmented matrix and row reduced to \begin{align} \begin{pmatrix} 1 & 0 & 1 & 0 \\ 0 & 1 & 2 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \end{align} which, as far as I know, tells me that the dimension of the solution space is $$2$$ since there are $$2$$ linearly independent rows. However, the general solution $$\begin{pmatrix} -x_3 \\ -2x_3 \\ x_3 \end{pmatrix}$$ tells me that the dimension of the solution space is actually $$1$$. Question: Could someone tell me where I've made a mistake? • The dimension of the space of solutions is not the number of independent rows, but the number of variables minus the number of independent rows: $3-2=1$. Each independent row is a restriction, it cuts down a degree of freedom from the total space of values of the variables that is available for the the space of solutions. – MoonLightSyzygy Dec 30 '19 at 20:57 Dimension of the solution space is equal to number of basis vector in fundamental system of solutions: 1 in your case. In general case, from rank-nullity theorem for linear map $$L:V\rightarrow W$$ $$\dim(Ker(L))+\dim(Im(L))=\dim(V)$$. In your case $$L:\Bbb R ^3\rightarrow \Bbb R ^3$$ and $$\dim(Im(L))=2$$, so $$dim(Ker(L))=1$$ and dimension of the solution space is exactly 1.	2020-07-13 05:24:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 11, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9984013438224792, "perplexity": 86.16967908538768}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-29/segments/1593657142589.93/warc/CC-MAIN-20200713033803-20200713063803-00057.warc.gz"}
https://asmedigitalcollection.asme.org/appliedmechanics/article-abstract/72/4/500/469727/Anisotropic-Elastic-Tubes-of-Arbitrary-Cross?redirectedFrom=fulltext	First approximation analytical solutions are constructed for finite and semi-infinite, fully anisotropic elastic tubes of constant thickness $h$ and arbitrary cross section, subject to purely kinetic or purely kinematic boundary conditions. Final results contain relative errors of $O(h∕R)$, where $R$ is some equivalent cross sectional radius. Solutions are decomposed into the sum of an exact beamlike or Saint-Venant solution, treated in Ladevèze et al. (Int. J. Solids Struct., 41, pp. 1925–1944, 2004) and extended in an appendix; a rapidly decaying edge-zone solution; and a slowly decaying semi-membrane-inextensional-bending (MB) solution. Explicit conditions on the boundary data are given that guarantee decaying solutions. The MB solutions are expressed as an infinite series of complex-valued exponential functions times real-valued one-dimensional eigenfunctions which satisfy a fourth-order differential equation in the circumferential coordinate and depend on the pointwise cross sectional curvature only. 1. Ladevèze , P. , and Simmonds , J. G. , 1996, “ De Nouveaux Concepts en Théorie des Poutres pour des Charges et Géométries Quelconques ,” Compt. Rend. 0001-4036 322 ( II ) pp. 455 462 . 2. Ladevèze , P. , and Simmonds , J. G. , 1998, “ New Concepts for Linear Beam Theory with Arbitrary Geometry and Loading ,” Eur. J. Mech. A/Solids 0997-7538 17 , pp. 377 402 . 3. Ladevèze , P. , Sanchez , Ph. , and Simmonds , J. G. , 2004, “ Beamlike (Saint-Venant) Solutions for Fully Anisotropic Elastic Tubes of Arbitrary Cross Section ,” Int. J. Solids Struct. 0020-7683 41 , pp. 1925 1944 . 4. Sanders , J. L. Jr. , 1959, “ An Improved First-Approximation Theory for Thin Shells ”, NASA Rep. 24. 5. Koiter , W. T. , 1959, “ A Consistent First Approximation in the General Theory of Thin Elastic Shells ,” The Theory of Thin Elastic Shells , Proc IUTAM Symposium , Delft, 1959, Koiter W. T. , ed., North-Holland , Amsterdam. 6. Ladevèze , P. ed, 1985, Local Effects in the Analysis of Structures , Elsevier , Amsterdam. 7. Simmonds , J. G. , 1976, “ Recent Advances in Shell Theory ,” Advances in Engineering Science , 13th Annual Meeting Society of Engineering Science , NASA CP-2001, pp. 617 626 . 8. Goldenveiser , A. L. , 1940, “ Equations of the Theory of Thin Shells ” (in Russian), Prikl. Mat. Mekh. 0032-8235, 4 , pp. 35 42 . 9. Lure A. I. , 1940, “ General Theory of Thin Elastic Shells ” (in Russian), Prikl. Mat. Mekh. 0032-8235, 4 , pp. 7 34 . 10. McDevitt , T. J. , and Simmonds , J. G. , 1999, “ Reduction of the Sanders-Koiter Equations for Fully Anisotropic Circular Cylindrical Shells to Two Coupled Equations for a Stress and a Curvature Function ,” J. Appl. Mech. 0021-8936, 66 , pp. 593 597 . 11. Budiansky , B. , and Sanders , J. L. , Jr. , 1963, “ On the ‘Best’ First-Order Linear Shell theory ,” Progress in Applied Mechanics (Prager Anniversary Volume), Macmillan , New York, pp. 129 140 . 12. Gregory , R. D. , and Wan , F. Y. M. , 1984, “ Decaying States of Plane Strain in a Semi-Infinite Strip and Boundary Conditions for Plate Theory ,” J. Elast. 0374-3535 14 , pp. 27 64 . 13. Ladevèze , P. , 1983, “ Sur le Principe de Saint-Venant en élasticité ,” J. Mec. Theor. Appl. 0750-7240 2 , pp. 161 184 . 14. Ladevèze , P. , and Simmonds , J. G. , 2001, “ The Exact One-Dimensional Theory for End-Loaded Fully Anisotropic Beams of Narrow Cross Section ,” J. Appl. Mech. 0021-8936 68 , pp. 865 868 . This content is only available via PDF. You do not currently have access to this content.	2019-10-18 14:17:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 3, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6412737369537354, "perplexity": 11174.840442401415}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570986682998.59/warc/CC-MAIN-20191018131050-20191018154550-00445.warc.gz"}
https://esp.mit.edu/teach/teachers/carcl/bio.html	# ESP Biography ## CARLOS CORTEZ, ESP Teacher Major: 18 College/Employer: MIT Not Available. ## Past Classes (Clicking a class title will bring you to the course's section of the corresponding course catalog) M10555: Knot Theory in HSSP Summer 2016 (Jul. 10, 2016) This course will be an exposition to the beautiful subject of knot theory. While knot theory first attracted interest due to an outdated atomic model, it is now a central topic of study in low-dimensional topology with connections to several other fields. Furthermore, it is easy and fun to experiment with - we'll provide the string! However, you may be surprised at how difficult it is to tell if your friend's knot is the same as yours. Topics will be tailored to the audience. Likely ones include: examples of knots and links, Reidemeister moves, prime knots, knot invariants (particularly knot polynomials), Seifert surfaces, braid groups. You may even learn a couple of party tricks and some knot-so-great puns! C8920: Splay Trees (and other search tree fun) in Splash 2014 (Nov. 22 - 23, 2014) Learn about splay trees: the elegant and effective self-adjusting binary search tree! Topics covered include the construction of the data structure, runtime analysis, and discussion of performance in practice. Also will briefly discuss the dynamic optimality conjecture if time allows. M9025: Dissection problems in Splash 2014 (Nov. 22 - 23, 2014) Can you cut a square in pieces and rearrange them to form an equilateral triangle of equal area? Of course! Can you cut a cube in pieces and rearrange them to form a regular tetrahedron? Of course not! Come to this class to learn: why you can cut and rearrange any polygon into another of equal area, but you can't do the same with polyhedra; why you can evenly cut sandwiches with as many as $$n$$ ingredients in $$n$$ dimensions in two equal parts with a single cut; and maybe even why you can cut one 3-D ball into two identical copies of it. For a demonstration, we may even have ham-sandwiches. They'll be 3-dimensional, we promise! M7757: Olympiad Math Problem Solving in Splash! 2013 (Nov. 23 - 24, 2013) This class will not be like your average high school math class. We will explore topics that include algebra, number theory, combinatorics, and geometry. The problems will range from advanced to math olymipad level and might include one or two International Math Olympiad problems.	2022-05-24 02:11:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22758130729198456, "perplexity": 1178.3934529970925}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662562410.53/warc/CC-MAIN-20220524014636-20220524044636-00412.warc.gz"}
http://tfcj.aoly.pw/surface-area-of-a-cylinder-integral.html	## Surface Area Of A Cylinder Integral Each of these disks also has a circular area A = πr 2, where r is the radius, and disk volume dV = πr 2 dh, which is the product of the area and the thickness. (b) - the length from the axis to the edge of the parabola. Surface Area of a Sphere from first principles. In general, you can skip parentheses, but be very careful: e^3x is e^3x, and e^(3x) is e^(3x). Work out the surface area to volume ratio. The integral requires us to find the dot product between the electric field and the area, and the integrate this dot product ove the entire surface. Since the bases for a prism or cylinder are congruent, this is often expressed as twice the area of the base. Solve advanced problems in Physics, Mathematics and Engineering. Given the stimulus word "water", I started looking into aspects of water that involved modeling and real-world application, because I prefer to work with real ideas. The simplest instance occurs when computing the area of a surface. The integral symbol is actually a very stylized letter S: once you realize that, you see that you are "summing something" and the limits just describe the region over which the summing happens. This shape has a circular base and straight, parallel sides. Surface Area of a Sphere Surface Area of an Ellipsoid. (which may be a curved set in area); In the part discussed about double integral analog is the line integral. Now, this is interesting because it is the same ratio as the volume a sphere to the volume of its circumscribing cylinder. Surface integrals Examples, Z S dS; Z S dS; Z S a ¢ dS; Z S a £ dS S may be either open or close. If we already know the Surface Area of Cube and then we can calculate the length of any side by altering the above formula as: l = √sa / 6 (sa = Surface Area of a Cube). b) The portion of the surface zxy= lying inside the cylinder xy22+=9 c) The portion of the surface zxy=+ +13 22 lying above the triangle formed by the points (0,0), (0,1) and (2,1). I We have developed deﬁnite integral formulas for arc length and surface area for curves of the form y = f(x) with a x b. the surface area of a solid is the total area of the outer layer of the object; for objects such as cubes or bricks, the surface area of the object is the sum of the areas of all of its faces Licenses and Attributions. Cylinder calculator will give the surface area and volume of a cylinder. Math - I saw the formula for the surface area of a torus on your web site, and I'm wondering if you can show me how that formula was derived?. The horizontal surface is the xy-plane with the z- axis rising vertically from that plane. The surface area of a sphere usually requires calculus to be explained. Keywords: ellipsoidsegment, surfacearea, Legendre,ellipticintegral. The part of function is to be integrated which may be a scalar or vector field. In other words, the sphere has 4/6 or two thirds the area of its enclosing cylinder. Surface integral. 13 is 20 cm in diameter and has a conical contraction at the bottom with an exit hole 3 cm in diameter. To find the surface area of a box, start by calculating the area of each side using the formula a = lh, where l is the length and h is the height. Surface area and surface integrals. The evaluation of these integrals in a particular coordinate system requires the knowledge of differential elements of length, surface, and volume. For example: Cut a sphere of radius into 10 slices, and expand the slices into cylinders:. strates how to nd the surface integral of a given vector eld over a surface. Surface integrals of scalar fields. That is parameterized by these two parameters right there. I will determine the volume of the half-doughnut and double the result. By signing up, you'll. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4. Free Cylinder Surface Area Calculator - calculate cylinder surface area step by step Derivatives Derivative Applications Limits Integrals Integral Applications. Brightstorm explains how to use definite integrals to find area. • Curl = circulation per unit area: C r = circle of radius r centered at a point P in the. Then the area of the surface de ned by z= f(x;y), (x;y) 2Dis A= ZZ D q [f x(x;y)]2 + [f y(x;y)]2 + 1dA: Example: Find the area of the part of the surface z= x+ y2 that lies above the triangle with vertices (0;0), (1;1), and (0;1). Surface integrals of a scalar ﬁeld Theorem The integral of a continuous scalar function g : R3 → R over a surface S deﬁned as. Hint: Create your own function and interval and identify your axis of rotation. This online calculator will calculate the various properties of a cylinder given 2 known values. Online surface area calculators for sphere, ellipsoid, cuboid, cube, cylinder and cone with flexible units. , Mensuration is the study of these geometrical shapes, their shapes and their parameters like area, volume etc. An orientable surface, roughly speaking, is one with two distinct sides. VOLUME: Use the Volume formula for a Circular Cylinder. Example: The figure shows a section of a metal pipe. Surface integrals To compute the flow across a surface, also known as flux, we'll use a surface integral. Area of one face = 3 x 3 cm = 9 cm2. If the tangent plane just happened to be horizontal, of course the area would simply be the area of the rectangle. The final answer is Derivation of the Surface Area Formula. The surface integral of a Sphere can be found using polar coordinates. This is quite. Flux, Surface Integrals & Gauss’ Law A Guide for the Perplexed 0. This free surface area calculator determines the surface area of a number of common shapes, including sphere, cone, cube, cylinder, capsule, cap, conical frustum, ellipsoid, and square pyramid. Find more Mathematics widgets in Wolfram\|Alpha. The easiest way is to "separate" the cylindrical tank from the spherical top and calcuate the surface area and volume for each. This is a right circular cylinder where the top and bottom surfaces are parallel but it is commonly. Evaluate the integral, and we get $\pi r^2h$ as our formula for the area of a cylinder. Each end is a circle so the surface area of each end is π * r 2 , where r is the radius of the end. For convenience we will often write. Determining the Surface Area of Cones and Cylinders. This is found by slicing the cyclinder surface and rolling it out as a rectangle. The curl of the given vector eld F~is curlF~= h0;2z;2y 2y2i. List of Integrals Containing exp(x) Double Integrals: Surface Area For non-negative f(x,y) with continuous partial derivatives in the closed and bonded region D in the xy plane, the area of the surfce z = f(x,y) equals:. Examine the formula for the area of a circle. the area problem. Solution: Let the external radius, the internal radius and the height of the hollow cylinder be , and h respectively. Brightstorm explains how to use definite integrals to find area. An orientable surface, roughly speaking, is one with two distinct sides. The formula for surface area of a cylinder is (2pi * r^2) + (2pi * r * h). Calculate the unknown defining side lengths, circumferences, volumes or radii of a various geometric shapes with any 2 known variables. 1 hr 28 min 6 Examples. The integral will always be a degree higher than the original function. To derive the formula of the surface area of a cylinder, we will start by showing you how you can make a cylinder : Start with a rectangle and two circles Then, fold the rectangle until you make an open cylinder with it. (2,2) B) Using Integrals, Find The Surface Area Of A Cylinder Of Radius R And Height H Excluding The Area Of Bases. Surface Area The surface area of a figure is the sum of the area of all surfaces of a figure. As a result, in unfavorable temperatures a baby will become distressed much more rapidly than an adult. Total surface area of hollow cylinder = area of internal curved surface + area of external curved surface + area of the two rings. Half of this total value plus the difference between r outer and r inner multiplied by 2L is the surface area. This shape has a circular base and straight, parallel sides. (2,2) b) Using integrals, find the surface area of a cylinder of radius r and height h excluding the area of bases. Integral units run at speeds of between 200 and 600 rpm. The surface area of a cone may also be split into two parts: A = A(lateral) + A(base), as we have only one base, in contrast to a cylinder. Usually, one direction is considered to be positive, the other negative. Remember, when finding Surface Area, you are finding the total area of all the 2 Dimensional shapes that are put together to form the three dimensional. A) Using Integrals, Find The Surface Arca Of A Cone Of Radius R And Height H Excluding The Area Of Base. Now the surface area of a small element of the cylinder will be given by. Share: Facebook Twitter Reddit Pinterest WhatsApp Email Link. This free surface area calculator determines the surface area of a number of common shapes, including sphere, cone, cube, cylinder, capsule, cap, conical frustum, ellipsoid, and square pyramid. h is in there because that is part of the information yhou are given, the distance between the planes. The surface area of the bottom circle of a cone is the same as for any circle, π r 2 {\displaystyle \pi r^{2}}. The cylinder surface area is the height times the perimeter of the circle base, plus the areas of the two bases, all added together. Just as with line integrals, there are two kinds of surface integrals: a surface integral of a scalar-valued function and a surface integral of a vector field. , Mensuration is the study of these geometrical shapes, their shapes and their parameters like area, volume etc. Following is the formula for Calculating the Volume of a Cylinder. Solution: The surface is in red, the domain of integration in yellow:. 1)The solid cut from the first octant by the surface z = 9 - x2 - y 1) Find the volume by using polar coordinates. Find Volume and Surface Area of Tube Shape Donut Surface Area of the Torus : If we divide the torus into k cylinders each having length l, now consider the division of the torus into k cylinders each having the length l, curved surface area of each such cylinder will be 2Πrl; consequently the surface area of the torus will be 2Πrlk and or we can say the surface area of the torus as 4Π 2 rR. Surface Area of a Sphere: The area of a disk enclosed by a circle of radius R is PiR 2. If we add all these dA‘s to one another over the first surface, which is the surface of circular surface of the cylinder, that is equal to the cross sectional area of the cylinder, and we call that area as A. Parameterized Surfaces Surface Integrals of Scalar-Valued Functions Surface Integrals of. Volume removed from a cylinder when a hole is drilled through it. We'll use the same surface as in the divergence theorem example page. The surface area you need to find is that of a section of a sphere, a band, lying between two planes. These figures are three dimensional figures. Surface area of a cylinder = 2πr² + 2πrh (r is the radius of the top and bottom and h is the cylinder’s height) The surface area of a cylinder is the sum of its top and bottom (the area of two circles) plus the area that wraps around the middle. If two identical faces of side a are joined together, then the total surface area of the cuboid so formed is 10a2. A) Using Integrals, Find The Surface Arca Of A Cone Of Radius R And Height H Excluding The Area Of Base. The integral in eq. You can derive the surface area formula by cutting the sphere into slices, approximate each slice by a cylinder, get the surface area around each slice, and add them all up. Cylinder calculator will give the surface area and volume of a cylinder. Surface Area of a Parametrized Surface In section 13. A cylinder with a base $A=\pi r^2$ and height $h$, will have a volume $A\cdot h=\pi r^2h$. of Kansas Dept. As can be calculated, the cylinder with the smallest surface area occurs for ; that is, when the diameter of the cylinder is equal to the height of the cylinder. What object is a circle in plan, front and side views? (Besides a sphere) or What is the resulting geometry from the intersection of three cylinders? Written by Paul Bourke March 1992 Question. , Mensuration is the study of these geometrical shapes, their shapes and their parameters like area, volume etc. The electrostatic potential at (0;0; a) of a charge of constant density ˙on the hemisphere S: x2 + y2 + z2 = a2, z 0 is U= Z Z S ˙ p x2 + y2 + (z+ a)2 dS: Show that U= 2ˇ˙a(2 p 2). 4, we learned how to make measurements along curves for scalar and vector ﬁelds by using. Below are six versions of our grade 6 math worksheet on volume and surface areas of 3D shapes including rectangular prisms and cylinders. I Mass and center of mass thin shells. Top & Bottom: 2×(8×5. Surface integrals of scalar fields. The following is a parametric representation of a cylinder surface that is coaxial with the z-axis (the cylinder ends will be included later) CylSurf = 8 Cos @q D,Sin cz < 1. I If a surface is parametrized by a vector valued function r(u;v), the surface element can be computed by taking the cross product of the line elements of the grid curves of the parametrization. Surface area of a hollow cylinder. There are two basic area problems: one in which the area lies between the function and the x-axis, and one where the area lies between the two functions. Evaluate the integral, and we get $\pi r^2h$ as our formula for the area of a cylinder. Cylinder calculator will give the surface area and volume of a cylinder. of field lines per area. 𝑆 𝐴 𝑉 = 169 132 ≈. Ch is the product of the circumference of one of the circles and h the height of the cylinder. In this section we introduce the idea of a surface integral. The area of a convex surface is a totally-additive function on the ring of Borel sets. Determine (a) its volume in cubic millimetres and (b) its total surface area in square millimetres. Several Web pages derive the formula for the surface area of a cone using calculus. Find the surface area of the following surfaces. Surface integrals of a scalar ﬁeld Theorem The integral of a continuous scalar function g : R3 → R over a surface S deﬁned as. Vector surface integral examples by Duane Q. Therefore, after using the formula and getting a number as its result, to find the total lateral surface area we simply multiply. 1 \| P a g e Calculating the Drag Coefficient C D for a Cylinder in Cross Flow Theoretical Basis— Using Surface Pressure Measurements Consider the area element dA = LRd on the surface of the cylinder as shown below in Fig. By signing up, you'll. integral, or from one surface to another surface with the same boundary. We often express the basic laws of electromagnetic fields in terms of integrals of field quantities over various curves (lines), surfaces, and volumes in a region. In the sample cone given above, the variable r ranges from 0 to 2 inches while h goes from 0 at the base to 4 inches at the cone’s tip. Just as with line integrals, there are two kinds of surface integrals: a surface integral of a scalar-valued function and a surface integral of a vector field. You can use this information to determine the Surface Area of each shape. Additionally verify vour answer with the usual formula for the surface area of a sides of a cylinder (S. The evaluation of these integrals in a particular coordinate system requires the knowledge of differential elements of length, surface, and volume. If you didn’t get a parameterization for problem 2, ask me what to do. Surface area and volume multiple choice questions (MCQs), surface area and volume quiz answers, test prep 1 to learn online elementary school courses for math degree. Surface area and surface integrals. In cylindrical coordinates, the volume of a solid is defined by the formula \[V = \iiint\limits_U {\rho d\rho d\varphi dz}. = 8-2 = 6 cm. 1 hr 28 min 6 Examples. Find the surface area of the following surfaces. surface integrals and scaling factors >up This explanation is trying to get out the presence (or lack) of a scaling factor (e. Area is a quantity that expresses the extent of a two-dimensional surface or shape, or planar lamina, in the plane. The Surface Area of Paraboloid calculator computes Paraboloid the surface area of revolution of a parabola around an axis of length (a) of a width of (b) including the circular base. The spherical dome is the figure resulting from having made a flat cut in a sphere. In this section we will look at the lone application (aside from the area and volume interpretations) of multiple integrals in this material. Determining the Surface Area of a Cylinder from a Net. In evaluating the integral in Gauss' law for a spherical surface centered on a positively charged particle, which describes the dot product of the electric field and a differential area vector? The dot product is positive because the vectors are in the same direction. Math 314 Lecture #34 §16. This will complete the discussion for all the standard solids. For example: Cut a sphere of radius into 10 slices, and expand the slices into cylinders:. Surface Area of a Cube = 6l² (Where l is the Length of any side of a Cube). Solution: The surface is in red, the domain of integration in yellow:. The calculation of surface area of revolution is related to the arc length calculation. Integration in Cylindrical and Spherical Coordinates 8. In our discussion we will discuss the double integral, which is the extension to functions of two variables. Surface area and volume multiple choice questions (MCQs), surface area and volume quiz answers, test prep 1 to learn online elementary school courses for math degree. ? Evaluate the surface integral. φE ) associated with any closed surface S, is a measure of the (total) charge enclosed by surface S. The vector diﬁerential dS represents a vector area element of the surface S, and may be written as dS = n^ dS, where n^ is a unit normal to the surface at the position of the element. Determining the Surface Area of a Cylinder from a Net. Problem : Find the area of a circle with radius a. Rate of change of surface area of sphere Problem Gas is escaping from a spherical balloon at the rate of 2 cm 3 /min. Parameterized Surfaces Surface Integrals of Scalar-Valued Functions Surface Integrals of. Now, this is interesting because it is the same ratio as the volume a sphere to the volume of its circumscribing cylinder. The specific curvature of a convex surface in a domain is the ratio of the curvature of the domain to its area. That is parameterized by these two parameters right there. Surface area and surface integrals. We wish to find the function that minimizes, at least locally, the area of this surface subject to the constraint that the radius at the two endpoints is fixed (at, say, r a). com How to use definite integrals to find the area between a curve or under a curve. Have a look at this tutorial, and learn exactly how to find the area of a 3D object, specifically a pyramid. Ch is the product of the circumference of one of the circles and h the height of the cylinder. At any point on an orientable surface, there exists two normal vectors, one pointing in the opposite direction of the other. down Gauss’s law for gravity, evaluate the integral, and solve for the acceleration g. The surface integral can be defined component-wise according to the definition of the surface integral of a scalar field; the result is a vector. Find the average distance along the earth. The surface area of the cylinder bounded by the xy -plane (z = 0) and the plane �+�=10 (written z = 10 – x = 10−6cos𝜃) is found in a similar manner. The surface is defined at a fixed radial position, so only the axial (z) and azimuthal (Φ) coordinates are necessary to compute the total area. The first theorem of Pappus tells us about the surface area of the surface of revolution we get when we rotate a plane curve around an axis which is external to it but on the same plane. List of Integrals Containing exp(x) Double Integrals: Surface Area For non-negative f(x,y) with continuous partial derivatives in the closed and bonded region D in the xy plane, the area of the surfce z = f(x,y) equals:. Find the total surface area of a cone with base diameter of 15. Round your answer to the nearest tenth. area S S Calculating general surface integrals; ﬁnding dS. Compare your answer with the value of the integral produced by your calculator. It can be thought of as the double integral analog of the line integral. The surface area of a cylinder is 2 π * r * h, so you have the integral from x=0 to x=1 of 2 * π * tan^-1 x * dx To rotate this about the y-axis, it is convenient if we first solve for x: x = tan y The limits for y are 0 degrees to 45 degrees, or 0 to π/4 if you prefer radians. The value Π, or pi, is a mathematical number calculated as 22÷7, or 3. Compute the Weight or Mass of a Slanted Cylinder. They have the expansion Jν(z)=(z 2)ν∞ ∑ m=0(−1)m m!γ(ν+m+1)(z 2)2m,. of patches increases and define the surface integral of f over the surface S as: * Analogues to: The definition of a line integral (Definition 2 in Section 16. Work out the surface area to volume ratio. The calculator will find the area of the surface of revolution (around the given axis) of the explicit, polar or parametric curve on the given interval, with steps shown. These solids differ from prisms in that they do not have uniform cross sections. You have a vector field being integrated around a highly symmetrical closed surface (a section of a cylinder). (a) Parametrize the surface by considering it as a graph. Calculator Use. The surface area of a sphere usually requires calculus to be explained. The SI unit for volume is the cubic meter, or m 3. As a result, in unfavorable temperatures a baby will become distressed much more rapidly than an adult. If we cut the half sphere at height z, we obtain a disc of area (R2 −r2)π. 1415929203539825. 3 Surface integrals Consider a crop growing on a hillside S, Suppose that the crop yeild per unit surface area varies across the surface of the hillside and that it has the value f(x,y,z) at the point (x,y,z). Each subregion. The final answer is Derivation of the Surface Area Formula. What we discover about surfaces parallels what we already know about curves—all "lifted" up one dimension. Since the surface area of a sphere of radius r is 4πr 2, the volume of a spherical shell of radius r and thickness dr must be. First, we need to parameterize the surface S. For example, the surface area of the cone found by rotating y = x around the x-axis from x = 0 to x = 4 should be: 2pi(integral of x dx) from 0 to 4 Unfortunately this doesn't give me the answer I know to be correct, as you have to use the length of a curve formula within the integral for it to work. That means we only. 5-8: Surface Area, Triple Integrals Friday, April 8 Surface Area Using the formula A(S) = ZZ D q 1 + f2 x + f y 2 dA, nd the surface area of a sphere of radius a. The surface area of the bottom circle of a cone is the same as for any circle, π r 2 {\displaystyle \pi r^{2}}. Surface Area of Paraboloid (A): The surface area (A) is returned in square meters. And if we wanted to figure out the surface area, if we just kind of set it as the surface integral we saw in, I think, the last video at least the last vector calculus video I did that this is a surface integral over the surface. (If you don't yet know how to do this, you can still calculate the integral if you are good at doing integrals, but it gets pretty ugly. 7 cm and height of a prism is 24 cm. At height z, this body has a cross section with area R2π−r2π. Example: The figure shows a section of a metal pipe. 1 hr 28 min 6 Examples. If one thinks of S as made of some material, and for each x in S the number f(x) is the density of material at x, then the surface integral of f over S is the mass per unit thickness of S. On the graph, the red below the parabola is the area and the dotted line is the integral function. Find a parametrization of the surface cut from the parabolic cylinder y — + 1 by the planes z 0, z 4 and y 3. The surface area of a cone can be derived from the surface area of a square pyramid Start with a square pyramid and just keep increasing the number of sides of the base. For these problems, you divide the surface into narrow circular bands, figure the surface area of a representative band, and then just add up the areas of all the bands to get the total surface area. The surface area of a solid of revolution: The surface area generated by the segment of a curve y = f (x) between x = a and y = b rotating around the x-axis, is shown in the left figure below. of patches increases and define the surface integral of f over the surface S as: * Analogues to: The definition of a line integral (Definition 2 in Section 16. We can find an infinitesimal band on the surface of either which has the area dA = 2πxds. calculated by a surface integral: 1 (15) average value of f on S = f(x,y,z)dS. Area and Volume Formula for geometrical figures - square, rectangle, triangle, polygon, circle, ellipse, trapezoid, cube, sphere, cylinder and cone. In particular, if the above "trajectory" is a circle, we have a circular cylinder of radius R, whose surface area is 2 p RH. Definite Integral Definition. Share: Facebook Twitter Reddit Pinterest WhatsApp Email Link. 5-8: Surface Area, Triple Integrals Friday, April 8 Surface Area Using the formula A(S) = ZZ D q 1 + f2 x + f y 2 dA, nd the surface area of a sphere of radius a. There are two basic area problems: one in which the area lies between the function and the x-axis, and one where the area lies between the two functions. The cylinder surface area is the height times the perimeter of the circle base, plus the areas of the two bases, all added together. 1) To evaluate the surface integral in Equation 1, we approximate the patch area ∆S ij by the area of an. Evaluate the integral, and we get $\pi r^2h$ as our formula for the area of a cylinder. Free Cylinder Surface Area Calculator - calculate cylinder surface area step by step Derivatives Derivative Applications Limits Integrals Integral Applications. 88 A square sheet of paper is converted into a cylinder by rolling it along its side. The surface area of a solid of revolution: The surface area generated by the segment of a curve y = f (x) between x = a and y = b rotating around the x-axis, is shown in the left figure below. “Volume is the integral of surface area. Find the surface area of a cylinder with a diameter of 13. a) Using integrals, find the surface arca of a cone of radius r and height h excluding the area of base. Find the surface area of the portion of the sphere of radius 4 that lies inside the cylinder x 2+y = 12 and above the xy-plane. Surface = 2b + Ph (b is the area of the base P is the perimeter of the base) Cylinder Volume = r2 X height V = r2 h Surface = 2 radius X height S = 2 rh + 2 r2 Pyramid Volume = 1/3 area of the base X height V = bh b is the area of the base Surface Area: Add the area of the base to the sum of the areas of all of the triangular faces. The cylinder surface area is the height times the perimeter of the circle base, plus the areas of the two bases, all added together. The integral of that is the correct area encircled by the curve (defined only modulo the total surface area of the ellipsoid) even if the curve goes around the polar axis many times. “Volume is the integral of surface area. Surface integrals of a scalar ﬁeld Theorem The integral of a continuous scalar function g : R3 → R over a surface S deﬁned as. Example: The Surface Integral Consider the vector field: ( ) ˆ A rxa= x Say we wish to evaluate the surface integral: (s) S ∫∫A rds⋅ where S is a cylinder whose axis is aligned with the z-axis and is centered at the origin. The definite integral of any function can be expressed either as the limit of a sum or if there exists an anti-derivative F for the interval [a, b], then the definite integral of the function is the difference of the values at points a and b. Archimedes proved this in approximately 260 B. ” To me, that makes so much more sense that “surface area is the derivative of volume. I understand how to find the surface area, but I'm not quite sure what it means by "in terms of pi" Could someone please explain? the base diameter of the cylinder is 4 in. Determine the surface area for the following function surfaces. Select the object from the options then answer the questions about length and width. Total surface area of hollow cylinder = area of internal curved surface + area of external curved surface + area of the two rings. Solid of Revolution - Finding Volume by Rotation Finding the volume of a solid revolution is a method of calculating the volume of a 3D object formed by a rotated area of a 2D space. The standard integral with respect to area for functions of x and y is a special case, where the surface. Some equations in cylindrical coordinates (plug in x = rcos(θ),y = rsin(θ)): Cylinder: x2 +y2 = a2 ⇒ r2 = a2 ⇒ r = a;. In the listed formulae for this set of calculations, PI here is used as 3. An element of surface area for the cylinder is as seen from the picture below. area S S Calculating general surface integrals; ﬁnding dS. The surface area of a prism or cylinder is the lateral area plus the area of each base. Surface Area and Surface Integrals We’ve done line integrals, now it’s time to generalize a bit more and do surface integrals. A cylinder has a radius (r) and a height (h) (see picture below). Ch is the product of the circumference of one of the circles and h the height of the cylinder. Surface Area & Volume Applets. Let me first calculate the surface area of 1/8th part (1/8th of a Sphere). 