Last modified: March 23, 2026

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Lagrange Polynomial Interpolation

Lagrange Polynomial Interpolation is a widely used technique for determining a polynomial that passes exactly through a given set of data points. Suppose we have a set of $(n+1)$ data points $(x_0, y_0), (x_1, y_1), \ldots, (x_n, y_n)$ where all $x_i$ are distinct. The aim is to find a polynomial $L(x)$ of degree at most $n$ such that:

$$L(x_i) = y_i, \quad \text{for} \; i=0,1,\ldots,n$$

Instead of solving a system of linear equations (as would be required if we used a general polynomial form), Lagrange interpolation provides a direct formula for the interpolating polynomial in terms of Lagrange basis polynomials. This approach is conceptually straightforward and does not require forming and solving large linear systems.

Conceptual Illustration:

Imagine that you have three points $(x_0, y_0), (x_1, y_1), (x_2, y_2)$. The Lagrange form builds a polynomial that goes exactly through these points. Each Lagrange basis polynomial $\ell_i(x)$ is constructed to be zero at all $x_j$ except $x_i$. By taking a suitable linear combination of these basis polynomials with weights given by $y_i$, we get an interpolating polynomial $L(x)$.

Lagrange Polynomial Plot

The Lagrange polynomial passing through all these points is unique and matches every given data point exactly.

Mathematical Formulation

Given $(n+1)$ distinct points $(x_0, y_0), (x_1, y_1), \ldots, (x_n, y_n)$, the Lagrange interpolation polynomial is constructed as follows:

I. Lagrange Basis Polynomials:

For each $i$ in ${0,1,\ldots,n}$, define the $i$-th Lagrange basis polynomial $\ell_i(x)$ by:

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

Notice that $\ell_i(x_k) = \delta_{ik}$, where $\delta_{ik}$ is the Kronecker delta. In other words:

$$\ell_i(x_k) = \begin{cases} 1 & \text{if } i=k,\\ 0 & \text{if } i \neq k. \end{cases}$$

II. Lagrange Interpolating Polynomial:

Once we have the $\ell_i(x)$, the interpolating polynomial $L(x)$ is given by:

$$L(x) = \sum_{i=0}^{n} y_i \ell_i(x)$$

By construction, $L(x_j) = y_j$ for all $j$. The degree of $L(x)$ is at most $n$.

Derivation

Starting from the requirement that $L(x)$ matches all data points:

$$L(x_i) = y_i \quad \text{for } i=0,1,\ldots,n$$

Consider polynomials $\ell_i(x)$ defined as:

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

This construction ensures that for each fixed $i$:

Hence $\ell_i(x)$ acts like a "selector" polynomial that equals 1 at $x_i$ and 0 at every other $x_j$.

To construct $L(x)$ that passes through all points, we form:

$$L(x) = \sum_{i=0}^{n} y_i \ell_i(x)$$

Evaluating at $x = x_k$:

$$L(x_k) = \sum_{i=0}^{n} y_i \ell_i(x_k) = y_k,$$

since $\ell_k(x_k)=1$ and $\ell_i(x_k)=0$ for $i \neq k$.

Uniqueness: The Lagrange interpolating polynomial is the unique polynomial of degree at most $n$ that passes through the given $(n+1)$ points. To see why, suppose two polynomials $L(x)$ and $M(x)$, both of degree at most $n$, agree at $n+1$ distinct points. Their difference $D(x) = L(x) - M(x)$ is a polynomial of degree at most $n$ with $n+1$ roots. By the Fundamental Theorem of Algebra, $D(x)$ must be identically zero, so $L(x) = M(x)$. Therefore, regardless of the method used to construct it, the interpolating polynomial of degree at most $n$ is unique.

Algorithm Steps

I. Input:

A set of $(n+1)$ points $(x_i,y_i)$ with all $x_i$ distinct.

II. Initialization:

Set $L(x)=0$.

III. Compute Lagrange Basis Polynomials:

For each $i=0,1,\ldots,n$:

$$\ell_i(x) = \ell_i(x) \cdot \frac{x - x_j}{x_i - x_j}$$

IV. Form the Interpolating Polynomial:

Compute:

$$L(x) = \sum_{i=0}^{n} y_i \ell_i(x)$$

Result:

The polynomial $L(x)$ is the desired Lagrange interpolating polynomial. To interpolate at any $x$, just evaluate $L(x)$.

Example

Given Points:

Let’s consider three points:

$$A(-1,1), \quad B(2,3), \quad C(3,5)$$

We have $n=2$ (since there are 3 points), and thus the polynomial $L(x)$ will be of degree at most 2.

