Questions: Lagrange Polynomial Interpolation

Question 1

You have 5 distinct data points. What is the maximum degree of the Lagrange interpolating polynomial that passes through all of them?

• A. 5
• B. 4
• C. 3
• D. It depends on whether the y-values are linearly independent

Answer: (B)

Given n+1 distinct points, the unique interpolating polynomial has degree *at most* n. With 5 points, n+1 = 5 so n = 4, and the polynomial has degree at most 4. It may be lower if the points happen to lie on a lower-degree polynomial. A common error is confusing the number of points (5) with the degree (at most 4).

Question 2

What is the defining property of the Lagrange basis polynomial L_i(x)?

• A. L_i(x_i) = 0 for every interpolation node x_i
• B. L_i(x_j) = 1 for all nodes x_j in the data set
• C. L_i(x_j) = 0 for j ≠ i and L_i(x_i) = 1
• D. L_i(x) is the derivative of the interpolating polynomial at x_i

Answer: (C)

L_i(x) is constructed to equal 1 at node x_i and 0 at all other nodes x_j (j ≠ i). This property is what makes P(x) = Σ y_i L_i(x) work: at any node x_k, every term vanishes except y_k · L_k(x_k) = y_k · 1 = y_k, confirming the polynomial passes through all data points. The numerator product ∏_{j≠i}(x − x_j) gives zeros at all other nodes; dividing by ∏_{j≠i}(x_i − x_j) normalizes it to 1 at x_i.

Question 3

Given n+1 distinct data points, there exists exactly one polynomial of degree ≤ n that passes through all of them.

• T. True
• F. False

Answer: True

This is the uniqueness theorem for polynomial interpolation. Existence is provided by the Lagrange (or Newton) construction. Uniqueness follows because if two polynomials of degree ≤ n agree at n+1 points, their difference is a polynomial of degree ≤ n with n+1 roots, which forces it to be identically zero.

Question 4

The Lagrange basis polynomial L_i(x) equals 1 at nearly every interpolation node x_j in the data set.

• T. True
• F. False

Answer: False

L_i(x) equals 1 *only* at x_i and equals 0 at all other nodes x_j (j ≠ i). This selective property — being 1 at exactly one node and 0 at all others — is precisely what allows the Lagrange sum P(x) = Σ y_i L_i(x) to reproduce the correct y-value at every data point.

Question 5

Why does the Lagrange form become computationally inconvenient when a new data point is added to an existing interpolation set?

Model answer: Each Lagrange basis polynomial L_i(x) = ∏_{j≠i} (x − x_j)/(x_i − x_j) depends on all n+1 interpolation nodes. When a new node x_{n+1} is added, every existing basis polynomial must be updated because its numerator product must now include the new factor (x − x_{n+1}), and its denominator must include (x_i − x_{n+1}). All n+1 existing basis functions must be completely recomputed, making updates cost O(n²) rather than incremental.

This contrasts with Newton's divided-difference form, where adding a new point requires only one new divided difference and appending one new term. The Lagrange form is mathematically elegant and useful for theoretical analysis, but Newton's form is preferred when incremental updates are needed.