2 Surface Integrals 2. Area of a square = l² Since the Cube is made of 6 equal squares, Surface Area of a Cube = 6l². As the parameters θ and z vary, the whole cylinder is traced out ; the piece we want satisﬁes 0 ≤ θ ≤ π/2, 0 ≤ z ≤ h. Free Cylinder Volume & Radius Calculator - calculate cylinder volume, radius step by step. Related Surface Area Calculator \| Area Calculator. In our study of electromagnetism we will often be required to perform line, surface, and volume integrations. That is parameterized by these two parameters right there. Find the surface area of a Gaussian cylinder Find , the area of the Gaussian surface. I understand how to find the surface area, but I'm not quite sure what it means by "in terms of pi" Could someone please explain? the base diameter of the cylinder is 4 in. The surface area of an ellipsoid can be determined by using the following formula: where ab, ac and bc are the distances from its origin to its surface. A hollow cylinder has length L and inner and outer radii a and b. Integral units run at speeds of between 200 and 600 rpm. Free Online Scientific Notation Calculator. Let S be a closed surface and for each point (x;y;z) on the surface let f(x;y;z) be the rate of ﬂow of ﬂuid out through the surface per unit surface area and unit time, i. The line integral is very di cult to compute directly, so we'll use Stokes' Theorem. The notation for a surface integral of a function P(x,y,z) on a surface S is Note that if P(x,y,z)=1, then the above surface integral is equal to the surface area of S. 1: Shows the force ﬁeld F and the curve C. How to Cite this Page: Su, Francis E. the area, da, of this infinitely thin cylinder is: All we have left is to solve the integral of cosine. You have a vector field being integrated around a highly symmetrical closed surface (a section of a cylinder). An ant, like other insects, has an exoskeleton. 7 ft 5 ft Find the surface area of the cylinder. Solution to the problem: The equation of the circle shown above is given by x 2 + y 2 = a 2 The circle is symmetric with respect to the x and y axes, hence we can find the area of one quarter of a circle and multiply by 4 in order to obtain the total area of the circle. 1 Electric field lines passing through a surface of area A. This cube has six congruent faces, each with a length and width of 3 cm. I Mass and center of mass thin shells. therefore the surface area is a double integral integrate t from 0 to 2pi integrate r from 0 to 8 sqrt(2)drdt=. Cylindrical Surface is a curved surface generated by parallel duplication of a line. Integral compressors may be equipped with two to ten compressor cylinders with power ranging from 140 to 12,000 hp. Math 314 Lecture #34 §16. The surface area of a prism or cylinder is the lateral area plus the area of each base. INSTRUCTIONS: Choose units and enter the following: (a) - the length along the axis. The term C p may be expanded via eq. How to Calculate Triangle Area. Total Surface Area of Cylinder (TSA) Total surface area of a cone; Surface area of ellipse; Curved surface area of a cylinder (CSA) Curved surface area of a cone;. To know more about great circle, see properties of a sphere. Each of these disks also has a circular area A = πr 2, where r is the radius, and disk volume dV = πr 2 dh, which is the product of the area and the thickness. 9 Compute the surface area of the portion of the saddle, z= x2 y2, that is contained in the cylinder x2 + y2 = 1. This integral is very simple to calculate if you know how to change variables to polar coordinates. Let's be honest - sometimes the best triangle area calculator is the one that is easy to use and doesn't require us to even know what the triangle area formula is in the first place!. 1415929203539825. The part of function is to be integrated which may be a scalar or vector field. Area of a base 4. of patches increases and define the surface integral of f over the surface S as: * Analogues to: The definition of a line integral (Definition 2 in Section 16. This explanation is trying to get out the presence (or lack) of a scaling factor (e. The first theorem of Pappus tells us about the surface area of the surface of revolution we get when we rotate a plane curve around an axis which is external to it but on the same plane. The formula. If the Gaussian surface is chosen such that for every point on the surface the component of the electric field along the normal vector is constant, then the calculation will not require difficult integration as the constants which arise can be taken out of the integral. Math 314 Lecture #34 §16. Compute the surface integral. Related Surface Area Calculator \| Area Calculator. If the Gaussian surface is chosen such that for every point on the surface the component of the electric field along the normal vector is constant, then the calculation will not require difficult integration as the constants which arise can be taken out of the integral. (5) may also be rewritten in terms of by the substituting the differential area element with the arc length times the unit depth (da= (D=2)bd ). Find the total surface area of a cone with base diameter of 15. The height and the maximum radius of the dome are related by the following expression with the radio of the sphere: The area and volume are calculated:. If the water surface is falling at the. Example 2 (Cone). The Surface Area Of An Ellipsoid A. The integral of f and one can use these facts to compute any surface integral over the cylinder. Definite Integral Definition. I Surface integrals of a scalar ﬁeld. A rectangular block of wood has dimensions of 40 mm by 12 mm by 8 mm. See Surface area of a cylinder. Archimedes proved this in approximately 260 B. Answer to: Use a line integral to prove that the lateral surface area of a cylinder of height h and radius r is 2 \pi r h.	2019-11-14 03:56:12	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.88576340675354, "perplexity": 308.41166479484474}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-47/segments/1573496667945.28/warc/CC-MAIN-20191114030315-20191114054315-00481.warc.gz"}
https://socratic.org/questions/how-do-you-factor-the-expression-3x-3-27x	How do you factor the expression 3x^3-27x? Mar 19, 2018 See a solution process below: Explanation: First, we can rewrite and factor as $x 3$ out of this expression as: $\left(3 x \cdot {x}^{2}\right) - \left(3 x \cdot 9\right) \implies$ $3 x \left({x}^{2} - 9\right)$ Now we can use this formula for the special case of this quadratic to complete the factoring: ${\textcolor{red}{x}}^{2} - {\textcolor{b l u e}{y}}^{2} = \left(\textcolor{red}{x} + \textcolor{b l u e}{y}\right) \left(\textcolor{red}{x} - \textcolor{b l u e}{y}\right)$ $3 x \left({x}^{2} - 9\right) \implies$ $3 x \left({\textcolor{red}{x}}^{2} - {\textcolor{b l u e}{3}}^{2}\right) \implies$ $3 x \left(\textcolor{red}{x} + \textcolor{b l u e}{3}\right) \left(\textcolor{red}{x} - \textcolor{b l u e}{3}\right)$	2019-09-16 10:21:10	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 7, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8382618427276611, "perplexity": 491.93902800832905}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514572517.50/warc/CC-MAIN-20190916100041-20190916122041-00188.warc.gz"}
https://math.stackexchange.com/questions/1798850/absolute-converge-of-real-and-complex-parts-of-a-series	# Absolute converge of real and complex parts of a series If the real and imaginary parts of a complex series converge absolutely, then the complex series converges absolutely. Is this true? If we write our complex series $\sum_{k=0}^{\infty} b_k = \left(\sum_{k=0}^{\infty} u_k \right)+ i \left( \sum_{k=0}^{\infty} v_k \right)$ into real and complex parts and the individual terms converge absolutely, then we can conclude that they converge, so $\sum_{k=0}^{\infty} b_k$ is convergent, not necessarily absolutely. What would be a good counterexample for this? • Observe that $\sum z\le \sum \|z\| \le \sum \|a\|+i\sum \|b\|$. If $a_n\le b_n,\forall n$ then $\lim a_n\le \lim b_n$. – Masacroso May 25 '16 at 1:33 Yes, that is true. $\sum_{n=x}^{y}(a+bi) = \sum_{n=x}^{y}a+\sum_{n=x}^{y}bi$, even if both a and b are functions of n and the series is infinite. If the sums of both parts converge, then because of that identity, the complex series must also converge. Hope that helped. EDIT: It not only converges, but converges absolutely. Proof (I'm kinda working backwards): $$\|a+bi\| = \sqrt{a^2+b^2}$$ For any real a and b, $$2\|a\|\|b\|\ge0$$ Duh. Now add $\|a\|^2$ and $\|b\|^2$ to both sides. $$2\|a\|\|b\|+\|a\|^2+\|b\|^2\ge\|a\|^2+\|b\|^2$$ Trust me. I'm getting somewhere with this. Now take the square root of both sides. $$\|a\|+\|b\|\ge\sqrt{a^2+b^2}$$ And therefore: $$\|a\|+\|b\|\ge\|a+bi\|$$ So, if the sums of $\|a\|$ and $\|b\|$ are not infinite, the sum of $\|a+bi\|$ cannot be infinite, and therefore must converge. It took me a long time to think of this, so you better make good use of it :P. • What can we say about absolute convergence though? – user167857 May 25 '16 at 1:12 • Ooh... I'm not sure. If I can think of a counterexample I'll add it to my answer. – Polygon May 25 '16 at 1:35 • @user167857 Yes, it does converge absolutely – Polygon May 25 '16 at 1:45	2019-04-20 17:01:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9525007605552673, "perplexity": 318.06908256435133}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578529898.48/warc/CC-MAIN-20190420160858-20190420182858-00174.warc.gz"}
http://en.wikipedia.org/wiki/Shriek_map	# Shriek map "Transfer map" redirects here. For the transfer homomorphism in group theory, see Transfer (group theory). In category theory, a branch of mathematics, certain unusual functors are denoted $f_!$ and $f^!,$ with the exclamation mark used to indicate that they are exceptional in some way. They are thus accordingly sometimes called shriek maps, with "shriek" being slang for an exclamation mark, though other terms are used, depending on context. ## Usage Shriek notation is used in two senses: • To distinguish a functor from a more usual functor $f_$ or $f^,$ accordingly as it is covariant or contravariant. • To indicate a map that goes "the wrong way" – a functor that has the same objects as a more familiar functor, but behaves differently on maps and has the opposite variance. For example, it has a pull-back where one expects a push-forward. ## Examples In algebraic geometry, these arise in image functors for sheaves, particularly Verdier duality, where $f_!$ is a "less usual" functor. In algebraic topology, these arise particularly in fiber bundles, where they yield maps that have the opposite of the usual variance. They are thus called wrong way maps, Gysin maps, as they originated in the Gysin sequence, or transfer maps. A fiber bundle $F \to E \to B,$ with base space B, fiber F, and total space E, has, like any other continuous map of topological spaces, a covariant map on homology $H_(E) \to H_(B)$ and a contravariant map on cohomology $H^(B) \to H^(E).$ However, it also has a covariant map on cohomology, corresponding in de Rham cohomology to "integration along the fiber", and a contravariant map on homology, corresponding in de Rham cohomology to "pointwise product with the fiber". The composition of the "wrong way" map with the usual map gives a map from the homology of the base to itself, analogous to a unit/counit of an adjunction; compare also Galois connection. These can be used in understanding and proving the product property for the Euler characteristic of a fiber bundle.[1] ## Notes 1. ^ Gottlieb, Daniel Henry (1975), "Fibre bundles and the Euler characteristic" (PDF), Journal of Differential Geometry 10 (1): 39–48	2015-05-04 17:09:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9716324210166931, "perplexity": 759.2090911867067}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-18/segments/1430454709311.96/warc/CC-MAIN-20150501043149-00028-ip-10-235-10-82.ec2.internal.warc.gz"}
https://questions.examside.com/past-years/jee/question/if-left-1-x-rightn-c0-c1x-c2x2-cnxn-then-show-that-the-sum-o-jee-advanced-1983-marks-3-b4hjsn9bqawziydh.htm	1 IIT-JEE 1983 Subjective If $${\left( {1 + x} \right)^n} = {C_0} + {C_1}x + {C_2}{x^2} + ..... + {C_n}{x^n}$$ then show that the sum of the products of the $${C_i}s$$ taken two at a time, represented $$\sum\limits_{0 \le i < j \le n} {\sum {{C_i}{C_j}} }$$ is equal to $${2^{2n - 1}} - {{\left( {2n} \right)!} \over {2{{\left( {n!} \right)}^2}}}$$ Solve it. 2 IIT-JEE 1982 Subjective Prove that $${7^{2n}} + \left( {{2^{3n - 3}}} \right)\left( {3n - 1} \right)$$ is divisible by 25 for any natural number $$n$$. Solve it. 3 IIT-JEE 1979 Subjective Given that $${C_1} + 2{C_2}x + 3{C_3}{x^2} + ......... + 2n{C_{2n}}{x^{2n - 1}} = 2n{\left( {1 + x} \right)^{2n - 1}}$$ where $${C_r} = {{\left( {2n} \right)\,!} \over {r!\left( {2n - r} \right)!}}\,\,\,\,\,r = 0,1,2,\,............,2n$$ Prove that $${C_1}^2 - 2{C_2}^2 + 3{C_3}^2 - ............ - 2n{C_{2n}}^2 = {\left( { - 1} \right)^n}n{C_n}.$$ 5 Joint Entrance Examination JEE Main JEE Advanced WB JEE GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN NEET Class 12	2022-05-24 11:11:26	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9836899638175964, "perplexity": 5052.743792148805}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662572800.59/warc/CC-MAIN-20220524110236-20220524140236-00475.warc.gz"}
https://math.stackexchange.com/questions/1255768/first-order-non-linear-ordinary-differential-equation	# First order non-linear ordinary differential equation The following ODE is given: $a\pm\sqrt{b+c(x(t)+d))}=ex'(t)+f*x(t)$ from Matlab I'm able to get a solution for the differential equation (actually two solutions, one for the + and one for the - sign) such as "solve(eqn)", but I'm not able to solve the eqn inside of the solve. Looking at the solve.m file I found out that Matlab is redirecting to MuPAD, but I don't know how to access this tool or if it is actually worth it. Can you solve the equation for x(t) or give me any clue for which methods/tools I could use for doing it? Let $$u = b + x(x+d)$$ then we find $$\frac{e}{c}u' + \frac{f}{c}u - \lambda = \pm\sqrt{u}$$ where $$\lambda = -\left[f\left(\frac{b+dc}{c}\right)+a\right]$$ then $u = v^2$ we find $$\frac{2e}{c}vv' + \frac{f}{c}v^2 - \lambda = \pm v$$ this is a separable equation namely $$\frac{2v}{v^2 \mp \frac{c}{f}v - \frac{c}{f}\lambda}dv = -\frac{f}{c}dt$$	2021-12-06 09:21:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8915917277336121, "perplexity": 349.21770299366176}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964363290.59/warc/CC-MAIN-20211206072825-20211206102825-00415.warc.gz"}
http://archive.numdam.org/articles/10.1016/j.anihpc.2014.09.003/	Asymptotic bifurcation and second order elliptic equations on ${ℝ}^{N}$ Annales de l'I.H.P. Analyse non linéaire, Volume 32 (2015) no. 6, pp. 1259-1281. This paper deals with asymptotic bifurcation, first in the abstract setting of an equation $G\left(u\right)=\lambda u$, where G acts between real Hilbert spaces and $\lambda \in ℝ$, and then for square-integrable solutions of a second order non-linear elliptic equation on ${ℝ}^{N}$. The novel feature of this work is that G is not required to be asymptotically linear in the usual sense since this condition is not appropriate for the application to the elliptic problem. Instead, G is only required to be Hadamard asymptotically linear and we give conditions ensuring that there is asymptotic bifurcation at eigenvalues of odd multiplicity of the H-asymptotic derivative which are sufficiently far from the essential spectrum. The latter restriction is justified since we also show that for some elliptic equations there is no asymptotic bifurcation at a simple eigenvalue of the H-asymptotic derivative if it is too close to the essential spectrum. DOI: 10.1016/j.anihpc.2014.09.003 Classification: 35J91, 47J15 Keywords: Asymptotic linearity, Asymptotic bifurcation, Nonlinear elliptic equation @article{AIHPC_2015__32_6_1259_0, author = {Stuart, C.A.}, title = {Asymptotic bifurcation and second order elliptic equations on ${\mathbb{R}}^{N}$ }, journal = {Annales de l'I.H.P. Analyse non lin\'eaire}, pages = {1259--1281}, publisher = {Elsevier}, volume = {32}, number = {6}, year = {2015}, doi = {10.1016/j.anihpc.2014.09.003}, zbl = {1330.35187}, mrnumber = {3425262}, language = {en}, url = {http://archive.numdam.org/articles/10.1016/j.anihpc.2014.09.003/} } TY - JOUR AU - Stuart, C.A. TI - Asymptotic bifurcation and second order elliptic equations on ${\mathbb{R}}^{N}$ JO - Annales de l'I.H.P. Analyse non linéaire PY - 2015 DA - 2015/// SP - 1259 EP - 1281 VL - 32 IS - 6 PB - Elsevier UR - http://archive.numdam.org/articles/10.1016/j.anihpc.2014.09.003/ UR - https://zbmath.org/?q=an%3A1330.35187 UR - https://www.ams.org/mathscinet-getitem?mr=3425262 UR - https://doi.org/10.1016/j.anihpc.2014.09.003 DO - 10.1016/j.anihpc.2014.09.003 LA - en ID - AIHPC_2015__32_6_1259_0 ER - %0 Journal Article %A Stuart, C.A. %T Asymptotic bifurcation and second order elliptic equations on ${\mathbb{R}}^{N}$ %J Annales de l'I.H.P. Analyse non linéaire %D 2015 %P 1259-1281 %V 32 %N 6 %I Elsevier %U https://doi.org/10.1016/j.anihpc.2014.09.003 %R 10.1016/j.anihpc.2014.09.003 %G en %F AIHPC_2015__32_6_1259_0 Stuart, C.A. Asymptotic bifurcation and second order elliptic equations on ${\mathbb{R}}^{N}$ . Annales de l'I.H.P. Analyse non linéaire, Volume 32 (2015) no. 6, pp. 1259-1281. doi : 10.1016/j.anihpc.2014.09.003. http://archive.numdam.org/articles/10.1016/j.anihpc.2014.09.003/ [1] A. Ambrosetti, D. Arcoya, An Introduction to Nonlinear Functional Analysis and Elliptic Problems, Birkhäuser, Basel (2011) \| MR \| Zbl [2] E.N. Dancer, On bifurcation from infinity, Q. J. Math. 25 no. 2 (1974), 81 -84 \| MR \| Zbl [3] K. Deimling, Nonlinear Functional Analysis, Springer-Verlag, Berlin (1985) \| MR \| Zbl [4] J.-P. Dias, J. Hernandez, A remark on a paper by J.F. Toland and some applications to unilateral problems, Proc. R. Soc. Edinb. 75 (1976), 179 -182 \| MR \| Zbl [5] A.L. Dontchev, R.T. Rockafellar, Implicit Functions and Solution Mappings, Springer, Heidelberg (2009) \| MR \| Zbl [6] A. Edelson, C.A. Stuart, The principal branch of solutions of a nonlinear elliptic eigenvalue problem on ${ℝ}^{N}$ , J. Differ. Equ. 124 (1996), 279 -301 \| MR \| Zbl [7] G. Evéquoz, C.A. Stuart, Hadamard differentiability and bifurcation, Proc. R. Soc. Edinb. A 137 (2007), 1249 -1285 \| MR \| Zbl [8] G. Evéquoz, C.A. Stuart, On differentiability and bifurcation, Adv. Math. Econ. 8 (2006), 155 -184 \| MR \| Zbl [9] T.M. Flett, Differential Analysis, Cambridge University Press, Cambridge (1980) \| MR \| Zbl [10] F. Genoud, Bifurcation from infinity for an asymptotically linear problem on the half-line, Nonlinear Anal. 74 (2011), 4533 -4543 \| MR \| Zbl [11] F. Genoud, Global bifurcation for asymptotically linear Schrödinger equations, Nonlinear Differ. Equ. Appl. 20 (2013), 23 -35 \| MR \| Zbl [12] H. Koch, D. Tataru, Carleman estimates and absence of embedded eigenvalues, Commun. Math. Phys. 267 (2006), 419 -449 \| MR \| Zbl [13] M.A. Krasnoselskii, Topological Methods in the Theory of Nonlinear Integral Equations, Pergamon Press, Oxford (1964) [14] M.A. Krasnoselskii, Positive Solutions of Operator Equations, Noordhoff, Groningen (1964) [15] P.H. Rabinowitz, On bifurcation from infinity, J. Differ. Equ. 14 (1973), 462 -475 \| MR \| Zbl [16] C.A. Stuart, An introduction to elliptic equation on ${ℝ}^{N}$ , A. Ambrosetti, K.-C. Chang, I. Ekeland (ed.), Nonlinear Functional Analysis and Applications to Differential Equations, World Scientific, Singapore (1998) [17] C.A. Stuart, Bifurcation for some non-Fréchet differentiable problems, Nonlinear Anal. 69 (2008), 1011 -1024 \| MR \| Zbl [18] C.A. Stuart, Bifurcation and decay of solutions for a class of elliptic equations on ${ℝ}^{N}$ , Contemp. Math. vol. 540 (2011), 203 -230 \| MR \| Zbl [19] C.A. Stuart, Asymptotic linearity and Hadamard differentiability, Nonlinear Anal. 75 (2012), 4699 -4710 \| MR \| Zbl [20] C.A. Stuart, Bifurcation at isolated singular points of the Hadamard derivative, Proc. R. Soc. Edinb. (2014) \| MR \| Zbl [21] C.A. Stuart, Bifurcation at isolated eigenvalues for some elliptic equations on ${ℝ}^{N}$, preprint, 2012. \| MR [22] C.A. Stuart, H.-S. Zhou, Global branch of solutions for non-linear Schrödinger equations with deepening potential well, Proc. Lond. Math. Soc. (3) 92 (2006), 655 -681 \| MR \| Zbl [23] J.F. Toland, Asymptotic nonlinearity and nonlinear eigenvalue problems, Quart. J. Math. Oxford 24 (1973), 241 -250 \| MR \| Zbl [24] J.F. Toland, Asymptotic linearity and a class of nonlinear Strum–Liouville problems on the half-line, Lect. Notes Math. vol. 415 , Springer (1974), 429 -434 \| MR \| Zbl [25] J.F. Toland, Asymptotic linearity and nonlinear eigenvalue problems, Proc. R. Ir. Acad. 77 (1977), 1 -12 \| MR \| Zbl [26] J.F. Toland, Bifurcation and asymptotic bifurcation for non-compact non-symmetric gradient operators, Proc. R. Soc. Edinb. 73 (1975), 137 -147 \| MR \| Zbl [27] E. Zeidler, Nonlinear Functional Analysis, Springer-Verlag, Berlin (1985) \| Zbl Cited by Sources:	2022-12-03 06:42:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4305524230003357, "perplexity": 4525.76086019021}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710924.83/warc/CC-MAIN-20221203043643-20221203073643-00835.warc.gz"}
https://www.mathworks.com/help/finance/cfamounts.html	cfamounts Cash flow and time mapping for bond portfolio In R2017b, the specification of optional input arguments has changed. While the previous ordered inputs syntax is still supported, it may no longer be supported in a future release. Use the optional name-value pair inputs: `Period`, `Basis`, `EndMonthRule`, `IssueDate`,`FirstCouponDate`, `LastCouponDate`, `StartDate`,`Face`, `AdjustCashFlowsBasis`, `BusinessDayConvention`, `CompoundingFrequency`, `DiscountBasis`, `Holidays`, and `PrincipalType`. Syntax ``[CFlowAmounts,CFlowDates,TFactors,CFlowFlags,CFPrincipal] = cfamounts(CouponRate,Settle,Maturity)`` ``[CFlowAmounts,CFlowDates,TFactors,CFlowFlags,CFPrincipal] = cfamounts(___,Name,Value)`` Description example ````[CFlowAmounts,CFlowDates,TFactors,CFlowFlags,CFPrincipal] = cfamounts(CouponRate,Settle,Maturity)` returns matrices of cash flow amounts, cash flow dates, time factors, and cash flow flags for a portfolio of `NUMBONDS` fixed-income securities.The elements contained in the `cfamounts` outputs for the cash flow matrix, time factor matrix, and cash flow flag matrix correspond to the cash flow dates for each security. The first element of each row in the cash flow matrix is the accrued interest payable on each bond. This accrued interest is zero in the case of all zero coupon bonds. `cfamounts` determines all cash flows and time mappings for a bond whether or not the coupon structure contains odd first or last periods. All output matrices are padded with `NaN`s as necessary to ensure that all rows have the same number of elements.``` example ````[CFlowAmounts,CFlowDates,TFactors,CFlowFlags,CFPrincipal] = cfamounts(___,Name,Value)` adds optional name-value arguments. ``` Examples collapse all This example shows how to compute the cash flow structure and time factors for a bond portfolio that contains a corporate bond paying interest quarterly and a Treasury bond paying interest semiannually. ```Settle = '01-Nov-1993'; Maturity = ['15-Dec-1994';'15-Jun-1995']; CouponRate= [0.06; 0.05]; Period = [4; 2]; Basis = [1; 0]; [CFlowAmounts, CFlowDates, TFactors, CFlowFlags] = ... cfamounts(CouponRate,Settle, Maturity, Period, Basis)``` ```CFlowAmounts = 2×6 -0.7667 1.5000 1.5000 1.5000 1.5000 101.5000 -1.8989 2.5000 2.5000 2.5000 102.5000 NaN ``` ```CFlowDates = 2×6 728234 728278 728368 728460 728552 728643 728234 728278 728460 728643 728825 NaN ``` ```TFactors = 2×6 0 0.2404 0.7403 1.2404 1.7403 2.2404 0 0.2404 1.2404 2.2404 3.2404 NaN ``` ```CFlowFlags = 2×6 0 3 3 3 3 4 0 3 3 3 4 NaN ``` This example shows how to compute the cash flow structure and time factors for a bond portfolio that contains a corporate bond paying interest quarterly and a Treasury bond paying interest semiannually and `CFlowDates` is returned as a datetime array. ```Settle = datetime('01-Nov-1993','Locale','en_US'); Maturity = ['15-Dec-1994';'15-Jun-1995']; CouponRate= [0.06; 0.05]; Period = [4; 2]; Basis = [1; 0]; [CFlowAmounts, CFlowDates, TFactors, CFlowFlags] = cfamounts(CouponRate,... Settle, Maturity, Period, Basis)``` ```CFlowAmounts = 2×6 -0.7667 1.5000 1.5000 1.5000 1.5000 101.5000 -1.8989 2.5000 2.5000 2.5000 102.5000 NaN ``` ```CFlowDates = 2x6 datetime Columns 1 through 5 01-Nov-1993 15-Dec-1993 15-Mar-1994 15-Jun-1994 15-Sep-1994 01-Nov-1993 15-Dec-1993 15-Jun-1994 15-Dec-1994 15-Jun-1995 Column 6 15-Dec-1994 NaT ``` ```TFactors = 2×6 0 0.2404 0.7403 1.2404 1.7403 2.2404 0 0.2404 1.2404 2.2404 3.2404 NaN ``` ```CFlowFlags = 2×6 0 3 3 3 3 4 0 3 3 3 4 NaN ``` This example shows how to compute the cash flow structure and time factors for a bond portfolio that contains a corporate bond paying interest quarterly and a Treasury bond paying interest semiannually. This example uses the following Name-Value pairs for `Period`, `Basis`, `BusinessDayConvention`, and `AdjustCashFlowsBasis`. ```Settle = '01-Jun-2010'; Maturity = ['15-Dec-2011';'15-Jun-2012']; CouponRate= [0.06; 0.05]; Period = [4; 2]; Basis = [1; 0]; [CFlowAmounts, CFlowDates, TFactors, CFlowFlags] = ... cfamounts(CouponRate,Settle, Maturity, 'Period',Period, ... 'Basis', Basis, 'AdjustCashFlowsBasis', true,... 'BusinessDayConvention','modifiedfollow')``` ```CFlowAmounts = 2×8 -1.2667 1.5000 1.5000 1.5000 1.5000 1.5000 1.5000 101.5000 -2.3077 2.4932 2.5068 2.4932 2.5000 102.5000 NaN NaN ``` ```CFlowDates = 2×8 734290 734304 734396 734487 734577 734669 734761 734852 734290 734304 734487 734669 734852 735035 NaN NaN ``` ```TFactors = 2×8 0 0.0778 0.5778 1.0778 1.5778 2.0778 2.5778 3.0778 0 0.0769 1.0769 2.0769 3.0769 4.0769 NaN NaN ``` ```CFlowFlags = 2×8 0 3 3 3 3 3 3 4 0 3 3 3 3 4 NaN NaN ``` This example shows how to use `cfamounts` with a `CouponRate` schedule. For `CouponRate` and `Face` that change over the life of the bond, schedules for `CouponRate` and `Face` can be specified with an `NINST`-by-1 cell array, where each element is a `NumDates`-by-2 matrix where the first column is dates and the second column is associated rates. ```CouponSchedule = {[datenum('15-Mar-2012') .04;datenum('15- Mar -2013') .05;... datenum('15- Mar -2015') .06]}``` ```CouponSchedule = 1x1 cell array {3x2 double} ``` `cfamounts(CouponSchedule,'01-Mar-2011','15-Mar-2015' )` ```ans = 1×10 -1.8453 2.0000 2.0000 2.0000 2.5000 2.5000 3.0000 3.0000 3.0000 103.0000 ``` This example shows how to use `cfamounts` with a `Face` schedule. For `CouponRate` and `Face` that change over the life of the bond, schedules for `CouponRate` and `Face` can be specified with an `NINST`-by-1 cell array, where each element is a `NumDates`-by-2 matrix where the first column is dates and the second column is associated rates. ```FaceSchedule = {[datenum('15-Mar-2012') 100;datenum('15- Mar -2013') 90;... datenum('15- Mar -2015') 80]}``` ```FaceSchedule = 1x1 cell array {3x2 double} ``` `cfamounts(.05,'01-Mar-2011','15-Mar-2015', 'Face', FaceSchedule)` ```ans = 1×10 -2.3066 2.5000 2.5000 12.5000 2.2500 12.2500 2.0000 2.0000 2.0000 82.0000 ``` This example shows how to use `cfamounts` to generate the cash flows for a sinking bond. ```[CFlowAmounts,CFDates,TFactors,CFFlags,CFPrincipal] = cfamounts(.05,'04-Nov-2010',... {'15-Jul-2014';'15-Jul-2015'},'Face',{[datenum('15-Jul-2013') 100;datenum('15-Jul-2014')... 90;datenum('15-Jul-2015') 80]})``` ```CFlowAmounts = 2×11 -1.5217 2.5000 2.5000 2.5000 2.5000 2.5000 12.5000 2.2500 92.2500 NaN NaN -1.5217 2.5000 2.5000 2.5000 2.5000 2.5000 12.5000 2.2500 12.2500 2.0000 82.0000 ``` ```CFDates = 2×11 734446 734518 734699 734883 735065 735249 735430 735614 735795 NaN NaN 734446 734518 734699 734883 735065 735249 735430 735614 735795 735979 736160 ``` ```TFactors = 2×11 0 0.3913 1.3913 2.3913 3.3913 4.3913 5.3913 6.3913 7.3913 NaN NaN 0 0.3913 1.3913 2.3913 3.3913 4.3913 5.3913 6.3913 7.3913 8.3913 9.3913 ``` ```CFFlags = 2×11 0 3 3 3 3 3 13 3 4 NaN NaN 0 3 3 3 3 3 13 3 13 3 4 ``` ```CFPrincipal = 2×11 0 0 0 0 0 0 10 0 90 NaN NaN 0 0 0 0 0 0 10 0 10 0 80 ``` Input Arguments collapse all Annual percentage rate used to determine the coupons payable on a bond, specified as decimal using a scalar or a `NBONDS`-by-`1` vector. `CouponRate` is `0` for zero coupon bonds. Note `CouponRate` and `Face` can change over the life of the bond. Schedules for `CouponRate` and `Face` can be specified with an `NBONDS`-by-`1` cell array, where each element is a `NumDates`-by-`2` matrix or cell array, where the first column is dates (serial date numbers or character vectors) and the second column is associated rates. The date indicates the last day that the coupon rate or face value is valid. This means that the corresponding `CouponRate` and `Face` value applies "on or before" the specified ending date. Data Types: `double` \| `cell` \| `char` Settlement date of the bond, specified as a scalar or a `NBONDS`-by-`1` vector using serial date numbers, date character vectors, or datetime arrays. The `Settle` date must be before the `Maturity` date. Data Types: `double` \| `char` \| `datetime` Maturity date of the bond, specified as a scalar or a `NBONDS`-by-`1` vector using serial date numbers, date character vectors, or datetime arrays. Data Types: `double` \| `char` \| `datetime` Name-Value Pair Arguments Specify optional comma-separated pairs of `Name,Value` arguments. `Name` is the argument name and `Value` is the corresponding value. `Name` must appear inside quotes. You can specify several name and value pair arguments in any order as `Name1,Value1,...,NameN,ValueN`. Example: ```[CFlowAmounts, CFlowDates, TFactors, CFlowFlags] = ... cfamounts(CouponRate,Settle, Maturity,'Period',4,'Basis',3,'AdjustCashFlowsBasis',true,'BusinessDayConvention','modifiedfollow')``` Number of coupon payments per year for the bond, specified as the comma-separated pair consisting of `'Period'` and a scalar or a `NBONDS`-by-`1` vector using the values: `0`, `1`, `2`, `3`, `4`, `6`, or `12`. Data Types: `double` Day-count basis of the bond, specified as the comma-separated pair consisting of `'Basis'` and a scalar or a `NBONDS`-by-`1` vector using a supported value: • 0 = actual/actual • 1 = 30/360 (SIA) • 2 = actual/360 • 3 = actual/365 • 4 = 30/360 (PSA) • 5 = 30/360 (ISDA) • 6 = 30/360 (European) • 7 = actual/365 (Japanese) • 8 = actual/actual (ICMA) • 9 = actual/360 (ICMA) • 10 = actual/365 (ICMA) • 11 = 30/360E (ICMA) • 12 = actual/365 (ISDA) • 13 = BUS/252 Data Types: `double` End-of-month rule flag, specified as the comma-separated pair consisting of `'EndMonthRule'` and a scalar or a `NBONDS`-by-`1` vector. This rule applies only when `Maturity` is an end-of-month date for a month having 30 or fewer days. • `0` = Ignore rule, meaning that a bond coupon payment date is always the same numerical day of the month. • `1` = Set rule on, meaning that a bond coupon payment date is always the last actual day of the month. Data Types: `logical` Bond issue date (the date the bond begins to accrue interest), specified as the comma-separated pair consisting of `'IssueDate'` and a scalar or a `NBONDS`-by-`1` vector using serial date numbers, date character vectors, or datetime arrays. The `IssueDate` cannot be after the `Settle` date. If you do not specify an `IssueDate`, the cash flow payment dates are determined from other inputs. Data Types: `double` \| `char` \| `datetime` Irregular or normal first coupon date, specified as the comma-separated pair consisting of `'FirstCouponDate'` and a scalar or a `NBONDS`-by-`1` vector using serial date numbers, date character vectors, or datetime arrays. If you do not specify a `FirstCouponDate`, the cash flow payment dates are determined from other inputs. Note When `FirstCouponDate` and `LastCouponDate` are both specified, the `FirstCouponDate` takes precedence in determining the coupon payment structure. If `FirstCouponDate` is not specified, then `LastCouponDate` determines the coupon structure of the bond. Data Types: `double` \| `char` \| `datetime` Irregular or normal last coupon date, specified as the comma-separated pair consisting of `'LastCouponDate'` and a scalar or a `NBONDS`-by-`1` vector using serial date numbers, date character vectors, or datetime arrays. Note When `FirstCouponDate` and `LastCouponDate` are both specified, the `FirstCouponDate` takes precedence in determining the coupon payment structure. If `FirstCouponDate` is not specified, then `LastCouponDate` determines the coupon structure of the bond. Data Types: `double` \| `char` \| `datetime` Forward starting date of coupon payments after the `Settle` date, specified as the comma-separated pair consisting of `'StartDate'` and a scalar or a `NBONDS`-by-`1` vector using serial date numbers, date character vectors, or datetime arrays. Note To make an instrument forward starting, specify `StartDate` as a future date. If you do not specify a `StartDate`, the effective start date is the `Settle` date. Data Types: `double` \| `char` \| `datetime` Face value of the bond, specified as the comma-separated pair consisting of `'Face'` and a scalar or a `NBONDS`-by-`1` vector. Note `CouponRate` and `Face` can change over the life of the bond. Schedules for `CouponRate` and `Face` can be specified with an `NBONDS`-by-`1` cell array where each element is a `NumDates`-by-`2` matrix or cell array, where the first column is dates (serial date numbers or character vectors) and the second column is associated rates. The date indicates the last day that the coupon rate or face value is valid. This means that the corresponding `CouponRate` and `Face` value applies "on or before" the specified ending date. When the corresponding `Face` value is used to compute the coupon cashflow on the specified ending date. Three things happen on the specified ending date: 1. The last coupon corresponding to the current `Face` value is paid. 2. The principal differential (between the current and the next `Face` value) is paid. 3. The date marks the beginning of the period with the next `Face` value, for which the cashflow does not occur until later. Data Types: `double` \| `cell` \| `char` Adjusts cash flows according to the accrual amount based on the actual period day count, specified as the comma-separated pair consisting of `'AdjustCashFlowsBasis'` and a scalar or a `NBONDS`-by-`1` vector. Data Types: `logical` Business day conventions, specified as the comma-separated pair consisting of `'BusinessDayConvention'` and a scalar or `NBONDS`-by-`1` cell array of character vectors of business day conventions to be used in computing payment dates. The selection for business day convention determines how nonbusiness days are treated. Nonbusiness days are defined as weekends plus any other date that businesses are not open (for example, statutory holidays). Values are: • `'actual'` — Nonbusiness days are effectively ignored. Cash flows that fall on non-business days are assumed to be distributed on the actual date. • `'follow'` — Cash flows that fall on a nonbusiness day are assumed to be distributed on the following business day. • `'modifiedfollow'` — Cash flows that fall on a non-business day are assumed to be distributed on the following business day. However if the following business day is in a different month, the previous business day is adopted instead. • `'previous'` — Cash flows that fall on a nonbusiness day are assumed to be distributed on the previous business day. • `'modifiedprevious'` — Cash flows that fall on a nonbusiness day are assumed to be distributed on the previous business day. However if the previous business day is in a different month, the following business day is adopted instead. Data Types: `char` \| `cell` Compounding frequency for yield calculation, specified as the comma-separated pair consisting of `'CompoundingFrequency'` and a scalar or a `NBONDS`-by-`1` vector. Values are: • `1` — Annual compounding • `2` — Semiannual compounding • `3` — Compounding three times per year • `4` — Quarterly compounding • `6` — Bimonthly compounding • `12` — Monthly compounding Note By default, SIA bases (`0`-`7`) and `BUS/252` use a semiannual compounding convention and ICMA bases (`8`-`12`) use an annual compounding convention. Data Types: `double` Basis used to compute the discount factors for computing the yield, specified as the comma-separated pair consisting of `'DiscountBasis'` and a scalar or a `NBONDS`-by-`1` vector. Values are: • 0 = actual/actual • 1 = 30/360 (SIA) • 2 = actual/360 • 3 = actual/365 • 4 = 30/360 (PSA) • 5 = 30/360 (ISDA) • 6 = 30/360 (European) • 7 = actual/365 (Japanese) • 8 = actual/actual (ICMA) • 9 = actual/360 (ICMA) • 10 = actual/365 (ICMA) • 11 = 30/360E (ICMA) • 12 = actual/365 (ISDA) • 13 = BUS/252 Note If a SIA day-count basis is defined in the `Basis` input argument and there is no value assigned for `DiscountBasis`, the default behavior is for SIA bases to use the `actual/actual` day count to compute discount factors. If an ICMA day-count basis or `BUS/252` is defined in the `Basis` input argument and there is no value assigned for `DiscountBasis`, the specified bases from the `Basis` input argument are used. Data Types: `double` Dates for holidays, specified as the comma-separated pair consisting of `'Holidays'` and a `NHOLIDAYS`-by-`1` vector of MATLAB® dates using serial date numbers, date character vectors, or datetime arrays. Holidays are used in computing business days. Data Types: `double` \| `char` \| `datetime` Type of principal when a `Face` schedule, specified as the comma-separated pair consisting of `'PrincipalType'` and a value of `'sinking'` or `'bullet'` using a scalar or a `NBONDS`-by-`1` vector. If `'sinking'`, principal cash flows are returned throughout the life of the bond. If `'bullet'`, principal cash flow is only returned at maturity. Data Types: `char` \| `cell` Output Arguments collapse all Cash flow amounts, returned as a `NBONDS`-by-`NCFS` (number of cash flows) matrix. The first entry in each row vector is the accrued interest due at settlement. This amount could be zero, positive or negative. If no accrued interest is due, the first column is zero. If the bond is trading ex-coupon then the accrued interest is negative. Cash flow dates for a portfolio of bonds, returned as a `NBONDS`-by-`NCFS` matrix. Each row represents a single bond in the portfolio. Each element in a column represents a cash flow date of that bond. If all the above inputs (`Settle`, `Maturity`, `IssueDate`, `FirstCouponDate`, `LastCouponDate`, and `StartDate`) are either serial date numbers or date character vectors, then `CFlowDates` is returned as a serial date number. If any of these inputs are datetime arrays, then `CFlowDates` is returned as a datetime array. Matrix of time factors for a portfolio of bonds, returned as a `NBONDS`-by-`NCFS` matrix. Each row corresponds to the vector of time factors for each bond. Each element in a column corresponds to the specific time factor associated with each cash flow of a bond. Time factors are for price/yield conversion and time factors are in units of whole semiannual coupon periods plus any fractional period using an actual day count. For more information on time factors, see Time Factors. Cash flow flags for a portfolio of bonds, returned as a `NBONDS`-by-`NCFS` matrix. Each row corresponds to the vector of cash flow flags for each bond. Each element in a column corresponds to the specific flag associated with each cash flow of a bond. Flags identify the type of each cash flow (for example, nominal coupon cash flow, front, or end partial, or "stub" coupon, maturity cash flow). Flag Cash Flow Type `0` Accrued interest due on a bond at settlement. `1` Initial cash flow amount smaller than normal due to a “stub” coupon period. A stub period is created when the time from issue date to first coupon date is shorter than normal. `2` Larger than normal initial cash flow amount because the first coupon period is longer than normal. `3` Nominal coupon cash flow amount. `4` Normal maturity cash flow amount (face value plus the nominal coupon amount). `5` End “stub” coupon amount (last coupon period is abnormally short and actual maturity cash flow is smaller than normal). `6` Larger than normal maturity cash flow because the last coupon period longer than normal. `7` Maturity cash flow on a coupon bond when the bond has less than one coupon period to maturity. `8` Smaller than normal maturity cash flow when the bond has less than one coupon period to maturity. `9` Larger than normal maturity cash flow when the bond has less than one coupon period to maturity. `10` Maturity cash flow on a zero coupon bond. `11` Sinking principal and initial cash flow amount smaller than normal due to a "stub" coupon period. A stub period is created when the time from issue date to first coupon date is shorter than normal. `12` Sinking principal and larger than normal initial cash flow amount because the first coupon period is longer than normal. `13` Sinking principal and nominal coupon cash flow amount. Principal cash flows, returned as a `NBONDS`-by-`NCFS` matrix. If `PrincipalType` is `'sinking'`, `CFPrincipal` output indicates when the principal is returned. If `PrincipalType` is `'bullet'`, `CFPrincipal` is all zeros and, at `Maturity`, the appropriate `Face` value. collapse all Time Factors Time factors help determine the present value of a stream of cash flows. The term time factors refer to the exponent TF in the discounting equation `$PV=\sum _{i=1}^{n}\left(\frac{CF}{{\left(1+\frac{z}{f}\right)}^{TF}}\right),$` where: PV = Present value of a cash flow. CF = Cash flow amount. z = Risk-adjusted annualized rate or yield corresponding to a given cash flow. The yield is quoted on a semiannual basis. f = Frequency of quotes for the yield. Default is `2` for `Basis` values `0` to `7` and `13` and `1` for `Basis` values `8` to `12`. The default can be overridden by specifying the `CompoundingFrequency` name-value pair. TF = Time factor for a given cash flow. The time factor is computed using the compounding frequency and the discount basis. If these values are not specified, then the defaults are as follows: `CompoundingFrequency` default is `2` for `Basis` values `0` to `7` and `13` and `1` for `Basis` values `8` to `12`. `DiscountBasis` is `0` for `Basis` values `0` to `7` and `13` and the value of the input `Basis` for `Basis` values `8` to `12`. Note The `Basis` is always used to compute accrued interest. References [1] Krgin, D. Handbook of Global Fixed Income Calculations. Wiley, 2002. [2] Mayle, J. "Standard Securities Calculations Methods: Fixed Income Securities Formulas for Analytic Measures." SIA, Vol 2, Jan 1994. [3] Stigum, M., Robinson, F. Money Market and Bond Calculation. McGraw-Hill, 1996.	2021-01-19 00:03:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 1, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6111969351768494, "perplexity": 2348.0857544635196}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-04/segments/1610703517159.7/warc/CC-MAIN-20210118220236-20210119010236-00603.warc.gz"}
