Lagrange Interpolation Calculator

Reminder: what Lagrange interpolation means for your data

Suppose you have n + 1 distinct data points . There exists a unique polynomial P(x) of degree at most n such that:

P(x_i) = y_i for every index i = 0, 1, …, n.

Lagrange interpolation expresses this polynomial as a sum of Lagrange basis polynomials built directly from the x-values in your data. Each basis polynomial is equal to 1 at one data point and 0 at all the others, so the final weighted sum reproduces every y-value exactly.

Lagrange interpolation formulas

The interpolating polynomial for this calculator is written as

P(x) = ∑i=0n yi Li(x),

where the Lagrange basis polynomials are

Li(x) = ∏j=0, j ≠ in (x − xj)/(xi − xj).

Each Li(x) satisfies Li(xi) = 1 and Li(xk) = 0 for all k ≠ i, so the sum P(x) = ∑ yiLi(x) passes exactly through all points.

The same Lagrange interpolation formula in MathML form is:

The calculator evaluates these products numerically when you click Interpolate. That is convenient for a moderate number of points and avoids setting up a separate coefficient system.

How to enter data into the Lagrange interpolation calculator

To use this Lagrange interpolation calculator with your own data, follow these steps:

1. List your data points.
   • Enter one point per line in the format x,y.
   • You may use integers or decimals, for example 0,1 or 1.5,2.75.
   • The x-values must all be distinct, because duplicate x-values make a single interpolating polynomial impossible.
2. Choose an evaluation x-value.
   • Enter the value of x where you want the interpolating polynomial evaluated.
   • This can be inside the range of your data (interpolation) or outside it (extrapolation), but the limitations below still matter.
3. Click the Interpolate button.
   • The calculator parses your points and builds the Lagrange basis polynomials.
   • It forms P(x) = ∑ yiLi(x) and evaluates it at the x-value you entered.
   • If any point is invalid or the x-values repeat, an error message is shown instead of a result.

For small interpolation problems, a compact set of well-spaced points is usually easier to interpret than a long list. As the polynomial degree rises, the curve can become more sensitive to tiny changes in the inputs.

Interpreting the Lagrange interpolation results

After running the Lagrange interpolation calculator, you will typically see:

• An expression for the interpolating polynomial. This may appear in Lagrange form or as a simplified polynomial. You can copy the expression into plotting software or a symbolic algebra system if you want to inspect the curve further.
• The interpolated value at your chosen x. This numeric value is P(xeval), where xeval is the x-value you entered. If that x matches one of your original x-values, the result should match the corresponding y-value apart from rounding error.

You can sanity-check a Lagrange interpolation result in a few practical ways:

• Substitute each original x-value into the polynomial and verify that the original y-values come back.
• Plot the polynomial and the source points in an external graphing tool to confirm that the curve passes through every sample.
• Compare the interpolated value with known or theoretical values whenever you have them.

Worked example: a quadratic through three interpolation points

This worked example uses three Lagrange interpolation points:

• (0, 1)
• (1, 2)
• (2, 0)

We want the polynomial P(x) of degree at most 2 that passes through all three points, and we will evaluate it at x = 0.5.

Step 1: Enter the interpolation points

In the data points field, type:

0,1
1,2
2,0
  

Then set the evaluation x-value to 0.5 and click Interpolate.

Step 2: Form the Lagrange basis polynomials

For three interpolation points we have indices i = 0, 1, 2:

• L0(x) = (x - 1)(x - 2) / ((0 - 1)(0 - 2)) = (x - 1)(x - 2) / 2
• L1(x) = (x - 0)(x - 2) / ((1 - 0)(1 - 2)) = -x(x - 2)
• L2(x) = (x - 0)(x - 1) / ((2 - 0)(2 - 1)) = x(x - 1) / 2

The polynomial is then

P(x) = 1 × L0(x) + 2 × L1(x) + 0 × L2(x).

Substituting the expressions for L0 and L1 and simplifying gives:

P(x) = -½x2 + &frac32;x + 1.

Step 3: Evaluate the Lagrange polynomial at x = 0.5

Now compute:

P(0.5) = -½(0.5)2 + &frac32;(0.5) + 1 = -0.125 + 0.75 + 1 = 1.625.

The calculator will display this value as the interpolated result at x = 0.5. It will also show the symbolic polynomial, which you can compare with the analytic expression above.

How to use Lagrange interpolation vs. other curve-fitting approaches

Lagrange interpolation is one exact way to pass a polynomial through known points. Depending on how many points you have and how smooth you want the final curve to be, another interpolation method may be a better fit.

This calculator focuses on the Lagrange form, which is especially useful when you want the explicit polynomial through a small set of points and do not want to solve for coefficients separately.

Assumptions, limitations, and good practices for Lagrange interpolation

The Lagrange interpolation implemented here rests on several important assumptions:

• Distinct x-values. All input data points must have different x-coordinates. If two points share the same x but different y-values, no single-valued function can pass through both.
• Exact polynomial fit. The method assumes that your data are samples of, or can be reasonably approximated by, a single polynomial of degree at most n.
• Finite, moderate number of points. The implementation is designed for a moderate number of points, for example up to a few dozen. Very large sets can be slow and numerically unstable.

Keep these practical limitations in mind when you interpret a Lagrange interpolation result:

• Runge’s phenomenon and oscillations. Using a high-degree polynomial through many points, especially evenly spaced ones, can cause large oscillations between points. A visually smooth curve is not guaranteed.
• Extrapolation risk. Evaluating the polynomial far outside the range of your data can give unreliable results. Interpolation, meaning values inside the span of your x-values, is usually much safer than extrapolation.
• Sensitivity to noise. If your data are noisy measurements, a polynomial that passes exactly through every point may amplify measurement error instead of smoothing it. Curve fitting or regression may be more appropriate in that case.
• Floating-point arithmetic. Results are computed with standard double-precision floating-point arithmetic. For extreme scales or very ill-conditioned point sets, round-off error can appear in the coefficients or evaluated values.

Some guidelines for using this calculator responsibly:

• Prefer a small, representative set of points rather than many closely spaced samples.
• Use interpolation primarily inside the span of your x-values.
• Cross-check important results with analytical formulas, experimental data, or alternative numerical methods when possible.

Typical uses for the Lagrange interpolation calculator

This Lagrange interpolation calculator can support a range of tasks:

• Physics and engineering labs. Interpolate between a few measured operating points to estimate a value that was not measured directly, and compare the result with a theoretical prediction.
• Numerical analysis coursework. Demonstrate how polynomial interpolation works, verify manual computations, and explore the impact of adding or removing data points on the interpolating polynomial.
• Computer graphics and animation. Fit a simple curve through control points to prototype trajectories, then compare it with more advanced spline-based methods.
• General data exploration. Obtain a smooth analytical curve through a small set of key samples when you need a symbolic expression for further calculus operations.

Used with care, the Lagrange Interpolation Calculator is a precise way to turn tabulated values into an explicit polynomial, inspect the shape of that polynomial, and estimate intermediate values.