Solutions of Equations in One Variable. FixedPoint Iteration II


 Jocelin Woods
 3 years ago
 Views:
Transcription
1 Solutions of Equations in One Variable FixedPoint Iteration II 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
2 Outline 1 Functional (FixedPoint) Iteration Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 2 / 54
3 Outline 1 Functional (FixedPoint) Iteration 2 Convergence Criteria for the FixedPoint Method Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 2 / 54
4 Outline 1 Functional (FixedPoint) Iteration 2 Convergence Criteria for the FixedPoint Method 3 Sample Problem: f (x) = x 3 + 4x 2 10 = 0 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 2 / 54
5 Outline 1 Functional (FixedPoint) Iteration 2 Convergence Criteria for the FixedPoint Method 3 Sample Problem: f (x) = x 3 + 4x 2 10 = 0 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 3 / 54
6 Functional (FixedPoint) Iteration Now that we have established a condition for which g(x) has a unique fixed point in I, there remains the problem of how to find it. The technique employed is known as fixedpoint iteration. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 4 / 54
7 Functional (FixedPoint) Iteration Now that we have established a condition for which g(x) has a unique fixed point in I, there remains the problem of how to find it. The technique employed is known as fixedpoint iteration. Basic Approach To approximate the fixed point of a function g, we choose an initial approximation p 0 and generate the sequence {p n } n=0 by letting p n = g(p n 1 ), for each n 1. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 4 / 54
8 Functional (FixedPoint) Iteration Now that we have established a condition for which g(x) has a unique fixed point in I, there remains the problem of how to find it. The technique employed is known as fixedpoint iteration. Basic Approach To approximate the fixed point of a function g, we choose an initial approximation p 0 and generate the sequence {p n } n=0 by letting p n = g(p n 1 ), for each n 1. If the sequence converges to p and g is continuous, then ( ) p = lim p n = lim g(p n 1 ) = g lim p n 1 = g(p), n n n and a solution to x = g(x) is obtained. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 4 / 54
9 Functional (FixedPoint) Iteration Now that we have established a condition for which g(x) has a unique fixed point in I, there remains the problem of how to find it. The technique employed is known as fixedpoint iteration. Basic Approach To approximate the fixed point of a function g, we choose an initial approximation p 0 and generate the sequence {p n } n=0 by letting p n = g(p n 1 ), for each n 1. If the sequence converges to p and g is continuous, then ( ) p = lim p n = lim g(p n 1 ) = g lim p n 1 = g(p), n n n and a solution to x = g(x) is obtained. This technique is called fixedpoint, or functional iteration. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 4 / 54
10 Functional (FixedPoint) Iteration y p 2 5 g(p 1 ) p 3 5 g(p 2 ) p 1 5 g(p 0 ) (p 1, p 1 ) (p 1, p 2 ) (p 2, p 2 ) y 5 x (p 0, p 1 ) y p 3 5 g(p 2 ) p 2 5 g(p 1 ) p 1 5 g(p 0 ) (p 0, p 1 ) y 5 x (p 2, p 3 ) y 5 g(x) (p 2, p 2 ) (p 1, p 1 ) y 5 g(x) p 1 p 3 p 2 p 0 x p 0 p 1 p 2 x (a) (b) Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 5 / 54
11 Functional (FixedPoint) Iteration Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 6 / 54
12 Functional (FixedPoint) Iteration FixedPoint Algorithm To find the fixed point of g in an interval [a, b], given the equation x = g(x) with an initial guess p 0 [a, b]: Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 6 / 54
13 Functional (FixedPoint) Iteration FixedPoint Algorithm To find the fixed point of g in an interval [a, b], given the equation x = g(x) with an initial guess p 0 [a, b]: 1. n = 1; Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 6 / 54
14 Functional (FixedPoint) Iteration FixedPoint Algorithm To find the fixed point of g in an interval [a, b], given the equation x = g(x) with an initial guess p 0 [a, b]: 1. n = 1; 2. p n = g(p n 1 ); Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 6 / 54
15 Functional (FixedPoint) Iteration FixedPoint Algorithm To find the fixed point of g in an interval [a, b], given the equation x = g(x) with an initial guess p 0 [a, b]: 1. n = 1; 2. p n = g(p n 1 ); 3. If p n p n 1 < ɛ then 5; Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 6 / 54
16 Functional (FixedPoint) Iteration FixedPoint Algorithm To find the fixed point of g in an interval [a, b], given the equation x = g(x) with an initial guess p 0 [a, b]: 1. n = 1; 2. p n = g(p n 1 ); 3. If p n p n 1 < ɛ then 5; 4. n n + 1; go to 2. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 6 / 54
17 Functional (FixedPoint) Iteration FixedPoint Algorithm To find the fixed point of g in an interval [a, b], given the equation x = g(x) with an initial guess p 0 [a, b]: 1. n = 1; 2. p n = g(p n 1 ); 3. If p n p n 1 < ɛ then 5; 4. n n + 1; go to End of Procedure. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 6 / 54
18 A Single Nonlinear Equation Example 1 The equation x 3 + 4x 2 10 = 0 has a unique root in [1, 2]. Its value is approximately Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 7 / 54
19 f (x) = x 3 + 4x 2 10 = 0 on [1, 2] y 14 y = f(x) = x 3 + 4x x 5 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 8 / 54
20 f (x) = x 3 + 4x 2 10 = 0 on [1, 2] Possible Choices for g(x) Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 9 / 54
21 f (x) = x 3 + 4x 2 10 = 0 on [1, 2] Possible Choices for g(x) There are many ways to change the equation to the fixedpoint form x = g(x) using simple algebraic manipulation. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 9 / 54
22 f (x) = x 3 + 4x 2 10 = 0 on [1, 2] Possible Choices for g(x) There are many ways to change the equation to the fixedpoint form x = g(x) using simple algebraic manipulation. For example, to obtain the function g described in part (c), we can manipulate the equation x 3 + 4x 2 10 = 0 as follows: 4x 2 = 10 x 3, so x 2 = 1 4 (10 x 3 ), and x = ± 1 2 (10 x 3 ) 1/2. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 9 / 54
23 f (x) = x 3 + 4x 2 10 = 0 on [1, 2] Possible Choices for g(x) There are many ways to change the equation to the fixedpoint form x = g(x) using simple algebraic manipulation. For example, to obtain the function g described in part (c), we can manipulate the equation x 3 + 4x 2 10 = 0 as follows: 4x 2 = 10 x 3, so x 2 = 1 4 (10 x 3 ), and x = ± 1 2 (10 x 3 ) 1/2. We will consider 5 such rearrangements and, later in this section, provide a brief analysis as to why some do and some not converge to p = Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 9 / 54
24 Solving f (x) = x 3 + 4x 2 10 = 0 5 Possible Transpositions to x = g(x) x = g 1 (x) = x x 3 4x x = g 2 (x) = x = g 3 (x) = x 4x 10 x 3 x = g 4 (x) = x x = g 5 (x) = x x 3 + 4x x 2 + 8x Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 10 / 54
