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03 truncation errors

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03 truncation errors

  1. 1. Part 3Truncation Errors 1
  2. 2. Key Concepts• Truncation errors• Taylors Series – To approximate functions – To estimate truncation errors• Estimating truncation errors using other methods – Alternating Series, Geometry series, Integration 2
  3. 3. IntroductionHow do we calculate sin( x), cos( x), e x , x y , x , log( x), ...on a computer using only +, -, x, ÷?One possible way is via summation of infiniteseries. e.g., x2 x3 xn x n +1 ex = 1 + x + + + ... + + + ... 2! 3! n! ( n + 1)! 3
  4. 4. Introduction x2 x3 xn x n +1 e =1+ x + x + + ... + + + ... 2! 3! n! ( n + 1)!• How to derive the series for a given function?• How many terms should we add? or• How good is our approximation if we only sum up the first N terms? 4
  5. 5. A general form of approximation is interms of Taylor Series. 5
  6. 6. Taylors TheoremTaylors Theorem: If the function f and its first n+1derivatives are continuous on an interval containing aand x, then the value of the function at x is given by f " (a) f ( 3) ( a ) f ( x) = f (a) + f (a)( x − a ) + ( x − a) 2 + ( x − a )3 + ... 2! 3! f ( n ) (a) + ( x − a ) n + Rn n!where the remainder Rn is defined as x ( x − t ) n ( n +1) Rn = ∫a n! f (t )dt (the integral form) 6
  7. 7. Derivative or Lagrange Form of the remainderThe remainder Rn can also be expressed as ( n +1) f (c ) (the Lagrange form) Rn = ( x − a ) n +1 (n + 1)! for some c between a and x The Lagrange form of the remainder makes analysis of truncation errors easier. 7
  8. 8. Taylor Series f " (a ) f ( 3) ( a )f ( x ) = f ( a ) + f ( a )( x − a ) + ( x − a)2 + ( x − a ) 3 + ... 2! 3! f ( n ) (a ) + ( x − a ) n + Rn n!• Taylor series provides a mean to approximate any smooth function as a polynomial.• Taylor series provides a mean to predict a function value at one point x in terms of the function and its derivatives at another point a.• We call the series "Taylor series of f at a" or "Taylor series of f about a". 8
  9. 9. Example – Taylor Series of ex at 0f ( x) = e x => f ( x) = e x => f " ( x) = e x => f ( k ) ( x) = e x for any k ≥ 0Thus f ( k ) (0) = 1 for any k ≥ 0.With a = 0, the Taylor series of f at 0 becomes f " ( 0) f ( n ) (0)f (0) + f (0)( x − 0) + ( x − 0) 2 + ... + ( x − 0) n + ... 2! n! x 2 x3 xn= 1 + x + + + ... + + ... 2! 3! n! Note: Taylor series of a function f at 0 is also known as the Maclaurin series of f. 9
  10. 10. Exercise – Taylor Series of cos(x) at 0 f ( x ) = cos( x ) => f (0) = 1 f ( x ) = − sin( x ) => f (0) = 0 f " ( x ) = − cos( x ) => f " (0) = −1 f ( 3) ( x ) = sin( x ) => f ( 3) (0) = 0 f ( 4 ) ( x ) = cos( x ) => f ( 4 ) (0) = 1 f ( 5) ( x ) = − sin( x ) => f ( 5) (0) = 0With a = 0, the Taylor series of f at 0 becomes f " ( 0) f ( n ) ( 0)f (0) + f (0)( x − 0) + ( x − 0) + ... + 2 ( x − 0) n + ... 2! n! x2 x4 x6=1+ 0 − +0 + +0 − + ... 2! 4! 6! ∞ 2n n x= ∑( −1) n =0 ( 2n )! 10
