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) ?
Solution:
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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