25 Numerical Results for f (x) = x 3 + 4x 2 10 = 0 n g 1 g 2 g 3 g 4 g ( 8.65) 1/ Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 11 / 54
26 Solving f (x) = x 3 + 4x 2 10 = 0 x = g(x) with x 0 = 1.5 x = g 1 (x) = x x 3 4x Does not Converge x = g 2 (x) = x = g 3 (x) = x 4x Does not Converge 10 x 3 Converges after 31 Iterations x = g 4 (x) = x x = g 5 (x) = x x 3 + 4x x 2 + 8x Converges after 12 Iterations Converges after 5 Iterations Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 12 / 54
27 Outline 1 Functional (FixedPoint) Iteration 2 Convergence Criteria for the FixedPoint Method 3 Sample Problem: f (x) = x 3 + 4x 2 10 = 0 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 13 / 54
28 Functional (FixedPoint) Iteration A Crucial Question How can we find a fixedpoint problem that produces a sequence that reliably and rapidly converges to a solution to a given rootfinding problem? Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 14 / 54
29 Functional (FixedPoint) Iteration A Crucial Question How can we find a fixedpoint problem that produces a sequence that reliably and rapidly converges to a solution to a given rootfinding problem? The following theorem and its corollary give us some clues concerning the paths we should pursue and, perhaps more importantly, some we should reject. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 14 / 54
30 Functional (FixedPoint) Iteration Convergence Result Let g C[a, b] with g(x) [a, b] for all x [a, b]. Let g (x) exist on (a, b) with g (x) k < 1, x [a, b]. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 15 / 54
31 Functional (FixedPoint) Iteration Convergence Result Let g C[a, b] with g(x) [a, b] for all x [a, b]. Let g (x) exist on (a, b) with g (x) k < 1, x [a, b]. If p 0 is any point in [a, b] then the sequence defined by p n = g(p n 1 ), n 1, will converge to the unique fixed point p in [a, b]. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 15 / 54
32 Functional (FixedPoint) Iteration Proof of the Convergence Result Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 16 / 54
33 Functional (FixedPoint) Iteration Proof of the Convergence Result By the Uniquenes Theorem, a unique fixed point exists in [a, b]. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 16 / 54
34 Functional (FixedPoint) Iteration Proof of the Convergence Result By the Uniquenes Theorem, a unique fixed point exists in [a, b]. Since g maps [a, b] into itself, the sequence {p n } n=0 is defined for all n 0 and p n [a, b] for all n. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 16 / 54
35 Functional (FixedPoint) Iteration Proof of the Convergence Result By the Uniquenes Theorem, a unique fixed point exists in [a, b]. Since g maps [a, b] into itself, the sequence {p n } n=0 is defined for all n 0 and p n [a, b] for all n. Using the Mean Value Theorem MVT and the assumption that g (x) k < 1, x [a, b], we write p n p = g(p n 1 ) g(p) Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 16 / 54
36 Functional (FixedPoint) Iteration Proof of the Convergence Result By the Uniquenes Theorem, a unique fixed point exists in [a, b]. Since g maps [a, b] into itself, the sequence {p n } n=0 is defined for all n 0 and p n [a, b] for all n. Using the Mean Value Theorem MVT and the assumption that g (x) k < 1, x [a, b], we write p n p = g(p n 1 ) g(p) g (ξ) p n 1 p Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 16 / 54
37 Functional (FixedPoint) Iteration Proof of the Convergence Result By the Uniquenes Theorem, a unique fixed point exists in [a, b]. Since g maps [a, b] into itself, the sequence {p n } n=0 is defined for all n 0 and p n [a, b] for all n. Using the Mean Value Theorem MVT and the assumption that g (x) k < 1, x [a, b], we write where ξ (a, b). p n p = g(p n 1 ) g(p) g (ξ) p n 1 p k p n 1 p Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 16 / 54
38 Functional (FixedPoint) Iteration Proof of the Convergence Result (Cont d) Applying the inequality of the hypothesis inductively gives Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 17 / 54
39 Functional (FixedPoint) Iteration Proof of the Convergence Result (Cont d) Applying the inequality of the hypothesis inductively gives p n p k p n 1 p Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 17 / 54
40 Functional (FixedPoint) Iteration Proof of the Convergence Result (Cont d) Applying the inequality of the hypothesis inductively gives p n p k p n 1 p k 2 p n 2 p Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 17 / 54
41 Functional (FixedPoint) Iteration Proof of the Convergence Result (Cont d) Applying the inequality of the hypothesis inductively gives p n p k p n 1 p k 2 p n 2 p k n p 0 p Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 17 / 54
42 Functional (FixedPoint) Iteration Proof of the Convergence Result (Cont d) Applying the inequality of the hypothesis inductively gives Since k < 1, and {p n } n=0 converges to p. p n p k p n 1 p k 2 p n 2 p k n p 0 p lim p n p lim k n p 0 p = 0, n n Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 17 / 54
43 Functional (FixedPoint) Iteration Corrollary to the Convergence Result If g satisfies the hypothesis of the Theorem, then p n p k n 1 k p 1 p 0. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 18 / 54
44 Functional (FixedPoint) Iteration Corrollary to the Convergence Result If g satisfies the hypothesis of the Theorem, then p n p k n 1 k p 1 p 0. Proof of Corollary (1 of 3) For n 1, the procedure used in the proof of the theorem implies that p n+1 p n = g(p n ) g(p n 1 ) Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 18 / 54
45 Functional (FixedPoint) Iteration Corrollary to the Convergence Result If g satisfies the hypothesis of the Theorem, then p n p k n 1 k p 1 p 0. Proof of Corollary (1 of 3) For n 1, the procedure used in the proof of the theorem implies that p n+1 p n = g(p n ) g(p n 1 ) k p n p n 1 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 18 / 54
46 Functional (FixedPoint) Iteration Corrollary to the Convergence Result If g satisfies the hypothesis of the Theorem, then p n p k n 1 k p 1 p 0. Proof of Corollary (1 of 3) For n 1, the procedure used in the proof of the theorem implies that p n+1 p n = g(p n ) g(p n 1 ) k p n p n 1 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 18 / 54
47 Functional (FixedPoint) Iteration Corrollary to the Convergence Result If g satisfies the hypothesis of the Theorem, then p n p k n 1 k p 1 p 0. Proof of Corollary (1 of 3) For n 1, the procedure used in the proof of the theorem implies that p n+1 p n = g(p n ) g(p n 1 ) k p n p n 1 k n p 1 p 0 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 18 / 54
48 Functional (FixedPoint) Iteration Proof of Corollary (2 of 3) p n+1 p n k n p 1 p 0 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 19 / 54
49 Functional (FixedPoint) Iteration p n+1 p n k n p 1 p 0 Proof of Corollary (2 of 3) Thus, for m > n 1, p m p n = p m p m 1 + p m 1 p m p n+1 p n Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 19 / 54
