Interpolation Formula

Interpolation Formula

The method of finding new values for any function using the set of values is done by interpolation. The unknown value on a point is found out using this formula. If linear interpolation formula is concerned then it should be used to find the new value from the two given points. If compared to Lagrange’s interpolation formula, the “n” set of numbers should be available and Lagrange’s method is to be used to find the new value.

The following is  Linear Interpolation Formula

\[\large y=y_{1}+\frac{(x-x_{1})}{(x_{2}-x_{1})} \times (y_{2}-y_{1})\]

The La-grange’s Interpolation Polynomial is given as,

\(\begin{array}{l}P(x) = \begin{array}{c} \frac{\left(x-x_{2}\right)\left(x-x_{3}\right) \cdots\left(x-x_{n}\right)}{\left(x_{1}-x_{2}\right)\left(x_{1}-x_{3}\right) \cdots\left(x_{1}-x_{n}\right)} y_{1}+\frac{\left(x-x_{1}\right)\left(x-x_{3}\right) \cdots\left(x-x_{n}\right)}{\left(x_{2}-x_{1}\right)\left(x_{2}-x_{3}\right) \cdots\left(x_{2}-x_{n}\right)} \\ y_{2}+\cdots+\frac{\left(x-x_{1}\right)\left(x-x_{2}\right) \cdots\left(x-x_{n-1}\right)}{\left(x_{n}-x_{1}\right)\left(x_{n}-x_{2}\right) \cdots\left(x_{n}-x_{n-1}\right)} y_{n} \end{array}\end{array} \)

Solved Examples

Question 1: Using the interpolation formula, find the value of y at x = 8 given some set of values (2, 6), (5, 9) ?


The known values are,

\(\begin{array}{l}x_{0}=8, x_{1}=2, x_{2}=5, y_{1}=6, y_{2}=9\end{array} \)
\(\begin{array}{l}y=y_{1}+\frac{(x-x_{1})}{(x_{2}-x_{1})} \times (y_{2}-y_{1}) \end{array} \)
\(\begin{array}{l}y=6+(\frac{(8-2)}{(5-2)} \times (9-6)\end{array} \)

y = 6 + 6

y = 12

More topics in Interpolation Formula
Quadratic Interpolation Formula Linear Interpolation Formula
Lagrange Interpolation Formula


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