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xStefx
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11 Jun 2013, 19:33

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gauss
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10 Jun 2013, 21:46

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Am mai raspuns la exact aceeasi problema cu cateva luni mai inainte.
In orice caz, este de preferat sa se incerce pe propriile puteri, de exemplu incepand cu calculul pentru valorile urmatoare:

(1/2)
(1/2) * (1/3)
(1/2) * (1/3) * (1/4)
(1/2) * (1/3) * (1/4) * (1/5)

Daca lucrurile par complicate, macar primele trei numere din lista...
Iata-le pe urmatoarele:

(20:44) gp > a = 1/2
%11 = 1/2
(20:44) gp > for( k=3, 100, a = f(a, 1/k); print( "(1/2) * .
(1/2) * ... * (1/3) = 5/7
(1/2) * ... * (1/4) = 9/11
(1/2) * ... * (1/5) = 7/8
(1/2) * ... * (1/6) = 10/11
(1/2) * ... * (1/7) = 27/29
(1/2) * ... * (1/8) = 35/37
(1/2) * ... * (1/9) = 22/23
(1/2) * ... * (1/10) = 27/28
(1/2) * ... * (1/11) = 65/67
(1/2) * ... * (1/12) = 77/79
(1/2) * ... * (1/13) = 45/46
(1/2) * ... * (1/14) = 52/53
(1/2) * ... * (1/15) = 119/121
(1/2) * ... * (1/16) = 135/137
(1/2) * ... * (1/17) = 76/77
(1/2) * ... * (1/18) = 85/86
(1/2) * ... * (1/19) = 189/191
(1/2) * ... * (1/20) = 209/211
(1/2) * ... * (1/21) = 115/116
(1/2) * ... * (1/22) = 126/127
(1/2) * ... * (1/23) = 275/277
(1/2) * ... * (1/24) = 299/301
(1/2) * ... * (1/25) = 162/163
(1/2) * ... * (1/26) = 175/176
(1/2) * ... * (1/27) = 377/379
(1/2) * ... * (1/28) = 405/407
(1/2) * ... * (1/29) = 217/218
(1/2) * ... * (1/30) = 232/233
(1/2) * ... * (1/31) = 495/497
(1/2) * ... * (1/32) = 527/529
(1/2) * ... * (1/33) = 280/281
(1/2) * ... * (1/34) = 297/298
(1/2) * ... * (1/35) = 629/631
(1/2) * ... * (1/36) = 665/667
(1/2) * ... * (1/37) = 351/352
(1/2) * ... * (1/38) = 370/371
(1/2) * ... * (1/39) = 779/781
(1/2) * ... * (1/40) = 819/821
(1/2) * ... * (1/41) = 430/431
(1/2) * ... * (1/42) = 451/452
(1/2) * ... * (1/43) = 945/947
(1/2) * ... * (1/44) = 989/991
(1/2) * ... * (1/45) = 517/518
(1/2) * ... * (1/46) = 540/541
(1/2) * ... * (1/47) = 1127/1129
(1/2) * ... * (1/48) = 1175/1177
(1/2) * ... * (1/49) = 612/613
(1/2) * ... * (1/50) = 637/638
(1/2) * ... * (1/51) = 1325/1327
(1/2) * ... * (1/52) = 1377/1379
(1/2) * ... * (1/53) = 715/716
(1/2) * ... * (1/54) = 742/743
(1/2) * ... * (1/55) = 1539/1541
(1/2) * ... * (1/56) = 1595/1597
(1/2) * ... * (1/57) = 826/827
(1/2) * ... * (1/58) = 855/856
(1/2) * ... * (1/59) = 1769/1771
(1/2) * ... * (1/60) = 1829/1831
(1/2) * ... * (1/61) = 945/946
(1/2) * ... * (1/62) = 976/977
(1/2) * ... * (1/63) = 2015/2017
(1/2) * ... * (1/64) = 2079/2081
(1/2) * ... * (1/65) = 1072/1073
(1/2) * ... * (1/66) = 1105/1106
(1/2) * ... * (1/67) = 2277/2279
(1/2) * ... * (1/68) = 2345/2347
(1/2) * ... * (1/69) = 1207/1208
(1/2) * ... * (1/70) = 1242/1243
(1/2) * ... * (1/71) = 2555/2557
(1/2) * ... * (1/72) = 2627/2629
(1/2) * ... * (1/73) = 1350/1351
(1/2) * ... * (1/74) = 1387/1388
(1/2) * ... * (1/75) = 2849/2851
(1/2) * ... * (1/76) = 2925/2927
(1/2) * ... * (1/77) = 1501/1502
(1/2) * ... * (1/78) = 1540/1541
(1/2) * ... * (1/79) = 3159/3161
(1/2) * ... * (1/80) = 3239/3241
(1/2) * ... * (1/81) = 1660/1661
(1/2) * ... * (1/82) = 1701/1702
(1/2) * ... * (1/83) = 3485/3487
(1/2) * ... * (1/84) = 3569/3571
(1/2) * ... * (1/85) = 1827/1828
(1/2) * ... * (1/86) = 1870/1871
(1/2) * ... * (1/87) = 3827/3829
(1/2) * ... * (1/88) = 3915/3917
(1/2) * ... * (1/89) = 2002/2003
(1/2) * ... * (1/90) = 2047/2048
(1/2) * ... * (1/91) = 4185/4187
(1/2) * ... * (1/92) = 4277/4279
(1/2) * ... * (1/93) = 2185/2186
(1/2) * ... * (1/94) = 2232/2233
(1/2) * ... * (1/95) = 4559/4561
(1/2) * ... * (1/96) = 4655/4657
(1/2) * ... * (1/97) = 2376/2377
(1/2) * ... * (1/98) = 2425/2426
(1/2) * ... * (1/99) = 4949/4951
(1/2) * ... * (1/100) = 5049/5051
(20:44) gp >


Putem "ghici" in orice caz o relatie pe care sa o verificam inductiv.
Care este deci raspunsul bun?
De ce?

Pe de alta parte, avem si un izomorfism de grupuri...
http://www.pro-didactica.ro/forum/index.php?forumID=8&ID=42733


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df (gauss)
xStefx
Grup: membru
Mesaje: 110
11 Jun 2013, 15:30

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Multumesc. Raspunsul corect este B. Deci sa inteleg ca respectiva lege de compozitie nu poate fi rescrisa favorabil?

gauss
Grup: Administrator
Mesaje: 6933
11 Jun 2013, 19:33

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O solutie stucturala este aici:
http://www.pro-didactica.ro/forum/index.php?forumID=21&ID=6724

"Alta" (de fapt aceeasi) solutie structurala foloseste morfismele F, G de aici
http://www.pro-didactica.ro/forum/index.php?forumID=8&ID=42733
pentru a muta operatia * de pe (-1,1) pe (0,oo) in operatia de inmultire,
"intamplator putem face inmultirea",
si venim inapoi.

Putem insa gasi si formula generala pentru
(1/2) * (1/3) * ... * (1/n)
care depinde de n,
pe care sa o demonstram prin inductie.


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df (gauss)
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