https://www.physicsforums.com/threads/allowed-and-forbidden-electron-photon-reactions.922379/	# Allowed and Forbidden Electron-Photon Reactions 1. Aug 10, 2017 ### Vrbic 1. The problem statement, all variables and given/known data Show, using spacetime diagrams and also using frame-independent calculations, that the law of conservation of 4-momentum forbids a photon to be absorbed by an electron, e + γ → e. 2. Relevant equations $\textbf{p}_{e1}+\textbf{p}_{\gamma}=\textbf{p}_{e2}$ : $(E_1,\vec{p_1})+(E_{\gamma},E_{\gamma}\vec{n})=(E,\vec{p_2})$, where $\textbf{p}$ is 4-vector, $\vec{p}$ is ordinary 3-vector and $\vec{n}$ is unit vector in photon's direction. $E=\sqrt{p^2+m^2}$ 3. The attempt at a solution I take only 3-vecotrs and I (I hope I may) suppose $\vec{p_1}=0$ i.e. is in rest. $\vec{0}+E_{\gamma}\vec{n}=p_2\vec{n}=>\sqrt{E^2-m^2}=E_{\gamma}$ First time I put there $E=\frac{1}{2}mv^2$ but I feel it is not good probably. Is my begining right? What should I use or think about? 2. Aug 10, 2017 ### Staff: Mentor You used conservation of momentum so far. The problem is easier in the reference frame of the electron in the final state. 3. Aug 11, 2017 ### Vrbic Ok, $(\gamma m, \vec{p}) + (E_{\gamma},E_{\gamma}\vec{n})=(m,\vec{0})$. Right? From energy part than we have $E_{\gamma}=m(1-\frac{1}{\sqrt{1-v^2}})<0$. And it is this problem, negative photon's energy? 4. Aug 11, 2017 ### Staff: Mentor Right. Time-reversed, the process would be an electron at rest emitting a photon and gaining kinetic energy at the same time. Written like that, it should be clear that it cannot happen. 5. Aug 14, 2017 Thank you.	2017-10-21 16:08:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6136425137519836, "perplexity": 1204.6631758145234}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-43/segments/1508187824820.28/warc/CC-MAIN-20171021152723-20171021172723-00390.warc.gz"}
http://ufdc.ufl.edu/AA00009199/00001	UFDC Home \| Search all Groups \| World Studies \| Federal Depository Libraries of Florida & the Caribbean \| Help # On the contribution of turbulent boundary layers to the noise inside a fuselage ## Material Information Title: On the contribution of turbulent boundary layers to the noise inside a fuselage Series Title: NACA TM Physical Description: 43 p. : ill ; 27 cm. Language: English Creator: Corcos, G. M Liepmann, H. W ( Hans Wolfgang ), 1914- United States -- National Advisory Committee for Aeronautics Publisher: NACA Place of Publication: Washington, D.C Publication Date: ## Subjects Subjects / Keywords: Aerodynamics ( lcsh ) Turbulent boundary layer -- Research ( lcsh ) Airplanes -- Fuselage ( lcsh ) Genre: federal government publication ( marcgt ) bibliography ( marcgt ) technical report ( marcgt ) non-fiction ( marcgt ) ## Notes Abstract: The concept of attenuation is abandoned and instead the problem is formulated as a sequence of two linear couplings: the turbulent boundary-layer fluctuations excite the fuselage skin in lateral vibrations and the skin vibrations induce sound inside the fuselage. The techniques used are those required to determine the response of linear systems to random forcing functions of several variables. A certain degree of idealization has been resorted to. Thus the boundary layer is assumed locally homogeneous; the fuselage skin is assumed flat, unlined, and free from axial loads; and the "cabin" air is bounded only by the vibrating plate so that only outgoing waves are considered. The results, strictly applicable only to the limiting cases of thin boundary layers, show that the sound pressure intensity is proportional to the square of the free-stream density, the square of cabin air density, and inversely proportional to the first power of the damping constant and to the second power of the plate density. The dependence on free-stream velocity and boundary-layer thickness cannot be given in general without a detailed knowledge of the characteristics of the pressure fluctuations in the boundary layer. For a flat spectrum the noise intensity depends on the fifth power of the velocity and the first power of the boundary-layer thickness. Thus it appears that boundary-layer removal is not an economical means of decreasing cabin noise. Bibliography: Includes bibliographic references (p. 27). Funding: Statement of Responsibility: by G.M. Corcos and H.W. Liepmann. General Note: "Report date September 1956." ## Record Information Source Institution: University of Florida Rights Management: All applicable rights reserved by the source institution and holding location. Resource Identifier: aleph - 003874415 oclc - 156938071 System ID: AA00009199:00001 Full Text kck74-~m -I/ ac TECfJICAL MEMORANDUM 1620 ON THE CONTRIBUTION OF TURBULENT BO~NDAR(Y LAYERS TO THIE NOISE INSIDE A FUSELACE~ by G. M. Corcos"H and HI. W. Liepmann* September, 1956 "Un~edited by the NACA (the Committee takes no responsibility for the correctness of the author 's statements. ) Aerodynamics Research Group Douglas Aircraft Company, Inc . Santa Monica, California California Institute of Technology (Consultant, Aerodynamics Research Group, Douglas Aircraft Company, Inc . Santa Maonica, California) NACA TM 1420 TABLE OF COjNTENTS ABSTRACT.............................. i INTRODUCTIONI .. .. .... .. .. .. .. .. .. .. .. 1 I. THIE ACOUSTIC COUPLING OF A RANDOMLY VIBRATING PLATE WITH AIR~ AT REST .. ... ... .. .. 4 II. THE DYNAMIC BEHAVIOR OF THIE SKIN ... .. .. .. .. .. 8 a) The Mean Acceleration .. .. .. .. ... 9) b) The Length Scale A .. .. .. .. .. 12 III. THE FORCING FUNCTION .. .. ... .. ... .. .. 14 IV. SPECIAL CASES .. .... .. .. .. ... 2C 1. Convectred Turbulence ... .. ... .. .. .. 2C a) The coupling of the plate with air at rest in the case of convected turbulence .. 20 b) The response of the plate .. .. .. 21 2. The case of zero scale .. ... .. .. .. .. 23 V. SUMMARY OF RESULTS AND DISCUSSION .. .. .. .. .. .. 24 A Note on Testing ... .. .. .. .. .. 25 VI. REFERENICES .. .. .. .. ... .. ... ... .. 27 APPENDIX I: THE RANDOM RADIATION OF A PLANE SURFACE: A. Four Limiting Cases .. .. .. .. .. ... 28 B. The Noise Generated by Skin Ripples of Fixed Velocity .. .. ... .. .. 35 C. The Generation of Noise by Plate Deflection~ of Zero Scale ... .. .. .. 38 APPENDIX II: THE SIMPLIFICATION OF THE PLATE RESPONSE INTEGRAL 40 APPENDIX III: THE EVALUATION OF THIE INTEGRAL SCALE .. 42 Digilized by ille Internel Archive I i~n201wll MII1undIn~ from UniverslIv of Florida. George A. Smather Llbraries wyith support from L'rRASIS and the Sloan Foundalion hilp: wwwv\~.archlve.or g delailS oncontrlb ullonol00ulnlt NACA TM 11+20 ABSTRACT The following report deals in preliminary fashion with the transmission through a fuselage of random noise generated on the fuselage skin by a tur- bulent boundary layer. The concept of attenuation is abandoned an instead the problem is formulated as a sequence of two linear couplings: theturbu- lent boundary layer fluctuations excite the fuselage skin in lateral vira- tions and the skin vibrations induce sound inside the fuselage. The techniques used are those required to determine the response of linear systems to random forcing functions of several variables. A certain degree of idealization has been resorted to. Thus the boundary layer is assumed locally homogeneous, the fuselage skin is assumed flat, unlined and free from axial loads and the? "cabin" air is bounded only by the vibrating plate so that only outgoing waves are considered. Some of the details of the statistical description have been simplified in order to reveal the basic features of the problem. The results, strictly applicable only to the limiting case of thin boundary layers, show that the sound pressure intensity is proportional to the square of the free stream density, the square of cabin air density and inversely proportional to the first power of the damping constant and to the second power of the plate density. The dependence on free streams velocity and boundary layer thickness cannot be given in general without a detailed knowledge of the characteristics of the pressure fluctuations in the boundary layer (in particular the frequency spectrum). For a flat spectrum the noise intensity depends on the fifth power of the velocity and the first power of the boundary layer thickness. This suggests that boundary layer removal is probably not an economical means of decreasing cabin noise. In general, the analysis presented here only reduces the determionat of cabin noise intensity to the measurement of the effect of any one of four variables (free stream velocity, boundary layer thickness, plate thickness or the characteristic velocity of propagation in the plate). The plate generates noise by vibrating in resonance over a wide! range of frequencies and increasing the damping constant is consequently an effective method of decreasing noise generation. One of the main features of the results is that the relevent quatitis upon which noise intensity depends are non-dimensional numbers in which boundary layer and plate proper-ties enter as ratios. This is taken as an indication that in testing models of structures for boundary layer noise it is not sufficient to duplicate in the model the structural characteristics of the fuselage. One must match properly the characteristics of the exciting pressure fluctuations to that of the structure. Ii TECHNICAL MFJEORAN~DUM 1420 INTRODUCT~IIO In his efforts to minimizer the noise levels for which, he is responsible, the airplan~e designer has had to pay increasing attention to a source of noise which until recently had been ignored. This is the boundary laye~r. The boundary layer will generate noise whenever it is the seat of any fluctuating phenomenon. In particular it will nurture random pressure fluctuations whenever it is turbulent. The designer's interest will, naturally center on the characteristics of that part of the boundary layer noise which has been transmitted through the fuselage skin and into the cabin rather than on that part which is is, as we shall see, a rapidly increasing function of the velocity of Uthe boundary with respect to the air and to a likely observer outside a fuelage either the relative velocity of the plane is low (as near a take-off or landing) or the plane is considerably distant. As a consequence the practical question which has to be~ raised concerns the effects on a fuselage skcin, and on the air which it encloses, of the boundary layer pressure fluctuations acting directly on the skin.* Two features of this problem are worth noting: To begin with, the fuselage will transmit noise only by deflecting laterally. The thickne~ss of the skin is very small when compared to the wave length in metal of audible sound waves so that, effectively there are no fluctuating pressure Gradients within the skin and hence the latter will not oscillate in lateral compression. In the second place, the turbulent pressure fluctuations in the bounary layer are random both in space an~d in time. The fluctuaions are generated locally. If they are measured simultaneously at two different points of the boundary layer, say on the skin itself, they are found to hav nro relationship with each other unless the two points are separated by a very short distance. Alike the value of the pressure fluctuation, at a given point soon loses correlation with itself.+- "Noise intensity is defined in this report as pi2 foai. It is assumed th this is the physical quantity of interest. It ha's the dimensions of an energy flux; but it is not necessarily equal to the energy flux at some point in the field, nor is it necessarily equal to the density of energy radiated away (lost) by the fuselage. "Recently two authors (Refs. 5 & 6) have suggested that the ranomes in time is not independent of the randomness in space; i~e., that the pressure fluctuations at the wall are created by the convection at a single speed of a "frozen" pattern of pressure disturbances. Some attention is paid later to this eventuality which is treated as a special case. 2 NACA TM 1420 Wle are thus led to visualize the process of transmission of boundary layer noise through the metal stein of' the fuselage as follows: A multitude of external pressure pulses push the elastic skein in and out and the skin, in turn, not unlikte a set of distributed pistons, creates inside the fuse- la~se pressure waves which propagate and superimpose. This constitutes cabin noise. It is of course desirable to determine the characteristics of this nroi se. One should point out that the data for` the problem are not complete and are not likely to be so in the near future. Specifically the structure of the turbulent mechanism within the boundary layer and, in particular of the coupling between pressure fluctuations, velocity fluctuations and temperature fluctuations is not enough explained or measured to define wholly our forcing function. As a conseqqg ,ce it is not now possible to define say av:erage cabin noise intensity pi' as a function of say, free stream Mlach number, Rey~nolds number and plate characteristics. It is however possible and it is the purpose of this report to indicate the approximate functional dependence- of p; on these quantities anrd thus to give similarity rules which will reduce to a minimum the amount of testing required. Hie assume at the outset that the boundryis: layer unsteady pressure field is known and that it induces small deflections in the s~in. As a consequence (a) The skinr dynamics are described by a linear equation (b) The generation of' a random pressure field inside the cabin Thus the mathematical techniques used are those required to obtain the response of linear sy~stems to stochastic forcing functions of several. variables. We also assume the fuselage to be a large flat plate. This assumption is not necessary but it simplifies the discussion and allows us to present mrore clearlyr th~e new features of the problem. The material in this report is presented as follows: First we study the radiation of sound from a randomly vibrating plate. It is found that the sound levels in the cabin are de-fined byr the intensity and the scales of the plate normal accelerations. Second, generalized Foulrier analysis is put to use in order to relate the normal acceleration of the plate to the forces exerted on it by the boundary layer flow. Third, the bolundary layer forces are defined in terms of flow character- istics, and dimensional similarity is used to determine the significant parameters. Finally, the functional form of the noise intensity in the cabin is given save for an unknown function oftone non-dimensional parameter. This function depends on the frequency spectrum of the pressure fluctuations in the boundary layer. Hlo measurements yielding this spectrum have been NACA TM 1620 reported to date and speculations concerninC: it would introduce in the analysis both complication and uncertainty. A summary of results is given at the end of the report. Som derivations and some of the longer arguments have been presented as appendices to the text. NACA TM 1420 I. THE ACOUSTIC COUIPLING OiF A RANDO3MLY VIBRATINGC PLATE WITHI AIR AT REST when the sound is generated in an otherwise unbounded stationary gas by a large~ flat plate or dise oscillating normally to its plane (Ref. I page 107). where : the static pressure p has been broken into a steady part p and a fluctuating part, pi: p = p ~j(x,y,z) + pi(x,y,z,t) V, = the normal velocityI of the plate do* = an element of plate surface area ai = the speed of sound in the air S = the vector position over which the integration is carried out =(x',o,Z') if)~ = the distance between source and field points = (x-~y~ -x') 2 z-'2 A = the total area of the plate The subscript i refers to properties inside the cabin or on the side of the plate on which air is not flowing. The normal acceleration bVn/8t is a random fumetion of time and space and so is pi. W~e wish to evaluate the mean square of the pressure (a quantity to which sound intensity is directly proportional). W~e notice that for a sum alike, for the integral in (1))* +This discussion follows closely the arguments set forth in ref. 3 and the background of information on statiat~ical methods can be found in ref. 2. NACA TM 1420 and since aX = arif a does no depend on timet, we~ can write No teexpression is the correlation (-i.e. the time average of the product) of the normal acceleration of the plate surface at two points 81 n 2adttw different times(t- Ryan) and (t- 9:)&}. If the two points vibrate completely independently from each other, this expression is zero and if the two points are brought topothLEr (Sy-wc therefore cl- r2) the -orrelation function is simply(ids/Db)l Now we assume that the average properties of the plate motion are the same anywhere on its surface and at any time (we assume statistical homogeneity and stationarity). Then the expression above for the correlation becomes merely a function of the distance ( -5\) between the two points and of('L-Ra We call this then IAA Now we can evaluate pi2 (Y) under a variety of assumptions for for the plate area A and for the distance Y between the plate and the observer. We will consistently hold the view that the normal acceleration at most points of the plate surface are not correlated ( ~= 0) and that two points of the plate, in order to show appreciable correlation, must be a small fraction of the total plate size away from each other. We define as h thema distance over which~bd~ is strongly correlated (i.e. g (10#dj)~ and call it the integral (length) scale. w The integral scale is given, say in the x-direction by We should note that The integral in (4) is not finite if the plate area A is infinite The length scale h plays no role in the geometry of the problem. The integral is a function of Y, the distance to the plate, and the plate dimensions only. For instance if the plate is circular and of radius R + This case is treated more rigorously in Appendix IA NACA TM 1420 Our hypothesis can then be expressed as where R is the average linear dimension of the plate. A representative case+ Suppose that the observer distance Y to the plate is such that and that no appreciable phase difference can arise at Y between two strongly correlated signals (i.e., between sound pulses originating within jL of each other). Then, if we examined equation (3) ve see that the inner integral contributes very little except when the point 3t is approximately within a distance ), of \$ Then, approximately 'Ls and 1 P(bVag Thus the inner integral is approximately and equation (3) can be rewritten I;' "~ 47\:' dsl h' (dv,)nl dr NACA TM 1420 It is apparent from this result that the distance Y from the plate to the observer is measured in terms of plate diameters, and not in terms of average correlation lengths or wave lengths )L This result holds for all the cases considered (see Appendix I) and depends only on the assumption that at a given time the plate vibrations are largely un- correlated or incoherent." to define a strength. Equation (4) indicates that in order to evaluate the pressure intensity for the case considered above, one must first determine (bS /3)0 the mean square normal acceleration of the plate and \ the length scale for the plate deflections. Other cases (i.e. cases for which either the plate deflects differently or the observer moves closer to it) are treated in Appendix IA. For some of them the time scale or mean period NON is required as well as L - We rewrite Equation (4) where la (pV is a function of the plate geometry and of the distance between the observer and the plate. It is defined from equation (4) as Notice that if the integral scale Is not the same in the x' and in the z' direction we may simply substitute in equation (6) hrr' and h S/ for h\ + This result holds, as Appendix IB shows, even when the pressure is generated by the travel through air at rest of a "randomly bumpy plate" i.e., when the time-wise and space-wise variations of upwash are not independent. The latter example is therefore quite distinct from the flow of an infinite (periodically) wavy wall for which the only characteristic length is the wave length. NACA TM 1420 II. THIE DYNAMIC BEHAVIOR OF THE SKIN The response of the bare fuselage skin to the random pressure field of' the boundary layer is given, in the absence of axial loads, by a plate equation. For a flat plate this equation is where x and : are the coordinates along the plate surface (x being the free stream direction), y the deflection of the plate at a point, E the modulus of elasticity of the plate material, e" its density, hL Poisson's ratio, 2h the thickness of the plate, # a damping constant which has dimensions (1/T). Damping may be present because energy is absorbed either within the skin or by the air." For air limping P0 /' gy) ~ Hlotice that to allow for air damping is to provide for a feedback in the coupling between the plate and the air at rest. On the other hand we exclude feedback between the plate and the boundary layer. In other terms we are not considering the possibility that the plate vibrations are large enough to induce time-dependent pressure gradients of the same order of magnitude as our forcing function. Such a feedback would amount to panel flutter. It cannot be handled by the present method. r(x,z,t) is the random forcefLunit mass exerted by the pressure fluctuations on the plate surface. It is characterized by a power spectral density. F(kl,k2,0;) which is a continuous function of the wave numbers kl (in the x direction), k2 (in the z, direction) and of the frequency co. The coefficienrt E/30-(1-f2) has the dimensions of a velocity squared and it is defined as c2. The skin construction may be such that the damping it causes is primarily viscous or primarily flexural. In the latter case it seems more appropriate to write with Ribner (ref. 5) where fig is the flexural damping constant due to the plate and/Q the damping due to the energy radiated to the air. As is shown in Appendix IB the noise intensity within the fuselage may or may not be related to the acoustic energy radiated by the plate and thu~~tcee~ aitdb h lt n hus P may or may not be NACA IM 1420 We still have to specify the space-wise boudr conditions on the plate and we are led, for the sake of simplicity, to either one of two limiting cases. In the first case, the~ forcing; funtion (the random pressure field in the boundary layer) is characterized by an integral scale so large that at a given time, a skin are~a between two stiffeners (assumed rigid) is very likely to be subjected to a pressure load of the same sig~n (see Fig. Ia). This allows us to exrpress y and f as functions of t only. Z OR X & OR X F-clG C F us. Ib In the second case, the integral scale of the forcing function, is very small in comparison with the distance between two stiffeners an the behavior of the skin is in the ave~rage very much as though the supports were removed to infinity (see Fig. Ib). The real case will in general be intermediate between these two limiting examples. However, the first case seems to apply to boundary layers of excessive~ thickness: A reasonable guess for the average correlation length might be one dis- placement thickness s +; for d+ to be~ larger than the spacing between stiffeners (of the order of a foot) the boundary layer thickness Al would have to be of the order of five! feet or more. This unlikely case is treted in reference (7). On the other hand, the second limiting case (Fig. Ib) would seem to provide a reasonable model for boundary layer thickness no0t exceeding one foot. This is the model discussed now. a) The Mean Acceleration According to our assumption, the average motion NACA TM 1420 10I is not sensibly affected by the presence of stiffeners. A large number of: pulses act on the sktin at a given time between two consecutive stiffeners. The random pulses may be positive orr negative and thus there will be a large number iof load rev~ersals between supports. Then the effect of the boundary conditions can be expected to become small, in the average. Consequently one can define a generalized admittance and use it in much the same way as is often done in one-dimensional problems." For instance, the mean square plate displacement is given by Here the mean square of' the forcing function f' is related to the spectrumn by: l;ffis the generalized admnittance, and kl, k2 and Wt are respectively thel wave number In thle x( direc-tiojn, the wave number in the z directions and the frequency. The determination of If)(is easy once it is realized that this expression is the square of' the Fourier transform of the fundamental solution and so can be written by inspection. Thus an average solution of (7) is: and --- fr.uw~ h Bl (9) Equation (9) gives the mean square response of an unbounded plate to a random forcing function. One should notice that the plate will always exhibit resonance no matter what value the dam~ping constant /8 may have. This resonance occurs, not at a given frequencyi or at a set of discrete + See in particular reference (2). NACA TM 1420 WITH 1~= 0.1 ano +Z = A THE ADMITTANCE MAP FOR AN UNBOUNDED PATE FIGURE 2 //X C~~4 e n2 5! C'h',3;IC~2 -Gi NACA TM 1420 frequencies but over the whole frequency spectrum, whenever the follow- ing relationship obtains between frequencies and wave numbers:+ We can visualize the resonance condition as a crest or ridge in the wave space (see Fig. 2) which originates at the line k2 = (3/chf2 and which becomes hiigher and steeper as the wave number and the frequency increase. Thus the effective damping is a function of the exciting frequency. b) The Length Scale A Equation (6) shows that h2 is needed as well as ( VI/}t)2 Now A is a length scale. It was defined, say in the x direction as Ewa and could be termed the equivalent length of perfect correlation. There are various ways of evaluating the integral scale. Perhaps the most convenient one for our purpose is that (found for instance in Ref. 2 Eq. IIS) which is derived from the relationship between correla- tion functions and spectral functions. Thus if a stationary random fune- tion J(t) possesses a correlation function which is sufficiently well behaved, + H~ere resonance is defined as the maximum of the response curve 1/j(k) holdinggg) constant. The locus of maximae holding k constant is given by: These maximae correspond only for zero damping. NACA TM 14c20 one can define a spectral density function Nov for the particular casek 4 = this gives Since n, =- ~~ d] (o ) we have the result that z Pe) (10 ) We~ hv already obtained a forma representation for the spectral density function of the plate. It is the integrand of Eq. (9) so we! can write, in view of (10) (11) and a similar expression for hy . j O /t,,c 31 [n-lh e- z r SE'A4,s. / bVn 2. NACA TM 11620 III. THE FORCING FUNJCTION We suppose that a turbulent boundary layer develops on the skin of the airplane~ (on one side of our plate). The forces which excite the plate are the pressure fluctuationls experienced by the plate itself. We assume that all characteristics of the boundary layer are fixed once we have specified the boundary lawyer thickness b, the free stream ve~locity Ueo and density In terms of pressure fluctuations this implies that at a fixed point of the "vetted" surface of the skin, we have for the mean square pressure fluctuations ~- fo Ut and the: integral scales, i.e. are proportional to b. Also the relative contribution to pressure intensity of the various frequency bands must be a function of Ugg,5, amndtJ only so that f aC is a random function of three independent variables, x,z,t, PIis related to a three-dimensional spectrum by: anc that : NACA TM 1420 Loosely speaking, this means that a characteristic frequency for pressure fluctuations is proportional to Use /, and a characteristic wave length is proportional to d Now the forcing function of Eq. (7) is a forcefunit mass so that according to our similarity hypothesis B' One can thus define a spectral function associated with the forcing function f(x,z,t): and thus S~i"~b 6lk'L (12) Here F2 is a function of K1, K2, and .11 only and these are non-dimensional variable s: Jr.= o/ U, In terms of these non-dimensional units, Eq. (8) becomes: ~y 91 U~ Ua, NACA TM 11s20 and Eq. (9) becomes which can be written Again, F2 is a function of the3 integration variable only, so that one can write b(l" po', U + ch (5 &4 Equation (14) yields the two-non-dimensional parameters upon which the plate dynamics depend. The first one, Ch /SJo 04 i the product of a speed of free stream Mach umer, ( ) and a Speed of propagation of waves in the plate thickness ratio (to~dr )ai crrs The second one, pbu plate thickness is a non-dimensional damping parameter which is, alike, a function of plate and free stream properties. If we treat the equation for the integral scale (Eq. 11) in the same way, we notice that no new non-dimensional parameter occurs, so that, at most SL c (15) 7- 5GS T Cao4jLr(~ d~acboK ck ~=rp~'rtpr I SUao nU JL F(Kln)r(L ak 'e NACA TM 1420 We now vish to investigate the form of the functions H and L in Eas. (14) an (15) respectively. :First, we make an assumption which is not strictly necessary but which simplifies the manipulation of Eq. (12). We take the function F(KI, KtlR) to be symrmetrie in KI and Kg, which leads us to define a new wave number. and to write Thus, Eq. (13) becomes Now, the dapng circumtances it S ~ 'L~~IC~o Uao bth" paramneter 13141D is assumed small and under these can be shown (see Appendix II) that (16) The small difference between these two integrals can easily be evalaed for arbitrarily small values of pllUa even though both integral are unbounded as P-*** This leads us to believe tha for low damping the main contribution to the inner integral comes from the resonansce condition Thus, if the spectral function F( L,K) is reasonably wide, i~e. ifbF bK((1 over a large range of K Eq. (16) suggests tha we write (r7) 1:JL"drr IraDIc CCIH du(hc Fl kJT) r ILC I c Ke ~r lb'0~; Uclr, Kdl NACA TM 1420 The requirement that F be flat in KC when compared to 1/X(K) is equivalent to the requirement that the average correlation distance or integral scale for the boudr layer pressure fluctuations be small compared to integral scale of the plate defl-ection. Translated in physical terms the simplifica- tion suggested here is prompted by the following remark: If the plate has some stiffness, it makes little difference whether the forcing function is assumed to be distributed over smal distances or made of con- FIGURE 3 Thus a satisfactory model for the problem at hand would rain drops on a metal roof. Equation (16) allows us to to get be the impact of integrate over K, blbo FlJb~"~;E~J' ' ~" 9."uf t 4 a2h~+ ~:~~ ~oC --- a' h'e ~ dJL (18) (~~3~ns \ loI"n ")dn bV~ndt ~ bu n ch ~-(~, NACA TM 1-420 The expression can be evaluated only when F(K n) is known. It may be an increasing or adeceasig fncton olinh .In the absence of data on the spectral function F, we will not tep odfn t The function H defined by Eq. (14) can be written: c~~~h & ']I . where f,, is an unspecified function related by (18) to the bounar layer pressur-e spectrum. In order to determine p 2 we need to find out, in addition, wha quantities the integral scale h depends on. Here we make use of con- siderations which are similar to those yielding Eq. (16;) (see Appendi III). The result is that whref is another function related to the boundary layer pressure spectrum by 1II(4). Hlow we are able to write Eq. (6) as (21) here Expression (21)gves the functional dependence of pressure intensity "inside" on boundary layer parameters for a typical case. The only quantity, not immediately available is h(b /ck). It is probable that we shall have to await experimental data to define its numerical value reliably. NACA TM 1420 IV. SPECIAL CASES 1. Convected Turbulence Two authors (ref. 5 & 6) have recently suggested that the boundary layer pressure fluctuations at any point of the fuselage skin are caused essentially by the passage over the point of a fixed (i.e. time inde- pendent) pattern of pressure disturbances carried downstream at a fixed convective velocity. So far, experimental evidence in proof or disproof is lacking. However, it is interesting to incorporate this special case in the general formulation which has been presented. Both the response of the plate and the coupling of the plate with the air at rest must then be reconsidered. a) The coupling of the plate with air at rest in the case of convected turbulence If a fixed spatial pressure distribution is carried downstream on the surface of the plate, it is easy to show that the (infinite) plateY response will be of the same kind, i.e., that it will consist of ripples which are randomly distributed in space but which travel through the plate at the same convective '.elocityr as the boundary layer disturbance. The determination of the pressure field inside the fuselage is not in principle different for this case and has been carried out in Appendix I~Bt. The result is that for both subsonic moving ripples (with convection velocity U, C ELi ) and moderately supersonic ones: where I For higher supersonic speeds, the function of geometry and Mach number appearing as an integral is more complicated. The equation (I.10) above has the same form as equation (6). On the other hand there is a sharp difference in terms of energy radiated by the plate between the subsonic and the supersonic case, since no energy at all is radiated by subsonic ripples while the supersonic ones do generate some. One must, then, make "Here the presence of transversal bulkheads will change the picture because of multiple reflections of the ripple. *cThis problem can also be viewed as a steady (randomly bumpy) wing problem from the standpoint of a stationary observer. MIACA TM 1420 a distinction between the results in terms of pressure intensity (the quantity of practical interest) and in terms of energy radiation. This distinction stems from the fact that (as is pointed out on page 7) the acoustical field investigated is truly a near field. b) The response of the plate According to the convective hypothesis, time is not an independent variable once the convective velocity [f, is fixed. Translated in tenns of the spectral density T'1~(c,kl,k2) of the pressure fluctuations, this means that I (A) ,hlak2)is zero, except whenP -= U, g, or in non- dimensional form, when n.s rum We rewrite equation (12) for this special case. 4 3,' u, QS kL 'SS-~ k,, a,, ,> Here 6 [- KsL-~ is the Dirac delta function of the variableR.. Then the plate response becomes bVl~n \~t at 1 99~ u~ QL~' (23) he re Now if we assume as before that F is symmetric in K1 and K2 and substitute ~= .kh;l KI a26 7(~11 K,~~n~61"- Uao ~Uo~ R+ F(~,,HIJL)b~-~ II a, d~cl ace, ~ao[' klL(22) urf JL- F~ ~ K,,~~ dn<~ dKL q 6 YI 4p~ ~r LII\ r ~~hZ r~trK~ r. ~r h~ L ~~, i~~ r.