50 Functional (FixedPoint) Iteration p n+1 p n k n p 1 p 0 Proof of Corollary (2 of 3) Thus, for m > n 1, p m p n = p m p m 1 + p m 1 p m p n+1 p n p m p m 1 + p m 1 p m p n+1 p n Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 19 / 54
51 Functional (FixedPoint) Iteration p n+1 p n k n p 1 p 0 Proof of Corollary (2 of 3) Thus, for m > n 1, p m p n = p m p m 1 + p m 1 p m p n+1 p n p m p m 1 + p m 1 p m p n+1 p n k m 1 p 1 p 0 + k m 2 p 1 p k n p 1 p 0 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 19 / 54
52 Functional (FixedPoint) Iteration p n+1 p n k n p 1 p 0 Proof of Corollary (2 of 3) Thus, for m > n 1, p m p n = p m p m 1 + p m 1 p m p n+1 p n p m p m 1 + p m 1 p m p n+1 p n k m 1 p 1 p 0 + k m 2 p 1 p k n p 1 p 0 k n 1 + k + k k m n 1) p 1 p 0. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 19 / 54
53 Functional (FixedPoint) Iteration p m p n k n ( 1 + k + k k m n 1) p 1 p 0. Proof of Corollary (3 of 3) Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 20 / 54
54 Functional (FixedPoint) Iteration p m p n k n ( 1 + k + k k m n 1) p 1 p 0. Proof of Corollary (3 of 3) However, since lim m p m = p, we obtain p p n = lim m p m p n Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 20 / 54
55 Functional (FixedPoint) Iteration p m p n k n ( 1 + k + k k m n 1) p 1 p 0. Proof of Corollary (3 of 3) However, since lim m p m = p, we obtain p p n = lim m p n m k n p 1 p 0 i=1 k i Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 20 / 54
56 Functional (FixedPoint) Iteration p m p n k n ( 1 + k + k k m n 1) p 1 p 0. Proof of Corollary (3 of 3) However, since lim m p m = p, we obtain p p n = lim m p n m k n p 1 p 0 = i=1 k i k n 1 k p 1 p 0. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 20 / 54
57 Functional (FixedPoint) Iteration Example: g(x) = g(x) = 3 x Consider the iteration function g(x) = 3 x over the interval [ 1 3, 1] starting with p 0 = 1 3. Determine a lower bound for the number of iterations n required so that p n p < 10 5? Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 21 / 54
58 Functional (FixedPoint) Iteration Example: g(x) = g(x) = 3 x Consider the iteration function g(x) = 3 x over the interval [ 1 3, 1] starting with p 0 = 1 3. Determine a lower bound for the number of iterations n required so that p n p < 10 5? Determine the Parameters of the Problem Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 21 / 54
59 Functional (FixedPoint) Iteration Example: g(x) = g(x) = 3 x Consider the iteration function g(x) = 3 x over the interval [ 1 3, 1] starting with p 0 = 1 3. Determine a lower bound for the number of iterations n required so that p n p < 10 5? Determine the Parameters of the Problem Note that p 1 = g(p 0 ) = = and, since g (x) = 3 x ln 3, we obtain the bound g (x) ln = k. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 21 / 54
60 Functional (FixedPoint) Iteration Use the Corollary Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 22 / 54
61 Functional (FixedPoint) Iteration Use the Corollary Therefore, we have p n p k n 1 k p 0 p 1 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 22 / 54
62 Functional (FixedPoint) Iteration Use the Corollary Therefore, we have p n p k n 1 k p 0 p n Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 22 / 54
63 Functional (FixedPoint) Iteration Use the Corollary Therefore, we have p n p k n 1 k p 0 p n n Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 22 / 54
64 Functional (FixedPoint) Iteration Use the Corollary Therefore, we have p n p k n 1 k p 0 p n n We require that n < 10 5 or n > Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 22 / 54
65 Footnote on the Estimate Obtained Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 23 / 54
66 Footnote on the Estimate Obtained It is important to realise that the estimate for the number of iterations required given by the theorem is an upper bound. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 23 / 54
67 Footnote on the Estimate Obtained It is important to realise that the estimate for the number of iterations required given by the theorem is an upper bound. In the previous example, only 21 iterations are required in practice, i.e. p 21 = is accurate to Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 23 / 54
68 Footnote on the Estimate Obtained It is important to realise that the estimate for the number of iterations required given by the theorem is an upper bound. In the previous example, only 21 iterations are required in practice, i.e. p 21 = is accurate to The reason, in this case, is that we used g (1) = whereas g ( ) = Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 23 / 54
69 Footnote on the Estimate Obtained It is important to realise that the estimate for the number of iterations required given by the theorem is an upper bound. In the previous example, only 21 iterations are required in practice, i.e. p 21 = is accurate to The reason, in this case, is that we used g (1) = whereas g ( ) = If one had used k = (were it available) to compute the bound, one would obtain N = 23 which is a more accurate estimate. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 23 / 54
70 Outline 1 Functional (FixedPoint) Iteration 2 Convergence Criteria for the FixedPoint Method 3 Sample Problem: f (x) = x 3 + 4x 2 10 = 0 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 24 / 54
71 A Single Nonlinear Equation Example 2 We return to Example 1 and the equation x 3 + 4x 2 10 = 0 which has a unique root in [1, 2]. Its value is approximately Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 25 / 54
72 f (x) = x 3 + 4x 2 10 = 0 on [1, 2] y 14 y = f(x) = x 3 + 4x x 5 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 26 / 54
73 Solving f (x) = x 3 + 4x 2 10 = 0 Earlier, we listed 5 possible transpositions to x = g(x) x = g 1 (x) = x x 3 4x x = g 2 (x) = x = g 3 (x) = x 4x 10 x 3 x = g 4 (x) = x x = g 5 (x) = x x 3 + 4x x 2 + 8x Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 27 / 54
74 Solving f (x) = x 3 + 4x 2 10 = 0 Results Observed for x = g(x) with x 0 = 1.5 x = g 1 (x) = x x 3 4x x = g 2 (x) = x = g 3 (x) = x 4x Does not Converge Does not Converge 10 x 3 Converges after 31 Iterations x = g 4 (x) = x x = g 5 (x) = x x 3 + 4x x 2 + 8x Converges after 12 Iterations Converges after 5 Iterations Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 28 / 54
75 Solving f (x) = x 3 + 4x 2 10 = 0 x = g(x) with x 0 = 1.5 x = g 1 (x) = x x 3 4x Does not Converge x = g 2 (x) = x = g 3 (x) = x 4x Does not Converge 10 x 3 Converges after 31 Iterations x = g 4 (x) = x x = g 5 (x) = x x 3 + 4x x 2 + 8x Converges after 12 Iterations Converges after 5 Iterations Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 29 / 54
76 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 1 (x) = x x 3 4x Iteration for x = g 1 (x) Does Not Converge Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 30 / 54
