Lagrange interpolation Please write a python code using the Lagrange interpolation method to find a polynomial to interpolate the following data Xi f(Xi) -1 0.56 3 2.68 5 9.01 8 19.23 10 45.74 use this polynomial to find f(x) at x=7 please upload the code

The Correct Answer and Explanation is:

Here’s the Python code implementing Lagrange interpolation for your data:

pythonCopyEditfrom sympy import symbols, simplify, lambdify

# Define symbol
x = symbols('x')

# Given data points
X = [-1, 3, 5, 8, 10]
F = [0.56, 2.68, 9.01, 19.23, 45.74]

# Number of points
n = len(X)

# Lagrange interpolation polynomial
lagrange_poly = 0

# Construct the polynomial
for i in range(n):
    term = F[i]
    for j in range(n):
        if i != j:
            term *= (x - X[j]) / (X[i] - X[j])
    lagrange_poly += term

# Simplify polynomial
lagrange_poly_simplified = simplify(lagrange_poly)

# Create function for evaluation
f_lagrange = lambdify(x, lagrange_poly_simplified, 'numpy')

# Evaluate at x = 7
f_at_7 = f_lagrange(7)

print("Lagrange Polynomial:")
print(lagrange_poly_simplified)
print(f"\nValue at x = 7: {f_at_7}")

Output

Interpolating Polynomial: f(x)=0.0289×4−0.4769×3+2.5639×2−1.8372x−4.3469f(x) = 0.0289x^4 – 0.4769x^3 + 2.5639x^2 – 1.8372x – 4.3469f(x)=0.0289×4−0.4769×3+2.5639×2−1.8372x−4.3469
Value at x = 7: f(7)≈14.2245f(7) \approx 14.2245f(7)≈14.2245

Explanation

Lagrange interpolation is a classical numerical method used to estimate a polynomial that exactly passes through a given set of points. Given n+1n+1n+1 data points (x0,f0),(x1,f1),…,(xn,fn)(x_0, f_0), (x_1, f_1), \dots, (x_n, f_n)(x0,f0),(x1,f1),…,(xn,fn), the Lagrange polynomial is defined as: L(x)=∑i=0nfi⋅ℓi(x)L(x) = \sum_{i=0}^{n} f_i \cdot \ell_i(x)L(x)=i=0∑nfi⋅ℓi(x)

where each ℓi(x)\ell_i(x)ℓi(x) is the Lagrange basis polynomial: ℓi(x)=∏j=0j≠inx−xjxi−xj\ell_i(x) = \prod_{\substack{j=0 \\ j \ne i}}^{n} \frac{x – x_j}{x_i – x_j}ℓi(x)=j=0j=i∏nxi−xjx−xj

Each basis polynomial ℓi(x)\ell_i(x)ℓi(x) equals 1 at x=xix = x_ix=xi and 0 at all other xjx_jxj, ensuring the polynomial matches the given function values exactly at the specified data points.

In this case, we were given five points:

xix_ixi	f(xi)f(x_i)f(xi)
-1	0.56
3	2.68
5	9.01
8	19.23
10	45.74

Using these, we constructed the interpolating polynomial of degree 4. The result is a fourth-degree polynomial explicitly defined in terms of xxx. To estimate the function value at x=7x = 7x=7, we substitute this value into the polynomial, yielding f(7)≈14.2245f(7) \approx 14.2245f(7)≈14.2245.

Lagrange interpolation is accurate for small datasets and smooth functions but can be unstable for large datasets due to Runge’s phenomenon. For such cases, piecewise methods (e.g., splines) may be preferable. However, for moderate-sized datasets like this, Lagrange interpolation is efficient and exact.

Output

Explanation

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