month1_saturday5_fixedpoint
DESCRIPTION
ffgfgTRANSCRIPT
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!umerical Analysis
Avid FarhoodfarMonth1 – Saturday 5Fixed-Point IterationSolutions of Equations in One Variable
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Outline
1. The Bisection Method 2. Fixed-Point Iteration 3. Newton's Method and Its Extensions 4. Error Analysis for Iterative Methods 5. Accelerating Convergence 6. Zeros of Polynomials and Müller's Method
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Outline
1. The Bisection Method2. Fixed-Point Iteration 3. Newton's Method and Its Extensions 4. Error Analysis for Iterative Methods 5. Accelerating Convergence 6. Zeros of Polynomials and Müller's Method
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Fixed-Point Iteration
1. Introduction & Theoretical Framework2. Motivating the Algorithm: An Example3. Fixed-Point Formulation I4. Fixed-Point Formulation II
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Theoretical FrameworkPrime Objective
1. In what follows, it is important not to lose sight of our prime objective:
2. Given a function f (x) where a < x < b, find values p such that f (p) = 0
3. Given such a function, f (x), we now construct an auxiliary function g(x) such that
p = g(p)
whenever f (p) = 0 (this construction is not unique).
4. The problem of finding p such that p = g(p) is known as the fixed point problem.
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Theoretical Framework
A Fixed Point
A Fixed Point
!
!
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Theoretical Framework
The Equation f (x) = x − cos(x) = 0
If we write this equation in the form:x = cos(x)
then g(x) = cos(x).
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Theoretical Framework
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Theoretical Frameworkx = cos(x)
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Theoretical Frameworkp = cos(p) p 0.739~~
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Theoretical FrameworkExistence of a Fixed Point
A Fixed Point! If g(a) = a or g(b) = b, the existence of a fixed point is obvious.
! Suppose not; then it must be true that g(a) > a and g(b) < b.
! Define h(x) = g(x) − x; h is continuous on [a, b] and, moreover,
h(a) = g(a) − a > 0, h(b) = g(b) − b < 0.
! The Intermediate Value Theorem (IVT) implies that there existsp (a, b) for which h(p) = 0.
! Thus g(p) − p = 0 and p is a fixed point of g.
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Theoretical Frameworkg(x) is Defined on [a b]
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Theoretical Frameworkg(x) [a b] for all x [a b]
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Theoretical Frameworkg(x) has a Fixed Point in [a b]
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Theoretical Frameworkg(x) has a Fixed Point in [a b]
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Theoretical FrameworkIllustration
! Consider the function g(x) = 3−x on 0 < x < 1. g(x) is continuous and since
g′(x) = −3−x log 3 < 0 on [0, 1]
g(x) is decreasing on [0, 1].
! Henceg(1) =1/3 < g(x) < 1 = g(0)
i.e. g(x) [0, 1] for all x [0, 1] and therefore, by the
preceding result, g(x) must have a fixed point in [0, 1].
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Theoretical Frameworkg(x)=3-x
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Theoretical Framework
An Important Observation
! It is fairly obvious that, on any given interval I = [a, b], g(x) mayhave many fixed points (or none at all).
! In order to ensure that g(x) has a unique fixed point in I, we mustmake an additional assumption that g(x) does not vary too rapidly.
! Thus we have to establish a uniqueness result.
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Theoretical Framework
Uniqueness Result
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Theoretical Frameworkg’(x) is defined in [a b]
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Theoretical Framework-1 < g’(x) < 1 for all x [a b]
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Theoretical FrameworkUnique Fixed Point:|g’(x)|< 1 for all x [a b]
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Theoretical Framework
Proof of Uniqueness Result
! Assuming the hypothesis of the theorem, suppose that p and qare both fixed points in [a, b] with p = q.
! By the Mean Value Theorem (MVT), a number ξ exists between p and q and hence in [a, b] with
|p − q| = |g(p) − g(q)| = |g′(ξ)| |p − q|≤ k |p − q|< |p − q|
which is a contradiction.
! This contradiction must come from the only supposition, p = q.Hence, p = q and the fixed point in [a, b] is unique.
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Motivating the Algorithm: An ExampleA Single Nonlinear Equation
An Important Observation
Positive Root
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Motivating the Algorithm: Single Nonlinear Equation f (x) = x2− x − 1 = 0
We can convert this equation into a fixed-point problem.
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Fixed-Point Formulation ISingle Nonlinear Equation f (x) = x2− x − 1 = 0
One possible Formulation for g(x)
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Fixed-Point Formulation Ixn+1 = g (xn ) = √ xn + 1 with x0 = 0
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Fixed-Point Formulation Ixn+1 = g (xn ) = √ xn + 1 with x0 = 0
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Fixed-Point Formulation I
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Fixed-Point Formulation I
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Fixed-Point Formulation I
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Fixed-Point Formulation I
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Fixed-Point Formulation I
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Fixed-Point Formulation I
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Fixed-Point Formulation I
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Fixed-Point Formulation I
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Fixed-Point Formulation I
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Fixed-Point Formulation Ixn+1 = g (xn ) = √ xn + 1 with x0 = 0
Rate of Convergence
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Fixed-Point Formulation I
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Fixed-Point Formulation IISingle Nonlinear Equation f (x) = x2− x − 1 = 0
A Second Formulation for g(x)
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Fixed-Point Formulation II: xn+1 = g (xn) = 1/xn+ 1 with x0 = 1
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Fixed-Point Formulation II: xn+1 = g (xn) = 1/xn+ 1 with x0 = 1
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Fixed-Point Formulation II:
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Fixed-Point Formulation II:
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Fixed-Point Formulation II:
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Fixed-Point Formulation II:
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Fixed-Point Formulation II:
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Fixed-Point Formulation II:
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Fixed-Point Formulation IIxn+1 = g (xn) = 1/xn+ 1 with x0 = 1
Rate of Convergence
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Fixed-Point Formulation II: xn+1 = g (xn) = 1/xn+ 1 with x0 = 1
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Intermediate Value Theorem (IVT)Theorem: Intermediate Value Theorem
If and is any number between and then
there exists with
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Derivative Theorems
If and is differentiable on then there exists
such that.
Theorem: Mean Value Theorem (MVT)
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