Position of Interpolated Point under Linear Interpolation

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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:

\(\ds \dfrac {y' - y_i} {x' - x_i}\) \(=\) \(\ds \dfrac {y_{i + 1} - y_i} {x_{i + 1} - x_i}\) Two-Point Form of Equation of Straight Line in Plane
\(\ds \leadsto \ \ \) \(\ds y' - y_i\) \(=\) \(\ds \dfrac {\paren {x' - x_i} \paren {y_{i + 1} - y_i} } {x_{i + 1} - x_i}\) 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

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