  11. 11. Question x2 x3 xn x n +1 ex = 1 + x + + + ... + + + ... 2! 3! n! ( n + 1)!What will happen if we sum up only the first n+1terms? 11
  12. 12. Truncation ErrorsTruncation errors are the errors that result fromusing an approximation in place of an exactmathematical procedure. Approximation Truncation Errors x2 x3 xn x n +1 ex = 1 + x + + + ... + + + ... 2! 3! n! ( n + 1)! Exact mathematical formulation 12
  13. 13. How good is our approximation? x2 x3 xn x n +1 ex = 1 + x + + + ... + + + ... 2! 3! n! ( n + 1)!How big is the truncation error if we only sum upthe first n+1 terms?To answer the question, we can analyze theremainder term of the Taylor series expansion. f " (a) f ( 3) ( a )f ( x) = f (a) + f (a)( x − a) + ( x − a) 2 + ( x − a)3 + ... 2! 3! f ( n ) (a) + ( x − a) n + Rn n! 13
  14. 14. Analyzing the remainder term of the Taylor series expansion of f(x)=ex at 0The remainder Rn in the Lagrange form is ( n +1) f (c ) Rn = ( x − a ) n +1 (n + 1)! for some c between a and xFor f(x) = ex and a = 0, we have f(n+1)(x) = ex. Thus ec Rn = x n +1 for some c in [0 , x] (n + 1)! x e We can estimate the largest possible ≤ x n +1 truncation error through analyzing Rn. (n + 1)! 14
  15. 15. ExampleEstimate the truncation error if we calculate e as 1 1 1 1 e = 1 + + + + ... + 1! 2! 3! 7!This is the Maclaurin series of f(x)=ex with x = 1 andn = 7. Thus the bound of the truncation error is ex 7 +1 e1 8 e −4R7 ≤ x = (1) = ≈ 0.6742 × 10 (7 + 1)! 8! 8!The actual truncation error is about 0.2786 x 10-4. 15
  16. 16. ObservationFor the same problem, with n = 8, the bound of the truncationerror is e R8 ≤ ≈ 0.7491 × 10−5 9!With n = 10, the bound of the truncation error is e R10 ≤ ≈ 0.6810 × 10−7 11!More terms used implies better approximation. 16
  17. 17. Example (Backward Analysis)This is the Maclaurin series expansion for ex x2 x3 xn e x = 1 + x + + + ... + + ... 2! 3! n!If we want to approximate e0.01 with an errorless than 10-12, at least how many terms areneeded? 17
  18. 18. With x = 0.01, 0 ≤ c ≤ 0.01, f ( x ) = e x => f ( n +1) ( x ) = e x ec n +1 e0.01 n +1 1.1 Rn = x ≤ (0.01) < (0.01) n +1 ( n + 1)! ( n + 1)! ( n + 1)! Note:1.1100 is about 13781 > eTo find the smallest n such that Rn < 10-12, we can findthe smallest n that satisfies 1.1 n +1 −12 With the help of a computer: (0.01) < 10 ( n + 1)! n=0 Rn=1.100000e-02 n=1 Rn=5.500000e-05 n=2 Rn=1.833333e-07 n=3 Rn=4.583333e-10 So we need at least 5 terms n=4 Rn=9.166667e-13 18
  19. 19. Same problem with larger step sizeWith x = 0.5, 0 ≤ c ≤ 0.5, f ( x ) = e x => f ( n +1) ( x ) = e x ec n +1 e0.5 n +1 1.7 Rn = x ≤ (0.5) < (0.5) n +1 ( n + 1)! ( n + 1)! ( n + 1)! Note:1.72 is 2.89 > eWith the help of a computer: n=5 Rn=3.689236e-05n=0 Rn=8.500000e-01 n=6 Rn=2.635169e-06n=1 Rn=2.125000e-01 n=7 Rn=1.646980e-07n=2 Rn=3.541667e-02 n=8 Rn=9.149891e-09n=3 Rn=4.427083e-03 n=9 Rn=4.574946e-10n=4 Rn=4.427083e-04 n=10 Rn=2.079521e-11 n=11 Rn=8.664670e-13 So we need at least 12 terms 19
