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1 Curve-Fitting Spline Interpolation

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2 Curve Fitting Regression Linear Regression Polynomial Regression Multiple Linear Regression Non-linear Regression Interpolation Newton's Divided-Difference Interpolation Lagrange Interpolating Polynomials Spline Interpolation

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3 For some cases, polynomials can lead to erroneous results because of round off error and overshoot. Alternative approach is to apply lower-order polynomials to subsets of data points. Such connecting polynomials are called spline functions.

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5 (a)Linear spline –Derivatives are not continuous –Not smooth (b) Quadratic spline –Continuous 1 st derivatives (c) Cubic spline –Continuous 1 st & 2 nd derivatives –Smoother

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6 Quadratic Spline

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7 Spline of Degree 2 A function Q is called a spline of degree 2 if –The domain of Q is an interval [a, b]. –Q and Q' are continuous functions on [a, b]. –There are points x i (called knots) such that a = x 0 < x 1 < … < x n = b and Q is a polynomial of degree at most 2 on each subinterval [x i, x i+1 ]. A quadratic spline is a continuously differentiable piecewise quadratic function.

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8 Exercise Which of the following is a quadratic spline?

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9 Exercise (Solution)

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10 Observations n+1 points n intervals Each interval is connected by a 2 nd -order polynomial Q i (x) = a i x 2 + b i x + c i, i = 0, …, n–1. Each polynomial has 3 unknowns Altogether there are 3n unknowns Need 3n equations (or conditions) to solve for 3n unknowns Quadratic Interpolation

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11 1.Interpolating conditions –On each sub interval [x i, x i+1 ], the function Q i (x) must satisfy the conditions Q i (x i ) = f(x i ) and Q i (x i+1 ) = f(x i+1 ) –These conditions yield 2 n equations Quadratic Interpolation ( 3n conditions)

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12 Quadratic Interpolation ( 3n conditions) 2.Continuous first derivatives –The first derivatives at the interior knots must be equal. –This adds n-1 more equations: We now have 2n + (n – 1) = 3n – 1 equations. We need one more equation.

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13 3.Assume the 2 nd derivatives is zero at the first point. –This gives us the last condition as Quadratic Interpolation ( 3n conditions) –With this condition selected, the first two points are connected by a straight line. –Note: This is not the only possible choice or assumption we can make.

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14 Example Fit quadratic splines to the given data points. i0123 xixi 34.579 f(xi)f(xi)2.51 0.5

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15 Example (Solution) 1. Interpolating conditions 2. Continuous first derivatives 3. Assume the 2nd derivatives is zero at the first point.

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16 Example (Solution) We can write the system of equations in matrix form as Notice that the coefficient matrix is sparse.

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17 Example (Solution) The system of equations can be solved to yield Thus the quadratic spline that interpolates the given points is

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18 Efficient way to derive quadratic spline

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19 Efficient way to derive quadratic spline

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20 Efficient way to derive quadratic spline

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21 Cubic Spline Spline of Degree 3 A function S is called a spline of degree 3 if –The domain of S is an interval [a, b]. –S, S' and S" are continuous functions on [a, b]. –There are points t i (called knots) such that a = t 0 < t 1 < … < t n = b and Q is a polynomial of degree at most 3 on each subinterval [t i, t i+1 ].

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22 Cubic Spline ( 4n conditions) 1.Interpolating conditions ( 2n conditoins). 2.Continuous 1 st derivatives ( n-1 conditions) The 1 st derivatives at the interior knots must be equal. 3.Continuous 2 nd derivatives ( n-1 conditions) The 2 nd derivatives at the interior knots must be equal. 4.Assume the 2 nd derivatives at the end points are zero ( 2 conditions). This condition makes the spline a "natural spline".

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23 Efficient way to derive cubic spline

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24 Summary Advantages of spline interpolation over polynomial interpolation The conditions that are used to derive the quadratic and cubic spline functions Characteristics of cubic spline –Overcome the problem of "overshoot" –Easier to derive (than high-order polynomial) –Smooth (continuous 2 nd -order derivatives)

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