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Add exercice 1 for intégrales généralisées
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\documentclass[a4paper,french,11pt]{article}
Add exercices Analyse
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\title{Analyse --- Exercices}
\author{}
\date{Dernière compilation~: \today{} à \currenttime}
\usepackage{../../cours}
\usepackage{enumitem}
\usepackage{xfrac}
\begin{document}
\maketitle
\section{Équations différentielles d'ordre 1}
\paragraph{$(E_1)$}
$y' - 2y = x^2$
\begin{enumerate}[label=\alph*)]
\item Solution homogène
\begin{align*}
r = 2 \implies y_0 = \lambda e^{2x}
\end{align*}
\item Solution particulière
\begin{align*}
\left\{
\begin{array}{l}
y_1 = ax^2 + bx + c \\
y_1' = 2ax + b \\
\end{array}
\right.
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\end{align*}
\begin{align*}
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\text{Dans } (E_1)
&\implies
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2ax + b - 2(ax^2 + bx + c) = x^2 \\
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&\iff
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\left\{
\begin{array}{l}
-2a = 1 \\
2a - 2b = 0 \\
b - 2c = 0 \\
\end{array}
\right.
Minor change
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\iff
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\left\{
\begin{array}{l}
a = \sfrac{-1}{2} \\
b = \sfrac{-1}{2} \\
c = \sfrac{-1}{4} \\
\end{array}
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\right.
\end{align*}
\begin{align*}
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\implies y_1 = -\frac{1}{2}x^2 -\frac{1}{2}x -\frac{1}{4}
\end{align*}
\item Solution générale
\begin{equation*}
y = y_0 + y_1 = \boxed{-\frac{1}{2}x^2 -\frac{1}{2}x -\frac{1}{4} + \lambda e^{2x}}
\end{equation*}
\end{enumerate}
\paragraph{$(E_2)$}
$3y' - 9y = 7e^{3x}$
\begin{enumerate}[label=\alph*)]
\item Solution homogène
\begin{align*}
r = 3 \implies y_0 = \lambda e^{3x}
\end{align*}
\item Solution particulière
\begin{align*}
\left\{
\begin{array}{l}
y_1 = ax e^{3x} \\
y_1' = ae^{3x} + 3axe^{3x} = (3ax + a)e^{3x} \\
\end{array}
\right.
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\end{align*}
\begin{align*}
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\text{Dans } (E_2)
&\implies 3(3ax + a)e^{3x} - 9axe^{3x} = 7e^{3x} \\
&\iff (9ax + 3a)e^{3x} - 9axe^{3x} = 7e^{3x} \\
&\iff 9ax + 3a - 9ax = 7 \\
&\iff 3a = 7 \\
&\iff a = \frac{7}{3}
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\end{align*}
\begin{align*}
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\implies y_1 = \frac{7}{3}xe^{3x}
\end{align*}
\item Solution générale
\begin{equation*}
y = y_0 + y_1 = \lambda e^{3x} + \frac{7}{3}xe^{3x} = \boxed{(\frac{7}{3}x + \lambda)e^{3x}}
\end{equation*}
\end{enumerate}
\paragraph{$(E_3)$}
$2y' - 4y = 5e^{3x}$
\begin{enumerate}[label=\alph*)]
\item Solution homogène
\begin{align*}
r = 2 \implies y_0 = \lambda e^{2x}
\end{align*}
\item Solution particulière
\begin{align*}
\left\{
\begin{array}{l}
y_1 = ae^{3x} \\
y_1' = 3ae^{3x} \\
\end{array}
\right.
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\end{align*}
\begin{align*}
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\text{Dans } (E_3)
&\iff 6ae^{3x} - 4ae^{3x} = 5e^{3x} \\
&\iff 2a = 5
\iff a = \frac{5}{2}
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\end{align*}
\begin{align*}
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\implies y_1 = \frac{5}{2}e^{3x}
\end{align*}
\item Solution générale
\begin{equation*}
y = y_0 + y_1 = \boxed{\lambda e^{2x} + \frac{5}{2}e^{3x}}
\end{equation*}
\end{enumerate}
\paragraph{$(E_4)$}
$y' + 4y = 3\cos(2x)$
\begin{enumerate}[label=\alph*)]
\item Solution homogène
\begin{align*}
r = -4 \implies y_0 = \lambda e^{-4x}
\end{align*}
\item Solution particulière
\begin{align*}
\left\{
\begin{array}{l}
y_1 = a\cos(2x) + b\sin(2x) \\
y_1' = -2a\sin(2x) + 2b\cos(2x) \\
\end{array}
\right.