  20. 20. To approximate e10.5 with an error less than 10-12,we will need at least 55 terms. (Not very efficient)How can we speed up the calculation? 20
  21. 21. Exercise If we want to approximate e10.5 with an error less than 10-12 using the Taylor series for f(x)=ex at 10, at least how many terms are needed?The Taylor series expansion of f ( x ) at 10 is f (10) f " (10) f ( n ) (10)f ( x ) = f (10) + ( x − 10) + ( x − 10) + ... + 2 ( x − 10) n + Rn 1! 2! n! ( x − 10) 2 ( x − 10) n = e (1 + ( x − 10) + 10 + ... + ) + Rn 2! n! f ( n +1) ( c )Rn = ( x − 10) n +1 for some c between 10 and x ( n + 1)! The smallest n that satisfy Rn < 10-12 is n = 18. So we need at least 19 terms. 21
  22. 22. Observation• A Taylor series converges rapidly near the point of expansion and slowly (or not at all) at more remote points. 22
  23. 23. Taylor Series Approximation Example:More terms used implies better approximation f(x) = 0.1x4 - 0.15x3 - 0.5x2 - 0.25x + 1.2 23
  24. 24. Taylor Series Approximation Example:Smaller step size implies smaller error Errors Reduced step size f(x) = 0.1x4 - 0.15x3 - 0.5x2 - 0.25x + 1.2 24
  25. 25. Taylor Series (Another Form)If we let h = x – a, we can rewrite the Taylor seriesand the remainder as (n) f " (a) 2 f (a) n f ( x) = f (a) + f (a)h + h + ... + h + Rn 2! n! f ( n +1) (c) n +1 When h is small, hn+1 is muchRn = h (n + 1)! smaller.h is called the step size.h can be +ve or –ve. 25
  26. 26. The Remainder of the Taylor Series Expansion f ( n +1) ( c) n +1 n +1 Rn = h = O (h ) ( n + 1)!SummaryTo reduce truncation errors, we can reduce h or/andincrease n.If we reduce h, the error will get smaller quicker (with less n).This relationship has no implication on the magnitude of theerrors because the constant term can be huge! It only giveus an estimation on how much the truncation error wouldreduce when we reduce h or increase n. 26
  27. 27. Other methods for estimating truncation errors of a series S = t0 + t1 + t 2 + t3 + ... + t n + t n +1 + t n + 2 + t n +3 + ...         Sn Rn 1. By Geometry Series 2. By Integration 3. Alternating Convergent Series TheoremNote: Some Taylor series expansions may exhibit certaincharacteristics which would allow us to use different methodsto approximate the truncation errors. 27
  28. 28. Estimation of Truncation Errors By Geometry SeriesIf |tj+1| ≤ k|tj| where 0 ≤ k < 1 for all j ≥ n, thenRn = tn +1 + tn +2 + tn +3 +... ≤ tn +1 + k tn +1 + k 2 tn +1 +... = tn +1 (1 + k + k 2 + k 3 +...) tn +1 = 1 −k k tn Rn ≤ 1− k 28
  29. 29. Example (Estimation of Truncation Errors by Geometry Series) What is |R6| for the following series expansion? S =1 +π −2 + 2π −4 + 3π −6 +... + jπ −2 j +... tj = jπ −2 jSolution: t j +1 j +1π−2 j −2 1 = = 1 + π−2Is there a k (0 ≤ k < 1) s.t. tj jπ−2 j j|tj+1| ≤ k|tj| or |tj+1|/|tj| ≤ k t j +1 1 ≤ 1 + π−2 ∀ ≥6 jfor all j ≤ n (n=6)? tj 6 < 0.11If you can find this k, then k = 0.11, t6 < 3 ×10 −6 k tn Rn ≤ k tn 0.11 1− k R6 ≤ < ×3 ×10 −6 1 −k 1 −0.11 29
