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MATH 441/541 - Numerical Analysis Sixth Meeting: Computer Arithmetic Continued & Root Finding Algorithms Thursday, September 27 th
M¨uller’s Method
- Given the continuous function f (x) and three reasonable guesses to the root, x 0 , x 1 , and x 2.
- A Visual Explanation:
- The Formula for the New Guess for the Root: Given three guesses, x 0 , x 1 , and x 2 generate the new approximation, x 3 as:
x 3 = x 2 − 2 c b + sgn(b)
b^2 − 4 ac where
c = f (x 2 )
b = (x^0 −^ x^2 )
(^2) (f (x 1 ) − f (x 2 )) − (x 1 − x 2 ) (^2) (f (x 0 ) − f (x 2 )) (x 0 − x 2 )(x 1 − x 2 )(x 0 − x 1 )
a =
(x 1 − x 2 )^2 (f (x 0 ) − f (x 1 )) − (x 0 − x 2 )^2 (f (x 1 ) − f (x 2 )) (x 0 − x 2 )(x 1 − x 2 )(x 0 − x 1 )
- Strengths and Weaknesses:
- Rates of Convergence for Methods^1 :
Bisection Secant Newton False Position Fixed Point M¨uller’s α = 2 α ≈ 1 α = 1 +^
≈ 1. 62 (for simple roots) α = 1 α ≈ 1. 839 α = 1 (for multiple roots)
(^1) For M¨uller’s Method rate in particular see Michael T. Heath. Scientific Computing: An Introductory Survey, 2nd (^) Ed.. McGraw Hill, 2002. Page 234
1
Interpolation And Polynomial Approximation
- Section 3.1: Interpolation and the Lagrange Polynomial
- The Concept:
- How is this different from / similar to regression?
- How is this different from / similar to using Taylor Polynomials?
- Reminder of Taylor Polynomials and Their Error Terms:
- Lagrange Polynomial (a) Deriving the Concept and Terms: i. Ln,k(x) = Lk(x) = (x^ −^ x^0 )(x^ −^ x^1 )^ · · ·^ (x^ −^ xk−^1 )(x^ −^ xk+1)^ · · ·^ (x^ +^ xn) (xk − x 0 )(xk − x 1 ) · · · (xk − xk− 1 )(xk − xk+1) · · · (xk + xn) ii. The nth^ degree Lagrange Polynomial, Pn(x), through the points (x 0 , f (x 0 )), · · · , (xn, f (xn)) is given by:
Pn(x) = Px 0 ,x 1 ,···,xn (x) = L 0 (x)f (x 0 ) + L 1 (x)f (x 1 ) + · · · + Ln(x)f (xn)
(b) An Example: Find the Second Order Lagrange Polynomial through the points (1, 3), (4, 2) and (6, 5), from the function f (x), and use it to approximate f (2). (c) The Error Term: (d) The Uniqueness of the Interpolating Polynomial: Theorem 3.2: If x 0 , x 1 , · · · , xn are n + 1 distinct numbers and f is a function whose values are given at these numbers, then a unique polynomial P (x) of degree at most n exists with
f (xk) = P (xk), k = 0, 1 , · · · n
where this polynomial is given by
P (x) =
∑^ n k=
f (xk)Ln,k(x)
- Section 3.2 Divided Differences
- Question: What if you had P 1 (x) (through points (x 0 , f (x 0 )) and (x 1 , f (x 1 ))) and you wanted to generate P 2 (x) (which also passes through (x 2 , f (x 2 ))? How would you do this?
- New Notations: (a) New way for writing the Lagrange Polynomial of degree n, Pn(x): (b) Divided Difference Notation:
- Connecting the Lagrange Polynomial and Divided Differences:
- An Excel Example (See Excel DividedDifferences.xls Handout)
