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Last updated on September 25, 2025
Interpolation is a mathematical technique used to estimate unknown values that fall between known values. In this topic, we will learn about interpolation formulas and how they are used in various applications.
Interpolation is a method to find new data points within the range of a discrete set of known data points. Let’s learn the formulas used in various interpolation techniques.
Linear interpolation is the simplest form of interpolation and is used to estimate the value of a function between two known values.
The formula for linear interpolation is: \[ f(x) = f(x_0) + \frac{(x - x_0)(f(x_1) - f(x_0))}{x_1 - x_0} \] where \( x_0 \) and \( x_1 \) are known data points, and \( f(x_0) \) and \( f(x_1) \) are the corresponding function values.
Polynomial interpolation is the process of estimating values between known data points using polynomials. The formula for polynomial interpolation can be expressed using Lagrange polynomials:
\[ P(x) = \sum_{i=0}^{n} f(x_i) \prod_{\substack{j=0 \\ j \neq i}}^{n} \frac{x - x_j}{x_i - x_j} \] where \( x_0, x_1, \ldots, x_n \) are the given data points.
Spline interpolation uses piecewise polynomials called splines to approximate the data. The most common type is cubic spline interpolation, which ensures that the first and second derivatives are continuous across the interval.
The formula for cubic spline interpolation is defined piecewise for each interval \([x_i, x_{i+1}]\).
Interpolation formulas are crucial for estimating unknown values in various fields. Here are some important aspects of interpolation:
Students may find interpolation formulas complex and confusing.
Here are some tips to master them:
Students often make errors when applying interpolation formulas. Here are some mistakes and tips to avoid them:
Estimate the value of a function at \( x = 2.5 \) using linear interpolation between points \( (2, 4) \) and \( (3, 6) \).
The estimated value is 5
Using linear interpolation:
[ f(x) = 4 + \frac{(2.5 - 2)(6 - 4)}{3 - 2} = 4 + 0.5 \times 2 = 5 \]
Find the value at \( x = 4 \) using polynomial interpolation with points \( (1, 1), (3, 9), (5, 25) \).
The estimated value is 16
Using Lagrange polynomial interpolation, the polynomial is found to be ( P(x) = x^2 ).
Thus, ( P(4) = 16 ).
Use cubic spline interpolation to estimate a value between \( x = 0 \) and \( x = 2 \) given points \( (0, 0), (1, 1), (2, 0) \).
The estimated value at \( x = 1.5 \) is approximately 0.5
Cubic spline interpolation results in a piecewise polynomial.
Evaluating at \( x = 1.5 \) yields an approximate value of 0.5.
Estimate the temperature at noon given temperatures at 10 AM (15°C) and 2 PM (25°C) using linear interpolation.
The estimated temperature at noon is 20°C
Using linear interpolation: \[ T = 15 + \frac{(12 - 10)(25 - 15)}{2} = 15 + 10 \times 0.5 = 20 \]
Determine the interpolated stock price at day 3 using polynomial interpolation with prices on days 1, 2, 4, and 5 as follows: \( (1, 100), (2, 110), (4, 130), (5, 150) \).
The interpolated stock price is approximately 120
Using polynomial interpolation, the estimated stock price at day 3 is approximately 120.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
: He loves to play the quiz with kids through algebra to make kids love it.