Lagrange Interpolation

The Lagrange Basis Polynomials

The interpolation polynomial p(x) is constructed as a linear combination of Lagrange basis polynomials, L_i(x). Each basis polynomial is designed to equal 1 at its corresponding node x_i and 0 at all other nodes x_j (where j ≠ i).

The formula for the i-th basis polynomial is the product of (x - x_j) / (x_i - x_j) for all j ≠ i. The full interpolating polynomial is simply the sum of y_i × L_i(x) across all points.

Worked example: Quadratic interpolation

Let's interpolate a quadratic polynomial through the nodes (-1, 1), (0, 0), and (1, 1). This is the "Quadratic" preset.

1. Calculate L₀(x) for x₀ = -1:
   L₀(x) = (x - 0) / (-1 - 0) × (x - 1) / (-1 - 1) = x(x - 1) / 2 = (x² - x) / 2
2. Calculate L₁(x) for x₁ = 0:
   L₁(x) = (x - (-1)) / (0 - (-1)) × (x - 1) / (0 - 1) = (x + 1)(x - 1) / -1 = 1 - x²
3. Calculate L₂(x) for x₂ = 1:
   L₂(x) = (x - (-1)) / (1 - (-1)) × (x - 0) / (1 - 0) = (x + 1)x / 2 = (x² + x) / 2
4. Multiply each by its corresponding y-value and sum:
   p(x) = 1 × L₀(x) + 0 × L₁(x) + 1 × L₂(x)
   p(x) = (x² - x) / 2 + 0 + (x² + x) / 2 = x²

To evaluate this interpolant at x = 0.5:
L₀(0.5) = (0.25 - 0.5) / 2 = -0.125 (-1/8)
L₁(0.5) = 1 - 0.25 = 0.75 (3/4)
L₂(0.5) = (0.25 + 0.5) / 2 = 0.375 (3/8)
The sum of these basis weights is -1/8 + 6/8 + 3/8 = 1.
p(0.5) = 1 × (-1/8) + 0 × (3/4) + 1 × (3/8) = 2/8 = 1/4 (0.25).

Verify the interpolation with the module

Open Lagrange Interpolation to explore how moving points affects the global polynomial and local basis weights.

1. Select the Quadratic preset. The nodes are (-1, 1), (0, 0), and (1, 1). Set the evaluation point to 0.5. Verify the basis weights are exactly -0.1250, 0.7500, and 0.3750, summing to 1.0000, and p(0.5) = 0.2500.
2. Move the evaluation point precisely to one of the nodes, for example x = 1.00. Notice that the weight for L₂ becomes 1.0000 and all other weights become 0.0000, yielding p(1.00) = 1.0000 exactly. This demonstrates that Lagrange polynomials perfectly interpolate the given nodes.
3. Switch to the Linear preset with nodes (0, 1) and (2, 5). The resulting polynomial is p(x) = 1 + 2x. Set the evaluation point to 1 and verify p(1) = 3.0000.
4. Switch to the Constant preset. Every interpolated value across the domain is 2. Evaluating outside the node interval (e.g., x < -1 or x > 1) is correctly labeled as extrapolation.

Limitations and Runge's Phenomenon

While Lagrange interpolation guarantees finding a polynomial that passes through all points, it does not guarantee a good approximation between those points, especially for higher degrees with uniformly spaced nodes. This severe oscillation at the edges of an interval is known as Runge's phenomenon.

Evaluating the polynomial outside the range of the given nodes is called extrapolation. As seen in the interface, extrapolation is explicitly labeled. High-degree polynomials can diverge wildly outside the interpolated interval, so an extrapolated value does not imply a guaranteed or safe approximation error.

Mathematically, the formula breaks down if two nodes share the same x-coordinate, which implies division by zero when calculating basis polynomials. The module explicitly rejects duplicate or near-coincident x values (separations below 1e-6) rather than fabricating a unique polynomial, avoiding NaN or Infinity results.

Primary References

For a formal definition of Lagrange interpolation and the remainder term for interpolation error, refer to the Digital Library of Mathematical Functions: NIST DLMF: Polynomial Interpolation. This section supports the construction of the basis polynomials used in this interactive module.