\end{align*}
\begin{align*}
Minor change
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\text{Dans } (E_4)
&\implies -2a\sin(2x) + 2b\cos(2x) + 4(a\cos(2x) + b\sin(2x)) = 3\cos(2x) \\
&\iff
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\left\{
\begin{array}{l}
4a + 2b = 3 \\
4b - 2a = 0 \\
\end{array}
\right.
Minor change
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\iff
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\left\{
\begin{array}{l}
4a + 2b = 3 \\
4a - 8b = 0 \\
\end{array}
\right.
Minor change
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\iff
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\left\{
\begin{array}{l}
10b = 3 \\
4b - 2a = 0 \\
\end{array}
\right. \\
Minor change
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&\iff
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\left\{
\begin{array}{l}
b = \sfrac{3}{10} \\
a = \sfrac{12}{20} = \sfrac{3}{5} \\
\end{array}
\right.
\end{align*}
\begin{align*}
\implies y_1 = \frac{3}{5}\cos(2x) + \frac{3}{10}\sin(2x)
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\end{align*}
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\item Solution générale
\begin{equation*}
y = y_0 + y_1 = \boxed{\lambda e^{-4x} + \frac{3}{5}\cos(2x) + \frac{3}{10}\sin(2x)}
\end{equation*}
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\end{enumerate}
\section{Équations différentielles d'ordre 2}
Start exercices second ordre
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\paragraph{$(E_5)$}
$y'' - 7y' + 10 y = (x + 3) e^{2x}$
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\begin{enumerate}[label=\alph*)]
\item Solution homogène
\begin{align*}
r^2 - 7r + 10 = 0
\implies \Delta = 9
\implies
\left\{
\begin{array}{l}
r_1 = \frac{7 - 3}{2} = 2 \\\\
r_2 = \frac{7 + 3}{2} = 5 \\
\end{array}
\right.
\end{align*}
\begin{align*}
\implies
y_0 = \lambda e^{2x} + \mu e^{5x}
\end{align*}
\item Solution particulière
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second membre~: $(x + 3)e^{\alpha x}$ avec $\alpha = 2$ $\implies \alpha$ racine de l'équation caractéristique.
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\begin{align*}
&\left\{
\begin{array}{l}
y_1 = xe^{2x} (ax + b) \\
y_1' = (2x + 1)e^{2x}(ax + b) + axe^{2x} \\
y_1'' = (4ax + 2a + 2b)e^{2x} + (4ax^2 + 4ax + 4bx + 2b)e^{2x} \\
\end{array}
\right. \\
\implies
&\left\{
\begin{array}{l}
y_1 = xe^{2x} (ax + b) \\
y_1' = (2ax^2 + 2ax + 2bx + b)e^{2x} \\
y_1'' = (4ax^2 + 8ax + 4bx + 2a + 4b)e^{2x} \\
\end{array}
\right.
\end{align*}
Dans $(E_5)$~:
$(4ax^2 + 8ax + 4bx + 2a + 4b)e^{2x} - 7(2ax^2 + 2ax + 2bx + b)e^{2x} + 10(ax + b)xe^{2x} = (x + 3) e^{2x}$
\begin{align*}
\implies
-6ax + 2a - 3b = x + 3
\implies
\left\{
\begin{array}{l}
-6a = 1 \\
2a - 3b = 3 \\
\end{array}
\right.
\implies
\left\{
\begin{array}{l}
a = \frac{-1}{6} \\\\
b = \frac{-10}{9} \\
\end{array}
\right.
\end{align*}
\begin{align*}
\implies
y_1 = xe^{2x}(-\frac{1}{6}x - \frac{10}{9})
\end{align*}
\item Solution générale
\begin{equation*}
y = y_0 + y_1 = \boxed{\lambda e^{2x} + \mu e^{5x} + xe^{2x}(-\frac{1}{6}x - \frac{10}{9})}
\end{equation*}
\end{enumerate}
Start exercices second ordre
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\paragraph{$(E_6)$}
$y'' - y = x^3$
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\begin{enumerate}[label=\alph*)]
\item Solution homogène
\begin{align*}
r^2 - 1 = 0
\implies \Delta = 4
\implies
\left\{
\begin{array}{l}
r_1 = \frac{0 - 2}{2} = -1 \\\\
r_2 = \frac{0 + 2}{2} = 1 \\
\end{array}
\right.