77 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 1 (x) = x x 3 4x Iteration for x = g 1 (x) Does Not Converge Since g 1 (x) = 1 3x 2 8x g 1 (1) = 10 g 1 (2) = 27 there is no interval [a, b] containing p for which g 1 (x) < 1. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 30 / 54
78 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 1 (x) = x x 3 4x Iteration for x = g 1 (x) Does Not Converge Since g 1 (x) = 1 3x 2 8x g 1 (1) = 10 g 1 (2) = 27 there is no interval [a, b] containing p for which g 1 (x) < 1. Also, note that g 1 (1) = 6 and g 2 (2) = 12 so that g(x) / [1, 2] for x [1, 2]. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 30 / 54
79 Iteration Function: x = g 1 (x) = x x 3 4x Iterations starting with p 0 = 1.5 n p n 1 p n p n p n p Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 31 / 54
80 g 1 Does Not Map [1, 2] into [1, 2] y g 1 (x) = x x 3 4x x 10 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 32 / 54
81 g 1 (x) > 1 on [1, 2] y 1 2 x 10 g 1 (x) = 1 3x2 8x 27 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 33 / 54
82 Solving f (x) = x 3 + 4x 2 10 = 0 x = g(x) with x 0 = 1.5 x = g 1 (x) = x x 3 4x Does not Converge x = g 2 (x) = x = g 3 (x) = x 4x Does not Converge 10 x 3 Converges after 31 Iterations x = g 4 (x) = x x = g 5 (x) = x x 3 + 4x x 2 + 8x Converges after 12 Iterations Converges after 5 Iterations Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 34 / 54
83 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 2 (x) = Iteration for x = g 2 (x) is Not Defined 10 x 4x Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 35 / 54
84 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 2 (x) = Iteration for x = g 2 (x) is Not Defined 10 x 4x It is clear that g 2 (x) does not map [1, 2] onto [1, 2] and the sequence {p n } n=0 is not defined for p 0 = 1.5. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 35 / 54
85 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 2 (x) = Iteration for x = g 2 (x) is Not Defined 10 x 4x It is clear that g 2 (x) does not map [1, 2] onto [1, 2] and the sequence {p n } n=0 is not defined for p 0 = 1.5. Also, there is no interval containing p such that g 2 (x) < 1 since g (1) 2.86 g (p) 3.43 and g (x) is not defined for x > Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 35 / 54
86 Iteration Function: x = g 2 (x) = 10 x 4x Iterations starting with p 0 = 1.5 n p n 1 p n p n p n Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 36 / 54
87 Solving f (x) = x 3 + 4x 2 10 = 0 x = g(x) with x 0 = 1.5 x = g 1 (x) = x x 3 4x Does not Converge x = g 2 (x) = x = g 3 (x) = x 4x Does not Converge 10 x 3 Converges after 31 Iterations x = g 4 (x) = x x = g 5 (x) = x x 3 + 4x x 2 + 8x Converges after 12 Iterations Converges after 5 Iterations Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 37 / 54
88 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 3 (x) = x 3 Iteration for x = g 3 (x) Converges (Slowly) Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 38 / 54
89 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 3 (x) = x 3 Iteration for x = g 3 (x) Converges (Slowly) By differentiation, g 3 (x) = 3x 2 4 < 0 for x [1, 2] 10 x 3 and so g=g 3 is strictly decreasing on [1, 2]. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 38 / 54
90 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 3 (x) = x 3 Iteration for x = g 3 (x) Converges (Slowly) By differentiation, g 3 (x) = 3x 2 4 < 0 for x [1, 2] 10 x 3 and so g=g 3 is strictly decreasing on [1, 2]. However, g 3 (x) > 1 for x > 1.71 and g 3 (2) Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 38 / 54
91 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 3 (x) = x 3 Iteration for x = g 3 (x) Converges (Slowly) By differentiation, g 3 (x) = 3x 2 4 < 0 for x [1, 2] 10 x 3 and so g=g 3 is strictly decreasing on [1, 2]. However, g 3 (x) > 1 for x > 1.71 and g 3 (2) A closer examination of {p n } n=0 will show that it suffices to consider the interval [1, 1.7] where g 3 (x) < 1 and g(x) [1, 1.7] for x [1, 1.7]. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 38 / 54
92 Iteration Function: x = g 3 (x) = x 3 Iterations starting with p 0 = 1.5 n p n 1 p n p n p n Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 39 / 54
93 g 3 Maps [1, 1.7] into [1, 1.7] y g 3 (x) = x x Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 40 / 54
94 g 3 (x) < 1 on [1, 1.7] y g 3 3x2 (x) = 4 10 x x 1 Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 41 / 54
95 Solving f (x) = x 3 + 4x 2 10 = 0 x = g(x) with x 0 = 1.5 x = g 1 (x) = x x 3 4x Does not Converge x = g 2 (x) = x = g 3 (x) = x 4x Does not Converge 10 x 3 Converges after 31 Iterations x = g 4 (x) = x x = g 5 (x) = x x 3 + 4x x 2 + 8x Converges after 12 Iterations Converges after 5 Iterations Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 42 / 54
96 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 4 (x) = x Iteration for x = g 4 (x) Converges (Moderately) Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 43 / 54
97 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 4 (x) = x Iteration for x = g 4 (x) Converges (Moderately) By differentiation, and it is easy to show that g 4 (x) = 10 4(4 + x) 3 < < g 4 (x) < 0.15 x [1, 2] Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 43 / 54
98 Solving f (x) = x 3 + 4x 2 10 = 0 x = g 4 (x) = x Iteration for x = g 4 (x) Converges (Moderately) By differentiation, and it is easy to show that g 4 (x) = 10 4(4 + x) 3 < < g 4 (x) < 0.15 x [1, 2] The bound on the magnitude of g 4 (x) is much smaller than that for g 3 (x) and this explains the reason for the much faster convergence. Numerical Analysis (Chapter 2) FixedPoint Iteration II R L Burden & J D Faires 43 / 54
Iterative Techniques in Matrix Algebra. Jacobi & GaussSeidel Iterative Techniques II
Iterative Techniques in Matrix Algebra Jacobi & GaussSeidel Iterative Techniques II Numerical Analysis (9th Edition) R L Burden & J D Faires Beamer Presentation Slides prepared by John Carroll Dublin
More informationNumerical methods for finding the roots of a function
Numerical methods for finding the roots of a function The roots of a function f (x) are defined as the values for which the value of the function becomes equal to zero. So, finding the roots of f (x) means
More informationLecture 3: Solving Equations Using Fixed Point Iterations
cs41: introduction to numerical analysis 09/14/10 Lecture 3: Solving Equations Using Fixed Point Iterations Instructor: Professor Amos Ron Scribes: Yunpeng Li, Mark Cowlishaw, Nathanael Fillmore Our problem,
More informationMath 317 HW #5 Solutions
Math 317 HW #5 Solutions 1. Exercise 2.4.2. (a) Prove that the sequence defined by x 1 = 3 and converges. x n+1 = 1 4 x n Proof. I intend to use the Monotone Convergence Theorem, so my goal is to show
More informationHOMEWORK 5 SOLUTIONS. n!f n (1) lim. ln x n! + xn x. 1 = G n 1 (x). (2) k + 1 n. (n 1)!