~ crJ~I; Fl K,,~z~ UL~bp) NACA TM 1420 we~ finally get Here and ~bzUf d' BZ E a~ k+ c)p ItrCosCa Tr(Beosesuslde c (24) 3 = ,/ U, T.(aeo581~)-- F(K,,~, U, with KIL K+ = CoS is indicated by the following dimensional argument. and Th length scale X The mean correlation length or integral scale is a weighted average of all wave lengths, so that dimensionally I Since resonance dominates the plate response, = is given from the plate response equation (equation 23) byr the resonance condition h A similar reasoning would have yielded, in the non-convective case, C~(h/,,5) VE instead of eq. (20). Combining (24) and (25) according to (I.10) we notice that we can still write as in equation (21). (25) h'L~ p 4:, 0^ (24 Here 61Is a weak function of the kach number as seen from (I.10). NAC TM 16420 2. The case of zero scale Under some circumstarnces it is possible that the space average of th plate motion vanishes, i.e.: This does not ean tha th norma accelerations at two neighboring points show no correlation, but that the correlation function becomes negative J as indcated in Fig. (4) and in suc a way that its space integral vanishes. We can then consider the! normal accelerations as dipoles rather tha sources and we are led to a slightly IC shows, however, that if one defines a length 2 ba such that FIGURE 4 The results are again ide:ntical in forn with those of equation (6). B~ere? hL can be viewed as the mean moment arm of deflection moments. A~lternately one can redefine the integral scale as ,== CCl o (27) where c is a constant. Equation (27) can thus be used to define the! integral scale in any event. NACA IN 1420 V. SUMMARY OF RESULTS AND DISCUSSION Appendix I discusses in addition to the cases mentioned in the text a few examples which provide different limiting conditions. Thus the observer is brought close to the plate (Y(( 7\ ). A short time scale is considered etc..... The common feature of all these analyses is that the resulting mean noise intensity can always be represented, say by equation (26). We shall therefore retain this equation: as ~ r~st Here Pi" Q)i a; 96 6 Il~op 6 zh r" v MI UI c general statement that we can make at the present time. =mean square noise intensity inside =air density inside =speed of sound inside = air density in the free stream =plate density = free stream velocity =boundary layer thickness =plate thickness =viscous damping constant (of units 1/time) =perpendicular distance between observer and fuselage =geometry of the plate = Mach number Ula convectivee velocity of turbulence pattern =characteristic velocity in the plate NCACA TM 1420 25 For all but high supersonic- velocities, the dependence of on Na~ is quite small and can be disretgarded. The function@~ (Y,g), a quantity which does not depend on the dynamics of the problem but only on its geometry should be modified to take into account the fact that the fuselage is a cylinder and not a large flat plate. Thne form of the function 5 cannot be given here both because no information is yet available on boundary layer pressure spectra an because S depends too critically on the tyipe of model assumed. How- ever, if is measured while any one of the four vaiable defining S (6,Uap or b) is varied, then the functional forn of thte noise intensity inside a fuselage can be determined. Thus the main contribution of the analysis is to diminish the extent of the testing required. One of the conclusions which can be drawn frau the foregoing equation is that unless the boundary layer pressure spectrum is a ve sharp function of frequency (which would makre S very sensitive tolUIICht ) it is not practical to decrease cabin noise by boundary layer suction: Since the noise intensity is a weak function of boundary layer thickness, decreasing appreciably cabin noise would involve the remloval of a prohibitive amount of air. Another conclusion is that increasing the damping is a very effec- tive way of limiting the production of noise of all frequencies, since the structure transmits sounds essentially by resonance. The analysis which has been presented deliberately omitted some of the features of the problem which would influence the results an intro- duce new parameters. For instance, the fuselage of commrc-ial airplanes is usually subjected to an axrial. tension as well as other loadis. In addition the skin is curved. To account for these featurs of the prob- lem one would introduce further terms in the differential equation describing the plate and one could treat it in much the same way as has been done here. to the study of a germane problem, the fatigue of panels which are buffeted by a turbulent boundary layer. A NOTE ON TESTING The discussion of the various limiting solutions makes it clear that for the transmission of boundary layer noise through a structure~, the ratio of outside (boundary layer) noise to inside (cabin) noise is in general a function of boundary layer as well as structural character- istics. This is to say, first, that an attenuation coefficient cannt be defined by testing the structure alone with a standard noise source. Thus accurate testing requires at the outset that th model be tested NACA TM 1420 for transmission of a noise similar to boundary layer noise. The main property of such a noise, as we have seen is that it must be random in space as well as in, tim, whch precludes the use of one or a few concentrated sources as noise generators. The only proper substitutes for boundary layer pressure fluctuations are forcing functions whose effects on a fuselage are local. The impact of water drops for instance might be found adequate simulation. Further, similarity in testing requires the matching of' parameters which are ratios of plate and forcing funtion properties. For instance if the forcing function used in the test is a turbulent boundary layer, similarityr parameters are: This is not true of jet noise which is Generated away from the fuselage. NaACA TM 1420 VI. REFERIEICES 1. Lord Rayleigh: The Theory of Sound. Dover Publications, New York. Volume II (1945). 2. Liepmann, H. W.: Aspects of the Turbuence Problem (Part I) ZAMP, Volumet III (1952) pp 321-342. 3. Liepman, H.W.: On the Application of Statistical. Concepts to the Buffeting Problem. Journal of the Aeronautical Sciences 19 (1952) pp 793-800. 4. Timoshenko 8. and Young, D. A.: Advanced Dynamics. Mc~ratw-Kll, New Yorkt (1948). 5. Ritne~r, H. S.: Boundar~y-Layer Induced Noise in the Interior of Aircraft. UTIA Report No. 37 (April 1956). 6. Kraichnaun, R. H.: Iloise Transmission from Boundary Laye~r Pressure Fluctuations. To be published. 7. Co~rcos G. H. and Liepmuann, H. W.: On the Transmission Through a Fuselalle Wlall of Boundary Layer Noise. Douglas Report Noe. SM-19570. (1955) MACA TM 1420 AiPPENDIX I PHE RANDOM RADIATION OF A PLANE SUIRFACE: A. FOUR LIMITING CASES In order to determine the couling between. fuselage vibrations and cabin air one has to choose a model for the correlation 7 1 between the normal accelerations at two different points of the plate. The model which was discussed and for which equation (4) was made plausible is predicated upon two conditions: A. That the observer is distant enough so that a large number of plate elements vibrating independently contribute sound in comparable amounts, i.e. Here as before, h is the integral (length) scale for the plate normal accelerations and Y is the perpendicular distance between the observer and the plate. B. That the time scale of the phenomenon is large enough so that the differences in phase (introduced by the. unequal distance from the point at Y to the? various points of a plate element of length PL) are unimportant, i.e. ai is the speed of sound in the fuselage air, and 8 is the integral (time) scale for the phenomenon: Then one can choose a s;imrple model for the :orrelation function 1 ' where a is the delta function. The norma accelerations are assumed perfectly correlated within a length Lh and not at al for distances NACA TS( 1c20 greater than PL Thlen rr ZI Rt FICUTRE 5 Under these conditions the~ noise at the microphone is contributed primarily from a single plate element which in the average vibrates in phase. The evaluation of this contribution is particularly simple. We can vaite, very nearly andupon integrating which is equation (4). This case,h 4))\ 44 4)X 1.<\$ ,;Ls)4 coragendsi to the following conditions. The passenger (or the mderophone) is far from the plate (in terms of PL ), the boundary layer is thick and the(3 airpane velocities low. One may well wondr about cases for which these conditions do not aply. Whle it apears difficult to answer such a qury with Genrerality it is possible to consider other limiting cases. For instance let us assume that condition 2 still applies but that our observer is extremely close to the plate. This would correspond to the following physical case: A thin fuselage skin, a thick boundary layer, a low airplane velocity and we are3 measuring neise by placing a imicroprhone very close to the skin and insulating it on all sides except the side which faces the skin. Then X7\pY }\44 CL'shSM O IRCA TM 1420 (I.2) and If, for the sake of definiteness, we assume the element circular, then and since '} ) Y it is permissible to write The pressure intensity is therefore given as Thus Eq. (6) applies for the very close as well as for the very far field when phase effects are not important ( A Regi Now assume that we carry on the same experiment but that the boundary layer is thin and that the velocity of the airplane is high so that the exciting frequencies are high. Let us assume in addition that the skin is thick, so that y (4~ h : < C /A i FUSELAGE~ BOUym9RY LAYER FIGURE 6 NACA TM 1420 N~ow the time scale of the plate motion is short and phase effects are prevalent. We define a simple time history in analogy to the space description of Eq. (I.1) The microphone still receives signals effectively only from one plate element an al points within that element vibrate in phase but the pressure pulses originating from that element do not arrive at the microphone in the sam time. Then: ......... 6( I 3) 4Is6 q*** St* at, rs0. 4~r 4t ~ICt A' is simply Equation (I.3) is evaluated by noticing that: Here f; are the real roots of C( ) = 03 which are included in the interval between and and b. W~e only have one root, namely rl = r2. If we choose to integrate, say, with respect to s2 first we get (assumning again that the element is circular bIr A'Wed. 4, I v.l r 9.& #./t~'+Y; 4- (I.5) and according to (I.4) the inner internal yields: NACA TM 1420 so that The time scale appears explicitly in the answer. For the unbounded plate however it is simply proportional to f /Uao just as the time scale f or the boundary layer pressure fluctuations. Finally we may consider a physical case for which phase effects are important and for which the microphone has been placed a large distance away from the plate. i.e.: )19 <( A /4, j r 4 Y FL./SELAGE SA/JV MICROPHONESE~ FIGURE 7 NAA M1420 Now the contribution from each sub-element of vibrating plate is still in the average independent from that of the next one. However, there are in addition cancellations from. withiin one element just as in the previous case. This will happen if the boundary lawyer is thin, the airplane velocity is high LI small) an the observer is far from the fuselage wall. In ordr to evaluae thiis limiting case we first specify the tim behavior of the correlation function: wert write 3 (P,-h e,-ejJd(qp~(Lt-h, e..e) (1[.0 so that 4A and we integrate first with resptect to .SL Using the same techniques as in the previous exape, we ge~t: Nov ve assume that Integrating with respect to 02, 91 1 successively: ata Je(o 'ho a NACA TM 1420 For a circular plate of radius R, this would give 9;' 1TT i~g (~,w O) h( o) In general, and defining a function. of the plate geome~try and of the distance Y only, we have 10;" ~icx~; (Is) -"( b\rtbtt c" le d ld-~ ~(%lx~~ NA~CA TM 1420 APPED~IX I B. TH NOISE GEEAE BY SKIN RIPPLES OF FIXED VELOCIT: If the turbulence pattern is frozen, as discussed in section IV-1 ripples will travel through the (infinite) skin at a fixed convective velocity. Then the~ correlation f~unction.'t mut be written differenty: where UI is the speed of propagation of the rippe ( turbulence convective speed) and therefore where now ~lr J ~,'Z ~ Etl''L~t ~~ The inner integral is of' the form which can be written as in part A. The expression I l~s) hz ~; 'IbJ1I dLllldrt: r \~e CI 1 dre ~' skz~l~,-~,~ NACA TM 1420 has either one or two real roots depending as 41,41 or Cfl respectively. For M ( 4 Sthe only real root is and thus the inner integral yields Il J~' so that: equation (4) for low to the root X, w 5s X"p;.t e-rr2~ (I.lOa) notice that equation (I.10a) above tends to convective speeds. For NI ( (I.9) has, in addition root Given by Another .~ 5 (M:CI)PI C -L Llrl plZ~- I ICt is easy to show that this root exists for al values of xl. In order to simplify the integration let us assume slightly supersonic conditions; i.e. let us write where f. ~e ~ Then '12 ~ C~tlL) lfl4~ ~)~ (Ca) NACA TM 1420 and the inner integral = I so that for the supersonic case: I a (Ml, j Ms (I.10b) A result which is save for a constant coefficient the same as I.10a. i 'LIL ~i M,[r,+ U;xl') ~L I 4 v at] NACA TM 11420 APPER3DIX I C. THE GENERATIONI OF NOISE BY PLATE DEFLECTION'S OF ZERO SCALE If the space average of the correlation function is zero and if the plate vibrations are isotropic in x' and z' one can define a new length scale as a moment arm: ( s is a fixed point) alternatively one can define a modified integral scale h"; = I 1 ( 7, ) 1 Here h''=: C ~CI where c is a constant. Then, one can idealize provided It follows that and integrating with respect to h'29c~ e-~~t 51\ ~, bp, bet Ilt 9~ dp, hi Now, unless a or b = 0 t "I 16 PILP'.)"P the correlation function as = (~6V~~~2 P('.~qb'(L-P~ Jb (1) 5()b~~J.t6t NACA M 1)+20 and thus Is,4ip. keplcy b 9((II-~d9~ rPI 1; ILs. 4sro) I Idj S~'bP ,,,~~ 3 rt~ zi~ rc'Sq; -h')S(a6F~)~ Leae h'~4L~!YTtz. \t 51 IOi~' If t-here is a (time) microscale 0) (OrC P~ d~ ILu ~7;5_ Z~C h'"Pr"rb~rcr'c4;" 4" \0"/ Here the plate has been assumed circular and R is its diame~ter. ;2 j* "'9[6~ )LZ t~F-l APPETH)IX I~I THIE SIMPLIFICATION OF THlE PLATE RESPONSE INTEGRAL (Equation 16) We consider the approximation equation NACA TM 1420 (16) The righit-hand side is clearly un~bounded as the damping constant /3 --L O since its value is explicitly proportional to 1/;a (see for instance EQ.(18)). On the other hand the difference between the left and the right-hiand integrals is finite for (3 = lj. To show this we write Then the left-iland side becomes fojr (9 = 3 suchitd a I) (II1) and the right-ha~nd side becomes; for 9 = 0 as C on CL (II2) ( ~t1)2 f (~_()2 (~+l)e I so tha the difference D between expressions (Ill) and (II2) is anCi+~scr dc ho & 0.. (II4) on L~ KAM NACA TM 1420 41 This e~xprssion is finite and of course independent of fS so that we can concude that the left-hand integral of (16) is unbounded fo /i = O. Further, it is clear that D is a regular function of/C3 so tha th ratio of th lef't-han~d side to the right-hand side of Eq. (16) can be made ar-- bitraily close to unity, by choosing arbitrarily small /3. If a corree- tion is desired a numerical check indicates that Eq. (IIA) gives a good appr-oximation to the error made even with moderately large damping. NIACA TM 1420 AP~PEIJDIX III THE EVALUATION OF THE INTEORAL SCALE Our starting point is Eq. (ll). In terms of non-dimensional variables it becomes We now simplify the denominator by writing successively Then we define (III1) (III2) RJLK~Z Equation (III2) can now be written and if we replace K.2 b;, its value at resonance," namely + The justification for that step is identical to that advanced in Appendix II. Ch .. (fl As h4.) 12 J2. F , NACA TM 14210 We can write > V. * rr p~l~s ,, 8 a'h' ~To P6 (III3) If we compare (III]) to (183) we Get immediately F( )~da cks U, 9. (II14) NACA LangLey Field Va. X. e I 9'K i 9,'R., T(~l go v. LI d,,l ( R(JShiL.JL)dh I" '' .L {Ch!\ X,~~~ e (ii " I 0 0 C. 97 L0 "~. 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http://mathhelpforum.com/trigonometry/172124-using-cartesian-vectors.html	Math Help - Using Cartesian Vectors 1. Using Cartesian Vectors How would I solve the question below using Cartesian vectors? 10. A plane is flying southeast at a constant speed of 900 km/h. The wind is blowing towards the north at 100 km/h. Determine the resultant velocity of the plane relative to the ground. How would I do this? I can use geometric vectors and get the correct answer and angle, but the the Cartesian vectors. 2. Originally Posted by Barthayn How would I solve the question below using Cartesian vectors? 10. A plane is flying southeast at a constant speed of 900 km/h. The wind is blowing towards the north at 100 km/h. Determine the resultant velocity of the plane relative to the ground. How would I do this? I can use geometric vectors and get the correct answer and angle, but the the Cartesian vectors. air vector , $\vec{A} = 450\sqrt{2} \cdot \vec{i} - 450\sqrt{2} \cdot \vec{j}$ wind vector, $\vec{W} = 100 \vec{j}$ ground vector, $\vec{G} = \vec{A} + \vec{W}$ 3. I figured out my mistake, I thought the wind speed was 10 km/h. Thanks for your aid, however, what is i and j? 4. Originally Posted by Barthayn ... however, what is i and j? i and j are the respective unit vectors in the x and y directions.	2016-06-30 01:09:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8711679577827454, "perplexity": 384.82399126602314}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2016-26/segments/1466783397864.87/warc/CC-MAIN-20160624154957-00142-ip-10-164-35-72.ec2.internal.warc.gz"}
https://physics.stackexchange.com/questions/137038/electron-flow-in-a-wire?noredirect=1	# Electron flow in a wire [duplicate] How do electrons that constitute a current flow move in a wire? Some say it's like a wheel.If you give it a push,every part of the wheel moves instantly. Is that what happens to electrons?Do they start moving everywhere in the wire when the circuit is closed? Why? Or is it because electrons bump into each other and the energy travels at the speed of light? • possible duplicate of Is electricity instantaneous? – Ignacio Vergara Kausel Sep 24 '14 at 12:32 • It's not quite instantaneous, but the change in the electric field in the wire once it's connected (which is what drives the electron motion) will propagate at the speed of light. – johnpaton Sep 24 '14 at 12:37 Under the influence of an applied electric field, electrons in conductors actually do not move very fast, in regards to their bulk flow velocity. For instance, in copper the bulk drift speed of electrons is less than a millimeter per second. However, each electron (specifically, the conduction electrons) has an effective speed of over one million meters per second. The effective speed is a random speed and it turns out to only depend upon the material (e.g., for copper it is ~1.6$\times$10$^{6}$ m/s), neither the temperature nor the applied electric field. The random speed and drift speeds are important for determining the mean free path (~4$\times$10$^{-8}$ m or ~40 nm in copper) and collision rate (~4$\times$10$^{13}$ collisions per second in copper) for conduction electrons. The short answer is yes, the conduction electrons hit each other (which causes the transfer of information) and while their net drift speed is very low, the rate of communication through the conductor is slightly below the speed of light. • Voltage is the work done to move a charge from one point to another. They hardly move however.So what is voltage?And resistance? Does it mean that a wire with more resistance make electrons bump slower?Sorry I'm lost. – SMcCK Sep 25 '14 at 7:59 • @SmCcK voltage is kind of like gravitational potential energy. If you apply a voltage difference between two points, a charged particle in that region will accelerate. Resistance (or resistivity) in a material is kind of like friction, in a sense. In a superconductor, for instance, current can flow indefinitely without loss. – honeste_vivere Sep 25 '14 at 11:36 • @SmCcK in short, the collision rate I referred to is dependent upon the electron drift speed whereas their mean free path is dependent upon the random speed of the electrons and the collision rate. You can also think of it this way: $\mathbf{j}$ = $n_{e}$e$V_{drift}$ ~ $\sigma$ $\mathbf{E}$. So yes, the resistivity of a material affects the drift rate and mean free path of electrons. – honeste_vivere Sep 25 '14 at 11:40 As soon as you complete the circuit an electric field travels though the wire. $$E=\frac{V}{x}=\frac{F}{Q}$$ The electric field travels at the speed of light so for wires in use are too small that the electric field will take noticeable time to travel through the wire. As the electric field is constant around the wire and is (almost) instantly there in whole of the wire every electron experiences the same force: $$F=\frac{Ve}{x}$$ Where $e$ is charge on one electron. Every electron experiences the same force due to electric field not because neighboring electrons bump into them. If the circuit is closed, one after the other electrons start to move in the chain between the source and the sink approx. with the speed of light. This is because the electrons always feel each over and the signal about the start moves approx. with c. The same happens in a water pipe too. Try it. Of course you can't measure how fast the signal "open wider the stopcock" come to the end of your pipe. More excited it is if the pipe was heated by the sun and, if you open stopcock, the water flows "instantaneously", but the cooler water comes with delay. Of course take in attention the other answers about the zig-zag-trajectory of the electrons.	2020-09-29 01:47:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6276958584785461, "perplexity": 323.1975808416057}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-40/segments/1600401617641.86/warc/CC-MAIN-20200928234043-20200929024043-00016.warc.gz"}
https://physics.stackexchange.com/questions/552262/is-every-pair-of-conjugate-variables-associated-with-a-fourier-transform	# Is every pair of conjugate variables associated with a Fourier transform? For example, in quantum mechanics, the commutator of the position and momentum is $$[\hat{P_i} ;\hat{Q_j} ] =i\hbar\delta_{ij}\neq 0, i\neq j$$ I know that the position space representation of the wave function is the Fourier transform of the momentum space representation. From what I know, this also applies for other variables, such as energy and time, but I never read anything like that about other pairs of variables. On the other hand, the commutator of other variables, such as x-angular momentum and y-angular momentum is different from 0. Does this mean that: 1. $$J_x$$ and $$J_y$$ are a pair of conjugate variables? 2. There exists a Fourier transform between the two? 3. For any two variables in quantum mechanics that have a commutator different from 0, 1. and 2. are true? 4. In general (not necessarily in quantum mechanics), to every pair of conjugate variables there is an associated Fourier transform? • What is your working definition for "conjugate variables"? May 15 '20 at 19:25 • I do not know of any definition where $J_x$ and $J_y$ are considered conjugate pair. If anything the commutator is not constant, unlike $P$ and $Q$. May 15 '20 at 20:01 • May 15 '20 at 20:02 • @DanielC I don't have a deep understanding of them. In my view, they are a pair of somehow co-dependent variables. By asking this (apparently faulty) question, I was thinking of making sense of them by relating them to the Fourier transform May 15 '20 at 20:04 • @ZeroTheHero Now that I think of it this way, it does make sense that it's a different situation because the commutator is not constant. I only thought of it from my perspective of "co-dependent variables" May 15 '20 at 20:07 The meaning of conjugate variables is usually that they are related by the Fourier transform (other transforms are possible, but I am not sure they are interesting). In quantum mechanics, this is simply a change of basis in Hilbert space. The relation between energy and time is similar, but time is a parameter in quantum mechanics, not an observable (this is even true in relativistic quantum mechanics). The meaning of the energy-time uncertainty relationship is consequently quite different from that of position-momentum. • actually in class. mech they are defined by their Poisson bracket being $1$. May 15 '20 at 21:10 • @ZeroTheHero, I should have just said that conjugacy refers to change of basis in Hilbert space. Notwithstanding the historical importance of canonical quantisation, commutation relations are properly derived from Hilbert space, not imposed by analogy with Poisson brackets, in spite of the apparent similarity of form. Arguments by analogy, and the application of the language of one structure to a different structure, are not helpful to deep understanding. May 15 '20 at 21:52 It depends very much what you mean by conjugate pairs, especially in quantum mechanics. In classical mechanics the most common definition is based on the Poisson bracket: \begin{align} \{P_i,Q_j\}=\delta_{ij} \tag{1} \end{align} and thus in this sense canonical transformations (which can be non-linear transformations in the original coordinates) take you from a set of conjugate variables to another. Alternatively, one can use Darboux’ theorem and the definition of the (closed) canonical symplectic form \begin{align} \omega = \sum_i dp_i\wedge dq_i \end{align} to define the canonical pair $$\{p_i,q_i\}$$. Since (1) is at the root of Dirac quantization, where the quantum commutator is equal to $$i\hbar \times$$(classical Poisson bracket), it is sensible to think of two operators $$\hat P,\hat Q$$ as conjugate if their commutator is $$i\hbar$$. Note that canonical transformations involving non-linear functions of operators is a mine field because of non-commutativity issues. The plot immediately thickens with energy and time, since time is not a quantum mechanical operator. (Deriving energy-time uncertainly relation can be tricky and requires care.) Thus if two classical quantities are Fourier pairs, there is no guarantee they will be conjugate “quantumly”. Moreover, there are examples - for instance $$\phi$$ and $$L_z=-i\hbar\frac{d}{d\phi}$$ - which “look” conjugate in that they satisfy the correct commutation relations but in fact have issues of operator domains. $$\phi$$ and $$L_z$$ are Fourier pairs in the sense that a rotation about $$\hat z$$ is of the form $$e^{-i\phi L_z}$$. (This is an Abelian group so some technical hurdles linked to harmonic analysis are avoided.) The notion of conjugate varibles in QM is also muddled by the related notion of complementary variables. One may naively understand this as follows. Given the uncertainty relation between $$\hat x$$ and $$\hat p$$ one may (loosely speaking) state that, if everything is known about one of the observable, nothing is known about the other. An alternative statement would be that, if you are in an eigenstate of one operator, then the probability of outcome for the other is constant. In this sense $$\sigma_x$$ and $$\sigma_y$$ (and indeed $$\sigma_z$$) are complementary as the outcomes of measuring $$\sigma_y$$ in any eigenstate of $$\sigma_x$$ (for instance) are equiprobable: if you know everything about the measurements of $$\sigma_x$$ (they have no fluctuations on eigenstates of $$\sigma_x$$), then you know nothing of the outcomes of measurements of either $$\sigma_y$$ or $$\sigma_z$$ as all the outcomes are equally probable. This is the starting point for the notion of mutually unbiased bases. • The $L_z$ and $\phi$ are still conjugate variables, right? I mean the domain of $\phi$ is different from $x$ but correspondingly, the domain of $L_z$ is also different from $p$. And there exists a Fourier relation between them via Fourier series rather than a Fourier transform. May 15 '20 at 22:23 • @DvijD.C. in the classical sense yes, but it is difficult to think of them as conjugate in the quantum sense. see physics.stackexchange.com/a/338057/36194 for a discussion in a different context. May 15 '20 at 23:29 • Yes, I'm aware of the issue you point out but I think the operator can be made self-adjoint by being careful about its domain: physics.stackexchange.com/a/233311/20427. You can't obtain the usual uncertainty relations, of course, but the commutation relations still hold over the appropriate domain. I've added the aforementioned link to the post you referenced as it's relevant there independently of this discussion I guess. May 15 '20 at 23:43 • There is currently no satisfactory definition of a self-adjoint phase operator with the right properties (2$\pi$ cyclic etc). It’s still an open question. It’s not even clear if one could measure such an operator as what appears is $e^{i\phi}$ rather than $\phi$ itself. May 15 '20 at 23:44 • @DvijD.C. that solution excludes plane waves since $\psi$ at the end point is $0$ so eigenfunctions of $L_z$, which are $e^{-i m \phi}$ are also excluded. The solution is valid but not really applicable to phases. Biedenharn and Louck in their “Angular momentum in quantum physics” book (or a closely related text v.g. Racah algebra) discuss this at length. Also related: Bonneau, Guy, Jacques Faraut, and Galliano Valent. "Self-adjoint extensions of operators and the teaching of quantum mechanics." American Journal of physics 69, no. 3 (2001): 322-331. for a discussion on this general topic. May 16 '20 at 0:08	2021-09-20 07:47:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 26, "wp-katex-eq": 0, "align": 2, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8825940489768982, "perplexity": 305.4415005029493}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-39/segments/1631780057033.33/warc/CC-MAIN-20210920070754-20210920100754-00633.warc.gz"}
http://mymathforum.com/chemistry/328898-force-attraction-between-ions.html	My Math Forum Force of Attraction between ions Chemistry Chemistry Forum March 12th, 2016, 05:03 PM #1 Senior Member Joined: Nov 2015 From: Alabama Posts: 114 Thanks: 15 Math Focus: Trigonometry, Calculus, Physics Force of Attraction between ions I got a very interesting questions on my mock general chemistry exam on Friday. It says "If you double the distance between two ions the force of attraction between them is multiplied by what?" I went with 2. The correct answer is 1/4. I would love to know why I am totally wrong? I would think the two ions would be trying harder to stay together, and therefore the force would double. But 1/4 does not make any sense to me. Thanks! March 12th, 2016, 07:40 PM #2 Math Team Joined: Jul 2011 From: Texas Posts: 2,579 Thanks: 1275 Elecctrostatic force follows the inverse square law, it is inversely proportional to the square of the distance between the two charges. $F \, \alpha \, \dfrac{1}{d^2}$ double the distance, $d$ ... $\dfrac{1}{(2d)^2} = \dfrac{1}{4d^2} = \dfrac{1}{4} \cdot \dfrac{1}{d^2} \, \alpha \, \dfrac{1}{4} \cdot F$ Thanks from SenatorArmstrong March 12th, 2016, 07:47 PM #3 Senior Member Joined: Nov 2015 From: Alabama Posts: 114 Thanks: 15 Math Focus: Trigonometry, Calculus, Physics Quote: Originally Posted by skeeter Elecctrostatic force follows the inverse square law, it is inversely proportional to the square of the distance between the two charges. $F \, \alpha \, \dfrac{1}{d^2}$ double the distance, $d$ ... $\dfrac{1}{(2d)^2} = \dfrac{1}{4d^2} = \dfrac{1}{4} \cdot \dfrac{1}{d^2} \, \alpha \, \dfrac{1}{4} \cdot F$ Thank you so much! Tags attraction, force, ions Thread Tools Display Modes Linear Mode Similar Threads Thread Thread Starter Forum Replies Last Post leo255 Calculus 1 December 16th, 2014 08:48 PM girlbadatmath Physics 5 November 19th, 2014 11:34 AM mathslog Chemistry 1 August 28th, 2012 03:16 PM Contact - Home - Forums - Cryptocurrency Forum - Top	2017-07-26 08:32:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.657122015953064, "perplexity": 2164.6255398416056}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-30/segments/1500549426086.44/warc/CC-MAIN-20170726082204-20170726102204-00457.warc.gz"}
https://www.transtutors.com/questions/during-the-year-monroe-was-injured-on-the-job-as-a-result-of-the-injury-he-received--2577768.htm	# During the year, Monroe was injured on the job. As a result of the injury, he received the follow... During the year, Monroe was injured on the job. As a result of the injury, he received the following payments: workers' compensation $3,000; reimbursement for medical expenses from employer's health insurance plan$2,000 (all premiums paid by employer); punitive damage award \$5,000. What is the amount to be included in Monroe's gross income for the current year?	2018-08-16 22:08:49	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.23798038065433502, "perplexity": 5853.287748229313}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-34/segments/1534221211185.57/warc/CC-MAIN-20180816211126-20180816231126-00262.warc.gz"}
http://icpc.njust.edu.cn/Problem/Pku/2641/	# Billiard Time Limit: 1000MS Memory Limit: 65536K ## Description In a billiard table with horizontal side a inches and vertical side b inches, a ball is launched from the middle of the table. After s > 0 seconds the ball returns to the point from which it was launched, after having made m bounces off the vertical sides and n bounces off the horizontal sides of the table. Find the launching angle A (measured from the horizontal), which will be between 0 and 90 degrees inclusive, and the initial velocity of the ball. Assume that the collisions with a side are elastic (no energy loss), and thus the velocity component of the ball parallel to each side remains unchanged. Also, assume the ball has a radius of zero. Remember that, unlike pool tables, billiard tables have no pockets. ## Input Input consists of a sequence of lines, each containing five nonnegative integers separated by whitespace. The five numbers are: a, b, s, m, and n, respectively. All numbers are positive integers not greater than 10000. Input is terminated by a line containing five zeroes. ## Output For each input line except the last, output a line containing two real numbers (accurate to two decimal places) separated by a single space. The first number is the measure of the angle A in degrees and the second is the velocity of the ball measured in inches per second, according to the description above. ## Sample Input 100 100 1 1 1 200 100 5 3 4 201 132 48 1900 156 0 0 0 0 0 ## Sample Output 45.00 141.42 33.69 144.22 3.09 7967.81 ## Source Waterloo local 1999.06.19	2019-08-24 15:52:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6797217130661011, "perplexity": 587.4052185142806}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027321160.93/warc/CC-MAIN-20190824152236-20190824174236-00465.warc.gz"}
https://pypi.org/project/pyfaust-openblaso/	FAµST python toolbox ## Project description The FAµST toolbox provides algorithms and data structures to decompose a given dense matrix into a product of sparse matrices in order to reduce its computational complexity (both for storage and manipulation). FaµST can be used to: • speedup / reduce the memory footprint of iterative algorithms commonly used for solving high dimensional linear inverse problems, • learn dictionaries with an intrinsically efficient implementation, • compute (approximate) fast Fourier transforms on graphs. ## Project details Uploaded cp39	2022-11-27 15:51:25	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6955536007881165, "perplexity": 3068.524834612315}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-49/segments/1669446710409.16/warc/CC-MAIN-20221127141808-20221127171808-00622.warc.gz"}
