Last modified: October 07, 2018

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Linear interpolation

Linear interpolation is one of the most basic and commonly used interpolation methods. The idea is to approximate the value of a function between two known data points by assuming that the function behaves linearly (like a straight line) between these points. Although this assumption may be simplistic, it often provides a reasonable approximation, especially when the data points are close together or the underlying function is relatively smooth.

Conceptual Illustration:

Imagine you have two points on a graph:

Linear interpolation draws a straight line between the two known data points $(x_i,y_i)$ and $(x_{i+1},y_{i+1})$, and then estimates the value at $x$ by following this line.

Mathematical Formulation

Given two known data points $(x_i,y_i)$ and $(x_{i+1},y_{i+1})$, and a target $x$-value with $x_i\le x\le x_{i+1}$, the line connecting these points has a slope $\alpha$ given by:

$$\alpha =\frac{y_{i+1}-y_i}{x_{i+1}-x_i}.$$

To find the interpolated value $y$ at $x$, start from $y_i$ and move along the line for the interval $(x-x_i)$:

$$y=y_i+\alpha (x-x_i).$$

Substituting $\alpha$:

$$y=y_i+(x-x_i)\frac{y_{i+1}-y_i}{x_{i+1}-x_i}.$$

This formula provides the interpolated $y$-value directly.

Derivation

I. Slope Calculation:

The slope $\alpha$ of the line passing through $(x_i,y_i)$ and $(x_{i+1},y_{i+1})$ is:

$$\alpha =\frac{y_{i+1}-y_i}{x_{i+1}-x_i}.$$

II. Linear Equation:

A line passing through $(x_i,y_i)$ with slope $\alpha$ is:

$$y-y_i=\alpha (x-x_i).$$

III. Substitution:

Replace $\alpha$ with its expression:

$$y-y_i=\frac{y_{i+1}-y_i}{x_{i+1}-x_i}(x-x_i).$$

IV. Final Formula:

Simplifying:

$$y=y_i+\frac{(y_{i+1}-y_i)}{x_{i+1}-x_i}(x-x_i).$$

Algorithm Steps

I. Identify the interval $[x_i,x_{i+1}]$ that contains the target $x$.

II. Compute the slope:

$$\frac{y_{i+1}-y_i}{x_{i+1}-x_i}.$$

III. Substitute into the linear interpolation formula:

$$y=y_i+\frac{(y_{i+1}-y_i)}{x_{i+1}-x_i}(x-x_i).$$

The result is the interpolated value $y$ at the desired $x$.

Example

Given Points: $A(-2,0)$ and $B(2,2)$. Suppose we want to find $y$ at $x=1$.

I. Compute the slope:

$$\alpha =\frac{2-0}{2-(-2)}=\frac{2}{4}=\mathrm{0.5.}$$

II. Substitute $x=1$:

$$y=0+0.5(1-(-2\left)\right)=0.5\times 3=\mathrm{1.5.}$$

So, the line passing through $(-2,0)$ and $(2,2)$ gives $y=1.5$ when $x=1$.

Advantages

• The method offers simplicity, as the calculation involves straightforward arithmetic, making it easy and quick to apply.
• Minimal data requirements make it practical, needing only two data points to estimate intermediate values.
• It provides a local approximation, working well when the function is nearly linear within the specified interval.

Limitations

• The linear assumption can lead to poor results if the actual relationship between points is not close to linear.
• Linear interpolation uses no derivative information, ignoring the slope or curvature of the function, which could enhance accuracy.
• Accuracy diminishes as the interval between points increases or as the function becomes more non-linear, leading to potential errors in approximation.