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本帖最后由 lin2009 于 2011-1-3 14:58 编辑
1) 下面无穷数列之和能否收敛，若能，收敛的确切值是多少？
}^2}\sin {{\left( {\frac{\pi }{6}\nu } \right)}^2}}}{{\sin {{\left( {\frac{\pi }{{18}}\nu } \right)}^2}{\nu ^2}}}$ "$\sum \limits_{\nu= 2}^\infty\frac{1}{9}\frac{{\sin {{\left( {\frac{\pi }{2}\nu } \right)}^2}\sin {{\left( {\frac{\pi }{6}\nu } \right)}^2}}}{{\sin {{\left( {\frac{\pi }{{18}}\nu } \right)}^2}{\nu ^2}}}$")
上式是- fabnv := proc (q, beta, nu) options operator, arrow; sin((1/2)*nu*beta*Pi)^2*sin((1/6)*nu*Pi)^2/(q^2*sin((1/6)*nu*Pi/q)^2*nu^2) end proc
maple不能得出该无穷数列的收敛值。但从下面的有限项之和可以看出
该无穷数列应该是收敛的，其值应在0.06879左右。
> evalf(subs({q = 3, beta = 1}, sum((sin((1/2)*nv*beta*Pi)*sin((1/6)*nv*Pi)/(q*sin((1/6)*nv*Pi/q)*nv))^2, nv = 2 .. 10000)));
print(`output redirected...`); # input placeholder
0.06878334966
> evalf(subs({q = 3, beta = 1}, sum((sin((1/2)*nv*beta*Pi)*sin((1/6)*nv*Pi)/(q*sin((1/6)*nv*Pi/q)*nv))^2, nv = 2 .. 20000)));
print(`output redirected...`); # input placeholder
0.06879168540
> evalf(subs({q = 3, beta = 1}, sum((sin((1/2)*nv*beta*Pi)*sin((1/6)*nv*Pi)/(q*sin((1/6)*nv*Pi/q)*nv))^2, nv = 2 .. 30000)));
print(`output redirected...`); # input placeholder
0.06879446199
> evalf(subs({q = 3, beta = 1}, sum((sin((1/2)*nv*beta*Pi)*sin((1/6)*nv*Pi)/(q*sin((1/6)*nv*Pi/q)*nv))^2, nv = 2 .. 40000)));
print(`output redirected...`); # input placeholder
0.06879585084
> evalf(subs({q = 3, beta = 1}, sum((sin((1/2)*nv*beta*Pi)*sin((1/6)*nv*Pi)/(q*sin((1/6)*nv*Pi/q)*nv))^2, nv = 2 .. infinity)));
原式输出
2）如何求出该无穷数列之和关于q和beta的一般表达式。
}^2} \times \sin {{\left( {\nu \times \frac{\pi }{6}} \right)}^2}}}{{{q^2} \times \sin {{\left( {\frac{\pi }{6} \times \frac{\nu }{q}} \right)}^2} \times {\nu ^2}}}} $ "$\sum\limits_{\nu = 2}^\infty {\frac{{\sin {{\left( {\nu \times \beta \times \frac{\pi }{2}} \right)}^2} \times \sin {{\left( {\nu \times \frac{\pi }{6}} \right)}^2}}}{{{q^2} \times \sin {{\left( {\frac{\pi }{6} \times \frac{\nu }{q}} \right)}^2} \times {\nu ^2}}}} $") - f := (q, beta, nu) -> sin((1/2)*nu*beta*Pi)^2*sin((1/6)*nu*Pi)^2/(q^2*sin((1/6)*nu*Pi/q)^2*nu^2)
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发表于 2011-1-3 14:29:43
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