Lecture Notes Course ÃMA 190 Numerical Mathematics, First ...

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or, in matrix form with the coefficients rounded to 6 decimal places:⎛0 −0.1 −0.2⎞ ⎛1.3⎞B = ⎝ −0.272727 0 −0.090909 ⎠ b = ⎝ 1.363636 ⎠ .−0.166667 −0.25 0 1.416667Jacobi iteration:We enter the starting vector x 1 = x 2 = x 3 = 0 into the system (4.2),(4.3)and (4.4) and get the first approximation, working with 6 decimal places: x 1 =1.3, x 2 = 15/11 = 1.363636, x 3 = 17/12 = 1.416667 This vector is entered intothe system to give the next approximation:x 1 = 0.1(13 − 1.363636 − 2 · 1.416667) = 0.880303x 2 = 1 (15 − 3 · 1.3 − 1.416667) = 0.88030311x 3 = 1 (17 − 2 · 1.3 − 3 · 1.363636) = 0.85909113This procedure is repeated generating the table below for Jacobi iterations.Each row of the table represents an approximation of the solution vector.Gauss-Seidel iterationIn the Gauss-Seidel scheme we use the latest calculated value of each componentof the solution vector by the updating. We start as before with the zero vectorand as before we update x 1 according to (4.2) and get x 1 = 1.3. (The latestvalues of x 2 and x 3 are 0) We next update x 2 with (4.3) and now we use thelatest value for x 1 , x 3 , namely 1.3 and 0. Hencex 2 = 1 (15 − 3 · 1.3) = 1.00909111We use (4.4) to update x 3 , using the latest values x 1 = 1.3, x 2 = 1.009091gettingx 3 = 1 (17 − 2 · 1.3 − 3 · 1.009091) = 0.94772712We continue in this way, always using the latest values of each component. Thusin the next iterationx 1 = 1 (13 − 1.009091 − 2 · 0.947727) = 1.00954610x 2 = 1 (15 − 3 · 1.009546 − 0.947727) = 1.00214911x 3 = 1 (17 − 2 · 1.009546 − 3 · 1.002149) = 0.99787012Continuing in this way we generate the Gauss-Seidel table.Remark 4.2.2 If we enter the vector x with x 1 = x 2 = x 3 = 1 into each of theiteration formulas we get the same vector back. It is hence the fix-point of thetwo iteration schemes and is also the solution of the given system of equations.35
We compare the results of Jacobi and Gauss-Seidel iterations. The followingoutput was obtained from simple Fortran programs implementing these methodsand using the null vector as starting vector.Jacobi iterationGiven system10.0000 1.0000 2.0000 13.00003.0000 11.0000 1.0000 15.00002.0000 3.0000 12.0000 17.0000Starting vector.000000 .000000 .000000Jacobi approximates1.300000 1.363636 1.416667.880303 .880303 .8590911.040151 1.045455 1.049874.985480 .984516 .9819441.005160 1.005601 1.006291.998182 .998021 .9977401.000650 1.000701 1.000798.999770 .999750 .9997161.000082 1.000088 1.000101.999971 .999969 .9999641.000010 1.000011 1.000013.999996 .999996 .9999951.000001 1.000001 1.0000021.000000 .999999 .9999991.000000 1.000000 1.000000Gauss-Seidel iterationGiven system10.000000 1.000000 2.000000 13.0000003.000000 11.000000 1.000000 15.0000002.000000 3.000000 12.000000 17.000000Starting vector.000000 .000000 .000000Gauss-Seidel approximates1.300000 1.009091 .9477271.009546 1.002149 .9978721.000211 1.000136 .9999311.000000 1.000006 .9999981.000000 1.000000 1.00000036
Page 1: Lecture NotesCourse ÅMA 190Numeric
Page 6 and 7: 13.8.4 Inverse linear interpolation
Page 8 and 9: AbstractNumerical analysts study th
Page 10 and 11: 1.1.1 Two standard inequalitiesLet
Page 12 and 13: Example 1.1.3 Examples with s = 3:N
Page 14 and 15: and the relative condition number b
Page 16 and 17: This general example may be special
Page 18 and 19: Theorem 1.5.2 Let A be a square mat
Page 20 and 21: 1.6.2 Arithmetic sequencen(n − 1)
Page 22 and 23: 1.6.7 Taylor’s formulaWe derive T
Page 24 and 25: )c)becauselimx→0x 2e x − 1 −
Page 26 and 27: 1.7.7 Locating rootsPutShow that th
Page 28 and 29: We next construct the general solut
Page 30 and 31: Definition 2.2.1 We distinguish bet
