展开$\frac{\cos (n x)-\cos (n y)}{\cos (x)-\cos (y)}$

hbghlyj

$f_n=\frac12\frac{\cos (n x)-\cos (n y)}{\cos (x)-\cos (y)}$
$f_0=0$
$f_1=\frac12$
$f_2=\cos (x)+\cos (y)$
$f_3-f_1=\cos (2 x)+\cos (x-y)+\cos (x+y)+\cos (2 y)$
$f_4-f_2=\cos (3 x)+\cos (2 x-y)+\cos (2 x+y)+\cos (x+2 y)+\cos (x-2 y)+\cos (3 y)$
$f_5-f_3=\cos (4 x)+\cos (3 x-y)+\cos (3 x+y)+\cos (2 x-2y)+\cos (2x+2y)+\cos (x+3 y)+\cos (x-3 y)+\cos (4 y)$
\begin{multline*}f_7-f_5=\cos (6 x)+\cos (5 x-y)+\cos (5 x+y)+\cos (4 x+2 y)+\cos (4 x-2 y)\\+\cos (3 x+3 y)+\cos (3 x-3 y)+\cos (2 x+4 y)+\cos (2 x-4 y)+\cos (x+5 y)+\cos (x-5 y)+\cos (6 y)\end{multline*}一般的公式\[f_{n+1}-f_{n-1}=\sum _{i=1}^n (\cos (ix+(n-i)y)+\cos ((n-i)x-i y))\]
验证:

1. 1/2*(Cos[(n+1)x]-Cos[(n+1)y]-Cos[(n-1)x]+Cos[(n-1)y])/(Cos[x]-Cos[y])-Sum[Cos[i x+(n-i)y]+Cos[(n-i)x-i y],{i,n}]//FullSimplify

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输出0
把$x,y$换成$\frac x2,\frac y2$, $n$换成$2n+1$得
\[\sum _{i=1}^{2 n+1} \left(\cos \left(i\frac x2+(2n+1-i)\frac y2\right)+\cos \left((2n+1-i)\frac x2-i\frac{y}{2}\right)\right)=\frac{\cos ((n+1) x)-\cos (n x)-\cos ((n+1) y)+\cos (n y)}{2 \left(\cos \left(\frac{x}{2}\right)-\cos \left(\frac{y}{2}\right)\right)}\]

hbghlyj

对于sin有类似等式
$g_n=\frac12\frac{\sin(nx)-\sin(ny)}{\sin x-\sin y}$
$g_1=\frac12$
$g_3+g_1=\cos (2 x)+ \cos (x+y)- \cos (x-y)+\cos (2 y)$
$g_5+g_3=\cos (4 x)+\cos (3 x+y)-\cos (3 x-y)+\cos (2 x+2 y)+\cos (2 x-2 y)+\cos (x+3 y)-\cos (x-3 y)+\cos (4 y)$
一般的公式: (可以从1#公式作代换$\frac\pi2-x,\frac\pi2-y$得到)
当$n$为偶数,\[\sum _{i=1}^n \left(\cos (ix+(n-i)y)+(-1)^i \cos ((n-i)x-i y)\right)=g_{n+1}+g_{n-1}\]
当$n$为奇数,\[\sum _{i=1}^n \left(\cos (ix+(n-i)y)+(-1)^i \cos ((n-i)x-i y)\right)=\frac{-2 \sin (n y+x)+\sin ((n-1) x)+\sin ((n+1) x)-\sin ((n-1) y)+\sin ((n+1) y)}{2 (\sin (x)-\sin (y))}\]当$n$为奇数$g_{n+1}+g_{n-1}$能否表示成cos的和式

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相关: Dirichlet kernel
$$D_{N}(t)=\frac{\sin \left(2 \pi\left(N+\frac{1}{2}\right) t\right)}{\sin (\pi t)}=1+2 \sum_{k=1}^{N} \cos (2 \pi k t)$$