Position of Interpolated Point under Linear Interpolation

Theorem

Let $f$ be a real function.

Let $y_1, y_2, \ldots, y_n$ be known values of $f$ corresponding to $x_1, x_2, \ldots, x_n$ respectively.

Let $x'$ be in the domain of $f$ such that $x_i < x' < x_{i + 1}$.

Let $y' = \map f {x'}$ be determined according to linear interpolation.

Then:

$y' = y_i + \dfrac {\paren {x' - x_i} \paren {y_{i + 1} - x_i} } {x_{i + 1} - x_i}$

Proof

By definition of linear interpolation, the three points $\tuple {x_i, y_i}$, $\tuple {x', y'}$ and $\tuple {x_{i + 1}, y_{i + 1} }$ are all collinear.

Then:

\(\ds \dfrac {y' - y_i} {x' - x_i}\)

\(=\)

\(\ds \dfrac {y_{i + 1} - y_i} {x_{i + 1} - x_i}\)

Equation of Straight Line in Plane through Two Points

\(\ds \leadsto \ \ \)

\(\ds y' - y_i\)

\(=\)

\(\ds \dfrac {\paren {x' - x_i} \paren {y_{i + 1} - y_i} } {x_{i + 1} - x_i}\)

multiplying both sides by $x' - x_i$

\(\ds \leadsto \ \ \)

\(\ds y'\)

\(=\)

\(\ds y_i + \dfrac {\paren {x' - x_i} \paren {y_{i + 1} - y_i} } {x_{i + 1} - x_i}\)

rearranging

$\blacksquare$

Sources

• 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): interpolation
• 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): interpolation