Page 32 and 33: Example 2.3.5 Consider a computer w
Page 34 and 35: We now determine c such that 2 + 3c
Page 36 and 37: e recorded) If there is no such col
Page 38 and 39: the left hand side and the right ha
Page 40 and 41: Chapter 4Iterative methods for Ax =
Page 44 and 45: Chapter 5Least squares fit5.1 Norma
Page 46 and 47: c 1,4 is the product of the first a
Page 48 and 49: PutandL i (t) =P (t) =5∏(t − t
Page 50 and 51: 6.2 On the choice of nodesIf t i in
Page 52 and 53: k arguments:f[t i1 , . . . , t ik ]
Page 54 and 55: Q(x) = 4.6 − 0.8(x + 2) + 0.6x(x
Page 56 and 57: a relative error bound if x ≠ 0.
Page 58 and 59: 7.2.2 MultiplicationPutwheres = x 1
Page 60 and 61: We now assume that f has a continuo
Page 62 and 63: functional values. Let s ∗ be an
Page 64 and 65: 7.5.2 Example: additionLetWe then h
Page 66 and 67: Chapter 8Numerical treatment ofnonl
Page 68 and 69: 8.3 Determination of approximate ro
Page 70 and 71: Proof:The case n = 0 is the stateme
Page 72 and 73: 8.6 The fixed-point methodTo use th
Page 74 and 75: Chapter 9On approximation offunctio
Page 76 and 77: 9.1.3 The square rootTo find √ a,
Page 78 and 79: which is achieved forTo gett i = 1/
Page 80 and 81: whereF (h, w) = h∞∑n=−∞e iw
Page 82 and 83: Proof PutThen we haveandM n = supr>
Page 84 and 85: 10.3 Numerical treatment of rapidly
Page 86 and 87: n a n−1 s n1 1.00000 1.000002 −
Page 88 and 89: phenomenon we needDefinition 10.4.1
Page 90 and 91: a r and b r are given bya r = exp(
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11.2 Constructing the trapezoidal r
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In the case of (11.2) it is suffici
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should give exact results for all l
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h=0.2 :f(0) + f(1.6)T (0.2) = 0.2
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h=0.5 :T (0.5) = 0.5 · f(0) + f(0.
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A system of ordinary differential e
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12.3.1 Euler’s methodx r+1 = x r
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It is directly verified that the so
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1.00 1.246834041.00 1.246818301.00
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13.1.3 Stirling’s formula13.1.4 B
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Power expansions for some elementar
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TranspositionThenA ∈ R n×m , B =
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13.5 Least squares solution of line
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Let f be a function which may be di
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13.9 Systems of nonlinear equations
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may be writtenx(t) = x h (t) + x p
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Chapter 14Laboratory exercises andc
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Start with x 0 = 1b) Use the bisect
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14.7 Case studies14.8 Evaluation of
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n piv,sing nonpiv,sing piv,dbl nonp
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We put c = 100 + u and solve the pr
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and hence we may consider minimisin
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x f(x)−1.0 0.3679−0.5 0.6065 .0
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14.12 Numerical integration problem
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14.12.5 Integrand with integrable s
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Lecture Notes Course ÃMA 190 Numerical Mathematics, First ...