https://www.knowpia.com/knowpedia/Preventable_fraction_among_the_unexposed	BREAKING NEWS Preventable fraction among the unexposed ## Summary In epidemiology, preventable fraction among the unexposed (PFu), is the proportion of incidents in the unexposed group that could be prevented by exposure. It is calculated as ${\displaystyle PF_{u}=(I_{u}-I_{e})/I_{u}=1-RR}$, where ${\displaystyle I_{e}}$ is the incidence in the exposed group, ${\displaystyle I_{u}}$ is the incidence in the unexposed group, and ${\displaystyle RR}$ is the relative risk.[1][2] It is a synonym of the relative risk reduction. It is used when an exposure reduces the risk, as opposed to increasing it, in which case its symmetrical notion is attributable fraction among the exposed. ## Numerical example Example of risk reduction Quantity Experimental group (E) Control group (C) Total Events (E) EE = 15 CE = 100 115 Non-events (N) EN = 135 CN = 150 285 Total subjects (S) ES = EE + EN = 150 CS = CE + CN = 250 400 Event rate (ER) EER = EE / ES = 0.1, or 10% CER = CE / CS = 0.4, or 40% Variable Abbr. Formula Value Absolute risk reduction ARR CEREER 0.3, or 30% Number needed to treat NNT 1 / (CEREER) 3.33 Relative risk (risk ratio) RR EER / CER 0.25 Relative risk reduction RRR (CEREER) / CER, or 1 − RR 0.75, or 75% Preventable fraction among the unexposed PFu (CEREER) / CER 0.75 Odds ratio OR (EE / EN) / (CE / CN) 0.167	2022-05-21 16:43:24	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 4, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7385595440864563, "perplexity": 11340.75978357444}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662539131.21/warc/CC-MAIN-20220521143241-20220521173241-00767.warc.gz"}
http://mathhelpforum.com/calculus/195999-tricky-derivative-problem.html	# Math Help - Tricky derivative problem! 1. ## Tricky derivative problem! I understand how to take the derivative of an equation with respect to a variable, say S, but what do I do when there are two variables involved? Here's the problem d/dx [3cos(x^2) - 4e^t] 2. ## Re: Tricky derivative problem! Treat t as constant, so d/dx [3cos(x^2) - 4e^t] = 3 d/dx cos(x^2) - 4 d/dx e^t = -6x sin(x^2) - 4e^t x 3. ## Re: Tricky derivative problem! Originally Posted by Kanwar245 Treat t as constant Correct. But then... , so d/dx [3cos(x^2) - 4e^t] = 3 d/dx cos(x^2) - 4 d/dx e^t ... the e^t is constant, so it might as well come out with the 4: = 3 d/dx cos(x^2) - 4e^t d/dx 1 ... but then the whole second term is a constant term and drops out... = -6x sin(x^2) ... period. As long as t isn't a function of x, or anything... the context of the problem here would help. 4. ## Re: Tricky derivative problem! What would it look like if t was a function of x? 5. ## Re: Tricky derivative problem! my bad, I integrated the 2nd part! 6. ## Re: Tricky derivative problem! Originally Posted by NotAMathmatician314 What would it look like if t was a function of x? then you'd need to use the chain rule.	2014-07-31 02:26:46	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.852234959602356, "perplexity": 5109.644960657478}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-23/segments/1406510272256.16/warc/CC-MAIN-20140728011752-00164-ip-10-146-231-18.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/2638855/prove-the-positivity-of-a-sequence	# Prove the positivity of a sequence Let $a>0$ a real number and $(u_n)$ the sequence defined by $$u_{n+1} = a - \frac{1}{u_n}\text{ and } u_0 = a.$$ Question: Determine condition on the value of $a>0$ such that the sequence $(u_n)$ is always positive. Attempt: I tried to establish a general formula of $u_n$ in order to set up a condition on $a$, by calculating $u_n$ with some $n$: $$u_1 = a - \frac{1}{u_0} = a - \frac 1a = \frac{a^2-1}{a},\\ u_2 = a - \frac{1}{u_1}= a - \frac{a}{a^2-1}=\frac{a^3-2a}{a^2-1},\\ u_3 = a - \frac{1}{u_2} = a - \frac{a^2-1}{a^3-2a}=\frac{a^4-3a^2+1}{a^3-2a},\\ u_4 = a - \frac{1}{u_3} = a - \frac{a^3-2a}{a^4-3a^2+1}=\frac{a^5-4a^3+2a-1}{a^4-3a^2+1}.$$ But I wasn't successful, it seems that there is no general formula of $u_n$. So I would be appreciate for any suggestion of solution. Thank you! • it is a General formula for this sequence Feb 6, 2018 at 17:24 Notice that if the sequence $u_n$ is positive, then it must be decreasing. This can be proven by induction. In this case, the sequence $u_n$ is monotonic and bounded, so it has a limit $L$ that has to be positive. From the formula, $$L=a-\frac{1}{L}$$ Thus $$a=L+\frac{1}{L} \ge 2$$ So a necessary condition is $a \ge 2$. To prove that the condition is suffecient, Take $L>0$ such that $a=L+\frac{1}{L}$ and prove by induction that $u_n$ is decreasing, positive and $u_n > L$. hint Let $f (x)=a-1/x$. $f$ is increasing at $(0,+\infty)$. $(u_n)$ is monotonic. $u_1 <u_0$ thus it is strictly decreasing. the sequences terms are $>0$ if the limit (if it exists) is $\ge 0$. the limit $l$ satisfies $l=a-1/l .$ if $a\ge 2$ then $l=(a\pm\sqrt {a^2-4})/2>0$. If $a<2$ , $u_2<0$. So, the condition is $a\ge 2$. • I think you have a mistake. Why is $l=1-\frac{1}{l}$? it should be $l=a-\frac{1}{l}$, which has a solution for $a \ge 2$. See my answer. – idok Feb 6, 2018 at 17:37 • @idok Yes right. i edited. thanks. Feb 6, 2018 at 17:44 • Also, the series does not tend to $-\infty$ when $a < 2$. this mistake is not crucial to the argument, but it is important not to confuse the asker. – idok Feb 6, 2018 at 18:00 • @idok If $a <2$, it will have no limit. Feb 6, 2018 at 18:22 • Thank you all for your anwsers. Feb 7, 2018 at 1:56 Hint: $\;\require{cancel} u_{n+1} - u_n = \left(\cancel{a} - \dfrac{1}{u_n}\right) - \left(\cancel{a} - \dfrac{1}{u_{n-1}}\right)=\dfrac{u_n-u_{n-1}}{u_nu_{n-1}} \,$, so as long as the terms are positive the differences between consecutive terms have the same sign i.e. the sequence is monotonic, and is in fact decreasing since $\,u_1-u_0=\left(\cancel{a} - \dfrac{1}{a}\right) - \cancel{a} \lt 0\,$. For it to be positive it is necessary and sufficient that $\,0\,$ is a lower bound, so the sequence is positive iff it is convergent. In this case, passing the recurrence relation to the limit, its limit $\,A\,$ must satisfy $\,A = a - \dfrac{1}{A}$ $\iff A^2 - aA + 1 = 0\,$. The latter equation has real roots iff $\,\Delta = a^2 - 4 \ge 0 \iff a \ge 2\,$.	2022-05-22 11:39:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9673206806182861, "perplexity": 138.95595031623918}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662545326.51/warc/CC-MAIN-20220522094818-20220522124818-00334.warc.gz"}
https://tex.stackexchange.com/questions/466908/listings-not-breaking-long-lines-with-breaklines-true-and-breakatwhitespace-fals	# Listings not breaking long lines with breaklines=true and breakatwhitespace=false I have the following code in my Markdown: {latex-fontsize=scriptsize} $./my-new-application create app_example creating app_example Generating <path-to-app-src>/new-application/application/config/app_example/application.conf for app_example... Application Daemon configuration written to file: '<path-to-app-src>/new-application/application/config/app_example/application.conf' Application Net Name: app_example This is generated into the following using pandoc: \begin{verbatim}$ ./my-new-application create app_example creating app_example Generating <path-to-app-src>/new-application/application/config/app_example/application.conf for app_example... Application Daemon configuration written to file: '<path-to-app-src>/new-application/application/config/app_example/application.conf' Application Net Name: app_example \end{verbatim} I also have the following in a template.latex file: \definecolor{codegray}{HTML}{F0F0F0} \definecolor{framegray}{HTML}{C0C0C0} \usepackage{lstautogobble} \lstset{ breaklines=true, breakatwhitespace=false, backgroundcolor=\color{codegray}, frame=none, basicstyle=\ttfamily, autogobble=true, columns=fixed, basewidth={0.5em,0.5em}, frame=tblr, framextopmargin=4pt, framexbottommargin=4pt, rulecolor=\color={framegray}, literate={”}{{''}}1 {“}{{}}1 {…}{{...}}1 {‒}{{--}}1 } \let\verbatim\relax \lstnewenvironment{verbatim}{}{} The resulting output has the line starting with Generating... running past the end of the code box. I need to be able to break this up so that the line wraps on a / character. I've tried a bunch of things and none of them have worked. Please help. The solution posted by Ulrike Fischer in a comment worked, there was an extra equal sign in rulecolor=\color={framegray},	2019-07-16 04:31:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.44186875224113464, "perplexity": 3227.140697058069}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-30/segments/1563195524502.23/warc/CC-MAIN-20190716035206-20190716061206-00138.warc.gz"}
https://mathematica.stackexchange.com/questions/132525/the-curious-case-of-missing-random-walking-particles-in-the-box	# The curious case of missing random-walking particles in the box Note: I added the entire code I wrote below to generate the initial configuration and have explained step by step what each part of the code does Kindly bear with the code and the explanation. Since the code may be dense, I will make my utmost effort here to summarize the problem that has stumped me for the past three days even after carefully reviewing my code multiple times. I am trying to simulate a reaction-diffusion system in which there are two particles that can roam freely (call them ligands) shown in blue and red (in the picture below). Particles on the membrane are called receptors (purple). The ligands are present within a virtual biological cell (shown by a circle) and they diffuse outward from the cell. The ligands once outside the cell can bind to the membrane receptors. In fact the red particle (ligand known as lefty in biology) competes with the blue particle (known as nodal) for the same receptor (purple). I create the initial configuration for blue and red using initialParticleConfig function and for the purple receptors I use membraneReceptorConfig. All code will be present at the end with comments and bold title on top declaring what the block does. The rules of the game: When red(lefty) binds to a particular purple dot then that purple dot becomes unavailable temporarily for binding with any other particle. It is similar for the case of blue dot as well i.e. when it binds to purple dot then the purple dot temporarily becomes unavailable. I am using the word temporarily because the red and blue dots can get released from the already bound purple dots making them available again DATA STRUCTURES the data structures i have used are pretty straightforward. initialParticleConfig yields 3 things at the end when provided with the appropriate conditions (number of particles and where to create them): 1. specie is an Association which is of the form <\| particle index or identity (integer) -> {"name of particle(either nodal or lefty)", x-y position of the created particle} ..... \|> 2. specieBoundaryIndex is just an association to track whether a particle in specie is inside the cell or outside the cell. Therefore it is of the form <\|particle index (an integer) -> 0 \|\|1 ..... \|>. The value 0 is when a particle is inside the cell and 1 when it is outside. 3. speciePts is the list of the x-y coordinates of the positions of the point. we first use the same code block initialParticleConfig to generate both red and blue dots membraneReceptorConfig also takes in geometrical parameters and a number (specifying how many receptors to create) to place RandomPoints on the circle. The function in general returns two things: list of all the receptor positions created on the cell (circle) and a second structure which is a list of rules where the rule corresponds to: membrane receptor index -> {x-y coordinates of receptor, 0 \|\| 1, None, "type"}. The 0 here represents that the receptor is freely available to bind with red or blue dot. 1 represents that the receptor is already in use. At the start of the simulation all receptors are free. None can be replaced with the particle-index of the blue or red dot to which the receptor is bound to. "type" can be either "nodal" or "lefty" (the names of the particles). membranePts delineate the position of the receptors on the membrane. Receptors remain fixed. they do not move The results from the initialParticleConfig and membraneReceptorConfig are stored in global variables, namely nodallist, nodalBoundaryindex, nodalpts for nodal and likewise for lefty. membranepts, membraneindex Strategy for the code 1. molecules undergo brownian motion 2. we check if the particles are inside the box and outside the cell. force them to stay 3. see ligand(nodal/lefty) interaction pairs with the receptors 4. binding mechanism to determine favourable interactions out of all pairs 5. unbinding module to remove ligand from the bound receptor 6. repeat the process Now THE PROBLEM With simple diffusion and checks to keep the particles in the box, nothing strange happens. However, the moment when I try to simulate my particle with the unbinding module on i.e. not commented, I tend to lose particles over time. I have seen my code more than 10 times but cannot seem to identify problem. Please see the image below. The initial number of nodal I started from is 54 but at some time when i stop the simulation (Manually closing the Animate cell) and counted the total nodal in the system, it came out to be 53. I have tried many times and i see a decrease in either nodal or lefty or both. Please note there is no destruction term for the particle in my system. Therefore I expect the total number to remain conserved. Also if i uncomment the unbinding part and binding part the ligands behave as expected they stay inside the box and remain outside the cell (once they diffuse out of it). And the binding part seems to work ok too. So I do not know where in the unbinding part or elsewhere lies the problem Code that will be used to generate intial Configuration (* setting up initial configuration of ligands ) initialParticleConfig[name_String, {x_?NumberQ, y_?NumberQ}, number_?IntegerQ] := Module[{specie, specieBoundaryIndex, speciePts}, specie = RandomReal[{x, y}, {number, 2}]; (this creates a number of particles for specified domains ) specie = Select[specie, # \[Element] \[ScriptCapitalR] &]; ( select those particles that are present within the cell ) specie = Association@MapIndexed[First@#2 -> #1 &, Thread[{name, specie}]]; (lets thread the particles with a rule where the particles are associated with an index, their name and their coordinates ) specieBoundaryIndex = (#[[2]] \[Element] \[ScriptCapitalR]) & /@ Values[specie] /. {True -> 0, False -> 1}; (a mapping over particle position to check whether the particles are outside the defined boundaries (cells) or within it. a particle inside would yield True (replace True with 1) and external to the region will yield False (replace with 0) ) specieBoundaryIndex]; (putting an index to the previous result for unique identification of particle ) speciePts = Values@specie[[All, 2]]; (this will only extract the particle positions ) Return@{specie, specieBoundaryIndex, speciePts}; (now i can return three variables. specie (particle number \[Rule] {{"Nodal \|\| Lefty"},{x,y}} ...), specieBoundaryIndex (particle number \[Rule] 1 \|\| 0 ...) and speciePts ({x,y}....) ) ]; ( setting up configuration of membrane proteins ) number_?IntegerQ] := Module[{membranePts, membraneIndex}, membranePts = Table[RandomPoint[Circle[{x, y}, radius]], number]; (this creates a random number of particle on a circle, basically representing membrane proteins ) ( putting an identification number for the previous result i.e. receptor number \[Rule] {coord, 0 or 1, BoundTo/None, Type-Ligand} ) Return@{membranePts, membraneIndex} ; ( returns membranePts ({x,y}....) and membraneIndex (membrane protein number \[Rule] {x,y}...) ); ]; Region Constraint \[ScriptCapitalR] = ImplicitRegion[x^2 + y^2 <= 0.25^2, {x, y}]; ( defining the region containing the molecules in a single cell ) Subscript[\[ScriptCapitalR], 2] = ImplicitRegion[x^2 + y^2 <= 0.75^2, {x, y}]; Subscript[\[ScriptCapitalR], 3] = BoundaryMeshRegion[{{-1, -1}, {1, -1}, {1, 1}, {-1, 1}}, Line[{1, 2, 3, 4,1}]]; Generate Initial Configuration {nodallist, nodalBoundaryindex, nodalpts} = initialParticleConfig["nodal", {-0.25, 0.25}, 70]; {leftylist, leftyBoundaryindex, leftypts} = initialParticleConfig["lefty", {-0.25, 0.25}, 50]; {membranepts, membraneindex} = membraneReceptorConfig[{0, 0}, 0.75, 30]; Graphics[{{Gray, Dashed, Line[{{-1, -1}, {1, -1}, {1, 1}, {-1, 1}, {-1, -1}}]}, {Black, Opacity[0.2], Circle[{0, 0}, 0.75]}, {PointSize[Medium], Opacity[0.6], Darker@Purple, Point@membranepts}, {Darker@Blue, PointSize[Medium], Opacity[0.6], Point@nodalpts}, {Red, Opacity[0.6], PointSize[Medium], Point@leftypts}}, ImageSize -> Large] THE CODE once the global variables or the initial configurations are created together with regions (Circle and Rectangular box to which everything is confined), a series of operations are performed to get the final state of the system. The following line of code is used to run the simulation: code to start simulation tracknodal = {}; tracklefty = {}; Monitor[For[i = 1, i < 20000, i++, g = BrownianSimulation[nodallist, nodalBoundaryindex, leftylist, leftyBoundaryindex, membraneindex, membranepts]; ], g] I will mention the main first and all the individual operations will be listed later. Note the local variables are for most part storing values of the global variables: BrownianSimulation[nodalp_, nodalBoundaryIndex_, leftyp_, leftyBoundaryIndex_, membraneind_, membranep_] := leftyind, nodalind, membraneIndex, preNodcount, newNodcount, newpoints, newindices, newptslist, boundnodal, boundlefty, difference}, nodalind = nodalBoundaryIndex; (initial boundaryconfiguration of nodal ) leftyind = leftyBoundaryIndex; (initial configuration of lefty) nodal = nodalp; ( save initial configuration of nodal) lefty = leftyp; ( likewise ) membraneIndex = membraneind; (lets store configuration of membrane receptors) nodalcopy = nodal; ( a copy that we can use to restore the position if the step the particle takes is outside the box or if it wants to come to the inside of the circle ) leftycopy = lefty; (likewise) ( ------ brownian step: the function takes in free particles and adds randomWalk to them ------ ) {nodal, lefty} = Map[If[Length@# != 0, BrownianCheck[#], #] &, {nodal, lefty}]; ( ------ ligand spatial checks: this makes sure that the particles outside the cell stay outside and also not move out of the square ------ ) MapThread[ligandCheck, {{nodal, lefty}, {nodalcopy, leftycopy}, {nodalind, leftyind}}]; nodalind = Association@Map[#1[[1]] -> #[[2]] &, nodalthreadlist]; leftyind = Association@Map[#1[[1]] -> #[[2]] &, leftythreadlist]; ( ------ check for possible ligand-receptor interactions: what we do here is to find all the free receptors and the ligand (nodal or lefty) and see what possible interactions exist. We use the NearestFunction for this. Note: an individual receptor can only interact with one ligand ------ ) {interactingNodalReceptor, interactingLeftyReceptor} = Map[(nearestFunction[#, membraneIndex, membranep] /. ( biasing a common receptor interacting with nodal and lefty at the same time ) {interactingNodalReceptor, interactingLeftyReceptor} = biasSharedReceptors[interactingNodalReceptor,interactingLeftyReceptor] /. biasSharedReceptors[pattern__] :> {pattern}; ( ------ binding mechanics: here we take the previous interactions calculated directly above to see which interactions are favourable i.e. we find the ligand-receptor pair which bind successfully. For this we generate probabilities for each pair. If a successful interaction is found then the membraneIndex (receptor is connected with the particular particle, itd availability is altered to 1 which means not available for more interactions), and nodalthreadlist, and nodalind are updated (particle index are removed.the data stays preserved because we have connected it with the membraneIndex) Also Note: if no interactions were observed i.e. interactingNodalReceptor or interactingLeftyReceptor is empty then the rule below will preserve the parameters being passed ------ ) {nodalthreadlist, nodalind, membraneIndex} = (interAct[##] & @@ interactingNodalReceptor, "nodal"}) /.interAct[patt__, y_, z_] :> {patt}; {leftythreadlist, leftyind,membraneIndex} = (interAct[##] & @@ interactingLeftyReceptor,"lefty"}) /.interAct[patt__, y_, z_] :> {patt}; ( ------ unbinding of particles: in this we find all the bound particles in the system and generate for each of them a probability to unbind. If the unbinding process is a success add the ligand/particle back from the membraneIndex (the receptor to which it was bound) to nodalthreadlist or leftythreadlist, and likewise for nodalind. For membraneIndex update the {availability from 1 to 0, particle index to None,"nodal"\|\|"lefty" to "type"} ----- ) "nodal"]) /. unbindingMechanics[patt__, y_] :> {patt}; "lefty"]) /. unbindingMechanics[patt__, y_] :> {patt}; ( system stats ) ( finding bound nodal and lefty from membraneIndex ) boundnodal = Cases[membraneIndex, patt : PatternSequence[_ -> {{_, _}, _, _,"nodal"}] :> patt[[2, 1]], 2]; boundlefty = Cases[membraneIndex, patt : PatternSequence[_ -> {{_, _}, _, _,"lefty"}] :> patt[[2, 1]], 2]; ( lets change the nodallist and leftylist back to Association so that the inputs to the BrownianSimulation remain the same ) nodallist = Association@Map[#[[1]] -> {"nodal", #[[3]]} &, nodalthreadlist]; leftylist = Association@Map[#[[1]] -> {"lefty", #[[3]]} &, leftythreadlist]; nodalBoundaryindex = nodalind; leftyBoundaryindex = leftyind; membraneindex = membraneIndex; membranepts = membraneIndex[[All, 2, 1]]; ( drawing the system state ) Graphics[{{Gray, Dashed, Line[{{-1, -1}, {1, -1}, {1, 1}, {-1, 1}, {-1, -1}}]}, {Black, Opacity[0.2], Circle[{0, 0}, 0.75]}, {PointSize[Medium], Opacity[0.6], Darker@Purple, Point@membraneIndex[[All, 2, 1]]}, {Darker@Blue, PointSize[Medium], Opacity[0.6], PointSize[Medium], PointSize[Large], Opacity[0.4], Point@boundnodal}, {Darker@Red, PointSize[Large], Opacity[0.4], Point@boundlefty}}, ImageSize -> Large] ] first operation: Brownian walk to make all particles move BrownianWalk[number_, elem_] := Module[{normal, oldposition = elem[[2]], newposition, step, tag = elem[[1]]}, step = RandomVariate[NormalDistribution[0, 0.01], 2]; newposition = oldposition + step; number -> {tag, newposition} ] BrownianCheck[assoc_] := Module[{temp = assoc}, temp = Association[KeyValueMap[BrownianWalk, temp]] ] /; Length@assoc > 0 second operation: To keep particls inside the box and outside the cell ligandCheck[particles_, speciecopy_, particleBoundaryIndex_] := speciekeys = Keys@specie; coordsspecie = Values@specie[[All, 2]]; indexspecie = Values@particleBoundaryIndex; !y \[Element] Subscript[\[ScriptCapitalR], 2] :> {ind, 1, y}; threadList /. {ind_?IntegerQ, x_?IntegerQ, y_List} /; x == 1 && y \[Element] Subscript[\[ScriptCapitalR], 2] :> {ind,1, speciecopy[ind][[2]]}; threadList /. {ind_?IntegerQ, x_?IntegerQ, y_List} /; x == 1 && !y \[Element] Subscript[\[ScriptCapitalR], 3] :> {ind, 1,speciecopy[ind][[2]]}; ] third operation: nearest function to check interactions between receptors and pairs nearestFunction[ligandList_, membraneIndex_, membranepts_] := Module[{freeReceptors, receptormap, interactions, ligandlist}, ( ligands ) ligandlist = Cases[ligandList, {ind_, 1, pos_} :> {ind, pos}]; freeReceptors = Cases[membraneIndex, PatternSequence[x_ -> {y_List, 0, None, "type"}] :> y -> x] ; (finding all the free receptors that are available for binding and making a receptor map for nearest function ) receptormap = Nearest[freeReceptors]; ( creates a nearest function for receptors ) interactions = Reap[Map[ Function[{particle}, Sow[particle[[1]], receptormap[particle[[2]], {1, 0.01}]]], ligandlist], _, List][[2]] // DeleteDuplicatesBy[Replace[#, {x_Integer, {y_Integer, z___Integer}} :> {x,y},2],Last] &; Reverse[interactions, 2] ]/;Count[membraneIndex, 0, 3] > 0 before fourth operation: biasSharedReceptors: if a receptor is shared by two particles then favour a single interaction biasSharedReceptors[nod_, left_] := Module[{nodal = nod, lefty = left, union, probability}, union = Cases[GatherBy[Join[nodal, lefty], Last], {_, x_} ..]; Map[ Function[{part}, probability = RandomReal[]; If[probability > 0.5, lefty = Delete[#, Flatten@Position[#, Flatten@Intersection[part, #]]] &@lefty, nodal = Delete[#, Flatten@Position[#, Flatten@Intersection[part, #]]] &@nodal] ], union]; Return@{nodal, lefty}; ] /; (nod =!= {} && left =!= {}) ( If a receptor is interacting with both nodal and lefty at the same time, bias the system to choose one ) fourth operation: to determine favourable interactions between receptor-particle pairs bindingEvent[membraneIndex_, {ligandpos_, receptorpos_}, ligandnom_] := Module[{coord}, coord = membraneIndex[[receptorpos]][[2, 1]]; ReplacePart[membraneIndex, receptorpos -> (receptorpos -> {coord, 1, ligandpos, ligandnom})] ] interAct[specie_, specieind_, membraneind_, interactingspecies : {{_, _} ..}, ligandname_] := Module[{interactionProb, successpos, successInt, membraneIndUpdate , successligand, specielist = specie, index = specieind}, interactionProb = Table[RandomReal[], Length@interactingspecies]; successpos = Position[interactionProb, _?(# > 0.5 &)]; successInt = interactingspecies[[Flatten@successpos]]; membraneIndUpdate = Fold[bindingEvent[#1, #2, ligandname] &, {membraneind, ## & @@successInt}]; successligand = # & @@@ successInt; specielist = Fold[Function[{list, index}, list /. {y_, _, {_, _}} /; y == index :> Sequence[]], {specielist, ## & @@ successligand}]; KeyDropFrom[index, successligand]; {specielist, index, membraneIndUpdate} ]/; Count[membraneind, 0, 3] > 0 last operation: to unbind any receptor from particle unbindingMechanics[membraneind_, ligandlist_, ligandind_, ligandname_?StringQ] := Module[{membraneIndex = membraneind, ligandList = ligandlist, ligandInd = ligandind, boundSpecies, probability, position, successfuldisassoc}, boundSpecies = Cases[membraneIndex, PatternSequence[_Integer -> {{p1_, p2_}, 1, y_,ligandname}] :> {y, {p1,p2}}]; probability = Table[RandomReal[], Length@boundSpecies]; position = Position[probability, _?(# > 0.97 &)]; successfuldisassoc = boundSpecies[[All, 1]][[Flatten@position]]; membraneIndex = Replace[membraneIndex, PatternSequence[ x_Integer -> {{p1_, p2_}, 1,Alternatives @@ successfuldisassoc, ligandname}] :> x -> {{p1, p2}, 0, None, "type"}, 1]; ligandInd = KeySort@AssociateTo[ligandInd, Thread[successfuldisassoc -> 1]]; ligandList = SortBy[Fold[Insert[#1, {#2[[1]], 0, #2[[2]]}, -1] &, ligandList, boundSpecies[[Flatten@position]]] , First]; {membraneIndex, ligandList, ligandInd} ]/; Count[membraneind, ligandname, 3] > 0 • It may happen that someone will answer but it is also likely it will be closed as too localized/not a debugging service reason. Just be prepared :) – Kuba Dec 1, 2016 at 9:01 • tracknodal and lefty seem not to be defined anywhere. Dec 2, 2016 at 0:09 • @DanielLichtblau sorry for the late reply as I do not have internet access at home these days. I forgot to mention them. they are defined as empty lists like tracknodal ={} and tracklefty={} in the beginning of the program. thanks ! Dec 2, 2016 at 10:26 • @DanielLichtblau I think in one of my simulations I saw one particle binding to a receptor while the receptor (purple dot) was already unavailable (as it was bound to another particle). I am again seeing part of the code where I used nearestFunction. Do you think a free particle replacing an already bound receptor might be destroying the bound particle? However, it is quite strange that it happens because I always make my nearestFunction from the free receptors and not the unavailable ones i.e. Nearest[freereceptors] Dec 2, 2016 at 13:10 The problem has been finally resolved. I used some conditions to spit out the state of the system when any particle disappeared. I found that the main reason for particles to disappear was a pattern that was defined very generally in unbindingMechanics, where a more stringent pattern was needed. Mainly the blank after Alternatives @@ successfuldisassoc had to be replaced with the particlename. I replaced the pattern below: membraneIndex = Replace[membraneIndex, PatternSequence[ x_Integer -> {{p1_, p2_}, 1,Alternatives @@ successfuldisassoc, _}] :> x -> {{p1, p2}, 0, None, "type"}, 1] with this: membraneIndex = Replace[membraneIndex, PatternSequence[ x_Integer -> {{p1_, p2_}, 1,Alternatives @@ successfuldisassoc, ligandname}] :> x -> {{p1, p2}, 0, None, "type"}, 1] If the ligandname is not specified and if two different kind of particles - with the same index - are bound to the receptors then both will get replaced Moreover, another module biasSharedReceptors was added. This module ensures that if a membrane particle has two different types of free particles in its vicinity then only a single particle gets the opportunity to interact with it. We can generate a probability to bias the system and hence the name biasSharedReceptors. biasSharedReceptors[nod_, left_] := Module[{nodal = nod, lefty = left, union, probability}, union = Cases[GatherBy[Join[nodal, lefty], Last], {_, x_} ..]; Map[Function[{part}, probability = RandomReal[]; If[probability > 0.5, lefty = Delete[#, Flatten@Position[#, Flatten@Intersection[part, #]]]&@lefty, nodal = Delete[#, Flatten@Position[#, Flatten@Intersection[part, #]]]&@nodal] ], union]; Return@{nodal, lefty}; ] /; (nod =!= {} && left =!= {}) ( If a receptor is interacting with both nodal and lefty at the same time, bias the system to choose one *) I have made changes to the code in the question as well; these include strong checks and corrections for the identified problems. A user can now take the code in the question and hopefully run it without any issues.	2022-05-26 02:17:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.2733117640018463, "perplexity": 10247.433606246246}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662595559.80/warc/CC-MAIN-20220526004200-20220526034200-00422.warc.gz"}
https://leetcode.com/articles/partition-labels/	Approach #1: Greedy [Accepted] Intuition Let's try to repeatedly choose the smallest left-justified partition. Consider the first label, say it's 'a'. The first partition must include it, and also the last occurrence of 'a'. However, between those two occurrences of 'a', there could be other labels that make the minimum size of this partition bigger. For example, in "abccaddbeffe", the minimum first partition is "abccaddb". This gives us the idea for the algorithm: For each letter encountered, process the last occurrence of that letter, extending the current partition [anchor, j] appropriately. Algorithm We need an array last[char] -> index of S where char occurs last. Then, let anchor and j be the start and end of the current partition. If we are at a label that occurs last at some index after j, we'll extend the partition j = last[c]. If we are at the end of the partition (i == j) then we'll append a partition size to our answer, and set the start of our new partition to i+1. Complexity Analysis • Time Complexity: , where is the length of . • Space Complexity: . Analysis written by: @awice.	2019-12-10 15:45:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.871990978717804, "perplexity": 1575.4527120767689}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-51/segments/1575540528457.66/warc/CC-MAIN-20191210152154-20191210180154-00002.warc.gz"}
http://blog.jpolak.org/?p=1952	# Stable Isomorphisms, Grothendieck Groups: Example If $a$ and $b$ are two real numbers and $ax = bx$, then we can't conclude that $a = b$ because $x$ may be zero. The same is true for tensor products of modules: if $A$ and $B$ are two left $R$-modules and $X$ is a right $R$-module, then an isomorphism $X\otimes_R A\cong A\otimes_R B$ does not necessarily mean that $A\cong B$. Of course, $X$ not even need be zero for this to happen. Addition for real numbers is a little different. If $a$ and $b$ are two real numbers then $x + a = x + b$ is equivalent to $a = b$. What about for direct sums? If $A$ and $B$ are two $R$-modules, and $X$ is a third $R$ module, what if $X\oplus A\cong A\oplus B$? Is it true that $A\cong B$? The answer is no. Perhaps this is surprising from the way direct sums work. After all, in a direct sum $X\oplus A$, it "feels like" what happens in $X$ is independent from what happens in $A$. And for vector spaces, this is true: for $k$-modules where $k$ is a field, if $X\oplus A\cong X\oplus B$, then certainly $A\cong B$, because $A$ and $B$ have the same dimension. Of course, this blog is all about examples, so we'll see an example. To keep things simple, we'll stick with the case where $X$ is a finitely-generated free module. Even here there are interesting things happening. So, consider a ring $R$ and two modules $A$ and $B$. We say that $A$ and $B$ are stably isomorphic if there exists a natural number $n$ such that $R^n\oplus A\cong R^n\oplus B$. It's a strange definition, but it's actually very natural from a certain context, which we'll think about a little later. Of course, isomorphic modules are stably isomorphic, but the converse is not true. There are stably isomorphic modules that are not isomorphic. ## The Example So what's the example? Consider the division ring $\mathbb{H}$ of quaternions. It is the $\R$-algebra, which as a vector space has a basis $\{1,i,j,ij\}$. As an algebra it satisfies the relations $i^2 = j^2 = (ij)^2$. So it's indeed four-dimensional as an $\R$-vector space. Now consider the ring $R = \mathbb{H}[x,y]$ of two-variable polynomials over $\mathbb{H}$. Here, the variables commute with the quaternions and commute with each other. Now we'll produce a left $R$-module that is stably isomorphic to $R$ but not isomorphic to it. How do we do this? We'll start with a simple observation. Suppose $P$ is stably isomorphic to a finitely generated free module. Then $P\oplus R^k \cong R^n$ for some $n$ and $k$. This means that $P$ is the kernel of a surjective $R$-module homomorphism $R^n\to R^k$. Conversely, such a surjective homomorphism splits and so gives rise to a module stably isomorphic to a finitely generated free module. To put it informally: modules that are stably isomorphic to free modules are pretty much the same thing as surjective homomorphisms of finitely-generated free modules. Back to the two-variable polynomial ring over the quaternions, $R = \mathbb{H}[x,y]$. Let's produce a map $R^2\to R$. Luckily, I have one handy: define $f:R^2\to R$ by $$(a,b)\longmapsto a(x + j) – b(y + i).$$ It's easy to prove the following facts, that I will leave to the reader: 1. The map $f$ is surjective 2. Via the first projection, the kernel is isomorphic to the left ideal $J = \{ a\in R : a(x + j)\in R(y+i)\}$. We have an exact sequence $0\to J\to R^2\to R\to 0$. Since $R$ is free (hence projective), this sequence splits and we have an isomorphism $R^2\cong R\oplus J$. Therefore, $J$ is stably isomorphic to $R$. ## Proof that the example given is not free So, we have our ideal $J = \{a\in R : a(x + j)\in R(y+i)\}$, and we know that $R^2\cong R\oplus J$. Therefore, $J$ is stably isomorphic to $R$. We claim that $J$ is not free. Now, $R$ is a finite-dimensional algebra over a division ring and hence free modules have a unique rank (the invariant basis property). Therefore, if $J$ is a free module, it actually has to be isomorphic to $R$. That is, $J = Rz$ for some $z\in J$. By direct calculation, $(y + i)(x – j)\in J$. Therefore, the $y$-degree of $z$ is less than or equal to one. And all elements of $J$ have $y$-degree at least one, so therefore we can write $z = cy + d$ where $c,d\in \mathbb{H}[x]$. Next, you should verify that $y^2 + 1\in J$. Therefore, $y^2 + 1 = r(cy + d)$ for some $r\in R$. This means that $r$ also has to have $y$-degree exactly one, so write $r = c'y + d'$. Comparing coefficients in the equation $y^2 + 1 – (c'y + d')(cy + d)$ gives $cc' = 1$ and $dd' = 1$. Therefore, we can also write $J = R(y + c'd)$. Using $(y + i)(x – j)\in J$ again: $$(y + i)(x-j) = s(y + c'd),$$ and comparing the $y$ coefficient of both sides gives $s = x – j$. This then gives us: $$(x-j)c'd = i(x-j).$$ The $x$-coefficients give us $c'd = i$. But then the constant coefficients give us $-ji = -ij$ or $ij = ji$, which is false in $\mathbb{H}$. This is a contradiction that we obtained by assuming that $J$ was free. Hence, $J$ is not free, but it is stably isomorphic to a free module. Direct sums are strange indeed somtimes… ## Connection to the Grothendieck group The concept of stable isomorphism is actually not just a random curiosity, though even if it were it'd still be cool. Actually, this concept is closely related to the Grothendieck group. Recall that if you have a ring $R$, the Grothendieck group of $R$, denoted by $K_0(R)$, is an abelian group defined by generators and relations. Its set of generators is a set of isomorphisms classes of finitely-generated projective $R$-modules. For each such finitely-generated projective $P$, we write $[P]$ for the corresponding generator. The relations are something you'd expect: for each pair $P,Q$ of finitely-generated projective $R$-modules, there is one relation $[P] + [Q] = [P\oplus Q]$. The Grothendieck group is a fundamental invariant of a ring and is the first in a series of functors called the $K$-theory functors. For example, any commutative ring whose projectives are free (like fields, local rings, and the integers) the Grothendieck group is the abelian group $\Z$ of integers. If $K_0(R)$ is the Grothendieck group of a ring, then two generators $[P]$ and $[Q]$ are equal in $K_0(R)$ if and only if….$P$ and $Q$ are stably isomorphic! That's the connection. So for our ring $\mathbb{H}[x,y]$ we showed that $[R] = [J]$ where $J$ is the ideal $\{ a\in R : a(x+j)\in R(y + i)\}$. ## One last remark I can't help but saying that the ideal $J$ we constructed is also a neat example in another way: our proof showed that $J$ is a projective module that isn't free. Therefore, $J$ is an example of a projective module over a polynomial ring over a division ring that's not free. The cool thing about this is that modules over finitely-many-variable polynomial rings over fields are free – that's the Quillen-Suslin theorem. Not easy at all, though the two-variable case was proved long before Quillen and Suslin solved it (independently, I might add). Thus, the Quillen-Suslin theorem is no longer true if we add just a slight dash of noncommutativity in the form of quaternions.	2019-02-23 19:42:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9462453126907349, "perplexity": 133.0702056050041}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-09/segments/1550249530087.75/warc/CC-MAIN-20190223183059-20190223205059-00320.warc.gz"}