Math 7 Fall 205 HOMEWORK 5 SOLUTIONS Problem. 2008 B2 Let F 0 x = ln x. For n 0 and x > 0, let F n+ x = 0 F ntdt. Evaluate n!f n lim n ln n. By directly computing F n x for small n s, we obtain the following
More informationSolving nonlinear equations in one variable
Chapter Solving nonlinear equations in one variable Contents.1 Bisection method............................. 7. Fixed point iteration........................... 8.3 NewtonRaphson method........................
More informationNumerical Analysis and Computing
Joe Mahaffy, mahaffy@math.sdsu.edu Spring 2010 #4: Solutions of Equations in One Variable (1/58) Numerical Analysis and Computing Lecture Notes #04 Solutions of Equations in One Variable, Interpolation
More informationIndiana State Core Curriculum Standards updated 2009 Algebra I
Indiana State Core Curriculum Standards updated 2009 Algebra I Strand Description Boardworks High School Algebra presentations Operations With Real Numbers Linear Equations and A1.1 Students simplify and
More information(Quasi)Newton methods
(Quasi)Newton methods 1 Introduction 1.1 Newton method Newton method is a method to find the zeros of a differentiable nonlinear function g, x such that g(x) = 0, where g : R n R n. Given a starting
More informationNonlinear Algebraic Equations. Lectures INF2320 p. 1/88
Nonlinear Algebraic Equations Lectures INF2320 p. 1/88 Lectures INF2320 p. 2/88 Nonlinear algebraic equations When solving the system u (t) = g(u), u(0) = u 0, (1) with an implicit Euler scheme we have
More information1 Fixed Point Iteration and Contraction Mapping Theorem
1 Fixed Point Iteration and Contraction Mapping Theorem Notation: For two sets A,B we write A B iff x A = x B. So A A is true. Some people use the notation instead. 1.1 Introduction Consider a function
More informationNumerical Analysis and Computing
Numerical Analysis and Computing Lecture Notes #3 Solutions of Equations in One Variable: ; Root Finding; for Iterative Methods Joe Mahaffy, mahaffy@math.sdsu.edu Department of Mathematics Dynamical Systems
More informationINTRODUCTION TO THE CONVERGENCE OF SEQUENCES
INTRODUCTION TO THE CONVERGENCE OF SEQUENCES BECKY LYTLE Abstract. In this paper, we discuss the basic ideas involved in sequences and convergence. We start by defining sequences and follow by explaining
More informationCS 221. Tuesday 8 November 2011
CS 221 Tuesday 8 November 2011 Agenda 1. Announcements 2. Review: Solving Equations (Text 6.16.3) 3. Rootfinding with Excel ( Goal Seek, Text 6.5) 4. Example Rootfinding Problems 5. The Fixedpoint
More informationAdaptive Online Gradient Descent
Adaptive Online Gradient Descent Peter L Bartlett Division of Computer Science Department of Statistics UC Berkeley Berkeley, CA 94709 bartlett@csberkeleyedu Elad Hazan IBM Almaden Research Center 650
More informationMath 2602 Finite and Linear Math Fall 14. Homework 9: Core solutions
Math 2602 Finite and Linear Math Fall 14 Homework 9: Core solutions Section 8.2 on page 264 problems 13b, 27a27b. Section 8.3 on page 275 problems 1b, 8, 10a10b, 14. Section 8.4 on page 279 problems
More informationReal Roots of Univariate Polynomials with Real Coefficients
Real Roots of Univariate Polynomials with Real Coefficients mostly written by Christina Hewitt March 22, 2012 1 Introduction Polynomial equations are used throughout mathematics. When solving polynomials
More informationBasics of Polynomial Theory
3 Basics of Polynomial Theory 3.1 Polynomial Equations In geodesy and geoinformatics, most observations are related to unknowns parameters through equations of algebraic (polynomial) type. In cases where
More informationItems related to expected use of graphing technology appear in bold italics.
 1  Items related to expected use of graphing technology appear in bold italics. Investigating the Graphs of Polynomial Functions determine, through investigation, using graphing calculators or graphing
More informationNonlinear Algebraic Equations Example
Nonlinear Algebraic Equations Example Continuous Stirred Tank Reactor (CSTR). Look for steady state concentrations & temperature. s r (in) p,i (in) i In: N spieces with concentrations c, heat capacities
More informationRolle s Theorem. q( x) = 1
Lecture 1 :The Mean Value Theorem We know that constant functions have derivative zero. Is it possible for a more complicated function to have derivative zero? In this section we will answer this question
More information4.3 Lagrange Approximation
206 CHAP. 4 INTERPOLATION AND POLYNOMIAL APPROXIMATION Lagrange Polynomial Approximation 4.3 Lagrange Approximation Interpolation means to estimate a missing function value by taking a weighted average
More information11. Nonlinear equations with one variable
EE103 (Fall 201112) 11. Nonlinear equations with one variable definition and examples bisection method Newton s method secant method 111 Definition and examples x is a zero (or root) of a function f
More informationNumerical Methods for the Root Finding Problem
Chapter 1 Numerical Methods for the Root Finding Problem Oct. 11, 2011 HG 1.1 A Case Study on the RootFinding Problem: Kepler s Law of Planetary Motion The rootfinding problem is one of the most important
More informationThe Mean Value Theorem
The Mean Value Theorem THEOREM (The Extreme Value Theorem): If f is continuous on a closed interval [a, b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some numbers
More informationDirect Methods for Solving Linear Systems. Matrix Factorization
Direct Methods for Solving Linear Systems Matrix Factorization Numerical Analysis (9th Edition) R L Burden & J D Faires Beamer Presentation Slides prepared by John Carroll Dublin City University c 2011
More informationFIXED POINT ITERATION
FIXED POINT ITERATION We begin with a computational example. solving the two equations Consider E1: x =1+.5sinx E2: x =3+2sinx Graphs of these two equations are shown on accompanying graphs, with the solutions
More informationx a x 2 (1 + x 2 ) n.