\end{align*}
\begin{align*}
\implies
y_0 = \lambda e^{-x} + \mu e^{x}
\end{align*}
\item Solution particulière
second membre~: $x^3e^{\alpha x}$ avec $\alpha = 0$ $\implies \alpha$ non racine de l'équation caractéristique.
\begin{align*}
&\left\{
\begin{array}{l}
y_1 = ax^3 + bx^2 + cx + d \\
y_1' = 3ax^2 + 2bx + c \\
y_1'' = 6ax + 2b \\
\end{array}
\right.
\end{align*}
Dans $(E_6)$~:
$6ax + 2b - ax^3 - bx^2 - cx - d = x^3$
\begin{align*}
\implies
-ax^3 - bx^2 + 6ax - cx + 2b - d = x^3
\implies
\left\{
\begin{array}{l}
-a = 1 \\
-b = 0 \\
6a - c = 0 \\
2b - d = 0 \\
\end{array}
\right.
\implies
\left\{
\begin{array}{l}
a = -1 \\
b = 0 \\
c = -6 \\
d = 0 \\
\end{array}
\right.
\end{align*}
\begin{align*}
\implies
y_1 = -x^3 - 6x
\end{align*}
\item Solution générale
\begin{equation*}
y = y_0 + y_1 = \boxed{\lambda e^{-x} + \mu e^{x} - x^3 - 6x}
\end{equation*}
\end{enumerate}
Start exercices second ordre
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\paragraph{$(E_7)$}
$y'' + y = \cos{x}$
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\begin{enumerate}[label=\alph*)]
\item Solution homogène
\begin{align*}
r^2 + 1 = 0
\implies \Delta = -4
\implies
\left\{
\begin{array}{l}
r_1 = \frac{0 - 2i}{2} \\\\
r_2 = \frac{0 + 2i}{2} \\
\end{array}
\right.
\implies
\left\{
\begin{array}{l}
\alpha = 0 \\
\beta = 1
\end{array}
\right.
\end{align*}
\begin{align*}
\implies
y_0 = e^{0}(\lambda \cos{x} + \mu \sin{x})
= \lambda \cos{x} + \mu \sin{x}
\end{align*}
\item Solution particulière
Correction
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second membre~: $e^{\alpha x}(P_0(x)\cos{\beta x} + 0 \times \sin{\beta x})$ avec $\alpha = 0, \beta = 1, P_0 = 1$ $\implies \alpha + i\beta$ racine de l'équation caractéristique.
Add E7 equa diff 2nd ordre
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\begin{align*}
y_1 &= xe^{\alpha x}(a\cos{\beta x} + b\sin{\beta x}) \\
&= x(a\cos{x} + b\sin{x}) \\
\implies
&\left\{
\begin{array}{l}
y_1 = x(a\cos{x} + b\sin{x}) \\
y_1' = a\cos{x} + b\sin{x} - x(a\sin{x} - b\cos{x}) \\
y_1'' = -a\sin{x} + b\cos{x} - a\sin{x} + b\cos{x} - x(a\cos{x} + b\sin{x}) \\
\end{array}
\right. \\
\implies
&\left\{
\begin{array}{l}
y_1 = x(a\cos{x} + b\sin{x}) \\
y_1' = a\cos{x} + b\sin{x} - x(a\sin{x} - b\cos{x}) \\
y_1'' = -2a\sin{x} + 2b\cos{x} - x(a\cos{x} + b\sin{x}) \\
\end{array}
\right.
\end{align*}
Dans $(E_7)$~:
\begin{align*}
-2a\sin{x} + 2b\cos{x} - x(a\cos{x} + b\sin{x}) + x(a\cos{x} + b\sin{x}) = \cos{x} \\
\iff
-2a\sin{x} + 2b\cos{x} = \cos{x} \\
\iff
2b\cos{x} - 2a\sin{x} = \cos{x} \\
\implies
\left\{
\begin{array}{l}
2b = 1 \\
-a = 0 \\
\end{array}
\right.
\implies
\left\{
\begin{array}{l}
a = 0 \\
Minor change
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b = \frac{1}{2} \\
Add E7 equa diff 2nd ordre
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\end{array}
\right.
\end{align*}
\begin{align*}
\implies
y_1 = \frac{x\sin{x}}{2} \\
\end{align*}
\item Solution générale
\begin{equation*}
y = y_0 + y_1 = \boxed{\lambda \cos{x} + \mu \sin{x} + \frac{x\sin{x}}{2}}
\end{equation*}
\end{enumerate}
Start exercices second ordre
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\paragraph{$(E_8)$}
$y'' - 4y = (-4x + 3) e^{2x}$
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\begin{enumerate}[label=\alph*)]
\item Solution homogène
\begin{align*}
r^2 - 4 = 0
\implies \Delta = 16
\implies
\left\{
\begin{array}{l}
r_1 = \frac{0 - 4}{2} = -2 \\\\
r_2 = \frac{0 + 4}{2} = 2 \\
\end{array}
\right.
