Math Problem Statement

Lagrange Polynomial answer me

Given the following data satisfying y = f(x). Find the Lagrange Polynomial. Hence, find f(12).

Solution

To find the Lagrange Polynomial f(x)f(x) for the given data points and calculate f(12)f(12), follow these steps:

1. Lagrange Polynomial Formula:

The Lagrange polynomial is given by:

f(x)=i=0nyiLi(x)f(x) = \sum_{i=0}^{n} y_i \cdot L_i(x)

where Li(x)L_i(x) is the Lagrange basis polynomial for xix_i:

Li(x)=j=0jinxxjxixjL_i(x) = \prod_{\substack{j=0 \\ j \neq i}}^{n} \frac{x - x_j}{x_i - x_j}

2. Given Data Points:

The data points are: (x0,y0)=(1,18),(x1,y1)=(3,8),(x2,y2)=(5,30),(x3,y3)=(6,88)(x_0, y_0) = (1, 18), \, (x_1, y_1) = (3, -8), \, (x_2, y_2) = (5, 30), \, (x_3, y_3) = (6, 88)

3. Calculate Li(x)L_i(x):

For each ii, compute Li(x)L_i(x):

For L0(x)L_0(x):

= \frac{(x - 3)(x - 5)(x - 6)}{-60}$$ #### For $$L_1(x)$$: $$L_1(x) = \frac{(x - 1)(x - 5)(x - 6)}{(3 - 1)(3 - 5)(3 - 6)} = \frac{(x - 1)(x - 5)(x - 6)}{30}$$ #### For $$L_2(x)$$: $$L_2(x) = \frac{(x - 1)(x - 3)(x - 6)}{(5 - 1)(5 - 3)(5 - 6)} = \frac{(x - 1)(x - 3)(x - 6)}{-24}$$ #### For $$L_3(x)$$: $$L_3(x) = \frac{(x - 1)(x - 3)(x - 5)}{(6 - 1)(6 - 3)(6 - 5)} = \frac{(x - 1)(x - 3)(x - 5)}{30}$$ ### 4. Form $$f(x)$$: Using the values of $$y_i$$, construct $$f(x)$$: $$f(x) = 18 \cdot L_0(x) - 8 \cdot L_1(x) + 30 \cdot L_2(x) + 88 \cdot L_3(x)$$ ### 5. Compute $$f(12)$$: Substitute $$x = 12$$ into the polynomial. To simplify this process and calculate the result, I will perform the calculation programmatically.from sympy import symbols, simplify, prod # Define the variables x = symbols('x') # Given data points points = [(1, 18), (3, -8), (5, 30), (6, 88)] # Calculate Lagrange basis polynomials and f(x) L = [] for i, (xi, yi) in enumerate(points): numerator = prod(x - xj for j, (xj, _) in enumerate(points) if j != i) denominator = prod(xi - xj for j, (xj, _) in enumerate(points) if j != i) Li = numerator / denominator L.append(yi * Li) # Sum up the terms to form the polynomial f_x = simplify(sum(L)) # Evaluate f(12) f_12 = f_x.subs(x, 12) f_x, f_12(x**3 - x**2 - 22*x + 40, 1360)(x**3 - x**2 - 22*x + 40, 1360)successThe Lagrange polynomial is: $$f(x) = x^3 - x^2 - 22x + 40$$ The value of $$f(12)$$ is: $$f(12) = 1360$$ Do you want me to explain any step in further detail or answer related questions? Here are some questions you might consider: 1. How do Lagrange basis polynomials work in interpolation? 2. Can you explain why Lagrange polynomials guarantee accuracy for given points? 3. What are some practical applications of Lagrange interpolation? 4. How does the polynomial behave between the given data points? 5. Can you calculate $$f(x)$$ at another specific value? **Tip:** The Lagrange polynomial is efficient for small data sets, but for larger ones, consider using piecewise or spline interpolation for better numerical stability!

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Math Problem Analysis

Mathematical Concepts

Lagrange Interpolation
Polynomial Interpolation

Formulas

f(x) = Σ(y_i * L_i(x))
L_i(x) = Π((x - x_j) / (x_i - x_j)) for j ≠ i

Theorems

Lagrange Interpolation Theorem

Suitable Grade Level

Undergraduate Level or Advanced High School