Duro suma con coseno

Question

Estoy tratando de calcular $\sum_{j=1}^{n} (-1)^{j-1} \cos^{2k} (\frac {j \pi}{2n+2}).$ $a = e^{\frac {jπ}{2n+2}}$ (y por tanto la satisfacción de $a^{2n+2}=e^{j \pi}=-1$) obtenemos que el lado izquierdo es el mismo

$ \sum_{j=1}^{n} (-1)^{j-1} (\frac{a^j + a^{-j}}{2})^{2k} = \frac {1}{2^{2k}} \sum_{j=1}^{n} (-1)^{j-1} (a^j + a^{-j})^{2k} $

y así nos queda probar

$S = \sum_{j=1}^{n} (-1)^{j-1} (a^j + a^{-j})^{2k} = 2^{2k-1} .$

Ahora denotar $f_n (x) = \sum_{j=1}^{n} (-x)^j.$ Tenemos $f_n (-1) = n$ $f_n (x) = \frac {1- (-x)^n}{1+x},$ todos los $x≠-1.$ Por el Teorema del Binomio tenemos

$S = \sum_{j=1}^{n} (-1)^{j-1} (a^j + a^{-j})^{2k} = \sum_{j=1}^{n} (-1)^{j-1} \sum_{m=0}^{2k} \binom {2k}m a^{2j(m-k)},$

o

$S = \sum_{j=1}^{n} \sum_{m=0}^{2k} (-1)^{j-1} \binom {2k}m a^{2j(m-k)},$

o

$S = \sum_{m=0}^{2k} \sum_{j=1}^{n} (-1)^{j-1} \binom {2k}m a^{2j(m-k)},$

o,

$S = \sum_{m=0}^{2k} \binom {2k}m \sum_{j=1}^{n} (-1)^{j-1} a^{2j(m-k)}.$

o,

$S = \sum_{m=0}^{2k} \binom {2k}m f(a^{2j(m-k)}).$

Me quedé allí. Estoy en lo cierto hasta el momento? Gracias

Answer 1

Tenga en cuenta que $$\cos ^{2k}\theta ={\frac {1}{2^{2k}}}\sum _{i=0}^{2k}{\binom {2k}{i}}\cos {{\big (}(2k-2i)\theta {\big )}}$$

Definir \begin{align} S_{n,k}&=\sum_{j=1}^{n-1} (-1)^{j-1} \cos^{2k} \frac {j \pi}{2n}\\ &=\sum_{j=1}^{n} (-1)^{j-1} \cos^{2k} \frac {j \pi}{2n}=\sum_{j=1}^{n}a_j \end{align} donde $$a_j=(-1)^{j-1}\cos^{2k} \frac {j \pi}{2n}=(-1)^{j-1}{\frac {1}{2^{2k}}}\sum _{l=0}^{2k}{\binom {2k}{l}}\cos {{\big (}(k-l)\frac{j\pi}{n} {\big )}}$$ Tenemos a continuación \begin{align} S_{n,k}&={\frac {1}{2^{2k}}}\sum _{l=0}^{2k}{\binom {2k}{l}}\sum_{j=1}^{n}(-1)^{j-1}\cos {{\big (}(k-l)\frac{j\pi}{n} {\big )}}\\ &=-{\frac {1}{2^{2k}}}\mathrm{Re}\;\sum _{l=0}^{2k}{\binom {2k}{l}}\sum_{j=1}^{n}(-1)^{j}\exp {{\big (}i(k-l)\frac{j\pi}{n} {\big )}} \end{align} Desde $$ \sum_{j=0}^{n}(-1)^{j}\exp {{\big (}i(k-l)\frac{j\pi}{n} {\big )}}=\frac{1+(-1)^{k-l+n}\exp {{\big (}i(k-l)\frac{\pi}{n} {\big )}}}{1+\exp {{\big (}i(k-l)\frac{\pi}{n} {\big )}}}=b_l= \begin{cases} 1\quad\text{if %#%#% is even}\\ b'_l\quad\text{if %#%#% is odd} \end{casos} $$ tenemos \begin{align} S_{n,k}=-{\frac {1}{2^{2k}}}\mathrm{Re}\;\sum _{l=0}^{2k}{\binom {2k}{l}}(b_l-1)&={\frac {1}{2^{2k}}}\sum _{l=0}^{2k}{\binom {2k}{l}}-{\frac {1}{2^{2k}}}\mathrm{Re}\;\sum _{l=0}^{2k}{\binom {2k}{l}}b_l\\ &=1-\frac12-{\frac {1}{2^{2k}}}\mathrm{Re}\;\sum _{l}{\binom {2k}{l}}b'_l\\ &=\frac12\tag{If %#%#%} \end{align} $k-l+n$ Asumiendo $k-l+n$, luego $n\ge k)\quad(*$, $(*):$, y $n\ge k$$ que los rendimientos de $\forall l$$