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Let's solve the given problem step-by-step.

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**1. Problem Breakdown:**

The function is \( f(x) = 2\cos(x) \). We are asked:

1. To find the Lagrange interpolation polynomial \( P_2(x) \) of degree 2 that interpolates \( f(x) \) at points \( 0 \), \( \pi \), and \( 2\pi \).
2. To compute \( P_2\left(\frac{3\pi}{4}\right) \).
3. To find an upper bound for the interpolation error.

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### **Step 1: Lagrange Interpolation Polynomial**
The general formula for the Lagrange interpolation polynomial is:
\[
P_2(x) = \sum_{i=0}^2 f(x_i) \ell_i(x),
\]
where \( \ell_i(x) \) is the Lagrange basis polynomial:
\[
\ell_i(x) = \prod_{\substack{j=0 \\ j \neq i}}^2 \frac{x - x_j}{x_i - x_j}.
\]
Here the interpolation points are:
\[
x_0 = 0, \quad x_1 = \pi, \quad x_2 = 2\pi.
\]

#### Step 1.1: Compute \( f(x_i) \)
\[
f(0) = 2\cos(0) = 2, \quad f(\pi) = 2\cos(\pi) = -2, \quad f(2\pi) = 2\cos(2\pi) = 2.
\]

#### Step 1.2: Basis Polynomials \( \ell_i(x) \)

- For \( \ell_0(x) \):
\[
\ell_0(x) = \frac{(x - \pi)(x - 2\pi)}{(0 - \pi)(0 - 2\pi)} = \frac{(x - \pi)(x - 2\pi)}{\pi(2\pi)} = \frac{(x - \pi)(x - 2\pi)}{2\pi^2}.
\]

- For \( \ell_1(x) \):
\[
\ell_1(x) = \frac{(x - 0)(x - 2\pi)}

Soit la fonction réelle f définie par f(x) = 2cos(x). Trouver le polynôme d'interpolation de Lagrange de degré 2, noté P2(x), qui interpole f aux points 0, π et 2π. Calculer P2(3π/4). Majorer l'erreur d'interpolation.

Solve the Lagrange interpolation problem for f(x) = 2cos(x) at 0, π, and 2π. Compute P2(3π/4) and estimate the interpolation error. Explore key concepts such as Lagrange polynomials, trigonometric values, and error bounds.

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