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[SPAM] [obm-l] Dúvidas
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Amigos da lista , na semana passada enviei a seguinte quest=E3o:
1) Calule a soma : S =3D 1/cos(pi/7) + 1/cos(3pi/7) + 1/cos(5pi/7) =
Resposta: S=3D 4 . Essa quest=E3o vi no forum internacional, hoje vi =
um internalta que envio a solu=E7=E3o. Mas n=E3o entendi nada, voc=EAs =
poderiam me ajudar na compreens=E3o da quest=E3o?=20
Solu=E7=E3o: Considerando que CHEBYSHEV POLYNOMIAL T_7 (x) =3D =
64x^7 -112x^5 56x^3 7x ( if have calculated correctly). the roots of [ =
1=B0 d=FAvida : o que =E9 CHEBYSHEV POLYNOMIAL , como ele chegou a =
esse polinomio? daria para explicar com detalhe ?] =20
=20
T_7(x)=3D cos(pi) =3D -1[2=B0 d=FAvida : como ele chegou nessa =
solu=E7=E3o?]
are cos(2k+1)pi/7, k =3D 0,1.....6. Note que 2S + 1/cos(pi) =3D [( =
somat=F3rio de K =3D 0 a 6) ]1/ cos(2k+1)pi/7. The polynomial With =
roots the reciprocal of T_7(x) + 1=3D 0 is the polinomial whose =
coefficients ae of above in reverse order, or
=20
x^7 - 7x^6 + STUFF( O QUE =C9 ISSO?)
and the sum of the roots of this polinomial is just 7, hence 2S -1 =3D =
7 implica S =3D4
2) Prove que tg(3pi/11) + tg(2pi/11) =3D raiz quadrada de 11
Qualquer ajuda me ajudarar a entender essas quest=F5es. desde j=E1 =
agrade=E7o pela aten=E7=E3o.
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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META http-equiv=3DContent-Type content=3D"text/html; =
charset=3Diso-8859-1">
<META content=3D"MSHTML 6.00.2900.2180" name=3DGENERATOR>
<STYLE></STYLE>
</HEAD>
<BODY bgColor=3D#ffffff>
<DIV><FONT face=3DArial size=3D2>Amigos da lista , na semana passada =
enviei a=20
seguinte quest=E3o:</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2> 1) Calule a soma : =
<FONT=20
face=3DArial size=3D2>S =3D 1/cos(pi/7) + 1/cos(3pi/7) =
+ =20
1/cos(5pi/7) Resposta: S=3D 4 . Essa quest=E3o vi =
no forum=20
internacional, hoje vi um internalta que envio a solu=E7=E3o. =
Mas n=E3o=20
entendi nada, voc=EAs poderiam me ajudar na compreens=E3o da=20
quest=E3o? </FONT></FONT></DIV>
<DIV><FONT face=3DArial size=3D2><FONT face=3DArial =
size=3D2></FONT></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2><FONT face=3DArial =
size=3D2> Solu=E7=E3o: Considerando=20
que CHEBYSHEV POLYNOMIAL T_7 (x) =3D 64x^7 =
-112x^5 56x^3=20
7x ( if have calculated correctly). the roots of [ <STRONG><FONT=20
color=3D#ff0000>1=B0 d=FAvida : o que =E9 CHEBYSHEV =
POLYNOMIAL , como=20
ele chegou a esse polinomio? daria para explicar com detalhe ?] =20
</FONT></STRONG></FONT></FONT></DIV>
<DIV><FONT face=3DArial size=3D2><FONT face=3DArial size=3D2> =
</FONT></FONT></DIV>
<DIV><FONT face=3DArial size=3D2> T_7(x)=3D cos(pi) =3D =
-1[<FONT=20
color=3D#ff0000>2=B0 d=FAvida : como ele chegou nessa =
solu=E7=E3o?]</FONT></FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2> are cos(2k+1)pi/7, k =
=3D=20
0,1.....6. Note que 2S + 1/cos(pi) =3D [( somat=F3rio de K =3D 0 a =
6) ]1/=20
cos(2k+1)pi/7. The polynomial With roots the reciprocal of T_7(x) + 1=3D =
0 is the=20
polinomial whose coefficients ae of above in reverse order, =
or</FONT></DIV>
<DIV><FONT face=3DArial size=3D2> </FONT></DIV>
<DIV><FONT face=3DArial size=3D2> x^7 - =
7x^6 + <FONT=20
color=3D#ff0000><FONT color=3D#000000>STUFF</FONT>( O QUE =C9=20
ISSO?)</FONT></FONT></DIV>
<DIV><FONT face=3DArial size=3D2><FONT face=3DArial size=3D2> and =
the sum of the=20
roots of this polinomial is just 7, hence 2S -1 =3D 7 implica S=20
=3D4</FONT></FONT></DIV>
<DIV><FONT face=3DArial size=3D2><FONT face=3DArial =
size=3D2></FONT></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2><FONT face=3DArial size=3D2>2) Prove =
que tg(3pi/11)=20
+ tg(2pi/11) =3D raiz quadrada de 11</FONT></FONT></DIV>
<DIV><FONT face=3DArial size=3D2><FONT face=3DArial =
size=3D2></FONT></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2><FONT face=3DArial =
size=3D2></FONT></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2><FONT face=3DArial =
size=3D2> Qualquer ajuda=20
me ajudarar a entender essas quest=F5es. desde j=E1 agrade=E7o pela =
aten=E7=E3o.</DIV></FONT></FONT></BODY></HTML>
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