\end{align*}
\begin{align*}
\implies
y_0 = \lambda e^{-2x} + \mu e^{2x}
\end{align*}
\item Solution particulière
second membre~: $(-4x + 3)e^{\alpha x}$ avec $\alpha = 2$ $\implies \alpha$ racine de l'équation caractéristique.
\begin{align*}
&\left\{
\begin{array}{l}
y_1 = xe^{2x} (ax + b) \\
y_1' = (2x + 1)e^{2x}(ax + b) + axe^{2x} \\
y_1'' = (4ax + 2a + 2b)e^{2x} + (4ax^2 + 4ax + 4bx + 2b)e^{2x} \\
\end{array}
\right. \\
\implies
&\left\{
\begin{array}{l}
y_1 = xe^{2x} (ax + b) \\
y_1' = (2ax^2 + 2ax + 2bx + b)e^{2x} \\
y_1'' = (4ax^2 + 8ax + 4bx + 2a + 4b)e^{2x} \\
\end{array}
\right.
\end{align*}
Dans $(E_8)$~:
$(4ax^2 + 8ax + 4bx + 2a + 4b)e^{2x} - 4x(ax + b)e^{2x} = (-4x + 3) e^{2x}$
\begin{align*}
\implies
8ax + 2a + 4b = -4x + 3
\implies
\left\{
\begin{array}{l}
8a = -4 \\
2a + 4b = 3 \\
\end{array}
\right.
\implies
\left\{
\begin{array}{l}
a = \frac{-1}{2} \\
b = 1 \\
\end{array}
\right.
\end{align*}
\begin{align*}
\implies
y_1 = xe^{2x} (\frac{-x}{2} + 1)
\end{align*}
\item Solution générale
\begin{equation*}
y = y_0 + y_1 = \boxed{\lambda e^{-2x} + \mu e^{2x} + xe^{2x} (\frac{-x}{2} + 1)}
\end{equation*}
\end{enumerate}
Add exercice 1 for intégrales généralisées
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\section{Intégrales généralisées}
\subsection{Exercice 1}
Étudier la convergence des intégrales généralisées suivantes.
\paragraph{$(I_1)$}
Make macro for d in integrals
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$\int_1^{+\infty} \frac{2x + x^3}{x^3 + x^4} \dif x$
Add exercice 1 for intégrales généralisées
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\begin{align*}
\forall x \in [1; +\infty[\quad \frac{2x + x^3}{x^3 + x^4} \geq 0 \\\\
\frac{2x + x^3}{x^3 + x^4} = \frac{x^3(\frac{2}{x^2} + 1)}{x^4(\frac{1}{x} + 1)} \sim \frac{x^3}{x^4} \sim \frac{1}{x} \\
Make macro for d in integrals
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\text{Or } \int_1^{+\infty}\frac{1}{x}\dif x \text{ diverge donc $I_1$ diverge par équivalence.} \\
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\end{align*}
\paragraph{$(I_2)$}
Make macro for d in integrals
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$\int_0^{+\infty} \frac{x\sqrt{x}}{x^2 \sqrt[3]{x} + 4} \dif x$
Add exercice 1 for intégrales généralisées
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\begin{align*}
\forall x \in [0; +\infty[\quad \frac{x\sqrt{x}}{x^2 \sqrt[3]{x} + 4} \geq 0 \\\\
\frac{x\sqrt{x}}{x^2 \sqrt[3]{x} + 4}
= \frac{x^{1+\frac{1}{2}}}{x^{2+\frac{1}{3}}(1 + \frac{4}{x^{\frac{7}{3}}})}
\sim \frac{x^{1+\frac{1}{2}}}{x^{\frac{7}{3}}}
\sim x^{\frac{3}{2} - \frac{7}{3}}
\sim x^{-\frac{5}{6}}
\sim \frac{1}{x^{\frac{5}{6}}} \\
Make macro for d in integrals
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\text{Or } \int_1^{+\infty}\frac{1}{x^{\frac{5}{6}}}\dif x \text{ diverge donc $I_2$ diverge par équivalence.} \\
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\end{align*}
\paragraph{$(I_3)$}
Make macro for d in integrals
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$\int_1^{+\infty} \frac{5x + x^2}{x^3 + x^3\sqrt{x}} \dif x$
Add exercice 1 for intégrales généralisées
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\begin{align*}
\forall x \in [1; +\infty[\quad \frac{5x + x^2}{x^3 + x^3\sqrt{x}} \geq 0 \\\\
\frac{5x + x^2}{x^3 + x^3\sqrt{x}}
= \frac{x^2(\frac{5x}{x^2} + 1)}{x^3\sqrt{x}(\frac{1}{\sqrt{x}} + 1)}
= \frac{x^2(\frac{5}{x} + 1)}{x^{3 + \frac{1}{2}}(\frac{1}{\sqrt{x}} + 1)}
\sim \frac{x^2}{x^{\frac{7}{2}}}
\sim \frac{1}{x^{\frac{3}{2}}} \\
Make macro for d in integrals
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\text{Or } \int_1^{+\infty}\frac{1}{x^{\frac{3}{2}}}\dif x \text{ converge donc $I_3$ converge par équivalence.} \\
Add exercice 1 for intégrales généralisées
