为与虚数单位i区分, 求和指标改成k.
比较直接的方法是用Euler公式e^(ix) = cos(x)+isin(x), 有2cos(x) = e^(ix)+e^(-ix).
因此2·∑{1 ≤ k} cos(kθ)cos(θ)^k
= ∑{1 ≤ k} e^(ikθ)·cos(θ)^k + ∑{1 ≤ k} e^(-ikθ)·cos(θ)^k
= ∑{1 ≤ k} (e^(iθ)·cos(θ))^k + ∑{1 ≤ k} (e^(-iθ)·cos(θ))^k
= e^(iθ)·cos(θ)/(1-e^(iθ)·cos(θ))+e^(-iθ)·cos(θ)/(1-e^(-iθ)·cos(θ))
= cos(θ)·((e^(iθ)-cos(θ))+(e^(-iθ)-cos(θ))/((1-e^(iθ)·cos(θ))(1-e^(-iθ)·cos(θ)))
= 0.
θ ∈ (0,π/2)的条件保证了上面各和式的收敛性.
比较直接的方法是用Euler公式e^(ix) = cos(x)+isin(x), 有2cos(x) = e^(ix)+e^(-ix).
因此2·∑{1 ≤ k} cos(kθ)cos(θ)^k
= ∑{1 ≤ k} e^(ikθ)·cos(θ)^k + ∑{1 ≤ k} e^(-ikθ)·cos(θ)^k
= ∑{1 ≤ k} (e^(iθ)·cos(θ))^k + ∑{1 ≤ k} (e^(-iθ)·cos(θ))^k
= e^(iθ)·cos(θ)/(1-e^(iθ)·cos(θ))+e^(-iθ)·cos(θ)/(1-e^(-iθ)·cos(θ))
= cos(θ)·((e^(iθ)-cos(θ))+(e^(-iθ)-cos(θ))/((1-e^(iθ)·cos(θ))(1-e^(-iθ)·cos(θ)))
= 0.
θ ∈ (0,π/2)的条件保证了上面各和式的收敛性.
