ECS 555 - INTERPOLATION
LEARNING OUTCOMES;
At the end of the week,
students will be able:
To evaluate the intermediate value between
precise data points using linear and polynomial
interpolation.
Adiza Jamadin
INTRODUCTION
• Interpolation is a procedure for
Interpolation estimating a value between known
values of data points.
The most common method is polynomial interpolation:
i) The Newton polynomial
ii) The Lagrange polynomial
f ( x )= a 0 +a1 x+a 2 x 2 +⋯+a n x n
Polynomial interpolation is used when
the point determined are very precise.
The curve representing the behavior has
to pass through every point.
INTRODUCTION
Polynomials are the most common choice of interpolation because
they are easy to:
Evaluate
Differentiate, and
Integrate.
Newton’s Divided-Difference
Interpolating Polynomials
Linear Interpolation – Example 18.1
Linear Interpolation
Quadratic Interpolation
Quadratic Interpolation
- Example 18.2
Quadratic Interpolation - Example 18.2
f2(x) = b0 + b1(x- x0) + b2(x- x0)(x- x1)
= 0 + 0.4620981(2-1) + (-0.0518731)(2-4)
= 0.56584
Newton’s Divided-Difference
Interpolating Polynomials
Quadratic Interpolation
- Example 18.3
x0 = 1 f(x0) = 0
x1 = 4 f(x1) = 1.386294
x2 = 6 f(x2) = 1.791759
x3 = 5 f(x3) = 1.609438
Quadratic Interpolation
- Example 18.3 x0 = 1
x1 = 4
f(x0) =
f(x1) =
0
1.386294
x2 = 6 f(x2) = 1.791759
x3 = 5 f(x3) = 1.609438
f[x1, x0]= 0.4620981
f[x2, x1]= 0.2027326
f[x3, x2]= 0.1823216
Quadratic Interpolation
- Example 18.3
Exercise
The following data defines the sea-level
concentration of dissolved oxygen for fresh water
as a function of temperature:
Estimate o(27) using;
(a) Linear interpolation
(b) Newton’s interpolating polynomial
Note the exact result is 7.986 mg/L
Answer;
(a) Linear Interpolation
(b) Polynomial Interpolation