Compute $\ell_0(x)$ for the point $A(-1, 1)$:

$$\ell_0(x) = \frac{(x - x_1)(x - x_2)}{(x_0-x_1)(x_0-x_2)} = \frac{(x - 2)(x - 3)}{(-1 -2)(-1 -3)} = \frac{(x - 2)(x - 3)}{(-3)(-4)} = \frac{(x - 2)(x - 3)}{12}$$

Compute $\ell_1(x)$ for the point $B(2,3)$:

$$\ell_1(x) = \frac{(x - x_0)(x - x_2)}{(x_1 - x_0)(x_1 - x_2)} = \frac{(x +1)(x - 3)}{(2 + 1)(2 - 3)} = \frac{(x+1)(x - 3)}{3 \cdot (-1)} = -\frac{(x+1)(x-3)}{3}$$

Compute $\ell_2(x)$ for the point $C(3,5)$:

$$\ell_2(x) = \frac{(x - x_0)(x - x_1)}{(x_2 - x_0)(x_2 - x_1)} = \frac{(x +1)(x - 2)}{(3 + 1)(3 - 2)} = \frac{(x+1)(x - 2)}{4}$$

Now, plug these into $L(x)$:

$$L(x) = y_0 \ell_0(x) + y_1 \ell_1(x) + y_2 \ell_2(x)$$

Substitute $(y_0, y_1, y_2) = (1,3,5)$:

$$L(x) = 1 \cdot \frac{(x - 2)(x - 3)}{12} + 3 \cdot \left(-\frac{(x+1)(x - 3)}{3}\right) + 5 \cdot \frac{(x+1)(x-2)}{4}$$

Simplify to standard polynomial form:

$$L(x) = \frac{1}{12}(x-2)(x-3) - (x+1)(x-3) + \frac{5}{4}(x+1)(x-2)$$

Expand each term:

$$\frac{1}{12}(x^2 - 5x + 6) = \frac{x^2}{12} - \frac{5x}{12} + \frac{1}{2}$$

$$-(x^2 - 2x - 3) = -x^2 + 2x + 3$$

$$\frac{5}{4}(x^2 - x - 2) = \frac{5x^2}{4} - \frac{5x}{4} - \frac{5}{2}$$

Sum the coefficients for each power of $x$:

$$x^2: \quad \frac{1}{12} - 1 + \frac{5}{4} = \frac{1}{12} - \frac{12}{12} + \frac{15}{12} = \frac{4}{12} = \frac{1}{3}$$

$$x: \quad -\frac{5}{12} + 2 - \frac{5}{4} = -\frac{5}{12} + \frac{24}{12} - \frac{15}{12} = \frac{4}{12} = \frac{1}{3}$$

$$\text{constant}: \quad \frac{1}{2} + 3 - \frac{5}{2} = \frac{1}{2} - \frac{5}{2} + 3 = -2 + 3 = 1$$

Therefore:

$$L(x) = \frac{1}{3}x^2 + \frac{1}{3}x + 1$$

Verify at the data points:

$$L(-1) = \frac{1}{3}(1) + \frac{1}{3}(-1) + 1 = \frac{1}{3} - \frac{1}{3} + 1 = 1 \; \checkmark$$

$$L(2) = \frac{1}{3}(4) + \frac{1}{3}(2) + 1 = \frac{4}{3} + \frac{2}{3} + 1 = 2 + 1 = 3 \; \checkmark$$

$$L(3) = \frac{1}{3}(9) + \frac{1}{3}(3) + 1 = 3 + 1 + 1 = 5 \; \checkmark$$

Evaluate at query points:

At $x = 0$:

$$L(0) = \frac{1}{3}(0) + \frac{1}{3}(0) + 1 = 1$$

At $x = 1$:

$$L(1) = \frac{1}{3}(1) + \frac{1}{3}(1) + 1 = \frac{2}{3} + 1 = \frac{5}{3} \approx 1.667$$

Error Bound:

If the data points are sampled from a function $f \in C^3[-1,3]$ (i.e., $f$ has a continuous third derivative on the interval), then the interpolation error at any point $x$ in that interval is bounded by:

$$|f(x) - L(x)| = \frac{|f'''(\xi)|}{3!} \cdot |(x+1)(x-2)(x-3)|$$

for some $\xi$ in the interval $[-1,3]$ that depends on $x$. More generally, for $n+1$ data points the error involves the $(n+1)$-th derivative and the product $\prod_{i=0}^{n}(x - x_i)$.

Advantages

I. Exact Fit:

The Lagrange interpolation polynomial passes through all given data points exactly. There is no approximation error at these nodes.

II. No Linear System Needed:

Unlike other polynomial interpolation techniques that require solving a system of equations, Lagrange interpolation provides a direct formula.

III. Simplicity of Form:

The formula for the interpolating polynomial is explicit and easy to implement.

IV. Flexibility:

Works for any set of points with distinct $x_i$.

Limitations

I. Runge’s Phenomenon:

For a large number of interpolation points, Lagrange interpolation may cause oscillations between the points, especially if the points are unevenly spaced. Using Chebyshev nodes instead of equally spaced points can mitigate this effect.

II. Recalculation for Added Points:

If a new point is added, the entire polynomial must be recomputed from scratch, unlike some other forms (e.g., Newton’s divided differences) that allow incremental updates more easily.

III. Computational Cost:

Direct evaluation of $L(x)$ at a single point costs $O(n^2)$ multiplications because each of the $n+1$ basis polynomials requires $O(n)$ work. The barycentric Lagrange form reduces this to $O(n)$ by precomputing a set of weights, making it the preferred approach in practice.

IV. Numerical Instability for Large $n$:

For high-degree interpolation, the individual basis polynomials $\ell_i(x)$ can attain very large values of alternating sign that nearly cancel when summed. This catastrophic cancellation leads to significant floating-point errors. The barycentric form of Lagrange interpolation avoids this issue and is numerically stable, making it the recommended implementation for practical use.

Verification

The implementation matches scipy.interpolate.lagrange to machine precision on every test case, including the worked example above. This serves as an independent confirmation that both the formula and the code are correct.