Polynomial Interpolation Calculator
Find the unique polynomial passing through given data points using Lagrange and Newton methods.
Data Points
Evaluate At
Lagrange Basis Polynomials
L0(1.5)
-0.062500
Contribution: 1 * -0.0625 = -0.062500
L1(1.5)
0.562500
Contribution: 2 * 0.5625 = 1.125000
L2(1.5)
0.562500
Contribution: 5 * 0.5625 = 2.812500
L3(1.5)
-0.062500
Contribution: 10 * -0.0625 = -0.625000
P(1.5)
3.250000
Polynomial degree: 3
Newton Divided Differences
| x | f[] | f[,] | f[,,] | f[,,,] |
|---|---|---|---|---|
| 0 | 1.0000 | 1.0000 | 1.0000 | 0.0000 |
| 1 | 2.0000 | 3.0000 | 1.0000 | |
| 2 | 5.0000 | 5.0000 | ||
| 3 | 10.0000 |
Newton Coefficients
a01.000000
a11.000000
a21.000000
a30.000000
Method Comparison
- Lagrange: Easy to understand, but adding points requires recalculation
- Newton: Efficient for adding new points, uses divided differences
- Both produce the same unique polynomial of degree n-1