linear_interpolation.md

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 $x_i \leq x \leq x_{i+1}$, 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:

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

II. Substitute $x=1$:

$$y = 0 + 0.5 (1 - (-2)) = 0.5 \times 3 = 1.5.$$

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.