Curve-Fitting Interpolation

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Curve-FittingInterpolation
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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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Polynomial Interpolation

• Objective
• Given n1 points, we want to find the polynomial
  of order n
• that passes through all the points.

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Polynomial Interpolation

• The nth-order polynomial that passes through n1
  points is unique, but it can be written in
  different mathematical formats
• The Newton's Form
• The Lagrange Form
• The conventional form

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Linear Interpolation (Newton's Form)

• Objective
• Connecting two points with a straight line.

f1(x) represents the first-order interpolating
polynomial.
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Two linear interpolations of f(x)ln(x) on two
different intervals.
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Quadratic Interpolation (Newton's Form)

• Connecting three points with a second-order
  polynomial or parabola.
• One way to form a 2nd-order polynomial is

• The advantage is that b0, b1, and b2 can be
  calculated conveniently.
• Only the format is different.
• There is till only one unique 2nd-order
  polynomial that passes through three points.
• Can be rewritten in the conventional form. i.e.,
  as

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Quadratic Interpolation Finding b0, b1, b2
Given three points (x0, f(x0)), (x1, f(x1)), and
(x2, f(x2)), we can create three equations with
three unknowns b0, b1, and b2 as
which can be solved for b0, b1, and b2
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Quadratic Interpolation Finding b0, b1, b2
Alternatively, we can also calculate b0, b1, and
b2 as
b1 Finite-divided difference for f'(x) b2
Finite-divided difference for f"(x)
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Comparing Linear and Quadratic Interpolation
The quadratic interpolation formula includes an
additional term which represents the 2nd-order
curvature.
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Fig 18.4
Linear vs. quadratic interpolation of ln(x)
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General Form of Newton's Interpolating Polynomials
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Graphical depiction of the recursive nature of
finite divided differences.
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Cubic interpolation of ln(x)
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Lagrange Interpolating Polynomials

• Simply a reformulation of the Newtons polynomial
  that avoids the computation of divided
  differences

e.g. 1st and 2nd-order polynomials in Lagrange
form
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• Second order case of Lagrange polynomial.
• Each of the three terms is a 2nd-order polynomial
  that passes through one of the data points and is
  zero at the other two.
• The summation of three terms must, therefore, be
  the unique 2nd-order polynomial that passes
  exactly through three points.

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Coefficients of an Interpolating Polynomial

• Newton and Lagrange polynomials are well suited
  for determining intermediate values between
  points.
• However, they do not provide a polynomial in the
  conventional form
• To calculate a0, a1, , an, we can use
  simultaneous linear systems of equations.

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Coefficients of an Interpolating Polynomial
Given n1 points, (x0, f(x0)), (x1, f(x1)), ,
(xn, f(xn)), we have n1 equations which can be
solved for n1 unknowns
Solve this system of linear equations for a0, a1,
, an.
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Coefficients of an Interpolating Polynomial

• Solving the system of linear equations directly
  is not the most efficient method.
• This system is typically ill-conditioned.
• The resulting coefficients can be highly
  inaccurate when n is large.

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Extrapolation

• Extrapolation is the process of estimating a
  value of f(x) that lies outside the range of the
  known base points, x0, x1, , xn.
• Extreme care should be exercised where one must
  extrapolate.

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Spline Interpolation

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

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Quadratic Spline
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Quadratic Interpolation

• Observations
• n1 points
• n intervals
• Each interval is connected by a 2nd-order
  polynomial fi(x) aix2bixci, i1, , n.
• Each polynomial has 3 unknowns
• Altogether there are 3n unknowns
• Need 3n equations (or conditions) to solve for 3n
  unknowns

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Quadratic Interpolation (3n conditions)

• The function values of adjacent polynomials must
  be equal at the interior knots.
• This condition can be represented as

• Since there are n-1 interior knots, this
  condition yields 2n-2 equations.

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Quadratic Interpolation (3n conditions)

• The first and last functions must pass through
  the end points.
• This adds 2 more equations

• The first derivatives at the interior knots must
  be equal.
• This adds n-1 more equations

We now have 2n - 2 2 n - 1 3n - 1
equations. We need one more equation.
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Quadratic Interpolation (3n conditions)

• Assume the 2nd derivatives is zero at the first
  point.
• This gives us the last condition as

• 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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Cubic Spline

• The function values must be equal at the
  interrior knots (2n-2 conditoins).
• The 1st and last functions must pass through the
  end points (2 conditions).
• The 1st derivatives at the interior knots must be
  equals (n-1 conditions).
• The 2nd derivatives at the interior knots must be
  equals (n-1 conditions).
• Assume the 2nd derivatives at the end points are
  zero (2 conditions).
• This condition makes the spline a "natural"
  spline.

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Efficient way to derive cubic spline

• The cubic equation on each interval can be
  expressed as

• There are only two unknowns in each equations
  the 2nd derivatives at the end of each interval

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Efficient way to derive cubic spline

• The unknowns can be evaluated using the following
  equation

• If this equation is written for all the interior
  knots, n-1 simultaneous equations result with n-1
  unknowns.

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Summary

• Polynomial interpolation for approximate
  complicated functions. (Data are exact)
• Newton's or Lagrange Polynomial interpolation are
  suitable for evaluating intermediate points.
• Cubic spline
• Overcome the problem of "overshoot"
• Easier to derive
• Smooth (continuous 2nd-order derivatives)