diff --git a/analyse/exercices/main.tex b/analyse/exercices/main.tex index 53cb4aa..db5f7a9 100644 --- a/analyse/exercices/main.tex +++ b/analyse/exercices/main.tex @@ -589,140 +589,40 @@ \subsection{Exercice 1} - \begin{enumerate} + Déterminer le développement en série de Fourier de la fonction $f$, 2-périodique, définie par~: \\ + $f(x) = |x| \quad \forall x \in [-1;1[$ - \item Déterminer le développement en série de Fourier de la fonction $f$, 2-périodique, définie par~: \\ - $f(x) = |x| \quad \forall x \in [-1;1[$ + $f(-x) = |-x| = |x| = f(x)$ \quad donc $f$ est paire $\implies b_n = 0$ - $f(-x) = |-x| = |x| = f(x)$ \quad donc $f$ est paire $\implies b_n = 0$ + \begin{tabularx}{\linewidth}{XX} - \begin{tabularx}{\linewidth}{XX} - - {\begin{align*} - a_0 &= \frac{1}{2} \int_{-1}^1 |x| \dif x \\ - &= \int_{0}^1 x \dif x \\ - &= \left[\frac{x^2}{2}\right]_0^1 \\ - a_0 &= \frac{1}{2} - \end{align*}} & - {\begin{align*} - a_n &= \frac{2}{T} \int_{-T/2}^{T/2} |x| \cos{\frac{2\pi nx}{T}} \dif x \\ - a_n &= \frac{2}{2} \int_{-1}^1 |x| \cos{\frac{2\pi nx}{2}} \dif x \\ - &= 2\int_0^1 x \cos(n\pi x) \dif x \\ - \text{IPP } - &\left\{ - \begin{array}{ll} - u = x & u' = 1 \\ - v' = \cos(n\pi x) & v = \frac{\sin{n\pi x}}{n\pi} \\ - \end{array} - \right. \\ - a_n &= 2 \left(\left[x\frac{\sin{n\pi x}}{n\pi}\right]_0^1 - \int_0^1 \frac{\sin{n\pi x}}{n\pi} \dif x \right) \\ - &= 2 \left(\frac{\sin{n\pi}}{n\pi} - 0 + \left[\frac{\cos{n\pi x}}{(n\pi)^2}\right]_0^1 \right) \\ - &= 2 \left(0 + \frac{\cos{n\pi}}{(n\pi)^2} - \frac{\cos{0}}{(n\pi)^2}\right) \\ - a_n &= 2\frac{(-1)^n - 1}{(n\pi)^2} - \end{align*}} \\ - - \end{tabularx} - - \begin{equation*} - \implies f(x) = \frac{1}{2} + \sum_{n=1}^{+\infty} 2\frac{(-1)^n - 1}{(n\pi)^2} \cos(n\pi x) - \end{equation*} - - \item En déduire que \quad $\frac{\pi^2}{8} = \sum_{n=0}^{+\infty} \frac{1}{(2n + 1)^2}$ - - \begin{tabularx}{\linewidth}{XX} - - {pour $x = 0$~: - \begin{align*} - f(0) &= \frac{1}{2} + \sum_{n=1}^{+\infty} 2\frac{(-1)^n - 1}{(n\pi)^2} \cos{0} \\ - \iff 0 &= \frac{1}{2} + \sum_{n=1}^{+\infty} 2\frac{(-1)^n - 1}{(n\pi)^2} \\ - \iff -\frac{1}{2} &= 2 \sum_{n=1}^{+\infty} \frac{(-1)^n - 1}{(n\pi)^2} \\ - \iff -\frac{1}{4} &= \sum_{n=1}^{+\infty} \frac{(-1)^n - 1}{(n\pi)^2} \\ - \iff \frac{1}{4} &= \sum_{n=1}^{+\infty} \frac{(-1)^{n+1} + 1}{(n\pi)^2} \\ - \iff \frac{\pi^2}{4} &= \sum_{n=1}^{+\infty} \frac{(-1)^{n+1} + 1}{n^2} \\ - \iff \frac{\pi^2}{8} &= \sum_{n=1}^{+\infty} \frac{(-1)^{n+1} + 1}{2n^2} - \end{align*}} & - - {pour $x = -1$~: - \begin{align*} - f(-1) &= \frac{1}{2} + \sum_{n=1}^{+\infty} 2\frac{(-1)^n - 1}{(n\pi)^2} \cos(-n\pi) \\ - \iff 1 &= \frac{1}{2} + 2\sum_{n=1}^{+\infty} \frac{(-1)^n - 1}{(n\pi)^2} \cos(n\pi) \\ - \iff \frac{1}{2} &= 2\sum_{n=1}^{+\infty} \frac{(-1)^n - 1}{(n\pi)^2} (-1)^n \\ - \iff \frac{1}{4} &= \sum_{n=1}^{+\infty} \frac{(-1)^{2n} - (-1)^n}{(n\pi)^2} \\ - \iff \frac{1}{4} &= \sum_{n=1}^{+\infty} \frac{1 - (-1)^n}{(n\pi)^2} \\ - \iff \frac{\pi^2}{4} &= \sum_{n=1}^{+\infty} \frac{1 + (-1)^{n+1}}{n^2} \\ - \iff \frac{\pi^2}{8} &= \sum_{n=1}^{+\infty} \frac{(-1)^{n+1} + 1}{2n^2} \\ - \end{align*}} \\ - - \end{tabularx} - - \item Appliquer l'identité de Parseval-Bessel. - En déduire la valeur de la somme \quad $\sum_{n=0}^{+\infty}\frac{1}{(2n + 1)^4}$ - - \begin{align*} - a_0^2 + \frac{1}{2}\sum_{n=1}^{+\infty}a_n^2 = \frac{1}{2}\int_{-1}^{1}f^2(x) \dif x \\ - \end{align*} - - \end{enumerate} - - \subsection{Exercice 2} - - \begin{enumerate} - - \item Déterminer la série de Fourier de la fonction $f$, $2\pi$-périodique, définie par~: \\ - \begin{align*} - \left\{ - \begin{array}{l} - f(x) = e^x \\ - f(\pi) = \cosh{\pi} - \end{array} - \right. - \quad \forall x \in \; ]{-\pi}, \pi[ - \end{align*} - - $f$ est ni paire, ni impaire, ni continue. - - La série de Fourier est donc de la forme~: - \begin{equation*} - S_f(x) = a_0 + \sum_{n=1}^{+\infty} ( a_n \cos(nx) + b_n \sin(nx) ) - \end{equation*} - - \begin{align*} - a_0 &= \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x) \dif x \\ - &= \frac{1}{2\pi} \int_{-\pi}^{\pi} e^x \dif x \\ - &= \frac{1}{2\pi} [e^x]_{-\pi}^{\pi} \\ - &= \frac{1}{2\pi} (e^{\pi} - e^{-\pi}) \\ - &= \frac{1}{2\pi} \left(e^{\pi} - \frac{1}{e^{\pi}}\right) \\ - &= \frac{1}{2\pi} \left(\frac{e^{2\pi}}{e^{\pi}} - \frac{1}{e^{\pi}}\right) \\ - &= \frac{1}{2\pi} \times \frac{e^{2\pi} - 1}{e^{\pi}} \\ - &= \frac{e^{2\pi} - 1}{2\pi e^{\pi}} \\ - \end{align*} - \begin{align*} - a_n &= \frac{2}{2\pi} \int_{-\pi}^{\pi} f(x) \cos{\frac{2\pi nx}{2\pi}} \dif x \\ - &= \frac{1}{\pi} \int_{-\pi}^{\pi} e^x \cos(nx) \dif x \\ - &\text{IPP } - \left\{ + {\begin{align*} + a_0 &= \frac{1}{2} \int_{-1}^1 |x| \dif x \\ + &= \int_{0}^1 x \dif x \\ + &= \left[\frac{x^2}{2}\right]_0^1 \\ + a_0 &= \frac{1}{2} + \end{align*}} & + {\begin{align*} + a_n &= \frac{2}{T} \int_{-T/2}^{T/2} |x| \cos{\frac{2\pi nx}{T}} \dif x \\ + a_n &= \frac{2}{2} \int_{-1}^1 |x| \cos{\frac{2\pi nx}{2}} \dif x \\ + &= 2\int_0^1 x \cos(n\pi x) \dif x \\ + \text{IPP } + &\left\{ \begin{array}{ll} - u = e^x & u' = e^x \\ - v' = \cos(nx) & v = \frac{\sin(nx)}{n} \\ + u = x & u' = 1 \\ + v' = \cos(n\pi x) & v = \frac{\sin{n\pi x}}{n\pi} \\ \end{array} \right. \\ - a_n &= \frac{1}{\pi} \left(\left[e^x \frac{\sin{nx}}{n} \right]_{-\pi}^{\pi} - \int_{-\pi}^{\pi} e^x \frac{\sin(nx)}{n} \dif x \right) \\ - &= \frac{1}{\pi} \left(\left[e^x \frac{\sin{nx}}{n} \right]_{-\pi}^{\pi} - \frac{1}{n} \int_{-\pi}^{\pi} e^x \sin(nx) \dif x \right) \\ - &\text{IPP } - \left\{ - \begin{array}{ll} - u = e^x & u' = e^x \\ - v' = \sin(nx) & v = \frac{-\cos(nx)}{n} \\ - \end{array} - \right. \\ - a_n &= \frac{1}{\pi} \left(\left[e^x \frac{\sin{nx}}{n} \right]_{-\pi}^{\pi} - \frac{1}{n} \left(\left[-e^x\frac{\cos(nx)}{n}\right]_{-\pi}^{\pi} - \int_{-\pi}^{\pi} -e^x \frac{\cos(nx)}{n} \dif x \right)\right) \\ - &= \frac{1}{\pi} \left(\left[e^x \frac{\sin{nx}}{n} \right]_{-\pi}^{\pi} - \frac{1}{n} \left(\left[-e^x\frac{\cos(nx)}{n}\right]_{-\pi}^{\pi} + \int_{-\pi}^{\pi} e^x \frac{\cos(nx)}{n} \dif x \right)\right) \\ - &= \frac{1}{\pi} \left(\left[e^x \frac{\sin{nx}}{n} \right]_{-\pi}^{\pi} + \frac{1}{n} \left(\left[e^x\frac{\cos(nx)}{n}\right]_{-\pi}^{\pi} - \frac{1}{n} \int_{-\pi}^{\pi} e^x \cos(nx) \dif x \right)\right) \\ - \end{align*} - \begin{align*} - b_n &= - \end{align*} + a_n &= 2 \left(\left[x\frac{\sin{n\pi x}}{n\pi}\right]_0^1 - \int_0^1 \frac{\sin{n\pi x}}{n\pi} \dif x \right) \\ + &= 2 \left(\frac{\sin{n\pi}}{n\pi} - 0 + \left[\frac{\cos{n\pi x}}{(n\pi)^2}\right]_0^1 \right) \\ + &= 2 \left(0 + \frac{\cos{n\pi}}{(n\pi)^2} - \frac{\cos{0}}{(n\pi)^2}\right) \\ + a_n &= 2\frac{(-1)^n - 1}{(n\pi)^2} + \end{align*}} \\ - \end{enumerate} + \end{tabularx} + + \begin{equation*} + \implies f(x) = \frac{1}{2} + \sum_{n=1}^{+\infty} 2\frac{(-1)^n - 1}{(n\pi)^2} \cos(n\pi x) + \end{equation*} \end{document}