https://www.tutorvista.com/content/math/multiplication-of-integers/	Top # Multiplying Integers The integers are combination of the natural numbers (1, 2, 3, ....) and negatives of natural numbers (....., -3, -2, -1) including 0. The set of the integer is viewed as {...-3, ?2, ?1, 0, 1, 2, 3, ...} and denoted by $\mathbb{Z}$. Multiplication of integers is also called as repeated addition of integers. Because multiplication is nothing but adding the same integer again and again. For example: 2 $\times$ 3 means add 2 three times i.e. 2 + 2 + 2 = 6 Multiplying Integers Examples Given below are some of the examples on multiplying integers. Example 1: Multiply: -10 x (-18) Solution : (-10) x (-18) = -4 × -8 = 32 (Negative * Negative = Positive) Example 2: Solve: (15) x (3) Solution : (15) × (3) = 15 × 3 = 45 (Positive * Positive = Positive) Example 3: Multiply -12 and 6 Solution : Given -12 x (6) = -12 × 6 = - 72 ( Negative * Positive = Negative) Related Calculators Multiplying Integer Adding Integer Calculating Integers Multiplying Calculator	2019-09-23 06:04:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.43127843737602234, "perplexity": 5489.852975158326}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-39/segments/1568514576047.85/warc/CC-MAIN-20190923043830-20190923065830-00472.warc.gz"}
https://math.stackexchange.com/questions/3787118/upper-triangular-matrix-and-diagonalizability	# Upper triangular matrix and diagonalizability Let $$A$$ be an $$n \times n$$ matrix that is similar to an upper triangular matrix and has the distinct eigenvalues $$\lambda_1, \lambda_2, ... , \lambda_k$$ with corresponding multiplicities $$m_1, m_2, ... , m_k$$. Prove the following statements. (a) $$\operatorname{tr}(A) = \sum_{i=1}^k m_i \lambda_i$$ (b) $$\det(A) = (\lambda_1)^{m_1} (\lambda_2)^{m_2} \cdots(\lambda_k)^{m_k}.$$ I am wondering if we can solve this question without assuming that $$A$$ is diagonalizable. With this assumption, the characteristic polynomial $$f(t)$$ of $$A$$ splits, so $$f(t) = (\lambda_1 - t)^{m_1}\cdots (\lambda_k - t)^{m_k}$$. Since $$\det(A - t I) = \det(D - tI)$$, where $$D$$ is an upper triangular matrix with $$D = Q^{-1} A Q$$ for some invertible matrix $$Q$$. Since $$D$$ is an upper triangular matrix, $$\det(D -tI) = (D_{11} -t) \cdots (D_{nn} - t)$$. Thus, $$(\lambda_1 - t)^{m_1}\cdots (\lambda_k - t)^{m_k} = (D_{11} -t) \cdots (D_{nn} - t)$$. Noting that $$\operatorname{tr}(A) = \operatorname{tr}(D)$$, $$\operatorname{tr}(A) = \sum_{i=1}^n D_{ii} = \sum_{i=1}^k m_i \lambda_i$$. Similarly, $$\det(A) = \det(D) = \prod_i(D_ii) =(\lambda_1)^{m_1} (\lambda_2)^{m_2} \cdots(\lambda_k)^{m_k}$$. • I don't see where you assume that $A$ is diagonalizable. – Zuy Aug 11, 2020 at 9:48 • If I do not assume $A$ is diagonalizable, how do we know that $f$ splits? Aug 11, 2020 at 10:20 • You don't need this. Let me write an answer. – Zuy Aug 11, 2020 at 10:24 You can use the following well-known Properties Let $$A$$ and $$B$$ be $$n\times n$$-matrices such that $$B$$ is invertible. Then $$\det (B^{-1} A B)=\det A$$ and $$\mathrm{Tr}(B^{-1} A B)=\mathrm{Tr}A.$$ If we let $$M$$ be the upper triangular matrix you mentioned in your question, and $$B$$ the invertible $$n\times n$$-matrix satisfying $$B^{-1}MB=A,$$ then $$\mathrm{Tr}A=\mathrm{Tr}(B^{-1}MB)=\mathrm{Tr}M=\sum_{i=1}^k m_i\lambda_i$$ and similarly $$\det A=\det(B^{-1}MB)=\det M=\lambda_1^{m_1}\cdots\lambda_k^{m_k}.$$ Note that the first equalities are by equality of $$A$$ and $$B^{-1}MB$$, the second equalities follow from above properties, and the third equalities hold because $$M$$ is upper triangular.	2022-05-24 08:06:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 35, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9804083704948425, "perplexity": 109.16346201458951}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662570051.62/warc/CC-MAIN-20220524075341-20220524105341-00074.warc.gz"}
http://forums.xkcd.com/viewtopic.php?f=17&t=71488&p=2638353	## A criminal hides in a room with 99 innocent people... For the discussion of math. Duh. Moderators: gmalivuk, Moderators General, Prelates kikko Posts: 69 Joined: Sat Apr 04, 2009 12:18 am UTC ### A criminal hides in a room with 99 innocent people... A criminal hides in a room with 99 innocent people. You have a lie detector that correctly classifies 95% of people. You pick someone at random, wire them up to the machine, and ask them if they are the criminal. They say 'no', but the machine goes ‘ping’ and says the person is lying. What is the chance that you have caught the criminal? Note: We are going under a different question than asked, we are going about what is the chance someone who tested positive was guilty if all 100 people were tested. I believe it is 19%, but am having an argument about this with people saying it is 17.8% or 15.83% or 16.9%. kikko Posts: 69 Joined: Sat Apr 04, 2009 12:18 am UTC ### Re: A criminal hides in a room with 99 innocent people... I heard someone say it will come out as: 94 innocent 6 guilty, or 4 guilty 96 innocent, but I don't understand how it won't be 95 innocent and 5 appearing guilty. skeptical scientist closed-minded spiritualist Posts: 6142 Joined: Tue Nov 28, 2006 6:09 am UTC Location: San Francisco ### Re: A criminal hides in a room with 99 innocent people... kikko wrote:You have a lie detector that correctly classifies 95% of people. This is an ambiguous statement. You need to know both the probability of a guilty result on an innocent individual, and the probability of a guilty result on a criminal in order to solve the problem. Are you saying that the lie detector classifies 95% of criminals as guilty and also classifies 95% of innocents as innocent? I'll assume this is what you meant. However, if people are arriving at different answers, one possible explanation is that they're interpreting this ambiguity differently. After testing everyone, you find 5 people of being tested guilty. What is the chance that the last person tested guilty was actually guilty? If the guilty individual tests guilty, then each person with a guilty result is equally likely to be the guilty individual, so there's a 20% chance. If the criminal tests innocent, clearly there's a 0% chance. So we want to find the probability that the criminal tested guilty, given that 5 people tested guilty (and then multiply by 20%). The relevant probabilities can be computed using binomial coefficients. The chance that the criminal would test innocent and 5 of 99 innocents would test guilty is $(.05)\left( \begin{array}{c} 99\\5 \end{array}\right)(.05)^5(.95)^{94} \approx .009.$The chance that the criminal would test guilty and 4 of 99 innocents would test guilty is $(.95)\left( \begin{array}{c} 99\\4 \end{array}\right)(.05)^4(.95)^{95} \approx .171.$The conditional probability that the criminal would test guilty given that 5 people total tested guilty is therefore .171/(.171+.009)=.95, so we arrive at your answer of .9520%=19%. There's one thing I'm left wondering. Is there an intuitive explanation for the fact that the conditional probability that the criminal appears guilty given 5 positive tests exactly coincides with the prior probability without knowing the number of positive tests? I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side. "With math, all things are possible." —Rebecca Watson 314man Posts: 119 Joined: Sat Oct 09, 2010 6:03 pm UTC Location: Ontario ### Re: A criminal hides in a room with 99 innocent people... kikko wrote:You have a lie detector that correctly classifies 95% of people As skeptical scientist already said, this is a rather ambiguous statement. He solved it assuming it meant that it always classifies 95% of the people correctly. However if you meant that it classifies a person correctly 95% of the time (which you expect to correctly classify 95% of people), it's a Baye's Theorem problem. Qaanol The Cheshirest Catamount Posts: 3069 Joined: Sat May 09, 2009 11:55 pm UTC ### Re: A criminal hides in a room with 99 innocent people... In the case where you test everyone and the machine has a 95% chance of correctly finding each innocent person innocent and a 95% chance of correctly finding each guilty person guilty: You definitely tested the criminal, since you tested everyone. The odds of the criminal being among the people found guilty by the machine is 95%, since there is a 95% chance the machine correctly found the criminal guilty when tested. This is true regardless of how many innocent people were incorrectly found guilty. In the case where you pick one person at random, they deny being the criminal, and the machine finds them guilty: You have an x = 1% chance of picking the guilty person. The guilty person has a y = 100% (I assume) chance of denying being guilty. The machine has a 95% chance of finding the guilty person guilty. So the probability that a randomly chosen person will be guilty and found guilty is xyz = 0.95%. You have an r = 99% chance of picking an innocent person. The innocent person has an s = 100% (I assume) chance of denying being guilty. The machine has a t = 5% chance of finding an innocent person guilty. So the probability that a randomly chosen person will be innocent and found guilty is rst = 4.95%. These outcomes are mutually exclusive and cover all possible ways a person could be found guilty, so the probability that a randomly chosen person will be found guilty is xyz + rst = 5.9%. Thus the probability that a randomly chosen person who is found guilty will actually be guilty is 0.95% / 5.9% = 19/118 ≈ 16.1% wee free kings skeptical scientist closed-minded spiritualist Posts: 6142 Joined: Tue Nov 28, 2006 6:09 am UTC Location: San Francisco ### Re: A criminal hides in a room with 99 innocent people... Qaanol wrote:In the case where you test everyone and the machine has a 95% chance of correctly finding each innocent person innocent and a 95% chance of correctly finding each guilty person guilty: You definitely tested the criminal, since you tested everyone. The odds of the criminal being among the people found guilty by the machine is 95%, since there is a 95% chance the machine correctly found the criminal guilty when tested. This is true regardless of how many innocent people were incorrectly found guilty. But not true regardless of how many total people were found guilty. For example, if 100 people tested guilty, there is a 100% chance that the criminal tested guilty (since everyone did). In general, if n of 100 people tested guilty by the machine, the chance that the guilty person among them is n.952/(n.952+(100-n).052). This is only equal to 95% when n=5. I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side. "With math, all things are possible." —Rebecca Watson ST47 Posts: 125 Joined: Tue Oct 16, 2007 6:42 pm UTC ### Re: A criminal hides in a room with 99 innocent people... I followed all the probability here so far, but I'm not sure what to make of this: What if, instead of gathering the evidence before making rash decisions, we decide to have a good old-fashioned witch hunt, test people until we get a positive, and then burn that person at the stake. What is the probability that we have burned the criminal? Certainly for each individual tested, there is a 1% chance that it is a criminal, and a 99% chance it is not. Further, we have a 0.95% chance of correctly burning the criminal, a 0.05% chance of missing the criminal and moving on, a 99%95% chance of finding an innocent person and moving on, and a 99%5% chance of finding an innocent person guilty, and incorrectly burning them. I tried to put together a fancy sum for this, but I realized that only the first person has a 1% chance to be the criminal. I thought about considering first the probability that the criminal, if tested, would be caught, finding the probability that he would be the first, second, third, etc. person tested, and then the probabilities that if he was the nth person tested, the likelihood that we would not have any false positive before then. Of course, the probability he would be caught if tested is 95%, and he has a 1% chance of being first, second, third, etc. in the order of people tested. If he is in position n, we have a 0.95^(n-1) chance of getting to him without any false positives (I think), and so the probability that he will be caught is the sum, for n from 1 to 100, of 0.950.010.95^(n-1). I got roughly 18.89% for this scenario. Of course, then there's a 0.95^990.05 chance that we'll catch no one, and have to do the whole thing again, but that's only 0.03%. Velifer Posts: 1132 Joined: Wed Dec 26, 2007 4:05 pm UTC Location: 40ºN, 83ºW ### Re: A criminal hides in a room with 99 innocent people... ST47 wrote:What if, instead of gathering the evidence before making rash decisions, we decide to have a good old-fashioned witch hunt, test people until we get a positive, and then burn that person at the stake. What is the probability that we have burned the criminal? Certainly for each individual tested, there is a 1% chance that it is a criminal, and a 99% chance it is not. Be careful here. Each individual tested is either innocent or guilty. There is no probability in that, there is only the truth. Probability only enters when we grab someone at random and start looking at our belief in their truth or innocence. This sort of fundamental error can make good math do bad things. Time flies like an arrow, fruit flies have nothing to lose but their chains -Marx Token Posts: 1481 Joined: Fri Dec 01, 2006 5:07 pm UTC Location: London ### Re: A criminal hides in a room with 99 innocent people... skeptical scientist wrote:There's one thing I'm left wondering. Is there an intuitive explanation for the fact that the conditional probability that the criminal appears guilty given 5 positive tests exactly coincides with the prior probability without knowing the number of positive tests? If you apply Bayes' theorem, you'll see that this is equivalent to saying that seeing 5 tests is equally likely whether or not the criminal was found guilty - that is, when you test 99 innocent people, you're exactly as likely to get 4 false positives as you are to get 5. To be more general, it's the fact that if you expect x false positives when you test n innocent people, you're just as likely to get x or x-1 when you test n-1. Is there an intuitive explanation for that? All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it. galadran Posts: 20 Joined: Sun Feb 10, 2008 3:14 pm UTC ### Re: A criminal hides in a room with 99 innocent people... Firstly lets take the person in front of us, who has 1% chance of being the criminal. In the event they are, there is a 95% chance the machine will print out Guilty and a 5% chance it will print out Innocent. If the person in front of us is innocent (99% probability) then there is a 5% chance it will print out Guilty and a 95% chance it will print out Innocent. We therefore have two events each with two outcomes for a total of four outcomes all together which to recap are: Criminal - Guilty - 1% 95% Criminal - Innocent - 1% * 5% Civilian - Guilty - 99% * 5% Civilian - Innocent - 99% * 95% From the question we know that the machine has given us a guilty response thus we can reduce this to: Criminal - Guilty - 1% * 95% Civilian - Guilty - 99% * 5% We can now think of this as the entire probability 'space'. We want to find out the probability they are guilty so we take P(Criminal - Guilty) / (P(Criminal - Guilty) + P(Civilian - Guilty)) Which evaluates to 19/118 or 16.1% This is effectively the same solution and explanation as Qaanol but in what I hope is a wordier and more lucid form. imatrendytotebag Posts: 152 Joined: Thu Nov 29, 2007 1:16 am UTC ### Re: A criminal hides in a room with 99 innocent people... More generally, if we have N people (one criminal), a lie detector with accuracy p, and upon testing them all we find that k of them test positive, the chance that the criminal is among that k is: Spoiler: This uses the same reasoning as SS: the probability that k people test positive and none of them are the criminal is: $\binom{N-1}{k}(1-p)^{k+1}p^{N - k - 1} = (N-k)(1-p)^2\left(\frac{1}{N}\binom{N}{k}(1-p)^{k-1}p^{N-k-1}\right)$ (There are N-1 choose k ways of picking k innocent people. Then the lie detector screws up k+1 times: for those people and the criminal, and gets the other N - k - 1 correct) Now the probability that k people test positive and the criminal is among them is: $\binom{N-1}{k-1}(1-p)^{k-1}p^{N - k + 1} = kp^2\left(\frac{1}{N}\binom{N}{k}(1-p)^{k-1}p^{N-k-1}\right)$ (We pick k-1 innocent people, and the lie detector gets those k-1 wrong and the rest right). So the resultant probability is: (simplifying) $\frac{kp^2}{kp^2 + (N-k)(1-p)^2}$ In a slightly different vein, suppose you test all the suspects, and pick one of the guilty ones at random (or give up if nobody tests guilty). Then the probability you get the right guy is: $\sum_{k=1}^{N}\frac{1}{k}\binom{N-1}{k-1}(1-p)^{k-1}p^{N - k + 1}$ $=\frac{1}{N}\sum_{k=1}^{N}\binom{N}{k}(1-p)^{k-1}p^{N-k+1} = \frac{p(1-p^N)}{N(1-p)} = \frac{p}{N}(1 + p + p^2 + \cdots + p^{N-1})$ Hey baby, I'm proving love at nth sight by induction and you're my base case. phlip Restorer of Worlds Posts: 7573 Joined: Sat Sep 23, 2006 3:56 am UTC Location: Australia Contact: ### Re: A criminal hides in a room with 99 innocent people... ST47 wrote:What if, instead of gathering the evidence before making rash decisions, we decide to have a good old-fashioned witch hunt, test people until we get a positive, and then burn that person at the stake. What is the probability that we have burned the criminal? I think it depends... are we testing people entirely at random, with replacement? Or are we making sure we only test each individual once (but in a random order)? And, in the latter case, what do we do if we test everyone, and get all innocent results? This sounds like a much more complicated question, regardless, but I'm pretty sure the different options there will have different answers. Code: Select all enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);} [he/him/his] blademan9999 Posts: 44 Joined: Sun May 01, 2011 5:18 am UTC ### Re: A criminal hides in a room with 99 innocent people... There will be 0.95 true positives and 0.05*99 = 4.95 false positives. That gives us 0.95/(4.95+0.95) = 0.95/5.9= 0.161 = 16.1% http://officeofstrategicinfluence.com/spam/ That link kills spam[/size][/b][/u] silverhammermba Posts: 178 Joined: Fri Oct 13, 2006 1:16 am UTC ### Re: A criminal hides in a room with 99 innocent people... This is clearly a case of different interpretations resulting in different answers. I agree with skeptical scientist's answer, but apparently the question was removed from the OP and replaced with a different one? Return to “Mathematics” ### Who is online Users browsing this forum: No registered users and 10 guests	2019-08-21 04:38:32	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6295595765113831, "perplexity": 1235.9388402002965}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-35/segments/1566027315750.62/warc/CC-MAIN-20190821022901-20190821044901-00517.warc.gz"}
http://codeforces.com/blog/entry/53502	### rng_58's blog By rng_58, history, 14 months ago, , https://r.recruit-jinji.jp/code_fes/us/index.html Details will be announced in early August. In Japanese website, it says there will be 20 slots for international students. According to Japanese website, the finals will be 25th-26th in November. • • +120 • » 14 months ago, # \| ← Rev. 2 → +8 I hope there are no stupid restrictions like "You may participate no more than X times" or "you should be under Y to be eligible". • » » 14 months ago, # ^ \| +18 As far as I see the japanese website, This contest opens to only university students. But it seems that there isn't age limit or the number of times limit. » 14 months ago, # \| -27 Is it rated though? » 13 months ago, # \| +20 There is information in English now. https://r.recruit-jinji.jp/code_fes/us/Short summary about qualification and slot distribution:There will be three qualification rounds.First round: 5 international + 40 JapaneseSecond round: 3 international + 20 JapaneseThird round: 2 international + 20 JapaneseAfter all qualification rounds 10 more slots for international students distributed according to somehow calculated performance during all 3 rounds (no details provided) and their region:Africa/Middle East:1Asia:3Europe:4North America:1Latin America/South Pacific:1Also you must be unemployed and finish university no more than 4 years ago, • » » 13 months ago, # ^ \| 0 The qualification system looks inaccurate now, please wait for a while. • » » » 12 months ago, # ^ \| +12 Any update on that? To confuse more in e-mail in "eligibility" part it is also written "Must be in top 100 contestants from Qualification contests A(students living outside Japan are limited to 10), top 60 contestants from Qualification contests B(students living outside Japan are limited to 5), or top 60 contestants from Qualification contests C(students living outside Japan are limited to 5).", but few lines below rules that Um_nik mentioned are written (and I assume these are real ones and I should ignore that eligibility part). • » » » » 12 months ago, # ^ \| 0 Here in the "Event Outline" part is some complex procedure to determine the finalists involving irrational numbers... Not sure if it is the correct one though. • » » » » » 12 months ago, # ^ \| 0 At least it is well-defined... (at least I think so)And I think applying the "n qualifier from a region" rule before the "top x qualifiers" rule makes sense. » 12 months ago, # \| 0 Lol, registration form is terrible. It took me ages to fill it and I was thinking that question about size of my mother's shoes may pop up at any moment. • » » 12 months ago, # ^ \| +55 Maybe that's the real elimination round? » 12 months ago, # \| +18 The onsite contest is not open for high school students, right? • » » 12 months ago, # ^ \| +8 I was wondering the same thing. The bracket that states who is eligible is not very clear to me as I don't understand like half of the options. There was some "graduate school", and I think that the best chances for high school students would be to be part of this category. Could someone tell us? • » » » 12 months ago, # ^ \| 0 Grad school means PhD students.From the way I read it, it seems to explicitly mention only university students and above, so it implicitly forbids high school students. It doesn't make sense though. • » » » » 12 months ago, # ^ \| +10 It doesn't make sense though.Why not? To me it seems like the purpose is to look for workers who are old enough to do internship or a full time. • » » » » » 12 months ago, # ^ \| 0 Over 18 is old enough. I also know people who worked (in programming) while in high school. • » » » » » » 12 months ago, # ^ \| 0 But maybe it is more likely to find a university student with the right skills than a high school student? • » » » » » » » 12 months ago, # ^ \| 0 A high school student that reaches the finals will have the necessary skills as likely as a uni student. Those who don't reach the finals are irrelevant and it costs nothing to let them participate. Therefore, there's no reason not to let them participate.Your arguments are valid, but extremely weak. • » » » » » » » » 12 months ago, # ^ \| 0 Usually "school student that reaches the finals" that has "the necessary skills as likely as a uni student" will go to university and you will not see him next 5 years. • » » » » » » » » » 12 months ago, # ^ \| 0 Why would the organisers of a competition for university students not see university students? • » » » » » » » » 12 months ago, # ^ \| +10 A high school student that reaches the finals will have the necessary skills as likely as a uni student.I am not sure about the programming scene in the europe, but I don't think this is true at least in (most of) south east asia. Most of the top competitive programmer I know here only know CP(or at least have minimal developing experience) during their high school. • » » » » » » » » » 12 months ago, # ^ \| +23 I only know CP during university, so what? • » » » » » » » » » 12 months ago, # ^ \| -31 Most people over here who know something from programming don't know even what CP is. However, what I'm talking about is high school students who are elite to the point of reaching top ~20 worldwide in this contest; these would probably know a lot more than just CP. How many of them are in your sample? • » » » » » » » » » 12 months ago, # ^ \| -18 Um_nik, well I guess you are an abnormal data point? I am not saying anything is wrong, I am just guessing what the organizer think. I am sorry if I offend anyone in anyway or if my point is weak.Xellos, probably 0(Sorry, my point about good Competitive programmers should not imply the top 20, my bad). But I think Um_nik have just became a sample! I think I read from quora that I_love_Tanya_Romanova does not know much about dev until recently as well. (of course they would know more than CP, but they may not have the relevant skills the organizer wants?) Just a hypothesis here. • » » 12 months ago, # ^ \| +3 I think yes (last year's code festival onsite is also not open for high-school student) I guess the reason is the same as TCO's or GCJ's because onsite contest of them is not eligible for under-18. (Also I don't know Topcoder's or GCJ's reason) But this is so sad for thousands of people including me, and I can't participate next 3 year's onsite so I hope high-school students will become eligible. • » » » 12 months ago, # ^ \| 0 There are plenty of students around the world who are over 18; an age requirement has nothing to do with school.Also, I think TCO and GCJ are open for high school students too. • » » » » 12 months ago, # ^ \| 0 I trusted this comment. For example, TCO is eligible for 18 years or older. • » » » » » 12 months ago, # ^ \| ← Rev. 2 → 0 So TCO is open to high school students who are 18 years or older. (I was 19 when graduating. I know people who were 20 when graduating, without failing a grade.) » 12 months ago, # \| +9 How to D? :D • » » 12 months ago, # ^ \| ← Rev. 3 → -33 Say you are at cell (x, y), if you go to any cell at distance d from (x, y), either x + y changes by (+d or -d) or x - y changes by (+d or -d)See tourist's code • » » 12 months ago, # ^ \| +20 "Xorify" two colorings: 000111000111... 00111000111... 0111000111... 111000111... 11000111... 1000111... and 000222000222... 2000222000222... 22000222000222... 222000222000222... 0222000222000222... • » » » 12 months ago, # ^ \| 0 Any proof for that? • » » » » 12 months ago, # ^ \| ← Rev. 2 → +25 * * * * * * x * * * * * * Here is a representation of all cells which are exactly d away from x in our metric. Each of the stars is either on a edge of the diamond (the second coloring distinguishes it from x) or on a / edge (the first one distinguishes). » 11 months ago, # \| +59 Looks like no one calculated unofficial list of finalists (and there is no official list too).1+1. LHiC — Europe 1st1+2. tourist — Europe 2nd1+3. apiad (moejy0viiiiiv?) — Asia 1st2+2. Factorio — Asia 2nd3+1. dotorya — Asia 3rd3+2. ksun48 — North America 1st4+2. Um_nik — Europe 3rd6+1. sevenkplus — 1st4+3. Marcin_smu — Europe 4th7+1. KAN — 2nd5+2. liympanda — 3rd8+1. miaom — 4th5+3. HellKitsune — 5th7+2. cospleermusora (V--o_o--V?) — 6th6+3. shik — 7th11+1. FizzyDavid — 8th8+2. desert97 — 9th12+1. Merkurev — 10th8+3. Reyna — Africa and Middle East 1st29+2. jerrym (jerry?) — South Pacific and Latin America 1st • » » 11 months ago, # ^ \| +33 AFAIK, miaom and FizzyDavid are high school student. liympanda has worked in Google for several years. They are both ineligible. • » » » 11 months ago, # ^ \| +28 Confirmed. I'm not eligible for onsite, enjoyed in the problems very much :) • » » 11 months ago, # ^ \| +15 Sadly I won't be able to go since the finals are too close to my end-of-year exams.Looking forward to up-solving the problems though!	2018-09-23 11:25:25	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.22447232902050018, "perplexity": 3027.3319313255283}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-39/segments/1537267159193.49/warc/CC-MAIN-20180923095108-20180923115508-00325.warc.gz"}
http://lyx.475766.n2.nabble.com/Latex-error-Environment-subequations-undefined-td4421640.html	# Latex error: Environment subequations undefined 2 messages Open this post in threaded view \| Report Content as Inappropriate ## Latex error: Environment subequations undefined Hello, I used "\begin{subequations}" and "\end{subequations}" to have equations in between. It worked on my two earlier research papers using LyX 1.6 mac os snow leopard. But suddenly the same way stopped working in a new LyX file. I just can't figure it out. No clue at all. Can anyone help. Thanks.	2017-06-27 08:58:04	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.919344961643219, "perplexity": 5884.501705561708}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-26/segments/1498128321306.65/warc/CC-MAIN-20170627083142-20170627103142-00116.warc.gz"}
https://dml.cz/handle/10338.dmlcz/119017	# Article Full entry \| PDF (0.2 MB) Keywords: free Abelian group; countable compactness; products; initially $\omega_1$-compact; Martin's Axiom Summary: It was known that free Abelian groups do not admit a Hausdorff compact group topology. Tkachenko showed in 1990 that, under CH, a free Abelian group of size ${\frak C}$ admits a Hausdorff countably compact group topology. We show that no Hausdorff group topology on a free Abelian group makes its $\omega$-th power countably compact. In particular, a free Abelian group does not admit a Hausdorff $p$-compact nor a sequentially compact group topology. Under CH, we show that a free Abelian group does not admit a Hausdorff initially $\omega_1$-compact group topology. We also show that the existence of such a group topology is independent of ${\frak C} = \aleph_2$. References: [1] Comfort W.W.: Topological groups. Handbook of Set-Theoretic Topology (K. Kunen and J. Vaughan, eds.), North-Holland, 1984, pp.1143-1263. MR 0776643 \| Zbl 1071.54019 [2] Comfort W.W.: Problems on topological groups and other homogeneous spaces. Open Problems in Topology (J. van Mill and G. M. Reed, eds.), North-Holland, 1990, pp.311-347. MR 1078657 [3] Comfort W.W., Remus D.: Imposing pseudocompact group topologies on Abelian groups. Fundamenta Mathematica 142 (1993), 221-240. MR 1220550 \| Zbl 0865.54035 [4] Dikranjan D., Shakhmatov D.: Pseudocompact topologies on groups. Topology Proc. 17 (1992), 335-342. MR 1255816 \| Zbl 0795.22001 [5] van Douwen E.K.: The product of two countably compact topological groups. Trans. Amer. Math. Soc. 262 (1980), 417-427. MR 0586725 \| Zbl 0453.54006 [6] Engelking R.: General Topology. Heldermann Verlag, 1989. MR 1039321 \| Zbl 0684.54001 [7] Hart K.P., van Mill J.: A countably compact $H$ such that $H\times H$ is not countably compact. Trans. Amer. Math. Soc. 323 (1991), 811-821. MR 0982236 [8] Hajnal A., Juhász I.: A separable normal topological group need not be Lindelöf. General Topology Appl. 6 (1976), 199-205. MR 0431086 [9] Kunen K.: Set Theory. North Holland, 1980. MR 0597342 \| Zbl 0960.03033 [10] Robbie D., Svetlichny S.: An answer to A.D. Wallace's question about countably compact cancellative semigroups. Proc. Amer. Math. Soc. 124 (1996), 325-330. MR 1328373 \| Zbl 0843.22001 [11] Tkachenko M.G.: Countably compact and pseudocompact topologies on free Abelian groups. Izvestia VUZ. Matematika 34 (1990), 68-75. MR 1083312 \| Zbl 0714.22001 [12] Tomita A.H.: The Wallace Problem: a counterexample from $M A_{countable}$ and $p$-compactness. Canadian Math. Bull. 39 (1996), 4 486-498. MR 1426694 [13] Tomita A.H.: On finite powers of countably compact groups. Comment. Math. Univ. Carolinae 37 (1996), 3 617-626. MR 1426926 \| Zbl 0881.54022 [14] Tomita A.H.: A group under $M A_{countable}$ whose square is countably compact but whose cube is not. to appear in Topology Appl. [15] Tomita A.H.: Countable compactness and related properties in groups and semigroups: free Abelian groups and the Wallace Problem. Ph.D Thesis, York University, June 1995. [16] Vaughan J.: Countably compact and sequentially compact spaces. Handbook of Set-Theoretic Topology (K. Kunen and J. Vaughan, eds.), North-Holland, 1984, pp.569-602. MR 0776631 \| Zbl 0562.54031 [17] Wallace A.D.: The structure of topological semigroups. Bull. Amer. Math. Soc. 61 (1955), 95-112. MR 0067907 \| Zbl 0065.00802 [18] Weiss W.: Versions of Martin's Axiom. Handbook of Set-Theoretic Topology (K. Kunen and J. Vaughan, eds.), North-Holland, 1984, pp.827-886. MR 0776638 \| Zbl 0571.54005 Partner of	2017-12-17 14:05:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9494341015815735, "perplexity": 2041.8040049015392}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-51/segments/1512948596051.82/warc/CC-MAIN-20171217132751-20171217154751-00433.warc.gz"}
https://www.gradesaver.com/textbooks/math/precalculus/precalculus-concepts-through-functions-a-unit-circle-approach-to-trigonometry-3rd-edition/chapter-3-polynomial-and-rational-functions-section-3-1-polynomial-functions-and-models-3-1-assess-your-understanding-page-208/19	## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition) $\text{Not a polynomial function because it has a term with a negative exponent/power.}$ $\text{A polynomial function is a function of the form}$ $$f(x)=a_n x^n +a_{n-1} x^{n-1}+...+a_1 x+ a_0$$ Where the coefficients $(a_n, a_{n-1}.....)$ are real numbers and $n$ is a non-negative integer $f \text{ can be rewritten as: }$ $$f(x)=1-x^{-1}$$ Since the term $x^{-1}$ has a negative power, the function isn't a polynomial.	2021-10-18 23:43:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8936677575111389, "perplexity": 293.12853057715245}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-43/segments/1634323585215.14/warc/CC-MAIN-20211018221501-20211019011501-00673.warc.gz"}
http://mathhelpforum.com/calculus/112084-double-integral-help-needed-print.html	# Double Integral help needed • November 2nd 2009, 07:16 PM Diggidy Double Integral help needed Ok if i am given the double integral (sorry about the format i dont knw how to enter integral signs and what not... i typed it as it looks) Integral from 1-3 Integral from 0-lnx of 2x dydx How do i write an equivalent integral with the order of integration reversed. I know that in order to reverse the order you simply switch the bounded integrals then switch dydx to dxdy and then integrate accordingly. but in my notes it says that you have to integrate the non-constant part first and that is where i get confused. If i switched the bounded integrals then the one iwth the lnx bound would have to integrated second.. How do i get around this? Also just to make sure if i am doing these right could somone please check these solutions. inegral from 0-4 integral from 0-7 of (x+y)dxdy For this i got 98 and Integral from 1-5 Integral from 0-lnx of e^y dydx for this one i got 8 Any help is greatly appreciated. Thank you, Diggidy • November 2nd 2009, 07:41 PM Prove It Quote: Originally Posted by Diggidy Ok if i am given the double integral (sorry about the format i dont knw how to enter integral signs and what not... i typed it as it looks) Integral from 1-3 Integral from 0-lnx of 2x dydx How do i write an equivalent integral with the order of integration reversed. I know that in order to reverse the order you simply switch the bounded integrals then switch dydx to dxdy and then integrate accordingly. but in my notes it says that you have to integrate the non-constant part first and that is where i get confused. If i switched the bounded integrals then the one iwth the lnx bound would have to integrated second.. How do i get around this? Also just to make sure if i am doing these right could somone please check these solutions. inegral from 0-4 integral from 0-7 of (x+y)dxdy For this i got 98 and Integral from 1-5 Integral from 0-lnx of e^y dydx for this one i got 8 Any help is greatly appreciated. Thank you, Diggidy Is this what you're trying to write? $\int_{x = 1}^{x = 3}{\int_{y = 0}^{y = \ln{x}}{2x\,dy}\,dx}$? If so, look at the terminals - reversing the order of integration simply involves rearranging the resulting inequalities. It should be pretty obvious that $1 \leq x \leq 3$ and $0 \leq y \leq \ln{x}$ Taking the logarithm of everything in the first inequality gives $\ln{1} \leq \ln{x} \leq \ln{3}$ $0 \leq \ln{x} \leq \ln{3}$. Exponentiating everything in the second inequality gives $e^0 \leq e^y \leq e^{\ln{x}}$ $1 \leq e^y \leq x$ Therefore $e^y \leq x \leq 3$ and $0 \leq y \leq \ln{3}$. So the double integral becomes $\int_{x = 1}^{x = 3}{\int_{y = 0}^{y = \ln{x}}{2x\,dy}\,dx} = \int_{y = 0}^{y = \ln{3}}{\int_{x = e^y}^{x = 3}{2x\,dx}\,dy}$. • November 2nd 2009, 07:56 PM Prove It Quote: Originally Posted by Diggidy Also just to make sure if i am doing these right could somone please check these solutions. inegral from 0-4 integral from 0-7 of (x+y)dxdy For this i got 98 and Integral from 1-5 Integral from 0-lnx of e^y dydx for this one i got 8 I get 42 for the first. I get the same as you for the second.	2014-12-27 08:59:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9358890056610107, "perplexity": 1108.962401488188}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-52/segments/1419447551125.41/warc/CC-MAIN-20141224185911-00096-ip-10-231-17-201.ec2.internal.warc.gz"}
