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Find values between data points using linear, Lagrange, and Newton interpolation methods. Our calculator provides step-by-step solutions for numerical analysis, data fitting, and polynomial interpolation with detailed mathematical explanations.
Last updated: February 2, 2026
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Using lagrange interpolation with 4 data points
For data points (0,1), (1,4), (2,9), (3,16) interpolating at x = 1.5:
f(1.5) = 6.25
Our interpolation calculator uses proven numerical methods to estimate values between known data points. The process involves constructing polynomials that pass through your data points and evaluating them at the target location. Each method offers different advantages for various types of data and accuracy requirements.
f(x) = y₀ + (y₁ - y₀) × (x - x₀) / (x₁ - x₀)P(x) = Σ yᵢ × Πⱼ≠ᵢ (x - xⱼ) / (xᵢ - xⱼ)P(x) = f[x₀] + f[x₀,x₁](x-x₀) + f[x₀,x₁,x₂](x-x₀)(x-x₁) + ...Shows how different methods fit curves through data points
Interpolation is based on the Weierstrass approximation theorem, which states that any continuous function can be uniformly approximated by polynomials. Each method constructs a unique polynomial of degree n-1 that passes through n data points, providing different computational advantages and numerical properties.
Need help with other numerical calculations? Check out our derivative calculator and quadratic formula calculator.
Get Custom Calculator for Your BusinessResult: f(1.5) = 6.25
This interpolated value estimates the function at x = 1.5 based on the given data points.
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