Interpolation & Polynomial Approximation Divided Differences: A Brief Introduction Numerical Analysis (9th Edition) R L Burden & J D Faires Beamer Presentation Slides prepared by John Carroll Dublin City University c 2011 Brooks/Cole, Cengage Learning
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Interpolation & Polynomial Approximation [0.125in]3.625in0 ... · IntroductionNotationNewton’s Polynomial Outline 1 Introduction to Divided Differences 2 The Divided Difference
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Interpolation & Polynomial Approximation
Divided Differences: A Brief Introduction
Numerical Analysis (9th Edition)R L Burden & J D Faires
Numerical Analysis 9 (Chapter 3) Divided Differences: A Brief Introduction John Carroll, DCU 3 / 16
Introduction Notation Newton’s Polynomial
Introduction to Divided Differences
A new algebraic representation for Pn(x)
Suppose that Pn(x) is the nth Lagrange polynomial that agreeswith the function f at the distinct numbers x0, x1, . . . , xn.Although this polynomial is unique, there are alternate algebraicrepresentations that are useful in certain situations.The divided differences of f with respect to x0, x1, . . . , xn are usedto express Pn(x) in the form
Numerical Analysis 9 (Chapter 3) Divided Differences: A Brief Introduction John Carroll, DCU 4 / 16
Introduction Notation Newton’s Polynomial
Introduction to Divided Differences
A new algebraic representation for Pn(x)
Suppose that Pn(x) is the nth Lagrange polynomial that agreeswith the function f at the distinct numbers x0, x1, . . . , xn.
Although this polynomial is unique, there are alternate algebraicrepresentations that are useful in certain situations.The divided differences of f with respect to x0, x1, . . . , xn are usedto express Pn(x) in the form
Numerical Analysis 9 (Chapter 3) Divided Differences: A Brief Introduction John Carroll, DCU 4 / 16
Introduction Notation Newton’s Polynomial
Introduction to Divided Differences
A new algebraic representation for Pn(x)
Suppose that Pn(x) is the nth Lagrange polynomial that agreeswith the function f at the distinct numbers x0, x1, . . . , xn.Although this polynomial is unique, there are alternate algebraicrepresentations that are useful in certain situations.
The divided differences of f with respect to x0, x1, . . . , xn are usedto express Pn(x) in the form
Numerical Analysis 9 (Chapter 3) Divided Differences: A Brief Introduction John Carroll, DCU 4 / 16
Introduction Notation Newton’s Polynomial
Introduction to Divided Differences
A new algebraic representation for Pn(x)
Suppose that Pn(x) is the nth Lagrange polynomial that agreeswith the function f at the distinct numbers x0, x1, . . . , xn.Although this polynomial is unique, there are alternate algebraicrepresentations that are useful in certain situations.The divided differences of f with respect to x0, x1, . . . , xn are usedto express Pn(x) in the form
To determine the first of these constants, a0, note that if Pn(x) iswritten in the form of the above equation, then evaluating Pn(x) atx0 leaves only the constant term a0; that is,
a0 = Pn(x0) = f (x0)
Similarly, when P(x) is evaluated at x1, the only nonzero terms inthe evaluation of Pn(x1) are the constant and linear terms,
f (x0) + a1(x1 − x0) = Pn(x1) = f (x1)
⇒ a1 =f (x1)− f (x0)
x1 − x0
Numerical Analysis 9 (Chapter 3) Divided Differences: A Brief Introduction John Carroll, DCU 5 / 16
To determine the first of these constants, a0, note that if Pn(x) iswritten in the form of the above equation, then evaluating Pn(x) atx0 leaves only the constant term a0; that is,
a0 = Pn(x0) = f (x0)
Similarly, when P(x) is evaluated at x1, the only nonzero terms inthe evaluation of Pn(x1) are the constant and linear terms,
f (x0) + a1(x1 − x0) = Pn(x1) = f (x1)
⇒ a1 =f (x1)− f (x0)
x1 − x0
Numerical Analysis 9 (Chapter 3) Divided Differences: A Brief Introduction John Carroll, DCU 5 / 16
To determine the first of these constants, a0, note that if Pn(x) iswritten in the form of the above equation, then evaluating Pn(x) atx0 leaves only the constant term a0; that is,
a0 = Pn(x0) = f (x0)
Similarly, when P(x) is evaluated at x1, the only nonzero terms inthe evaluation of Pn(x1) are the constant and linear terms,
f (x0) + a1(x1 − x0) = Pn(x1) = f (x1)
⇒ a1 =f (x1)− f (x0)
x1 − x0Numerical Analysis 9 (Chapter 3) Divided Differences: A Brief Introduction John Carroll, DCU 5 / 16