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Olá,
chame S = P'/P = cotgx + 2 . cotg(2x) + ... + 2^n .
cotg(2^nx)
S - 2S = cotgx + 2 [ cotg(2x) - cotgx ] + 4 [
cotg(4x) - cotg(2x) ] + ... + 2^n [ cotg(2^nx) - cotg(2^(n-1)x) ] - 2^(n+1) .
cotg(2^nx)
cotg(ax) - cotg(ax/2) = cos(ax)/sen(ax) -
cos(ax/2)/sen(ax/2) = [cos(ax) - 2(cos(ax/2))^2]/sen(ax) = (-1)/sen(ax) =
-1/sen(ax)
logo:
S - 2S = cotgx - 2/sen(2x) - 4/sen(4x) -
8/sen(8x) - ... - 2^n/sen(2^nx) - 2^(n+1) . cotg(2^nx)
-S = cotgx - 2/sen(2x) - 4/sen(4x) - 8/sen(8x) -
... - 2^n/sen(2^nx) - 2^(n+1) . cotg(2^nx)
S = -cotgx + 2/sen(2x) + 4/sen(4x) +
8/sen(8x) + ... + 2^n/sen(2^nx) + 2^(n+1) .
cotg(2^nx)
agora, quanto vale Sum(i = 1 até n, a/sen(ax))
?
hehe
parece q to travando toda hora :P
abraços,
Salhab
----- Original Message -----
Sent: Thursday, September 14, 2006 8:18
PM
Subject: Re: [obm-l] questao boa de
trigo.
Ae Vinicius, tudo bem?
P = senx . sen2x . sen4x . sen8x ...
sen2^nx
P' = dP/dx = cosx . sen2x . sen4x ... sen2^nx + 2
. senx . cos2x . sen4x ... sen2^nx + ... + 2^n . senx . sen2x . sen4x . sen8x
... cos2^nx
P'/P = cosx / senx + 2 cos2x / sen2x + ... + 2^n
. cos2^nx / sen2^nx
P'/P = cotgx + 2 . cotg(2x) + ... + 2^n .
cotg(2^nx)
d(lnP)/dx = P'/P
integrando, temos:
lnP = ln(senx) + ln(sen2x) + ... + ln sen(2^nx) +
c
P = k . senx . sen2x . sen4x ...
sen2^nx
usando um valor conhecido, temos que k =
1..
logo, voltei de onde comecei! hehehe
dps eu tento denovo!
vou estudar q tenho prova amanha..
abraços
Salhab
----- Original Message -----
Sent: Wednesday, September 13, 2006
10:42 PM
Subject: [obm-l] questao boa de
trigo.
qt vale:
Senx . sen2x . sen4x . sen8x .... sen2^nx
flw!
Vinicius Meireles Aleixo
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