$\cos(x) \; \cdot \; \cos(2x) \; \cdot \; \cos(3x) \; \cdots \cos(nx)=$?

Question

Aquí está mi intento,

$2 \cos(x) = e^{ix}+e^{-ix}$

$2 \cos(kx) = e^{kix}+e^{-kix}$

$\displaystyle\implies\prod_{k=1}^{n} \cos(kx) = \frac{1}{2^n} \prod_{k=1}^{n} \left(e^{kix}+e^{-kix}\right)$

Todas las ideas para tomar desde aquí? Gracias.

Por favor, no le entregue la totalidad de la solución :P

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\cos\pars{x}\cos\pars{2x}\cos\pars{3x}\ldots\cos\pars{nx} = \prod_{k = 0}^{n - 1}\cos\pars{\bracks{k + 1}x} = \prod_{k = 0}^{n - 1} {\expo{\ic\pars{k + 1}x} + \expo{-\ic\pars{k + 1}x} \over 2} \\[5mm] = &\ {1 \over 2^{n}} \prod_{k = 0}^{n - 1} \expo{\ic\pars{k + 1}x}\bracks{1 + \expo{-2\ic\pars{k + 1}x}} = {1 \over 2^{n}}\prod_{k = 0}^{n - 1} \pars{\expo{\ic x}}^{k + 1}\bracks{1 - \pars{-\expo{-2\ic x}} \pars{\expo{-2\ic x}}^{k}} \\[5mm] = &\ {1 \over 2^{n}}\,\expo{\ic n\pars{n + 1}x/2} \prod_{k = 0}^{n - 1}\ \bracks{1 - \pars{-\expo{-2\ic x}}\pars{\expo{-2\ic x}}^{k}} = \bbx{{\expo{\ic n\pars{n + 1}x/2} \over 2^{n}}\, \pars{-\expo{-2\ic x};\expo{-2\ic x}}_{n}} \end{align}

$\ds{\pars{a;b}_{m}}$ es el q-Símbolo de Pochhammer.