https://hurmanblirrikzpbv.firebaseapp.com/54787/91092.html	# Propanamine and N,N-dimethylmethanamine contain the same number of carbon atoms, even though Propanamine has higher boiling point than N,N-dimethylmethanamine. Why ? Post Answer. Answers (1) S Sumit. Propamine has higher boiling point than N,N diamethylamine because propamine is more associated than through H-bonding due to presence of 2 N-H b) The energy levels of atomic hydrogen are given by: En = −13.6 n2. eV, where n = 1,2,3,4, is the principal quantum number and each energy level has When this is done, the symbol N is used. For example, in referring to 1 mol of iodine atoms, we could write $$\text{n}_I = \text{1 mol} \text{ and } \text{N}_I = \text{6.022} \times \text{10}^{23}$$ Notice that N I is a pure number, rather than a quantity. To obtain such a pure number, we need a conversion factor which involves the number of 2020-3-27 · Needed: Number of atoms Equations: n = m / M n = # atoms/AC Solve for n in (1) and substitute it into (2) Substitute: (1) n = 0.551g (2) n = # atoms 2339.10g/mol 6.02x10 Molar Mass • The molar mass of a compound is found by adding together the molar masses of all of its elements, taking into account the number of moles of each element present The number of carbon atoms present in {eq}0.062 \ mol {/eq} acetic acid, {eq}HC_2H_3O_2 {/eq}, is a. {eq}7.5 \times 10^{22} {/eq} atoms b. Atoms are simply missing in the structure file provided to pdb2gmx; look for REMARK 465 and REMARK 470 entries in the .pdb file. The number of tetrahedral voids are formed if the number of atoms in a crystal is N/2. New questions in Chemistry every one please forgot me me nhi aa rahi yaha koi baat nhi krta 100 points khatam Number of atoms present in W g of a substance =MW×Na. Here, Na represents Avogadro's number. M represents the atomic mass of substance. A = Z + n° Example: 19K39 See below for the elements listed in Atomic Number Order or Name order. See also a copy of the 7, N, nitrogen, 14.006 43, 14.007 28. ## 2018-4-11 · The mathematical equation, N = n × N A, can be used to find the number of atoms, ions or molecules in any amount (in moles) of atoms, ions or molecules: 10 moles of helium atoms = 10 × (6.022 × 10 23) = 6.022 × 10 24 helium atoms. 10 moles of sodium … Anonymous. 10 years ago. 92.6 mg * 1 g / 10^3 mg = .0926 g.0926 g NH4NO2 The mole is the unit of measurement for amount of substance in the International System of Units. ### The name is of the form atomselect%d where '%d' is a non-negative number ( such as vmd> \$sel get {name backbone} {N 1} {H 0} {CA 1} {CB 0} {C 1} {O 1}. sv atomnummer n. en atomic number. de Atomnummer f. Resonant soft number n, which is associated with the energy level or the main shell the electron Enda Ni Atomen är direkt samordnas med en N atom. Revisor svärmor styrelseledamot [1,2]. By using the locality of N, n-disubstuffs alkenamider och phenylalkenamider, deras atoms with the ring nitrogen atoms and the substituent nitrogen atoms attached to the same carbon chain, which Publication number Priority date Publication date Assignee Title. Atomnumret, Z, skall inte förväxlas med masstalet A, som är antalet nukleoner, det totala antalet protoner och neutroner i en atoms kärna. Antalet neutroner, N av MP Savedoff · 1955 · Citerat av 48 — (20) \3crv/ 3 ` ° Here R0 is the radius of the Hii region for the density of the undisturbed gas, n~. Add to List. Given a chemical formula (given as a string), return the count of each atom. The atomic element always starts with an uppercase character, then zero or more lowercase letters, representing the name. Income per capita garantibelopp shb bendy toveks lastbilar uddevalla btb security services inc centre daily times nanny oil tanker ### 2013-03-15 · third, use the formula to calculate the number of atoms - atoms=mol x L. which L is a constant number of 6.02 x 10^23. 1) first step: Ar Na=23, Ar N=14 , Ar O=16. So the Mr of NaNO3 is= 23 + 14 + 3x16 = 85. second step: mol= grams/Mr = 0.72/85 = 0.0084. third step: number of atoms= mol x L = 0.0084 x 6.02 x 10^23 = 0.5 x 10^22 = 2.247 × 10 23 atoms of hydrogen. ∴ Number of N atoms in 5.6 g of urea = (2 × 0.0933) mol × 6.022 × 10 23 atoms/mol. The elements in the lower left hand corner of the periodic table have the largest atomic radius. Francium is in the absolute bottom left corner, so has the largest atomic radius. Hålla röd tråd anna wallander schauspielerin ### 2933 Heterocyclic compounds with nitrogen hetero-atom(s) only ring system have been substituted with the corresponding number of halogen atoms. Each atom of an element contains the same number of protons, which is the atomic number (Z). Neutral atoms have the same number of electrons and protons. Atoms of an element that contain different numbers of neutrons are called isotopes. ## So for the element of NITROGEN, you already know that the atomic number tells you the number of electrons. That means there are 7 electrons in a nitrogen Atomic number 7. 78% av jordens atmosfär betsår av kvävgas 600. H-N-H - H H NH3 C N Okay class, today we are going to be learning about H; C; The total number of valence electrons that will be used is 8. in the valence shell of the nitrogen atom, and 2 electrons for each hydrogen atom. N; av T Wallgren · 2019 · Citerat av 13 — Sweden has banned tail docking since 1988 and all pigs have intact tails, yet tail unless farmers are willing to reduce the actual number of sows in production Experimentally, the number of entities (atoms or molecules) in a mole is given by Avogadro's number: (Avogadro 1778-1856). 1. 95 nuclide i. Programmet genererar sedan en atommatris samt söker reda på och chemical reactions is 1 Name Symbol Atomic number Atomic weight (12 ) C=12 Do you want to give 1 linearly independent reaction/s ? n Reaction : 1 Stable nitrogen atoms that have the same atomic number as the element nitrogen, but differ in atomic weight. N-15 is a stable nitrogen isotope. av C Scheuner · 2017 · Citerat av 4 — Z 1 denotes the atomic number of the projectile ion which is 7 for nitrogen, Z 2 is in the planar case 8 for oxygen and n = 1/(7.28) Å−2 is the The degree of filtering of sputtered atoms because of the collimator depends on hν0 = quantum of energy; L = radiance; Lv = spectral radiance; N = number of (n – 2) f1 – 14(n – 1) d0 – 1 ns2.	2023-03-29 04:49:34	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7094910740852356, "perplexity": 4115.234459209489}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948932.75/warc/CC-MAIN-20230329023546-20230329053546-00458.warc.gz"}
https://papers.nips.cc/paper/1995/hash/818f4654ed39a1c147d1e51a00ffb4cb-Abstract.html	#### Authors Anders Krogh, Soren Riis #### Abstract Most current methods for prediction of protein secondary structure use a small window of the protein sequence to predict the structure of the central amino acid. We describe a new method for prediction of the non-local structure called ,8-sheet, which consists of two or more ,8-strands that are connected by hydrogen bonds. Since,8- strands are often widely separated in the protein chain, a network with two windows is introduced. After training on a set of proteins the network predicts the sheets well, but there are many false pos(cid:173) itives. By using a global energy function the ,8-sheet prediction is combined with a local prediction of the three secondary structures a-helix, ,8-strand and coil. The energy function is minimized using simulated annealing to give a final prediction.	2022-10-07 11:56:39	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8598373532295227, "perplexity": 1402.234132459318}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030338073.68/warc/CC-MAIN-20221007112411-20221007142411-00325.warc.gz"}
http://blog.logancyang.com/note/algo/2015/10/03/graphsearch.html	## Traits for using BFS: 1. Shortest path in a simple graph: given a initial state, a final state, a transition rule between states, ask how many (the least #) transitions from init to final. 2. Graph traversal. Only visit each node once. ## Traits for using DFS: (DFS method: build search tree, check conditions for recursing down a branch) 1. Enumerate subsets, permutations. 2. Find all possible solutions. Generally, when we do a search to enumerate cases using DFS recursion, there are 3 steps we should keep in mind, 1. Define the recursion. 2. Think about what to do in the base case. When should we return directly. 3. In general cases (other than base case), how to make the problem smaller and recurse down. ## Subsets I & II (DFS template) The thinking is to categorize cases by different head items. Enumerate the head item of a case (path) in a for loop in this way: 1. Append the item into the path. 2. DFS recurse down onto the next case (generally a smaller case, with advanced index and/or updated reference parameter). 3. Pop the item from the path, to iterate to a different head item on the next iteration. For this specific Subsets problem, the base case is just adding the current path. For Subsets II, the only difference is that the input can have duplicates and we don’t want the result subsets to be duplicate sets. Since the input is sorted (or we sort it by ourselves before DFS), when we encounter a number which is equal to the previous number in the for loop, we continue. Because the same number is taking the same place as the previous one, the resulting subsets with either of them are the same sets. ## Permutations I & II (DFS template) It’s quite similar to the Subsets problems. The thinking is also to categorize cases by different head items, and enumerate the head item of a case (path) in a for loop. The difference is that now we don’t want to keep track of the index as a parameter passed into DFS. Our base case is that when the path has the same length as the original input sequence, the current path is added. The for loop is now as such: 1. Append the item into the path. 2. DFS recurse down after appending the new head item. Avoid the same number by checking if it's already in path, if yes, continue. 3. Pop the item from the path, iterate to a different head item on the next iteration. For permutations II where we allow duplicates in the input list, we must sort it first and then do DFS. In the results, duplicate permutations must be avoided, but how? We introduce a new list, visited. We only add continuous same numbers to path, meaning if the previous same number is not visited, we continue. Check the code for details. ## Summary for Subsets and Permutations This kind of problems is not easy to understand. Recursion tree diagrams can help to clarify, but also keep in mind the code templates: inside the for loop, check condition to recurse down, then append in path, DFS down with path (with appropriate update in some parameter), pop from path. ### Subsets class Solution: """ @param S: The set of numbers. @return: A list of lists. See example. """ def subsets(self, S): if S is None or len(S) == 0: return [] S.sort() self.results = [] self.DFS([], 0, S) return self.results def DFS(self, path, ind, S): # base case, add each path of each recursion (sorted as required) # must make new list, list(path). If not, # res (path) points to the obj passed in, which is empty at the beginning res = list(path) self.results.append(res) # i is the first item's index in a path for i in xrange(ind, len(S)): path.append(S[i]) self.DFS(path, i+1, S) path.pop() ### Permutations: class Solution: """ @param nums: A list of Integers. @return: A list of permutations. """ def permute(self, nums): if nums is None or len(nums) == 0: return [] self.results = [] self.DFS([], nums) return self.results def DFS(self, path, nums): # base case if len(path) == len(nums): # must make new list, list(path). If not, # it points to the obj passed in, which is empty at # the beginning self.results.append(list(path)) return for i in xrange(len(nums)): # check if the ith number is already in path if nums[i] in path: continue path.append(nums[i]) self.DFS(path, nums) path.pop() Combination Sum is the Sum version of Subsets, with duplicates allowed. ### Palindrome Partitioning We deem the cuts to be the member of a subset, and this problem becomes finding all subsets of valid cuts. If there are N cuts, we can choose whether to include each cut, so there are 2^N ways to cut our string. For O(2^N) problems, it’s usually a Subsets problem. The thinking is that, we have a substring from start to i, s[start:i], called prefix. This is the next head item of the new node (path) in the DFS tree, we later append it to path, DFS down, and pop. But before that we should check if it is a valid palindrome. For a fixed start, we loop through all substrings starting there, check if it satisfies the condition (palindrome in this case), if yes DFS down starting at i (the next char after s[old_start:i]). Again the template of DFS: for i in range(old_start_ind, data.length): check if next_head satisfies our condition if not, continue DFS(path, i or i+1, data) # i or i+1 greater than old_start_ind path.pop() For "aab", we have start = 0: \|a\|ab, \|aa\|b, \|aab\|; start = 1: a\|a\|b, a\|ab\|; start = 2: aa\|b\|; start = 3 == len, aab\|, add one full path and return. start progresses by new recursion, i scans inside each recursion from start+1 to len+1. This is the method to enumerate all substrings that satisfies some condition. ### Factor Combinations For example, 8 -> [[2, 2, 2], [2, 4]]. Do not include 1 or n as a factor, and each sublist should be ascending. This is a typical DFS problem: list all solutions (combinations) or a certain decomposition problem, in this case, factorization. class Solution: def factors(self, n): self.result = [] if n <= 1: return self.result path = [] self.dfs(n, path) return self.result def dfs(self, n, path): # base case # len(path) == 1 is the case [n], which is not included # in this question if n == 1 and len(path) > 1: self.result.append(list(path)) return # recursion # factor must include n itself # or when it's down to the last factor, it's not added for factor in xrange(2, n+1): # check if it's a factor of not if n % factor != 0: continue # ensure ascending order if path == [] or path[-1] <= factor: path.append(factor) self.dfs(n/factor, path) path.pop() return 3 points that need special attention in this particular problem, • For n <= 3, result = [] • For the loop from 2 to n, must include n: for factor in xrange(2, n+1): .... Because if we don’t include n, the factors are not added to path. For example, If we use xrange(2, n)... Input: 8 Ex1: 1st level DFS: n = 8, factor = 2, dfs(n/2 = 4, [2]) 2nd level DFS: n = 4, factor = 2, dfs(n/2 = 2, [2, 2]) 3rd level DFS: n = 2, factor is in xrange(2, 2) which is nothing, abort Failed to add path [2, 2, 2] Ex2: 1st level DFS: n = 8, factor = 2, dfs(n/2 = 4, [2]) 2nd level DFS: n = 4, factor is in xrange(2, 4), factor cannot reach 4 Failed to add path [2, 4] • To ensure ascending order in paths, check for path == [] or path[-1] <= factor before append(factor). (Note that Python has short circuit evaluation in the conditionals. For or it means that if path == [], the latter part(s) won’t be checked. So there won’t be a case where path[-1] does not exist in this expression.)	2020-08-04 02:18:00	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.4701916575431824, "perplexity": 3645.8152702099937}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-34/segments/1596439735851.15/warc/CC-MAIN-20200804014340-20200804044340-00375.warc.gz"}
https://openstax.org/books/elementary-algebra-2e/pages/7-6-quadratic-equations	Elementary Algebra 2e ### Learning Objectives By the end of this section, you will be able to: • Solve quadratic equations by using the Zero Product Property • Solve applications modeled by quadratic equations ### Be Prepared 7.19 Before you get started, take this readiness quiz. Solve: $5y−3=05y−3=0$. If you missed this problem, review Example 2.27. ### Be Prepared 7.20 Solve: $10a=010a=0$. If you missed this problem, review Example 2.13. ### Be Prepared 7.21 Combine like terms: $12x2−6x+4x12x2−6x+4x$. If you missed this problem, review Example 1.24. ### Be Prepared 7.22 Factor $n3−9n2−22nn3−9n2−22n$ completely. If you missed this problem, review Example 7.32. We have already solved linear equations, equations of the form $ax+by=cax+by=c$. In linear equations, the variables have no exponents. Quadratic equations are equations in which the variable is squared. Listed below are some examples of quadratic equations: $x2+5x+6=03y2+4y=1064u2−81=0n(n+1)=42x2+5x+6=03y2+4y=1064u2−81=0n(n+1)=42$ The last equation doesn’t appear to have the variable squared, but when we simplify the expression on the left we will get $n2+nn2+n$. The general form of a quadratic equation is $ax2+bx+c=0,witha≠0ax2+bx+c=0,witha≠0$. An equation of the form $ax2+bx+c=0ax2+bx+c=0$ is called a quadratic equation. $a,b,andcare real numbers anda≠0a,b,andcare real numbers anda≠0$ To solve quadratic equations we need methods different than the ones we used in solving linear equations. We will look at one method here and then several others in a later chapter. ### Solve Quadratic Equations Using the Zero Product Property We will first solve some quadratic equations by using the Zero Product Property. The Zero Product Property says that if the product of two quantities is zero, it must be that at least one of the quantities is zero. The only way to get a product equal to zero is to multiply by zero itself. ### Zero Product Property If $a·b=0a·b=0$, then either $a=0a=0$ or $b=0b=0$ or both. We will now use the Zero Product Property, to solve a quadratic equation. ### Example 7.69 #### How to Use the Zero Product Property to Solve a Quadratic Equation Solve: $(x+1)(x−4)=0(x+1)(x−4)=0$. ### Try It 7.137 Solve: $(x−3)(x+5)=0(x−3)(x+5)=0$. ### Try It 7.138 Solve: $(y−6)(y+9)=0(y−6)(y+9)=0$. We usually will do a little more work than we did in this last example to solve the linear equations that result from using the Zero Product Property. ### Example 7.70 Solve: $(5n−2)(6n−1)=0(5n−2)(6n−1)=0$. ### Try It 7.139 Solve: $(3m−2)(2m+1)=0(3m−2)(2m+1)=0$. ### Try It 7.140 Solve: $(4p+3)(4p−3)=0(4p+3)(4p−3)=0$. Notice when we checked the solutions that each of them made just one factor equal to zero. But the product was zero for both solutions. ### Example 7.71 Solve: $3p(10p+7)=03p(10p+7)=0$. ### Try It 7.141 Solve: $2u(5u−1)=02u(5u−1)=0$. ### Try It 7.142 Solve: $w(2w+3)=0w(2w+3)=0$. It may appear that there is only one factor in the next example. Remember, however, that $(y−8)2(y−8)2$ means $(y−8)(y−8)(y−8)(y−8)$. ### Example 7.72 Solve: $(y−8)2=0(y−8)2=0$. ### Try It 7.143 Solve: $(x+1)2=0(x+1)2=0$. ### Try It 7.144 Solve: $(v−2)2=0(v−2)2=0$. ### Solve Quadratic Equations by Factoring Each of the equations we have solved in this section so far had one side in factored form. In order to use the Zero Product Property, the quadratic equation must be factored, with zero on one side. So we must be sure to start with the quadratic equation in standard form, $ax2+bx+c=0ax2+bx+c=0$. Then we can factor the expression on the left. ### Example 7.73 #### How to Solve a Quadratic Equation by Factoring Solve: $x2+2x−8=0x2+2x−8=0$. ### Try It 7.145 Solve: $x2−x−12=0x2−x−12=0$. ### Try It 7.146 Solve: $b2+9b+14=0b2+9b+14=0$. ### How To #### Solve a quadratic equation by factoring. 1. Step 1. Write the quadratic equation in standard form, $ax2+bx+c=0ax2+bx+c=0$. 2. Step 2. Factor the quadratic expression. 3. Step 3. Use the Zero Product Property. 4. Step 4. Solve the linear equations. 5. Step 5. Check. Before we factor, we must make sure the quadratic equation is in standard form. ### Example 7.74 Solve: $2y2=13y+452y2=13y+45$. ### Try It 7.147 Solve: $3c2=10c−83c2=10c−8$. ### Try It 7.148 Solve: $2d2−5d=32d2−5d=3$. ### Example 7.75 Solve: $5x2−13x=7x5x2−13x=7x$. ### Try It 7.149 Solve: $6a2+9a=3a6a2+9a=3a$. ### Try It 7.150 Solve: $45b2−2b=−17b45b2−2b=−17b$. Solving quadratic equations by factoring will make use of all the factoring techniques you have learned in this chapter! Do you recognize the special product pattern in the next example? ### Example 7.76 Solve: $144q2=25144q2=25$. ### Try It 7.151 Solve: $25p2=4925p2=49$. ### Try It 7.152 Solve: $36x2=12136x2=121$. The left side in the next example is factored, but the right side is not zero. In order to use the Zero Product Property, one side of the equation must be zero. We’ll multiply the factors and then write the equation in standard form. ### Example 7.77 Solve: $(3x−8)(x−1)=3x(3x−8)(x−1)=3x$. ### Try It 7.153 Solve: $(2m+1)(m+3)=12m(2m+1)(m+3)=12m$. ### Try It 7.154 Solve: $(k+1)(k−1)=8(k+1)(k−1)=8$. The Zero Product Property also applies to the product of three or more factors. If the product is zero, at least one of the factors must be zero. We can solve some equations of degree more than two by using the Zero Product Property, just like we solved quadratic equations. ### Example 7.78 Solve: $9m3+100m=60m29m3+100m=60m2$. ### Try It 7.155 Solve: $8x3=24x2−18x8x3=24x2−18x$. ### Try It 7.156 Solve: $16y2=32y3+2y16y2=32y3+2y$. When we factor the quadratic equation in the next example we will get three factors. However the first factor is a constant. We know that factor cannot equal 0. ### Example 7.79 Solve: $4x2=16x+844x2=16x+84$. ### Try It 7.157 Solve: $18a2−30=−33a18a2−30=−33a$. ### Try It 7.158 Solve: $123b=−6−60b2123b=−6−60b2$. ### Solve Applications Modeled by Quadratic Equations The problem solving strategy we used earlier for applications that translate to linear equations will work just as well for applications that translate to quadratic equations. We will copy the problem solving strategy here so we can use it for reference. ### How To #### Use a problem-solving strategy to solve word problems. 1. Step 1. Read the problem. Make sure all the words and ideas are understood. 2. Step 2. Identify what we are looking for. 3. Step 3. Name what we are looking for. Choose a variable to represent that quantity. 4. Step 4. Translate into an equation. It may be helpful to restate the problem in one sentence with all the important information. Then, translate the English sentence into an algebra equation. 5. Step 5. Solve the equation using good algebra techniques. 6. Step 6. Check the answer in the problem and make sure it makes sense. 7. Step 7. Answer the question with a complete sentence. ### Example 7.80 The product of two consecutive integers is 132. Find the integers. ### Try It 7.159 The product of two consecutive integers is $240240$. Find the integers. ### Try It 7.160 The product of two consecutive integers is $420420$. Find the integers. Were you surprised by the pair of negative integers that is one of the solutions to the previous example? The product of the two positive integers and the product of the two negative integers both give 132. In some applications, negative solutions will result from the algebra, but will not be realistic for the situation. ### Example 7.81 A rectangular garden has an area $1515$ square feet. The length of the garden is two feet more than the width. Find the length and width of the garden. ### Try It 7.161 A rectangular sign has an area of 30 square feet. The length of the sign is one foot more than the width. Find the length and width of the sign. ### Try It 7.162 A rectangular patio has an area of 180 square feet. The width of the patio is three feet less than the length. Find the length and width of the patio. In an earlier chapter, we used the Pythagorean Theorem $(a2+b2=c2)(a2+b2=c2)$. It gave the relation between the legs and the hypotenuse of a right triangle. We will use this formula to in the next example. ### Example 7.82 Justine wants to put a deck in the corner of her backyard in the shape of a right triangle, as shown below. The hypotenuse will be 17 feet long. The length of one side will be 7 feet less than the length of the other side. Find the lengths of the sides of the deck. ### Try It 7.163 A boat’s sail is a right triangle. The length of one side of the sail is 7 feet more than the other side. The hypotenuse is 13. Find the lengths of the two sides of the sail. ### Try It 7.164 A meditation garden is in the shape of a right triangle, with one leg 7 feet. The length of the hypotenuse is one more than the length of one of the other legs. Find the lengths of the hypotenuse and the other leg. ### Section 7.6 Exercises #### Practice Makes Perfect Use the Zero Product Property In the following exercises, solve. 315. $( x − 3 ) ( x + 7 ) = 0 ( x − 3 ) ( x + 7 ) = 0$ 316. $( y − 11 ) ( y + 1 ) = 0 ( y − 11 ) ( y + 1 ) = 0$ 317. $( 3 a − 10 ) ( 2 a − 7 ) = 0 ( 3 a − 10 ) ( 2 a − 7 ) = 0$ 318. $( 5 b + 1 ) ( 6 b + 1 ) = 0 ( 5 b + 1 ) ( 6 b + 1 ) = 0$ 319. $6 m ( 12 m − 5 ) = 0 6 m ( 12 m − 5 ) = 0$ 320. $2 x ( 6 x − 3 ) = 0 2 x ( 6 x − 3 ) = 0$ 321. $( y − 3 ) 2 = 0 ( y − 3 ) 2 = 0$ 322. $( b + 10 ) 2 = 0 ( b + 10 ) 2 = 0$ 323. $( 2 x − 1 ) 2 = 0 ( 2 x − 1 ) 2 = 0$ 324. $( 3 y + 5 ) 2 = 0 ( 3 y + 5 ) 2 = 0$ In the following exercises, solve. 325. $x 2 + 7 x + 12 = 0 x 2 + 7 x + 12 = 0$ 326. $y 2 − 8 y + 15 = 0 y 2 − 8 y + 15 = 0$ 327. $5 a 2 − 26 a = 24 5 a 2 − 26 a = 24$ 328. $4 b 2 + 7 b = −3 4 b 2 + 7 b = −3$ 329. $4 m 2 = 17 m − 15 4 m 2 = 17 m − 15$ 330. $n 2 = 5 n − 6 n 2 = 5 n − 6$ 331. $7 a 2 + 14 a = 7 a 7 a 2 + 14 a = 7 a$ 332. $12 b 2 − 15 b = −9 b 12 b 2 − 15 b = −9 b$ 333. $49 m 2 = 144 49 m 2 = 144$ 334. $625 = x 2 625 = x 2$ 335. $( y − 3 ) ( y + 2 ) = 4 y ( y − 3 ) ( y + 2 ) = 4 y$ 336. $( p − 5 ) ( p + 3 ) = −7 ( p − 5 ) ( p + 3 ) = −7$ 337. $( 2 x + 1 ) ( x − 3 ) = −4 x ( 2 x + 1 ) ( x − 3 ) = −4 x$ 338. $( x + 6 ) ( x − 3 ) = −8 ( x + 6 ) ( x − 3 ) = −8$ 339. $16 p 3 = 24 p 2 - 9 p 16 p 3 = 24 p 2 - 9 p$ 340. $m 3 − 2 m 2 = − m m 3 − 2 m 2 = − m$ 341. $20 x 2 − 60 x = −45 20 x 2 − 60 x = −45$ 342. $3 y 2 − 18 y = −27 3 y 2 − 18 y = −27$ Solve Applications Modeled by Quadratic Equations In the following exercises, solve. 343. The product of two consecutive integers is 56. Find the integers. 344. The product of two consecutive integers is 42. Find the integers. 345. The area of a rectangular carpet is 28 square feet. The length is three feet more than the width. Find the length and the width of the carpet. 346. A rectangular retaining wall has area 15 square feet. The height of the wall is two feet less than its length. Find the height and the length of the wall. 347. A pennant is shaped like a right triangle, with hypotenuse 10 feet. The length of one side of the pennant is two feet longer than the length of the other side. Find the length of the two sides of the pennant. 348. A reflecting pool is shaped like a right triangle, with one leg along the wall of a building. The hypotenuse is 9 feet longer than the side along the building. The third side is 7 feet longer than the side along the building. Find the lengths of all three sides of the reflecting pool. Mixed Practice In the following exercises, solve. 349. $( x + 8 ) ( x − 3 ) = 0 ( x + 8 ) ( x − 3 ) = 0$ 350. $( 3 y − 5 ) ( y + 7 ) = 0 ( 3 y − 5 ) ( y + 7 ) = 0$ 351. $p 2 + 12 p + 11 = 0 p 2 + 12 p + 11 = 0$ 352. $q 2 − 12 q − 13 = 0 q 2 − 12 q − 13 = 0$ 353. $m 2 = 6 m + 16 m 2 = 6 m + 16$ 354. $4 n 2 + 19 n = 5 4 n 2 + 19 n = 5$ 355. $a 3 − a 2 − 42 a = 0 a 3 − a 2 − 42 a = 0$ 356. $4 b 2 − 60 b + 224 = 0 4 b 2 − 60 b + 224 = 0$ 357. The product of two consecutive integers is 110. Find the integers. 358. The length of one leg of a right triangle is three feet more than the other leg. If the hypotenuse is 15 feet, find the lengths of the two legs. #### Everyday Math 359. Area of a patio If each side of a square patio is increased by 4 feet, the area of the patio would be 196 square feet. Solve the equation $(s+4)2=196(s+4)2=196$ for s to find the length of a side of the patio. 360. Watermelon drop A watermelon is dropped from the tenth story of a building. Solve the equation $−16t2+144=0−16t2+144=0$ for $tt$ to find the number of seconds it takes the watermelon to reach the ground. #### Writing Exercises 361. Explain how you solve a quadratic equation. How many answers do you expect to get for a quadratic equation? 362. Give an example of a quadratic equation that has a GCF and none of the solutions to the equation is zero. #### Self Check After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section. Overall, after looking at the checklist, do you think you are well-prepared for the next section? Why or why not? Do you know how you learn best? Order a print copy As an Amazon Associate we earn from qualifying purchases.	2022-09-29 10:35:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 184, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6843640804290771, "perplexity": 577.23765340372}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-40/segments/1664030335350.36/warc/CC-MAIN-20220929100506-20220929130506-00152.warc.gz"}
https://stats.stackexchange.com/questions/139290/a-psychology-journal-banned-p-values-and-confidence-intervals-is-it-indeed-wise	# A psychology journal banned p-values and confidence intervals; is it indeed wise to stop using them? On 25 February 2015, the journal Basic and Applied Social Psychology issued an editorial banning $p$-values and confidence intervals from all future papers. Specifically, they say (formatting and emphasis are mine): • [...] prior to publication, authors will have to remove all vestiges of the NHSTP [null hypothesis significance testing procedure] ($p$-values, $t$-values, $F$-values, statements about ‘‘significant’’ differences or lack thereof, and so on). • Analogous to how the NHSTP fails to provide the probability of the null hypothesis, which is needed to provide a strong case for rejecting it, confidence intervals do not provide a strong case for concluding that the population parameter of interest is likely to be within the stated interval. Therefore, confidence intervals also are banned from BASP. • [...] with respect to Bayesian procedures, we reserve the right to make case-by-case judgments, and thus Bayesian procedures are neither required nor banned from BASP. • [...] Are any inferential statistical procedures required? -- No [...] However, BASP will require strong descriptive statistics, including effect sizes. Let us not discuss problems with and misuse of $p$-values here; there already are plenty of excellent discussions on CV that can be found by browsing the p-value tag. The critique of $p$-values often goes together with an advice to report confidence intervals for parameters of interest. For example, in this very well-argued answer @gung suggests to report effect sizes with confidence intervals around them. But this journal bans confidence intervals as well. What are the advantages and disadvantages of such an approach to presenting data and experimental results as opposed to the "traditional" approach with $p$-values, confidence intervals, and significant/insignificant dichotomy? The reaction to this ban seems to be mostly negative; so what are the disadvantages then? American Statistical Association has even posted a brief discouraging comment on this ban, saying that "this policy may have its own negative consequences". What could these negative consequences be? Or as @whuber suggested to put it, should this approach be advocated generally as a paradigm of quantitative research? And if not, why not? PS. Note that my question is not about the ban itself; it is about the suggested approach. I am not asking about frequentist vs. Bayesian inference either. The Editorial is pretty negative about Bayesian methods too; so it is essentially about using statistics vs. not using statistics at all. Other discussions: reddit, Gelman. • There is a one-to-one mapping between p-values and confidence intervals in linear regression models, so I don't see a strong reason why banning p-values but keeping confidence intervals would make much sense. But banning both p-values and confidence intervals leaves a gap in description of results... I wonder if they allow reporting standard errors (that would be another measure of the same one-to-one mapping group). – Richard Hardy Feb 25 '15 at 19:07 • Everything could be misused so banning stuff on this condition is, well... strange. I am not the fan of p-values but this seems as a pretty naive approach to the problem. One thing is encouraging to use proper stuff, but banning things does not sound like a proper way to deal with the problem... – Tim Feb 25 '15 at 21:23 • Great idea. Using statistics just hides the unscientific nature of this field. – Aksakal Feb 25 '15 at 21:35 • This seems like a complete overreaction to the frustration over the misuse of p values. I would be much happier with a ban on the misuse of p values rather than P values in general. – TrynnaDoStat Feb 26 '15 at 1:17 • The 4th item in your list suggests they're not requiring point estimates, which would be inference, but effect sizes reported merely as descriptive statistics. (Nevertheless, a few lines down in the editorial, "we encourage the use of larger sample sizes than is typical in much psychology research, because as the sample size increases, descriptive statistics become increasingly stable and sampling error is less of a problem". I look forward to the 2016 editorial's calling for research into formalizing this notion of stability & accounting quantitatively for the effects of sampling error.) – Scortchi Feb 26 '15 at 18:55 The first sentence of the current 2015 editorial to which the OP links, reads: The Basic and Applied Social Psychology (BASP) 2014 Editorial emphasized that the null hypothesis significance testing procedure (NHSTP) is invalid... (my emphasis) In other words, for the editors it is an already proven scientific fact that "null hypothesis significance testing" is invalid, and the 2014 editorial only emphasized so, while the current 2015 editorial just implements this fact. The misuse (even maliciously so) of NHSTP is indeed well discussed and documented. And it is not unheard of in human history that "things get banned" because it has been found that after all said and done, they were misused more than put to good use (but shouldn't we statistically test that?). It can be a "second-best" solution, to cut what on average (inferential statistics) has come to losses, rather than gains, and so we predict (inferential statistics) that it will be detrimental also in the future. But the zeal revealed behind the wording of the above first sentence, makes this look -exactly, as a zealot approach rather than a cool-headed decision to cut the hand that tends to steal rather than offer. If one reads the one-year older editorial mentioned in the above quote (DOI:10.1080/01973533.2014.865505), one will see that this is only part of a re-hauling of the Journal's policies by a new Editor. Scrolling down the editorial, they write ...On the contrary, we believe that the p<.05 bar is too easy to pass and sometimes serves as an excuse for lower quality research. So it appears that their conclusion related to their discipline is that null-hypotheses are rejected "too-often", and so alleged findings may acquire spurious statistical significance. This is not the same argument as the "invalid" dictum in the first sentence. So, to answer to the question, it is obvious that for the editors of the journal, their decision is not only wise but already late in being implemented: they appear to think that they cut out what part of statistics has become harmful, keeping the beneficial parts -they don't seem to believe that there is anything here that needs replacing with something "equivalent". Epistemologically, this is an instance where scholars of a social science partially retract back from an attempt to make their discipline more objective in its methods and results by using quantitative methods, because they have arrived at the conclusion (how?) that, in the end, the attempt created "more bad than good". I would say that this is a very important matter, in principle possible to have happened, and one that would require years of work to demonstrate it "beyond reasonable doubt" and really help your discipline. But just one or two editorials and papers published will most probably (inferential statistics) just ignite a civil war. The final sentence of the 2015 editorial reads: We hope and anticipate that banning the NHSTP will have the effect of increasing the quality of submitted manuscripts by liberating authors from the stultified structure of NHSTP thinking thereby eliminating an important obstacle to creative thinking. The NHSTP has dominated psychology for decades; we hope that by instituting the first NHSTP ban, we demonstrate that psychology does not need the crutch of the NHSTP, and that other journals follow suit. • Yes...we have to be careful when writing tongue-in-cheek or sardonic replies on this site: they might be (completely) misunderstood! – whuber Feb 25 '15 at 21:53 • @naught101 ...that wouldn't be very diplomatic. Notice that the way the NHSTP is condemned, it spares the psychologists themselves that they have used it in all these decades. If it was written the way you propose, it would look much more like a direct attack on their colleagues as scientists. As it now stands essentially the text implies that psychologists full of good intentions have been unfortunately misled in using the approach, by "someone", which misused his "power of scientific authority" in the matter... Perhaps by evil statisticians driven by scientific imperialism? – Alecos Papadopoulos Feb 26 '15 at 1:09 • A bad workman blames his tools. – naught101 Feb 26 '15 at 1:39 • @BrianDHall I would suggest to look up more authoritative resources on the issues surrounding NHSTP (this site included), rather than the specific author's works on the issue. The matter is difficult and subtle -already from your comment one should discuss first the semantics around "accept" and "assert"... – Alecos Papadopoulos Feb 26 '15 at 3:51 • @naught101: If you notice that the workman can't handle the chainsaw properly, you might not blame the tool. But you would still take it away from the workman, to prevent further harm ;-) – nikie Feb 26 '15 at 11:33 I feel that banning hypothesis tests is a great idea except for a select few "existence" hypotheses, e.g. testing the null hypothesis that there is not extra-sensory perception where all one would need to demonstrate to have evidence that ESP exists is non-randomness. But I think the journal missed the point that the main driver of poor research in psychology is the use of a threshold on $P$-values. It has been demonstrated in psychology and most other fields that a good deal of gaming goes on to arrive at $P < 0.05$. This includes hypothesis substitution, removing of observations, and subsetting data. It is thresholds that should be banned first. The banning of confidence intervals is also overboard, but not for the reasons others have stated. Confidence intervals are useful only if one misinterprets them as Bayesian credible intervals (for suitable non-information priors). But they are still useful. The fact that their exact frequentist interpretation leads to nothing but confusion implies that we need to "get out of Dodge" and go Bayesian or likelihood school. But useful results can be obtained by misinterpreting good old confidence limits. It is a shame that the editors of the journal misunderstood Bayesian statistics and don't know of the existence of pure likelihood inference. What they are seeking can be easily provided by Bayesian posterior distributions using slightly skeptical priors. • +1, thanks. Let me clarify regarding confidence intervals. Confidence intervals are related to standard errors, so the suggestion is probably to stop using those as well. Let's consider the simplest case: some value is measured across a group of $n$ subjects/objects; let's say the mean is 3. As far as I understand this journal suggests to report it simply as 3. But would you not want to see standard error as well, e.g. $3 \pm 0.5$? This of course means that 95% confidence interval is $3 \pm 1$, which also means that $p<0.05$, so it's all related. I am not sure how you suggest to report it. – amoeba Feb 27 '15 at 12:00 • I think of standard errors are oversimplified (because they assume symmetric distributions) but useful measures of precision, like mean squared error. You can think of a precision interval based on root mean squared error without envisioning probability coverage. So I don't see where any of this discussion implies de-emphasis of standard errors. And I wasn't suggesting that we stop using CLs. But the difficulty with CLs comes mainly from attempts at probability interpretations. – Frank Harrell Feb 27 '15 at 12:48 • Hmmm. Interesting. To me it seems like there is such a small step from standard error to CI (a constant factor!), that treating them differently would be weird. But perhaps it is a semantic point; I guess what you mean is that people think about standard errors and CIs differently and tend to get more confused about CIs. I wonder what this particular journal policy says about standard errors (the Editorial doesn't mention them explicitly). – amoeba Feb 27 '15 at 22:26 • In symmetric situations, the standard error is a building block for a confidence interval. But in many cases the correct confidence interval is asymmetric so can't be based on a standard error at all. Some varieties of the bootstrap and back-transforming are two approaches of this type. Profile likelihood confidence intervals especially come to mind here. – Frank Harrell Feb 28 '15 at 12:11 • @Frank Harrell - As for "pure likelihood inference" I agree that an emphasis toward summarization of the data's likelihood without embellishing it with thresholds appears to be the answer the editors were grasping for. A. W. F. Edwards' book "Likelihood" (1972) speaks directly to the editor's concern: "We may defer consideration of these arguments (e.g. significance testing) until later chapters, and pass immediately to the description of a procedure, based on Fisher's concept of Likelihood, which is open to none of these objects which may be levelled at significance tests." – John Mark Apr 20 '15 at 15:43 I see this approach as an attempt to address the inability of social psychology to replicate many previously published 'significant findings.' 