Putnam Examen Integral

Question

Estoy tratando de evaluar$$ \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \cos^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big)dx_1 dx_2...dx_n. $$ Esto es de un viejo Putnam concurso de matemáticas. Cualquiera de 1965 o 1987 se me olvida. Debemos re-escribir la $\cos^2$ a plazo de primera o de cómo abordar? Gracias

Answer 1

El uso de $$ \cos^2(x)=\frac{1+\cos(2x)}{2} $$ tenemos que $$ \begin{align} &\int_0^1\int_0^1\cdots\int_0^1\cos^2\left(\frac{a\pi}{2n}(x_1+x_2+\dots+x_n)\right)\,\mathrm{d}x_1\,\mathrm{d}x_2\dots\,\mathrm{d}x_n\\ &=\frac12+\frac12\mathrm{Re}\left(\int_0^1\int_0^1\cdots\int_0^1e^{\frac{ia\pi}{n}(x_1+x_2+\dots+x_n)}\,\mathrm{d}x_1\,\mathrm{d}x_2\dots\,\mathrm{d}x_n\right)\\ &=\frac12+\frac12\mathrm{Re}\left(\left[\int_0^1e^{\frac{ia\pi}{n}x}\,\mathrm{d}x\right]^n\right)\\ &=\frac12+\frac12\mathrm{Re}\left(\left[\frac{n}{ia\pi}\right]^n\left[e^{\frac{ia\pi}{n}}-1\right]^n\right)\\ &=\frac12+\frac12\mathrm{Re}\left(\left[\frac{2n}{a\pi}\sin\left(\frac{a\pi}{2n}\right)\right]^ne^{\frac{ia\pi}{2}}\right)\\ &=\frac12+\frac12\left[\frac{2n}{a\pi}\sin\left(\frac{a\pi}{2n}\right)\right]^n\color{#C00000}{\cos\left(\frac{a\pi}{2}\right)}\\ &\to\frac12+\frac12\cos\left(\frac{a\pi}{2}\right)\\ &=\cos^2\left(\frac{a\pi}{4}\right) \end{align} $$ Si $a=1$, $\color{#C00000}{\cos\left(\frac{a\pi}{2}\right)}$ es $0$, por lo que la integral es $\frac12$ todos los $n$.

Answer 2

$\newcommand{\+}{^{\daga}} \newcommand{\ángulos}[1]{\left\langle\, nº 1 \,\right\rangle} \newcommand{\llaves}[1]{\left\lbrace\, nº 1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, nº 1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, nº 1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\piso}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\mitad}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\a la derecha\vert\,} \newcommand{\cy}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left (\, nº 1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\vphantom{\large Un}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, nº 1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} &\color{#c00000}{\lim_{n \to \infty}\int_{0}^{1}\int_{0}^{1}\ldots\int_{0}^{1} \cos^{2}\pars{{\pi \over 2n}\,\bracks{x_{1} + x_{2} + \cdots + x_{n}}} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}} \\[3mm]&=\half\bracks{% 1 + \color{#00f}{\lim_{n \to \infty}\int_{0}^{1}\int_{0}^{1}\ldots\int_{0}^{1} \cos\pars{{\pi \over n}\,\bracks{x_{1} + x_{2} + \cdots + x_{n}}} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}}}\tag{1} \end{align}

\begin{align} &\color{#00f}{\lim_{n \to \infty}\int_{0}^{1}\int_{0}^{1}\ldots\int_{0}^{1} \cos\pars{{\pi \over n}\,\bracks{x_{1} + x_{2} + \cdots + x_{n}}} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n}} \\[3mm]&=\lim_{n \to \infty}\Re\int_{0}^{1}\int_{0}^{1}\ldots\int_{0}^{1} \int_{-\infty}^{\infty}\expo{\ic\pi x/n}\delta\pars{x - \sum_{k = 1}^{n}x_{k}}\,\dd x \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n} \\[3mm]&=\lim_{n \to \infty}\Re\int_{0}^{1}\int_{0}^{1}\ldots\int_{0}^{1} \int_{-\infty}^{\infty}\expo{\ic\pi x/n} \int_{-\infty}^{\infty}\exp\pars{\ic q\bracks{x - \sum_{k = 1}^{n}x_{k}}} \,{\dd q \over 2\pi}\,\dd x\,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n} \\[3mm]&=\lim_{n \to \infty}\Re\int_{-\infty}^{\infty}\dd q\ \overbrace{\int_{-\infty}^{\infty}{\dd x \over 2\pi}\, \exp\pars{\ic\bracks{q + {\pi \over n}}x}}^{\ds{=\ \delta\pars{q + {\pi \over n}}}}\ \pars{\int_{0}^{1}\expo{-\ic q\xi}\,\dd\xi}^{n} \\[3mm]&=\lim_{n \to \infty} \Re\bracks{\pars{\int_{0}^{1}\expo{\ic\pi\xi/n}\,\dd\xi}^{n}} =\lim_{n \to \infty} \Re\bracks{\pars{\expo{\ic\pi/n} - 1 \over \ic\pi/n}^{n}} \\[3mm]&=\lim_{n \to \infty} \Re\bracks{% \expo{-\ic\pi/2}\pars{\expo{\ic\pi/2n} - \expo{-\ic\pi/2n} \over \ic\pi/n}^{n}} =\lim_{n \to \infty} \Re\braces{\expo{-\ic\pi/2}\bracks{\sin\pars{\pi/2n} \over \pi/2n}^{n}} \\[3mm]&=\color{#00f}{\large 0} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad\pars{2} \end{align}

Mediante la sustitución de $\pars{2}$ en la expresión $\pars{1}$ nos encontramos con: $$ \color{#00f}{\large\lim_{n \to \infty}\int_{0}^{1}\int_{0}^{1}\ldots\int_{0}^{1} \cos^{2}\pars{{\pi \más de 2n}\,\bracks{x_{1} + x_{2} + \cdots + x_{n}}} \,\dd x_{1}\,\dd x_{2}\ldots\dd x_{n} = \mitad} $$

Answer 3

$\displaystyle \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \cos^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big)dx_1 dx_2...dx_n $

$\displaystyle = \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \cos^2\big(\frac{\pi}{2n}(1-x_1+1-x_2+...+1-x_n)\big)dx_1 dx_2...dx_n $

$\displaystyle = \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \sin^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big)dx_1 dx_2...dx_n$

$\displaystyle = \frac12 \lim_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \sin^2\big(\frac{\pi}{2n}(x_1+x_2+...x_n)\big) + \cos^2\big(\frac{\pi}{2n}(x_1+x_2+...+x_n)\big)dx_1 dx_2...dx_n$

$=\dfrac12$

Answer 4

Sugerencia: $\displaystyle\frac1n\sum_1^nx_k$ es el valor de la media de $\bar x$, que, por $n\to\infty$, tiende a $\dfrac{a+b}2$$x_k\in(a,b)$.