Limits and continuity Suppose that we have a function f : R R. Let a R. We say that f(x) tends to the limit l as x tends to a; lim f(x) = l ; x a if, given any real number ɛ > 0, there exists a real number
More informationComputer Algorithms. Merge Sort. Analysis of Merge Sort. Recurrences. Recurrence Relations. Iteration Method. Lecture 6
Computer Algorithms Lecture 6 Merge Sort Sorting Problem: Sort a sequence of n elements into nondecreasing order. Divide: Divide the nelement sequence to be sorted into two subsequences of n/ elements
More informationCritical points of once continuously differentiable functions are important because they are the only points that can be local maxima or minima.
Lecture 0: Convexity and Optimization We say that if f is a once continuously differentiable function on an interval I, and x is a point in the interior of I that x is a critical point of f if f (x) =
More informationt := maxγ ν subject to ν {0,1,2,...} and f(x c +γ ν d) f(x c )+cγ ν f (x c ;d).
1. Line Search Methods Let f : R n R be given and suppose that x c is our current best estimate of a solution to P min x R nf(x). A standard method for improving the estimate x c is to choose a direction
More informationNonlinear Optimization: Algorithms 3: Interiorpoint methods
Nonlinear Optimization: Algorithms 3: Interiorpoint methods INSEAD, Spring 2006 JeanPhilippe Vert Ecole des Mines de Paris JeanPhilippe.Vert@mines.org Nonlinear optimization c 2006 JeanPhilippe Vert,
More information10.3 POWER METHOD FOR APPROXIMATING EIGENVALUES
58 CHAPTER NUMERICAL METHODS. POWER METHOD FOR APPROXIMATING EIGENVALUES In Chapter 7 you saw that the eigenvalues of an n n matrix A are obtained by solving its characteristic equation n c nn c nn...
More informationCONTINUED FRACTIONS AND PELL S EQUATION. Contents 1. Continued Fractions 1 2. Solution to Pell s Equation 9 References 12
CONTINUED FRACTIONS AND PELL S EQUATION SEUNG HYUN YANG Abstract. In this REU paper, I will use some important characteristics of continued fractions to give the complete set of solutions to Pell s equation.
More informationDouble Sequences and Double Series
Double Sequences and Double Series Eissa D. Habil Islamic University of Gaza P.O. Box 108, Gaza, Palestine Email: habil@iugaza.edu Abstract This research considers two traditional important questions,
More informationLecture 24: Series finale
Lecture 24: Series finale Nathan Pflueger 2 November 20 Introduction This lecture will discuss one final convergence test and error bound for series, which concerns alternating series. A general summary
More informationSolutions to Homework 10
Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x
More informationTaylor Polynomials and Taylor Series Math 126
Taylor Polynomials and Taylor Series Math 26 In many problems in science and engineering we have a function f(x) which is too complicated to answer the questions we d like to ask. In this chapter, we will
More informationNotes from Week 1: Algorithms for sequential prediction
CS 683 Learning, Games, and Electronic Markets Spring 2007 Notes from Week 1: Algorithms for sequential prediction Instructor: Robert Kleinberg 2226 Jan 2007 1 Introduction In this course we will be looking
More informationSeries Convergence Tests Math 122 Calculus III D Joyce, Fall 2012
Some series converge, some diverge. Series Convergence Tests Math 22 Calculus III D Joyce, Fall 202 Geometric series. We ve already looked at these. We know when a geometric series converges and what it
More informationSEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION
CHAPTER 5 SEQUENCES, MATHEMATICAL INDUCTION, AND RECURSION Alessandro Artale UniBZ  http://www.inf.unibz.it/ artale/ SECTION 5.2 Mathematical Induction I Copyright Cengage Learning. All rights reserved.
More informationLimit processes are the basis of calculus. For example, the derivative. f f (x + h) f (x)
SEC. 4.1 TAYLOR SERIES AND CALCULATION OF FUNCTIONS 187 Taylor Series 4.1 Taylor Series and Calculation of Functions Limit processes are the basis of calculus. For example, the derivative f f (x + h) f
More informationChapter 6 Finite sets and infinite sets. Copyright 2013, 2005, 2001 Pearson Education, Inc. Section 3.1, Slide 1
Chapter 6 Finite sets and infinite sets Copyright 013, 005, 001 Pearson Education, Inc. Section 3.1, Slide 1 Section 6. PROPERTIES OF THE NATURE NUMBERS 013 Pearson Education, Inc.1 Slide Recall that denotes
More informationMTH4100 Calculus I. Lecture notes for Week 8. Thomas Calculus, Sections 4.1 to 4.4. Rainer Klages
MTH4100 Calculus I Lecture notes for Week 8 Thomas Calculus, Sections 4.1 to 4.4 Rainer Klages School of Mathematical Sciences Queen Mary University of London Autumn 2009 Theorem 1 (First Derivative Theorem
More informationTHE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS
THE FUNDAMENTAL THEOREM OF ALGEBRA VIA PROPER MAPS KEITH CONRAD 1. Introduction The Fundamental Theorem of Algebra says every nonconstant polynomial with complex coefficients can be factored into linear
More informationLecture 1 Newton s method.
Lecture 1 Newton s method. 1 Square roots 2 Newton s method. The guts of the method. A vector version. Implementation. The existence theorem. Basins of attraction. The Babylonian algorithm for finding
More informationG.A. Pavliotis. Department of Mathematics. Imperial College London
EE1 MATHEMATICS NUMERICAL METHODS G.A. Pavliotis Department of Mathematics Imperial College London 1. Numerical solution of nonlinear equations (iterative processes). 2. Numerical evaluation of integrals.
More information10.3 POWER METHOD FOR APPROXIMATING EIGENVALUES
55 CHAPTER NUMERICAL METHODS. POWER METHOD FOR APPROXIMATING EIGENVALUES In Chapter 7 we saw that the eigenvalues of an n n matrix A are obtained by solving its characteristic equation n c n n c n n...
More information1. R In this and the next section we are going to study the properties of sequences of real numbers.
+a 1. R In this and the next section we are going to study the properties of sequences of real numbers. Definition 1.1. (Sequence) A sequence is a function with domain N. Example 1.2. A sequence of real
More informationSome Notes on Taylor Polynomials and Taylor Series
Some Notes on Taylor Polynomials and Taylor Series Mark MacLean October 3, 27 UBC s courses MATH /8 and MATH introduce students to the ideas of Taylor polynomials and Taylor series in a fairly limited
More informationFINAL EXAM SECTIONS AND OBJECTIVES FOR COLLEGE ALGEBRA
FINAL EXAM SECTIONS AND OBJECTIVES FOR COLLEGE ALGEBRA 1.1 Solve linear equations and equations that lead to linear equations. a) Solve the equation: 1 (x + 5) 4 = 1 (2x 1) 2 3 b) Solve the equation: 3x
More information(57) (58) (59) (60) (61)
Module 4 : Solving Linear Algebraic Equations Section 5 : Iterative Solution Techniques 5 Iterative Solution Techniques By this approach, we start with some initial guess solution, say for solution and
More information(Refer Slide Time: 00:00:56 min)
Numerical Methods and Computation Prof. S.R.K. Iyengar Department of Mathematics Indian Institute of Technology, Delhi Lecture No # 3 Solution of Nonlinear Algebraic Equations (Continued) (Refer Slide
More informationThe Derivative and the Tangent Line Problem. The Tangent Line Problem
The Derivative and the Tangent Line Problem Calculus grew out of four major problems that European mathematicians were working on during the seventeenth century. 1. The tangent line problem 2. The velocity
More informationSome Polynomial Theorems. John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.
