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