  30. 30. Estimation of Truncation Errors By IntegrationIf we can find a function f(x) s.t. |tj| ≤ f(j) ∀j ≥ nand f(x) is a decreasing function ∀x ≥ n, then ∞ ∞ Rn = t n +1 + t n +2 + t n +3 +... = ∑t j =n +1 j ≤ ∑ f ( j) j =n +1 ∞ Rn ≤ ∫ f ( x )dx n 30
  31. 31. Example (Estimation of Truncation Errors by Integration) Estimate |Rn| for the following series expansion. S =∑ j t where t j = ( j 3 +1) −1 j=1Solution: We can pick f(x) = x–3 because it would provide a tight bound for |tj|. That is 1 1 ≥ ∀ ≥1 j j 3 1+ j3 ∞ 1 1 So Rn ≤ ∫ 3 dx = n x 2n 2 31
  32. 32. Alternating Convergent SeriesTheorem (Leibnitz Theorem)If an infinite series satisfies the conditions – It is strictly alternating. – Each term is smaller in magnitude than that term before it. – The terms approach to 0 as a limit.Then the series has a finite sum (i.e., converge)and moreover if we stop adding the terms after thenth term, the error thus produced is between 0 andthe 1st non-zero neglected term not taken. 32
  33. 33. Alternating Convergent Series Theorem Example 1: Maclaurin series of ln(1 + x ) ∞ x2 x3 x4 xn S = x − + − + ... = ∑ ( −1) n −1 ( −1 < x ≤ 1) 2 3 4 n =1 n With n = 5, 1 1 1 1 S = 1 − + − + = 0.7833333340 2 3 4 5 Eerror ln 2 = 0.693 estimated 1 using the R = S − ln 2 ≈ 0.09 ≤ = 0.16666 althernating 6 convergent Actual error series theorem 33
  34. 34. Alternating Convergent Series Theorem Example 2: Maclaurin series of cos( x ) x2 x4 x6 ∞ x 2 n +1 S = 1 − + − + ... = ∑ ( −1) n 2! 4! 6! n =0 ( 2n + 1)! With n = 5, 12 14 16 18 S = 1 − + − + = 0.5403025793 2! 4! 6! 8! Eerror cos(1) = 0.5403023059 estimated using the −7 1 althernating S − cos(1) = 2.73 × 10 ≤ = 2.76 × 10−7 10! convergent series Actual error theorem 34
  35. 35. Exercise If the sine series is to be used to compute sin(1) with an error less than 0.5x10-14, how many terms are needed? 13 15 17 19 111 113 115 117 sin(1) = 1 − + − + − + − + − ... 3! 5! 7! 9! 11! 13! 15! 17! R0 R1 R2 R3 R4 R5 R6 R7 Solution: This series satisfies the conditions of the Alternating Convergent Series Theorem. Solving 1 1 Rn ≤ ≤ ×10 −14 ( 2n +3)! 2 for the smallest n yield n = 7 (We need 8 terms) 35
  36. 36. Exercise π4 1 1 1 =1 + 4 + 4 + 4 +... 90 2 3 4How many terms should be taken in order to computeπ4/90 with an error of at most 0.5x10-8? ∞ 1 ∞ Rn = tn +1 +tn +2 +... = ∑ < ∫ ( x +1) −4 dx j= + n 1 ( j +1) 4 n Solution −3 ∞integration): (by − ( x +1) ( n +1)3= = −3 n 3 1 1 < ×10 −8 = ( n +1) ≥ 406 = n ≥ 405 > >3( n +1) 3 2Note: If we use f(x) = x-3 (which is easier to analyze) insteadof f(x) = (x+1)-3 to bound the error, we will get n >= 406(just one more term). 36
  37. 37. Summary• Understand what truncation errors are• Taylors Series – Derive Taylors series for a "smooth" function – Understand the characteristics of Taylors Series approximation – Estimate truncation errors using the remainder term• Estimating truncation errors using other methods – Alternating Series, Geometry series, Integration 37

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