Some Polynomial Theorems by John Kennedy Mathematics Department Santa Monica College 1900 Pico Blvd. Santa Monica, CA 90405 rkennedy@ix.netcom.com This paper contains a collection of 31 theorems, lemmas,
More informationCHAPTER 3. Sequences. 1. Basic Properties
CHAPTER 3 Sequences We begin our study of analysis with sequences. There are several reasons for starting here. First, sequences are the simplest way to introduce limits, the central idea of calculus.
More informationMetric Spaces. Chapter 7. 7.1. Metrics
Chapter 7 Metric Spaces A metric space is a set X that has a notion of the distance d(x, y) between every pair of points x, y X. The purpose of this chapter is to introduce metric spaces and give some
More informationSection 3 Sequences and Limits, Continued.
Section 3 Sequences and Limits, Continued. Lemma 3.6 Let {a n } n N be a convergent sequence for which a n 0 for all n N and it α 0. Then there exists N N such that for all n N. α a n 3 α In particular
More informationLectures 56: Taylor Series
Math 1d Instructor: Padraic Bartlett Lectures 5: Taylor Series Weeks 5 Caltech 213 1 Taylor Polynomials and Series As we saw in week 4, power series are remarkably nice objects to work with. In particular,
More information1. Please write your name in the blank above, and sign & date below. 2. Please use the space provided to write your solution.
Name : Instructor: Marius Ionescu Instructions: 1. Please write your name in the blank above, and sign & date below. 2. Please use the space provided to write your solution. 3. If you need extra pages
More information1 if 1 x 0 1 if 0 x 1
Chapter 3 Continuity In this chapter we begin by defining the fundamental notion of continuity for real valued functions of a single real variable. When trying to decide whether a given function is or
More informationRootfinding for Nonlinear Equations
> 3. Rootfinding Rootfinding for Nonlinear Equations > 3. Rootfinding Calculating the roots of an equation is a common problem in applied mathematics. f(x) = 0 (7.1) We will explore some simple numerical
More informationA Slide Show Demonstrating Newton s Method
The AcroT E X Web Site, 1999 A Slide Show Demonstrating Newton s Method D. P. Story The Department of Mathematics and Computer Science The Universityof Akron, Akron, OH Now go up to the curve. Now go
More informationNumerisches Rechnen. (für Informatiker) M. Grepl J. Berger & J.T. Frings. Institut für Geometrie und Praktische Mathematik RWTH Aachen
(für Informatiker) M. Grepl J. Berger & J.T. Frings Institut für Geometrie und Praktische Mathematik RWTH Aachen Wintersemester 2010/11 Problem Statement Unconstrained Optimality Conditions Constrained
More informationAim. Decimal Search. An Excel 'macro' was used to do the calculations. A copy of the script can be found in Appendix A.
Aim The aim of this investigation is to use and compare three different methods of finding specific solutions to polynomials that cannot be solved algebraically. The methods that will be considered are;
More informationCONTINUED FRACTIONS, PELL S EQUATION, AND TRANSCENDENTAL NUMBERS
CONTINUED FRACTIONS, PELL S EQUATION, AND TRANSCENDENTAL NUMBERS JEREMY BOOHER Continued fractions usually get shortchanged at PROMYS, but they are interesting in their own right and useful in other areas
More informationRoots of Equations (Chapters 5 and 6)
Roots of Equations (Chapters 5 and 6) Problem: given f() = 0, find. In general, f() can be any function. For some forms of f(), analytical solutions are available. However, for other functions, we have
More informationMATH 2300 review problems for Exam 3 ANSWERS
MATH 300 review problems for Exam 3 ANSWERS. Check whether the following series converge or diverge. In each case, justify your answer by either computing the sum or by by showing which convergence test
More informationPUTNAM TRAINING POLYNOMIALS. Exercises 1. Find a polynomial with integral coefficients whose zeros include 2 + 5.
PUTNAM TRAINING POLYNOMIALS (Last updated: November 17, 2015) Remark. This is a list of exercises on polynomials. Miguel A. Lerma Exercises 1. Find a polynomial with integral coefficients whose zeros include
More informationProblem Set. Problem Set #2. Math 5322, Fall December 3, 2001 ANSWERS
Problem Set Problem Set #2 Math 5322, Fall 2001 December 3, 2001 ANSWERS i Problem 1. [Problem 18, page 32] Let A P(X) be an algebra, A σ the collection of countable unions of sets in A, and A σδ the collection
More informationMath 317 HW #7 Solutions
Math 17 HW #7 Solutions 1. Exercise..5. Decide which of the following sets are compact. For those that are not compact, show how Definition..1 breaks down. In other words, give an example of a sequence
More informationCHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e.
CHAPTER II THE LIMIT OF A SEQUENCE OF NUMBERS DEFINITION OF THE NUMBER e. This chapter contains the beginnings of the most important, and probably the most subtle, notion in mathematical analysis, i.e.,
More informationGeneral Framework for an Iterative Solution of Ax b. Jacobi s Method
2.6 Iterative Solutions of Linear Systems 143 2.6 Iterative Solutions of Linear Systems Consistent linear systems in real life are solved in one of two ways: by direct calculation (using a matrix factorization,
More informationMath 113 HW #9 Solutions
Math 3 HW #9 Solutions 4. 50. Find the absolute maximum and absolute minimum values of on the interval [, 4]. f(x) = x 3 6x 2 + 9x + 2 Answer: First, we find the critical points of f. To do so, take the
More informationFurther Study on Strong Lagrangian Duality Property for Invex Programs via Penalty Functions 1
Further Study on Strong Lagrangian Duality Property for Invex Programs via Penalty Functions 1 J. Zhang Institute of Applied Mathematics, Chongqing University of Posts and Telecommunications, Chongqing
More informationCHAPTER 5. Product Measures
54 CHAPTER 5 Product Measures Given two measure spaces, we may construct a natural measure on their Cartesian product; the prototype is the construction of Lebesgue measure on R 2 as the product of Lebesgue
More informationOfficial Math 112 Catalog Description
Official Math 112 Catalog Description Topics include properties of functions and graphs, linear and quadratic equations, polynomial functions, exponential and logarithmic functions with applications. A
More informationORDERS OF ELEMENTS IN A GROUP
ORDERS OF ELEMENTS IN A GROUP KEITH CONRAD 1. Introduction Let G be a group and g G. We say g has finite order if g n = e for some positive integer n. For example, 1 and i have finite order in C, since
More informationPolynomials Classwork
Polynomials Classwork What Is a Polynomial Function? Numerical, Analytical and Graphical Approaches Anatomy of an n th degree polynomial function Def.: A polynomial function of degree n in the vaiable
More informationPractice Problems Solutions
Practice Problems Solutions The problems below have been carefully selected to illustrate common situations and the techniques and tricks to deal with these. Try to master them all; it is well worth it!
