Applied Mathematics for Electrical Engineering (3130908)

Question-2c ( i )

BE | Semester-3 Winter-2019 | 26-11-2019

Q2) (c ( i ))

Using quadratic Lagrange interpolation, compute $\ln 9.2$ from $\ln 9.0 = 2.1972, \ln 9.5 = 2.2513, \ln 11 = 2.3979$ .

Here,

x_{0} = 9, x_{1} = 9.5, x_{2} = 11

y_{0} = 2.1972, y_{1} = 2.2513, y_{2} = 2.3979

Here,

x = 9.2

By Lagrange’s interpolation Formula,

f (x) = y = \frac{(x - x_{1}) (x - x_{2})}{(x_{0} - x_{1}) (x_{0} - x_{2})} y_{0} + \frac{(x - x_{0}) (x - x_{2})}{(x_{1} - x_{0}) (x_{1} - x_{2})} y_{1} + \frac{(x - x_{0}) (x - x_{1})}{(x_{2} - x_{0}) (x_{2} - x_{1})} y_{2}

f (x) = y = \frac{(9.2 - 9.5) (9.2 - 11)}{(9 - 9.5) (9 - 11)} (2.1972) + \frac{(9.2 - 9) (9.2 - 11)}{(9.5 - 9) (9.5 - 11)} (2.2513) + \frac{(9.2 - 9) (9.2 - 9.5)}{(11 - 9) (11 - 9.5)} (2.3979)

f (x) = y = 1.1865 + 1.0806 - 0.0480

f (x) = y = 2.2191

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