lin2009 发表于 2011-1-3 14:29:43

如何求出一无穷三角数列之和的收敛值?

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?
$\sum \limits_{\nu= 2}^\infty\frac{1}{9}\frac{{\sin {{\left( {\frac{\pi }{2}\nu } \right)}^2}\sin {{\left( {\frac{\pi }{6}\nu } \right)}^2}}}{{\sin {{\left( {\frac{\pi }{{18}}\nu } \right)}^2}{\nu ^2}}}$

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
在beta=1,q=3时的特例。

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。

$\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) 已知q为正整数,2/3 <= beta <= 1

zsq-w 发表于 2011-1-4 12:39:56

v=18的时候,式子是不是无意义呀?

lin2009 发表于 2011-1-4 17:08:05

2# zsq-w
v=18及18倍数时,f(q,beta,v)应为0;
单项计算时,会提示出错信息:
f(3, 1, 18)
Error, (in f) numeric exception: division by zero;

但在求和时Maple应该是按照极限值来计算的:
${\lim }\limits_{nv \to 18} f\left( {3,1,nv} \right) = 0$.
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