More informationANALYSIS I. How do we get hold of a real number? Answer, we look at successive approximations. E.g.: 1000, , 1414
ANALYSIS I 5 Real and complex sequences 5. Real numbers in practice How do we get hold of a real number? Answer, we look at successive approximations. E.g.: Our task is to make all this precise... 5. Sequences,
More informationFACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z
FACTORING POLYNOMIALS IN THE RING OF FORMAL POWER SERIES OVER Z DANIEL BIRMAJER, JUAN B GIL, AND MICHAEL WEINER Abstract We consider polynomials with integer coefficients and discuss their factorization
More informationZeros of Polynomial Functions
Zeros of Polynomial Functions The Rational Zero Theorem If f (x) = a n x n + a n1 x n1 + + a 1 x + a 0 has integer coefficients and p/q (where p/q is reduced) is a rational zero, then p is a factor of
More informationn k=1 k=0 1/k! = e. Example 6.4. The series 1/k 2 converges in R. Indeed, if s n = n then k=1 1/k, then s 2n s n = 1 n + 1 +...
6 Series We call a normed space (X, ) a Banach space provided that every Cauchy sequence (x n ) in X converges. For example, R with the norm = is an example of Banach space. Now let (x n ) be a sequence
More informationPaper 2 Revision. (compiled in light of the contents of paper1) Higher Tier Edexcel
Paper 2 Revision (compiled in light of the contents of paper1) Higher Tier Edexcel 1 Topic Areas 1. Data Handling 2. Number 3. Shape, Space and Measure 4. Algebra 2 Data Handling Averages Twoway table
More informationThe Expectation Maximization Algorithm A short tutorial
The Expectation Maximiation Algorithm A short tutorial Sean Borman Comments and corrections to: emtut at seanborman dot com July 8 2004 Last updated January 09, 2009 Revision history 2009009 Corrected
More informationAnalysis MA131. University of Warwick. Term
Analysis MA131 University of Warwick Term 1 01 13 September 8, 01 Contents 1 Inequalities 5 1.1 What are Inequalities?........................ 5 1. Using Graphs............................. 6 1.3 Case
More informationSection 3 Sequences and Limits
Section 3 Sequences and Limits Definition A sequence of real numbers is an infinite ordered list a, a 2, a 3, a 4,... where, for each n N, a n is a real number. We call a n the nth term of the sequence.
More informationS = k=1 k. ( 1) k+1 N S N S N
Alternating Series Last time we talked about convergence of series. Our two most important tests, the integral test and the (limit) comparison test, both required that the terms of the series were (eventually)
More information5. Convergence of sequences of random variables
5. Convergence of sequences of random variables Throughout this chapter we assume that {X, X 2,...} is a sequence of r.v. and X is a r.v., and all of them are defined on the same probability space (Ω,
More informationDifferentiability and some of its consequences Definition: A function f : (a, b) R is differentiable at a point x 0 (a, b) if
Differentiability and some of its consequences Definition: A function f : (a, b) R is differentiable at a point x 0 (a, b) if f(x)) f(x 0 ) x x 0 exists. If the it exists for all x 0 (a, b) then f is said
More informationContinued Fractions and the Euclidean Algorithm
Continued Fractions and the Euclidean Algorithm Lecture notes prepared for MATH 326, Spring 997 Department of Mathematics and Statistics University at Albany William F Hammond Table of Contents Introduction
More informationPseudocode Analysis. COMP3600/6466 Algorithms Lecture 5. Recursive Algorithms. Asymptotic Bounds for Recursions
COMP600/6466 Algorithms Lecture 5 Pseudocode Analysis Iterative Recursive S 0 Dr. Hassan Hijazi Prof. Weifa Liang Recursive Algorithms Asymptotic Bounds for Recursions Total running time = Sum of times
More informationMAT2400 Analysis I. A brief introduction to proofs, sets, and functions
MAT2400 Analysis I A brief introduction to proofs, sets, and functions In Analysis I there is a lot of manipulations with sets and functions. It is probably also the first course where you have to take
More informationPower Series. We already know how to express one function as a series. Take a look at following equation: 1 1 r = r n. n=0
Math 0 Calc Power, Taylor, and Maclaurin Series Survival Guide One of the harder concepts that we have to become comfortable with during this semester is that of sequences and series We are working with
More informationLecture 12 Basic Lyapunov theory
EE363 Winter 200809 Lecture 12 Basic Lyapunov theory stability positive definite functions global Lyapunov stability theorems Lasalle s theorem converse Lyapunov theorems finding Lyapunov functions 12
More informationk=1 k2, and therefore f(m + 1) = f(m) + (m + 1) 2 =
Math 104: Introduction to Analysis SOLUTIONS Alexander Givental HOMEWORK 1 1.1. Prove that 1 2 +2 2 + +n 2 = 1 n(n+1)(2n+1) for all n N. 6 Put f(n) = n(n + 1)(2n + 1)/6. Then f(1) = 1, i.e the theorem
More informationGeneral Theory of Differential Equations Sections 2.8, 3.13.2, 4.1
A B I L E N E C H R I S T I A N U N I V E R S I T Y Department of Mathematics General Theory of Differential Equations Sections 2.8, 3.13.2, 4.1 Dr. John Ehrke Department of Mathematics Fall 2012 Questions
More information2. Simplify. College Algebra Student SelfAssessment of Mathematics (SSAM) Answer Key. Use the distributive property to remove the parentheses
College Algebra Student SelfAssessment of Mathematics (SSAM) Answer Key 1. Multiply 2 3 5 1 Use the distributive property to remove the parentheses 2 3 5 1 2 25 21 3 35 31 2 10 2 3 15 3 2 13 2 15 3 2
More informationThis workshop will. Xintercepts and Zoom Box
This workshop will Arapahoe Community College MAT 111  Graphing Calculator Techniques for Survey of Algebra TI83 Graphing Calculator Workshop #8 Solving Equations  Graphical Approach 1) demonstrate
More information