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\end{align*}
\paragraph{$(I_4)$}
Make macro for d in integrals
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$\int_0^{+\infty} \frac{x \sqrt[3]{x}}{x^2\sqrt{x} + 5} e^{-x} \dif x$
Add exercice 1 for intégrales généralisées
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\begin{align*}
\forall x \in [0; +\infty[ \quad \frac{x \sqrt[3]{x}}{x^2\sqrt{x} + 5} e^{-x} \geq 0 \\\\
\frac{x \sqrt[3]{x}}{x^2\sqrt{x} + 5} e^{-x}
= \frac{x^{\frac{4}{3}}}{x^2\sqrt{x}(1 + \frac{5}{x^2\sqrt{x}})} e^{-x}
\sim \frac{x^{\frac{4}{3}}}{x^{\frac{5}{2}}} e^{-x}
\sim x^{\frac{-7}{6}} e^{-x}
\sim \frac{1}{x^{\frac{7}{6}}} e^{-x} \\
Make macro for d in integrals
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\text{Or } \int_0^{+\infty}\frac{1}{x^{\frac{7}{6}}} e^{-x}\dif x \text{ converge (l'exponentielle l'emporte) donc $I_4$ converge par équivalence.} \\
Add exercice 1 for intégrales généralisées
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(\sim e^{-kx} \text{ avec } k = 1 > 0)
\end{align*}
Add exercice 2 for intégrales généralisées
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\subsection{Exercice 2}
Étudier la convergence absolue de l'intégrale généralisée suivante.
\paragraph{$(I_5)$}
Make macro for d in integrals
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$\int_1^{+\infty} \frac{\sin x}{x^2} \dif x$
Add exercice 2 for intégrales généralisées
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$(I_5)$ n'est pas strictement positif sur $[1;+\infty[$.
Nous allons donc étudier sa valeur absolue.
\begin{align*}
-1 \leq \sin{x} \leq 1 \\
0 \leq |\sin{x}| \leq 1 \\
0 \leq \left|\frac{\sin{x}}{x^2}\right| \leq \frac{1}{x^2} \\
Make macro for d in integrals
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\text{Or } \int_1^{+\infty}\frac{1}{x^2}\dif x \text{ converge donc } \left|\frac{\sin{x}}{x^2}\right| \text{ converge par majoration.} \\
Add exercice 2 for intégrales généralisées
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\implies I_5 \text{ converge absolument.} \\
\end{align*}
Start exercices Fourier
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\section{Séries de Fourier}
\subsection{Exercice 1}
\begin{enumerate}
\item Déterminer le développement en série de Fourier de la fonction $f$, 2-périodique, définie par~: \\
$f(x) = |x| \quad \forall x \in [-1;1[$
More Fourier
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$f(-x) = |-x| = |x| = f(x)$ \quad donc $f$ est paire $\implies b_n = 0$
\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}
Start exercices Fourier
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\begin{equation*}
More Fourier
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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)
Start exercices Fourier
2021-10-16 18:20:03 +02:00
\end{equation*}
\item En déduire que \quad $\frac{\pi^2}{8} = \sum_{n=0}^{+\infty} \frac{1}{(2n + 1)^2}$
More Fourier
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\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}
Start exercices Fourier
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\item Appliquer l'identité de Parseval-Bessel.
En déduire la valeur de la somme \quad $\sum_{n=0}^{+\infty}\frac{1}{(2n + 1)^4}$
Update exo Fourier
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\begin{align*}
More Fourier
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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 \\
Update exo Fourier
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\end{align*}
Start exercices Fourier
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\end{enumerate}
Add exercices Analyse
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\end{document}
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