1. that it doesn't address many of the factors leading to spurious effects. E.g., • A) People can still peek at their data and stop running their studies when an effect size strikes them as being sufficiently large to be of interest. • B) Large effects sizes will still appear to have large power in retrospective assessments of power. • C) People will still fish for interesting and big effects (testing a bunch of hypotheses in an experiment and then reporting the one that popped up) or • D) pretend that an unexpected weird effect was expected all along. 2. As a field going forwards it will make a review of past findings pretty awful. There is no way to quantitatively assess the believability of different studies. If every journal implemented this approach, you'll have a bunch of social scientists saying there is evidence for X when it is totally unclear how believable X is and scientists arguing about how to interpret a published effect or arguing about whether it is important or worth talking about. Isn't this the point of having stats? To provide a consistent way to assess numbers. In my opinion, this new approach would cause a mess if it was widely implemented. 3. This change does not encourage researchers to submit the results of studies with small effect sizes so it doesn't really address the file-drawer effect (or are they going to publish findings with large n's regardless of effect size?). If we published all results of carefully designed studies, then even though the believability of results of individual studies may be uncertain, meta-analyses and reviews of studies that supplied statistical analysis would do a much better job at identifying the truth. • @captain_ahab Regarding point 3, we must mention that the previous editorial (2014) of the Editor explicitly encouraged the submission of "null-effect" studies. – Alecos Papadopoulos Feb 27 '15 at 11:48 • I can't seem to find a comment in the editorial discussing any criteria for publication except for the need have larger sample sizes than normal (how they are planning on identifying acceptable n's without inferential statistics is unclear to me). To me there is no emphasis in this editorial that they don't care what the effect size is. It seems to me that they will still be looking for interesting effects and interesting stories, which I think is the bigger problem in social science work (i.e., the post-hoc search for interesting effects and stories). – captain_ahab Feb 27 '15 at 19:34 • What seems like a better solution is that all scientists must log the hypothesis, basic rational, power of and analytics approach of a study in a PUBLIC place BEFORE running the study. And then being limited to publishing that study in the prescribe manner. If an unexpected interesting effect is found, they should publicly log then run a new study that examines that effect. This approach while controlling for false positives would also enable scientists to demonstrate their productivity without publishing new effects. – captain_ahab Feb 27 '15 at 19:39 I came across a wonderful quote that almost argues for the same point, but not quite -- since it is an opening paragraph in a textbook that is mostly about frequentist statistics and hypothesis testing. It is widely held by non-statisticians, like the author, that if you do good experiments statistics are not necessary. They are quite right. [...] The snag, of course, is that doing good experiments is difficult. Most people need all the help they can get to prevent them making fools of themselves by claiming that their favourite theory is substantiated by observations that do nothing of the sort. And the main function of that section of statistics that deals with tests of significance is to prevent people making fools of themselves. From this point of view, the function of significance tests is to prevent people publishing experiments, not to encourage them. Ideally, indeed, significance tests should never appear in print, having been used, if at all, in the preliminary stages to detect inadequate experiments, so that the final experiments are so clear that no justification is needed. -- David Colquhoun, Lectures on biostatistics, 1971 • Your post is really a comment, rather than an answer, so I am refraining from upvoting it, but I do wish to thank you for sharing the quotation. There are so many misunderstandings evident in this passage that it would take extensive effort (not to say space) to point out and debunk them all. In one word, though, the counter to these assertions is "efficiency." If everybody had unlimited time and budget we could at least aspire to perform "good experiments." But when resources are limited, it would be foolhardy (as well as costly) to conduct only "final, ... clear" experiments. – whuber Apr 8 '15 at 17:11 • Thanks for your comment, @whuber; I agree with what you are saying. Still, I must add that I do find it appealing to say that ideally experimental data should be so convincing as to render formal hypothesis tests redundant. This is not an unattainable ideal! In my field (where p-values are used a lot), I find that the best papers are convincing without them: e.g. because they present a sequence of several experiments supporting each other, which taken together, obviously cannot be a statistical fluke. Re comment: it was too long for a comment, and I figured it's okay as a CW answer. – amoeba Apr 8 '15 at 17:26 • Yes, I understand why it had to be posted as an answer, and therefore did not vote to move it into a comment (which would cut off the last part of the quote). I agree that the ideal is not unattainable in particular cases. I also agree it's a nice ideal to bear in mind. But as a guide to how to design experiments (which is, overall, a discipline of allocating resources), it could be a terrible mistake. (This is certainly debatable.) The suggestion that a "good" experiment would never require statistical methods is, however, one that does not stand up even to cursory examination. – whuber Apr 8 '15 at 18:52 • Perhaps one way of reading that is as saying the initial significance test that suggested a substance stimulates a certain physiological response is no longer relevant by the time you're publishing your investigations into the effects of different kinds of inhibitors on the dose-response curve. – Scortchi Apr 27 '16 at 9:16 ## protected by whuber♦Feb 26 '15 at 17:53 Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).	2019-04-23 22:07:27	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.5154492259025574, "perplexity": 1270.995387818521}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-18/segments/1555578613888.70/warc/CC-MAIN-20190423214818-20190423235832-00003.warc.gz"}
http://lists.gnu.org/archive/html/lilypond-user/2012-03/msg00437.html	lilypond-user [Top][All Lists] ## Re: separating design from pure score From: Hans Aikema Subject: Re: separating design from pure score Date: Wed, 14 Mar 2012 21:31:38 +0100 User-agent: Mozilla/5.0 (Windows NT 6.1; WOW64; rv:11.0) Gecko/20120312 Thunderbird/11.0 On 14-3-2012 19:16, Stjepan Horvat wrote: Hello, I'm working on a project and i want to separate the design from score because i want to have a clear score so that in the future when i maybe want to change the design i dont have to change the score (something like html and CSS :) ). So when i include my score into my design file variables are included but score block also. Now when i what to change, for example, \mark \default to have a preety box i dont know where to put it. Is there a way not to include a score block but all variables, or insert stuff into a score block that is invisible to me? If I understand your request correctly you're looking for a similar structure as what I'm working with for generating beamerslides of songs including their music. In that case the \include command is your friend. What I have in my projects: format/BaseBeamerslideDesign.ly contains - a \paper section containing lots of formatting settings - a \layout setting containing lots of formatting settings - an include for yet another lilypond file containing useful scheme functions such as the override-color-for-allgrobs scheme funtion I found in the LSR: http://lsr.dsi.unimi.it/LSR/Item?id=443 - variable definitions for a music staff (muziekbalk) and its associated lyrics (verstekst): muziekbalk = { \new Staff \new Voice = myMelody { \melodie } } verstekst = {\new Lyrics \lyricsto "myMelody" { \tekst }} format/BeamerRegular.ly contains - an include of format\BaseBeamerslideDesign.ly - a Score block combining the variables defined in BaseBeamerSlideDesign into an actual Score and override of the grob-colors to get colored output: === format/BeamerRegular.ly sample === \include "BaseBeamerslideDesign.ly" \score { << \applyContext #(override-color-for-all-grobs white) \muziekbalk \verstekst >> } === end of format/BeamerRegular.ly === Then any song requiring the beamer slide format: - a variable definition called 'melodie' (which is refered to in the format\BaseBeamerslideDesign.ly mentioned above) containing the melody of the song - a variable definition called 'tekst' which is refered to in the format\BaseBeamerslideDesign.ly mentioned above) containing the lyrics of the song - an include of the format/BeamerRegular.ly e.g.: === MySong.ly sample === \version "2.15.26" title = "Just some sample song" } melodie = \relative c'' { \clef treble \time 4/4 \key c \major c d e f \bar "\|." } tekst = \lyricmode { this is a song } \include format/BeamerGewoon.ly === end of MySong.ly sample === The reason for the BeamerRegular.ly is that I also have separate files for BeamerNoBars, BeamerNoTimeSignature, BeamerNoTimeSignatureNoBars for different score-tweaks on different songs. All basic formatting is done in BaseBeamerslideDesign, the tweaks for outputting different score types for different subsets of my songs are in these additional lilypond format/*.ly files. The songs than include the appropriate 'tweaked' design from the format folder. Any design change (e.g. font-size, fonts to use, paper-sizing) I can do on BaseBeamerslideDesign and then I can just regenerate all the songs of my library to get updated sheet-music using the new layout. This 'layering' of definitions can be extended as much as one wants of course, such as for different score setups (I only have need for a single-voice score with accompanying lyrics), someone else might find a need to define eg a set of SATB choral score designs including the baseDesign for the basic layout settings. Which might lead to for example: - baseDesign.ly (font setup, paper setup, staff-sizing setup) - SATBScoreBase.ly (variable setup for the 4 voices and accompanying lyrics) - SATBScore-S-A-T-B-SLyrics.ly including the SATBScoreBase and containing a score-block with each voice on its own staff and only lyrics accompanying the Soprano voice - SATBScore-SA-TB-STLyrics.ly including the SATBScoreBase and containg a score-block with the voices two-by-two on staffs and lyrics accompanying the Soprano and Tenor voices - SATBScore-S-A-T-B-LyricsAll.ly including the SATBScoreBase and containing a score-block with each voice on its own staff and lyrics accompanying each of the voices regards, Hans Aikema	2018-01-21 01:47:15	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.664409875869751, "perplexity": 10780.319043675403}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-05/segments/1516084889798.67/warc/CC-MAIN-20180121001412-20180121021412-00044.warc.gz"}
http://mathhelpforum.com/calculus/27885-fourier-sine-series.html	# Math Help - Fourier sine series... 1. ## Fourier sine series... Let f(x) = cos x on 0<= x <= pi. Calculate the Fourier sine series representation of f(x). Can anybody possibly start me off on this question or give me any help at all as I don't understand my lecture notes on this topic at all 2. Originally Posted by hunkydory19 Let f(x) = cos x on 0<= x <= pi. Calculate the Fourier sine series representation of f(x). Can anybody possibly start me off on this question or give me any help at all as I don't understand my lecture notes on this topic at all What do you mean by fourier sine? is it the sine part of the Fourier Series? if does, this should be a good hint: If $f(x)$ is an even function, then $b_n = 0$, i.e $f(x) \approx \frac{a_0}{2} + \sum a_n \, \cos nx$ If $f(x)$ is an odd function, then $a_n = 0$, i.e $f(x) \approx \frac{a_0}{2} + \sum b_n \, \sin nx$ 3. Thanks kalagota, So $a_n = 0$ and $b_n = \frac{2}{\pi}\int^\pi_0 cos x sin nx \, \mathrm{d}x$ But how do integrate this, I tried doing it by parts but realised that it will just go on forever! 4. You can turn a product of sine and cosine into a sum, do you know that formula? Applyin' that the rest follows. 5. You need to know the orthogonal properties of sine and cosine. (If L>0 we have): $\int_{-L}^L \sin \frac{\pi n x}{L} \sin \frac{\pi m x}{L} ~ dx = 0\mbox{ if }n\not =m , \ =L \mbox{ if }n=m$. $\int_{-L}^L \cos \frac{\pi n x}{L} \cos \frac{\pi m x}{L} ~ dx= \mbox{ same as above}$. $\int_{-L}^L \sin \frac{\pi n x}{L}\cos \frac{\pi m x}{L} ~ dx = 0$. Here $L=\pi$. 6. Originally Posted by ThePerfectHacker You need to know the orthogonal properties of sine and cosine. (If L>0 we have): $\int_{-L}^L \sin \frac{\pi n x}{L} \sin \frac{\pi m x}{L} ~ dx = 0\mbox{ if }n\not =m , \ =L \mbox{ if }n=m$. $\int_{-L}^L \cos \frac{\pi n x}{L} \cos \frac{\pi m x}{L} ~ dx= \mbox{ same as above}$. $\int_{-L}^L \sin \frac{\pi n x}{L}\cos \frac{\pi m x}{L} ~ dx = 0$. Here $L=\pi$. indeed! and the first two are 0 if $m \ne n$ you know i didn't know about these rules until after my differential equations class was over! i always had to do it the hard way (the product to sum formulas)	2014-10-31 15:09:05	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 17, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8643373250961304, "perplexity": 505.52285219342014}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-42/segments/1414637900024.23/warc/CC-MAIN-20141030025820-00170-ip-10-16-133-185.ec2.internal.warc.gz"}
https://math.stackexchange.com/questions/2727184/about-bounded-subspace-of-a-topological-vector-space	# About bounded subspace of a topological vector space A topological vector space $X$ is a vector space over a topological field $K$ (most often the real or complex numbers with their standard topologies) that is endowed with a topology such that vector addition $X \times X \to X$ and scalar multiplication $K \times X \to X$ are continuous functions (where the domains of these functions are endowed with product topologies). Now I want to show that in a topological vector space, the only bounded subspace is $\{0\}$. • What definition of bounded subspace used? [your question didn't specify a metric on $X$] – coffeemath Apr 8 '18 at 3:52	2019-05-23 17:20:07	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9274909496307373, "perplexity": 149.54058102462383}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-22/segments/1558232257316.10/warc/CC-MAIN-20190523164007-20190523190007-00090.warc.gz"}
https://www.fightfinance.com/?q=168,181,304,360,362,386,422,620,880,912	# Fight Finance #### CoursesTagsRandomAllRecentScores A four year bond has a face value of $100, a yield of 6% and a fixed coupon rate of 12%, paid semi-annually. What is its price? A stock pays annual dividends. It just paid a dividend of$5. The growth rate in the dividend is 1% pa. You estimate that the stock's required return is 8% pa. Both the discount rate and growth rate are given as effective annual rates. Using the dividend discount model, what will be the share price? Which one of the following is NOT usually considered an 'investable' asset for long-term wealth creation? Find Ching-A-Lings Corporation's Cash Flow From Assets (CFFA), also known as Free Cash Flow to the Firm (FCFF), over the year ending 30th June 2013. Ching-A-Lings Corp Income Statement for year ending 30th June 2013 $m Sales 100 COGS 20 Depreciation 20 Rent expense 11 Interest expense 19 Taxable Income 30 Taxes at 30% 9 Net income 21 Ching-A-Lings Corp Balance Sheet as at 30th June 2013 2012$m $m Inventory 49 38 Trade debtors 14 2 Rent paid in advance 5 5 PPE 400 400 Total assets 468 445 Trade creditors 4 10 Bond liabilities 200 190 Contributed equity 145 145 Retained profits 119 100 Total L and OE 468 445 Note: All figures are given in millions of dollars ($m). The cash flow from assets was: Three years ago Frederika bought a house for $400,000. Now it's worth$600,000, based on recent similar sales in the area. Frederika's residential property has an expected total return of 7% pa. She rents her house out for $2,500 per month, paid in advance. Every 12 months she plans to increase the rental payments. The present value of 12 months of rental payments is$29,089.48. The future value of 12 months of rental payments one year ahead is $31,125.74. What is the expected annual capital yield of the property? A risky firm will last for one period only (t=0 to 1), then it will be liquidated. So it's assets will be sold and the debt holders and equity holders will be paid out in that order. The firm has the following quantities: $V$ = Market value of assets. $E$ = Market value of (levered) equity. $D$ = Market value of zero coupon bonds. $F_1$ = Total face value of zero coupon bonds which is promised to be paid in one year. The risky corporate debt graph above contains bold labels a to e. Which of the following statements about those labels is NOT correct? Acquirer firm plans to launch a takeover of Target firm. The firms operate in different industries and the CEO's rationale for the merger is to increase diversification and thereby decrease risk. The deal is not expected to create any synergies. An 80% scrip and 20% cash offer will be made that pays the fair price for the target's shares. The cash will be paid out of the firms' cash holdings, no new debt or equity will be raised. Firms Involved in the Takeover Acquirer Target Assets ($m) 6,000 700 Debt ($m) 4,800 400 Share price ($) 40 20 Number of shares (m) 30 15 Ignore transaction costs and fees. Assume that the firms' debt and equity are fairly priced, and that each firms' debts' risk, yield and values remain constant. The acquisition is planned to occur immediately, so ignore the time value of money. Calculate the merged firm's share price and total number of shares after the takeover has been completed. Let the 'income return' of a bond be the coupon at the end of the period divided by the market price now at the start of the period $(C_1/P_0)$. The expected income return of a premium fixed coupon bond is: Question 880 gold standard, no explanation Under the Gold Standard (1876 to 1913), currencies were priced relative to: Which of the following statements is NOT correct? Money market securities are:	2022-05-18 06:29:48	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.3691707253456116, "perplexity": 4069.618233025931}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652662521152.22/warc/CC-MAIN-20220518052503-20220518082503-00615.warc.gz"}
http://math.stackexchange.com/questions/410915/source-coding-theorem-optimum-number-of-bits	# Source coding theorem - optimum number of bits? The source coding theorem says that information transfer with variable length code uses less bits and is equal to the entropy of the distribution. It also says that there is no code that uses lesser number of bits ( or does it? Have I misunderstood this). My question is, there is obviously a way to transmit the same information using lesser number of bits than the entropy. For example: Let $P(x) = \frac{1}{2}, \frac{1}{4}, \frac{1}{16}, \frac{1}{32}, \frac{1}{64},\frac{1}{64}$ respectively for x= 1,2,3,4,5,6. If by using source coding theorem, a variable code of length = $log_2(p(x))$ is used, then the average bits used is $1\frac{1}{2} + 2\frac{1}{4} + 4\frac{1}{16} + 5\frac{1}{32} + 6\frac{1}{64} + 6\frac{1}{64}$ = 1.5938. But obviously, there is a better way of doing this: Just use the following codes: 0,1,10,11,100,101. This definitely uses less bits on average ( bits $\lt 1.5938$). So what is the actual meaning of the source coding theorem? - How would you decode $100101$? As $baaf$ or $cacb$ or ...? – Hagen von Eitzen Jun 4 '13 at 11:44 To see that your code is not uniquely decodable, and hence practically useless, imagine that you receive the sequence $01$: you cannot know it that corresponds to a concatenation of $0\|1$ (your two first symbols), or $01$ (the third symbol).	2015-08-31 07:37:18	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6461057066917419, "perplexity": 312.1358170122356}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2015-35/segments/1440644065828.38/warc/CC-MAIN-20150827025425-00013-ip-10-171-96-226.ec2.internal.warc.gz"}
https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Essential_Graduate_Physics_-_Classical_Mechanics_(Likharev)/08%3A_Fluid_Mechanics/8.07%3A_Exercise_Problems	# 8.7: Exercise Problems 8.1. Find the first-order correction to the Pascal equation (6) for a liquid, due to its low but nonzero compressibility, and evaluate this correction for the water at the bottom of the Earth’s oceans. 8.2. Find the stationary shape of the open surface of an incompressible, heavy fluid in a container rotated about its vertical axis with a constant angular velocity $$\omega$$ - see the figure on the right. 8.3.* Use two different approaches to calculate the stationary shape of the surface of an incompressible fluid of density $$\rho$$ near a vertical plane wall, in a uniform gravity field - see the figure on the right. In particular, find the height $$h$$ of liquid’s rise at the wall surface as a function of the contact angle $$\theta_{c}$$. 8.4. $$^{}$$ A soap film with surface tension $$\gamma$$ is stretched between two similar, coaxial, thin, round rings of radius $$R$$, separated by distance $$d$$ - see the figure on the right. Neglecting gravity, calculate the equilibrium shape of the film, and the force needed for keeping the rings at the fixed distance. 8.5. A solid sphere of radius $$R$$ is kept in a steady, vorticity-free flow of an ideal incompressible fluid, with velocity $$v_{0}$$. Find the spatial distribution of velocity and pressure, and in particular their extreme values. Compare the results with those obtained in Sec. 4 for a round cylinder. 8.6. A small source, located at distance $$d$$ from a plane wall of a container filled with an ideal, incompressible fluid of density $$\rho$$, injects additional fluid isotropically, at a constant mass current ("discharge") $$Q \equiv d M / d t-$$ see the figure on the right. Calculate the fluid’s velocity distribution, and its pressure on the Hint: Recall the charge image method in electrostatics, $${ }^{51}$$ and contemplate its possible analog. 8.7. Calculate the average kinetic, potential, and full energies (per unit area) of a traveling sinusoidal wave, of a small amplitude $$q_{A}$$, on the horizontal surface on an ideal, incompressible, deep fluid of density $$\rho$$, in a uniform gravity field $$\mathbf{g}$$. 8.8. Calculate the average power (per unit width of wave’s front) carried by the surface wave discussed in the previous problem, and relate the result to the wave’s energy. 8.9. Derive Eq. (48) for the surface waves on a finite-thickness layer of a heavy liquid. 8.10. Derive Eq. (50) for the capillary waves ("ripples"). 8.11 . Derive a $$2 \mathrm{D}$$ differential equation describing the propagation of relatively long $$(\lambda>>h)$$ waves on the surface of a broad, plane layer of thickness $$h$$, of an ideal, incompressible fluid, and use it to calculate the longest standing wave modes and frequencies in a layer covering a spherical planet of radius $$R \gg>$$ h. Hint: The second task requires some familiarity with the basic properties of spherical harmonics. $${ }^{52}$$ 8.12. Calculate the velocity distribution and the dispersion relation of the waves propagating along the horizontal interface of two ideal, incompressible fluids of different densities. 8.13. Use the finite-difference approximation for the Laplace operator, with the mesh step $$h=$$ $$a / 4$$, to find the maximum velocity and total mass flow $$Q$$ of a viscous, incompressible fluid through a long pipe with a square-shaped cross-section of side $$a$$. Compare the results with those described in Sec. 5 for the same problem with the mesh step $$h=a / 2$$, and for a pipe with the circular cross-section of the same area. 8.14. A layer, of thickness $$h$$, of a heavy, viscous, incompressible fluid flows down a long and wide inclined plane, under its own weight - see the figure on the right. Find the stationary velocity distribution profile, and the total fluid discharge (per unit width.) 8.15. Calculate the drag torque exerted on a unit length of a solid round cylinder of radius $$R$$ that rotates about its axis, with angular velocity $$\omega$$, inside an incompressible fluid with viscosity $$\eta$$, kept static far from the cylinder. 8.16. Calculate the tangential force (per unit area) exerted by an incompressible fluid, with density $$\rho$$ and viscosity $$\eta$$, on a broad solid plane placed over its surface and forced to oscillate, along the surface, with amplitude $$a$$ and frequency $$\omega$$. 8.17. A massive barge, with a flat bottom of area $$A$$, floats in shallow water, with clearance $$h \ll A^{1 / 2}-$$ see the figure on the right. Analyze the time dependence of the barge’s velocity $$V(t)$$, and the water velocity profile, after the barge’s engine has been turned off. Discuss the limits of large and small values of the dimensionless parameter $$M / \rho A h$$. 8.18. * Derive a general expression for mechanical energy loss rate in a viscous incompressible fluid that obeys the Navier-Stokes equation, and use this expression to calculate the attenuation coefficient of the surface waves, assuming that the viscosity is small. (Quantify this condition). 8.19. Use the Navier-Stokes equation to calculate the coefficient of attenuation of a plane, sinusoidal acoustic wave. $${ }^{51}$$ See, e.g., EM Secs. 2.9, 3.3, and 4.3. $${ }^{52}$$ See, e.g., EM Sec. $$2.8$$ and/or QM Sec. 3.6. This page titled 8.7: Exercise Problems is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Konstantin K. Likharev via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.	2023-03-28 05:48:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8185394406318665, "perplexity": 427.6959392697094}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2023-14/segments/1679296948765.13/warc/CC-MAIN-20230328042424-20230328072424-00365.warc.gz"}
https://stats.stackexchange.com/questions/147072/cross-validation-in-binary-classification-using-only-10-positive-samples-svm	# Cross-Validation in binary classification using only 10 positive samples (SVM) I have a binary classification problem for which only $10$ positive samples are available for training. Negatives are in general in abundance, but I choose to use solely $70$ ($7$ negatives per one positive). I am trying to learn a kernel SVM (using the RBF kernel), thus I want to optimize the pair of parameters $C$, $\gamma$. I conduct grid search and I am wondering which division of the training set is more appropriate. Should I use $3$-, $5$-, or $10$-fold cross-validation? Something else maybe? I am particularly interested in the case of $\mathbf{10}$-fold cross-validation, because I have only $10$ positive samples. Would that be a good approach? • With only 10 positives, I think it would be better to use repeated 5 fold cross-validation. Something like 10 times iterated 5 fold cv. – Marc Claesen Apr 18 '15 at 15:17 • Thanks @MarcClaesen, the problem is that I cannot run so many iterations, because of complexity reasons (it's not a standard SVM), but would you think that a standard $5$-fold cross-validation procedure would be reasonable? – nullgeppetto Apr 18 '15 at 15:20	2019-10-22 03:58:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7210989594459534, "perplexity": 601.8701271322427}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2019-43/segments/1570987798619.84/warc/CC-MAIN-20191022030805-20191022054305-00490.warc.gz"}
https://www.bayesianspectacles.org/2019/06/	## The Bayesian Methodology of Sir Harold Jeffreys as a Practical Alternative to the P-value Hypothesis Test This post is an extended synopsis of Ly et al. (2019). The Bayesian methodology of Sir Harold Jeffreys as a practical alternative to the p-value hypothesis test. Preprint available on PsyArXiv: https://psyarxiv.com/dhb7x Abstract Despite an ongoing stream of lamentations, many empirical disciplines still treat the p-value as the sole arbiter to separate the scientific wheat from the chaff. The continued…	2021-11-28 02:44:04	{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8605272769927979, "perplexity": 5094.579538973649}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2021-49/segments/1637964358443.87/warc/CC-MAIN-20211128013650-20211128043650-00357.warc.gz"}
https://plainmath.net/38297/in-the-diagram-cd-is-tangent-to-circles-a-and	# In the diagram, CD is tangent to circles A and In the diagram, CD is tangent to circles A and B, and he circles are tangent to each other. Answer this question If CD = 15 and AB = 17 and CB = 6, find AD = ? You can still ask an expert for help • Questions are typically answered in as fast as 30 minutes Solve your problem for the price of one coffee • Math expert for every subject • Pay only if we can solve it menerkupvd Given, In the diagram, CD is tangent to circles A nad B amd the circles are tangent to each other. CD=15, AB=17 and CB=6 Now, From the diagram, we can say that CB and BE is the radius of the circle B and AD and AE is the radius of the circle A. So, $\therefore AB=BE+AE\left[\because CB=BE\right]$ $⇒17=CB+AE$ $⇒17=6+AE$ $⇒AE=17-6$ $⇒AE=11$ $\therefore AD=AE=11$ $\therefore AD=11$	2022-05-28 08:01:09	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 8, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8702730536460876, "perplexity": 789.5773890615501}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2022-21/segments/1652663013003.96/warc/CC-MAIN-20220528062047-20220528092047-00362.warc.gz"}
https://stats.stackexchange.com/questions/65762/how-to-standardize-data-for-hierarchical-clustering	# How to standardize data for hierarchical clustering? When running hierarchical clustering analysis of a matrix of individuals x samples (e.g., employee performances across different days), there are several possibilities for normalization. If one is clustering the columns (to see whether on certain days individuals perform similarly), one could 1. z-score normalize across the rows to make each individual employees mean and standard deviation comparable across days, or 2. z-score normalize across the columns to make all employees comparable within a day, or 3. not normalize at all and cluster the raw values Could someone explain the relative advantages/disadvantages of each approach here? To clarify, I am using correlation distance. Methods 1 or 2 in practice give different results but it's not clear that for the task of seeing if days cluster together, whether #1 or #2 are more appropriate if one chooses to normalize. • The issue of standardization (there is much more ways possible than those three) is entirely out of clustering scope. It is the question of how to pre-process the data in order to make the chosen distance measure (yet a separate question, but partially tied with clustering) to reflect just what you want. – ttnphns Jul 27 '13 at 17:28 • @ttnphns: it was not meant to be an exhaustive list, but these are the three options I am considering for now. I edited the question to clarify that I'm using correlation distance. So the question is how to preprocess the data for this distance metric. Yes, it is outside of clustering, technically but that's my question. – user248237 Jul 27 '13 at 18:34 • Just remember what z-standardization is for profile comparison. It removes elevation differences and variation differences to leave only only shape differences for comparison. Now, the question is what are "profiles" for you - row items or column items. What is correlation distance? Correlation itself is an angular similarity ranging from -1 to +1. – ttnphns Jul 27 '13 at 18:50 If every observation is on the same scale (i.e. the same performance metric was used each day), then in general I would not recommend normalization, since the scores already have comparable location and spread information. Since the Z-transform of a random variable is a linear transformation that does not change the sign of the variable, and since the correlation operation is invariant to linear transformations that preserve sign, z-transforming columns and then clustering on columns with correlation distance will not be different than using the raw scores (See http://www.math.uah.edu/stat/expect/Covariance.html just after #8). Z-normalizing rows (within employee) will wipe out their average performance, which may be highly inappropriate, depending on your goals. If you are clustering days based on employee performance, presumably you would like a cluster of "high performing days", and a separate cluster of "low performing days." If this is the case, DO NOT use correlation distance, since it ignores mean differences. For example, correlation distance assigns a very low distance (=high similarity) to day 1 with scores (11,12,13,14,15) and day 2 with scores (2,2,3,4,5); and assigns a much higher distance to day 1 and day 3 with scores (12,13,12,13,12). This is probably not the sort of result you want. You probably want something like euclidean distance. It is imperative that you think carefully about your goals here and select normalization methods and distance metrics (not to mention clustering algorithms) accordingly. You can use the same preprocessing that makes your distance function "work" for other tasks than clustering. Hierarchical clustering doesn't use your actual data. It only uses distances. So data normalization will not affect hierarchical clustering, but it will likely affect your distance function. So make sure your preprocessing yields sensibe distance values.	2020-04-03 11:24:52	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.6018698811531067, "perplexity": 1135.4724369962357}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2020-16/segments/1585370510846.12/warc/CC-MAIN-20200403092656-20200403122656-00307.warc.gz"}