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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 261 0 "" }{TEXT 
262 62 "Elements de correction du TD 4 sur la fonction zeta de Riemann
" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 38 "\nQuelques indication sur la
 partie 1 :" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "zeta := x-> s
um(1/n^x,n=1..infinity);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 
"limit(zeta(x),x=1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "plo
t(zeta,1..10);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "series(ze
ta(x),x=1,4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "Dzeta := p
roc(k) RETURN(x->sum((-1)^k*ln(n)^k/n^x,n=1..infinity)); end;" }}}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 256 "" 0 "" {TEXT 259 0 
"" }{TEXT 256 1 "\n" }{TEXT 258 61 "La correction de la question d\351
liquate (i.e. la question 4) :" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 315 "Newton := proc(A,p)\n  local S,i,j,tmp,n;\n  n := nops(A);\n  S
 := [seq(0,i=1..p)];\n  S[1] := A[1];\n  if not p = 1 then \n    for j
 from 2 to p do\n      tmp := (-1)^(j-1)*j*A[j];  \n      for i from 1
 to j-1 do\n        tmp := tmp + (-1)^(j-1-i)*A[j-i]*S[i];\n      od; \+
\n      S[j] := tmp;\n    od;  \n  fi;\nRETURN(S);\nend;\n" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT 257 36 "Un exemple de conclusion pour p=2 : " }
{TEXT -1 1 "\n" }{MPLTEXT 1 0 85 "A := [n*(2*n-1)/3,2*n*(2*n-1)*(2*n-2
)*(2*n-3)/(5!)];\nRES := simplify(Newton(A,2)[2]);" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT 260 12 "Conclusion :" }{TEXT -1 134 " le coefficient dom
inant 8/45 donne en cherchant l'\351quivalent (grace \340 l'in\351gali
t\351 de la question 5) la formule :\nzeta(4) = pi^4/90.  " }{MPLTEXT 
1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0 2" 62 }
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