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# Caims 2009

James talk in the Symbolic Computation Thread at the Canadian Applied and Industrial Mathematics Society\'s 2009 Meeting

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### Caims 2009

1. 1. The unreasonable eﬀectiveness of algebra (but don’t take it for granted) James Davenport University of Bath (visiting Waterloo) 11 June 2009
2. 2. What is the derivative of sin x?
3. 3. What is the derivative of sin x? sin(x + δ) − sin(x) f (x) = lim δ→0 δ
4. 4. What is the derivative of sin x? sin(x + δ) − sin(x) f (x) = lim δ→0 δ sin(x + δ) − sin(x) ∀ > 0∃D : δ < D ⇒ f (x) − < δ
5. 5. What is the derivative of sin x? sin(x + δ) − sin(x) f (x) = lim δ→0 δ sin(x + δ) − sin(x) ∀ > 0∃D : δ < D ⇒ f (x) − < δ sin(x + δ) − sin(x) = cos(x) sin(δ) − sin(x)(1 − cos(δ)) so we can make the choice f (x) = cos(x).
6. 6. What is the derivative of sin x? sin(x + δ) − sin(x) f (x) = lim δ→0 δ sin(x + δ) − sin(x) ∀ > 0∃D : δ < D ⇒ f (x) − < δ sin(x + δ) − sin(x) = cos(x) sin(δ) − sin(x)(1 − cos(δ)) so we can make the choice f (x) = cos(x). We can then take 1 √ D = 2 min( , 1), and some simple algebra shows the result.
7. 7. What is the derivative of sin x? sin(x + δ) − sin(x) f (x) = lim δ→0 δ sin(x + δ) − sin(x) ∀ > 0∃D : δ < D ⇒ f (x) − < δ sin(x + δ) − sin(x) = cos(x) sin(δ) − sin(x)(1 − cos(δ)) so we can make the choice f (x) = cos(x). We can then take 1 √ D = 2 min( , 1), and some simple algebra shows the result. Did any of you do that?
8. 8. What is the derivative of sin x? sin(x + δ) − sin(x) f (x) = lim δ→0 δ sin(x + δ) − sin(x) ∀ > 0∃D : δ < D ⇒ f (x) − < δ sin(x + δ) − sin(x) = cos(x) sin(δ) − sin(x)(1 − cos(δ)) so we can make the choice f (x) = cos(x). We can then take 1 √ D = 2 min( , 1), and some simple algebra shows the result. Did any of you do that? √ Did any of you take D = 2 max(| min( ,1) cos(x)|,| sin(x)|,1) to allow for x ∈ C?
9. 9. What is the derivative of sin x? sin(x + δ) − sin(x) f (x) = lim δ→0 δ sin(x + δ) − sin(x) ∀ > 0∃D : δ < D ⇒ f (x) − < δ sin(x + δ) − sin(x) = cos(x) sin(δ) − sin(x)(1 − cos(δ)) so we can make the choice f (x) = cos(x). We can then take 1 √ D = 2 min( , 1), and some simple algebra shows the result. Did any of you do that? √ Did any of you take D = 2 max(| min( ,1) cos(x)|,| sin(x)|,1) to allow for x ∈ C? Or did you “just know it”?
10. 10. π/2 What is the integral: 0 cos x?
11. 11. π/2 What is the integral: 0 cos x? I = lim S ∆ [cos(x)] = lim S ∆ [cos(x)] |∆|→0 |∆|→0 (∆ ranges over all dissections of [0, π/2])
12. 12. π/2 What is the integral: 0 cos x? I = lim S ∆ [cos(x)] = lim S ∆ [cos(x)] |∆|→0 |∆|→0 (∆ ranges over all dissections of [0, π/2]) Does anyone actually do this?
13. 13. π/2 What is the integral: 0 cos x? I = lim S ∆ [cos(x)] = lim S ∆ [cos(x)] |∆|→0 |∆|→0 (∆ ranges over all dissections of [0, π/2]) Does anyone actually do this? Can anyone actually do this?
14. 14. π/2 What is the integral: 0 cos x? I = lim S ∆ [cos(x)] = lim S ∆ [cos(x)] |∆|→0 |∆|→0 (∆ ranges over all dissections of [0, π/2]) Does anyone actually do this? Can anyone actually do this? Or do we say “well, we have proved that sin = cos, so cos = sin, and therefore the answer is sin π − sin 0 = 1? 2
15. 15. π/2 What is the integral: 0 cos x? I = lim S ∆ [cos(x)] = lim S ∆ [cos(x)] |∆|→0 |∆|→0 (∆ ranges over all dissections of [0, π/2]) Does anyone actually do this? Can anyone actually do this? Or do we say “well, we have proved that sin = cos, so cos = sin, and therefore the answer is sin π − sin 0 = 1? 2 We might say “therefore, by the Fundamental Theorem of Calculus, . . . ”
16. 16. What is the Fundamental Theorem of Calculus?
17. 17. What is the Fundamental Theorem of Calculus? Indeed, what is calculus?
18. 18. What is the Fundamental Theorem of Calculus? Indeed, what is calculus? . . . the deism of Liebniz over the dotage of Newton . . . [Babbage, chapter 4]
19. 19. What is the Fundamental Theorem of Calculus? Indeed, what is calculus? . . . the deism of Liebniz over the dotage of Newton . . . [Babbage, chapter 4] I claim that calculus is actually the interesting fusion of two, diﬀerent subjects.
20. 20. What is the Fundamental Theorem of Calculus? Indeed, what is calculus? . . . the deism of Liebniz over the dotage of Newton . . . [Babbage, chapter 4] I claim that calculus is actually the interesting fusion of two, diﬀerent subjects. What you learned in calculus, which I shall write as D δ : the “diﬀerentiation of –δ analysis”. Also d d x , and its inverse δ δ .
21. 21. What is the Fundamental Theorem of Calculus? Indeed, what is calculus? . . . the deism of Liebniz over the dotage of Newton . . . [Babbage, chapter 4] I claim that calculus is actually the interesting fusion of two, diﬀerent subjects. What you learned in calculus, which I shall write as D δ : the “diﬀerentiation of –δ analysis”. Also d d x , and its inverse δ δ . What is taught in diﬀerential algebra, which I shall write as d DDA : the “diﬀerentiation of diﬀerential algebra”. Also dDA x , and its inverse DA .
22. 22. D δ (for functions R → R)
23. 23. D δ (for functions R → R) Deﬁne CL(f , x0 ) (the “Cauchy Limit”) as f (x0 + h) − f (x0 ) CL(f , x0 ) = lim h→0 h and D δ (f ) = λx. CL(f , x).
24. 24. D δ (for functions R → R) Deﬁne CL(f , x0 ) (the “Cauchy Limit”) as f (x0 + h) − f (x0 ) CL(f , x0 ) = lim h→0 h and D δ (f ) = λx. CL(f , x). Then the following are theorems.
25. 25. D δ (for functions R → R) Deﬁne CL(f , x0 ) (the “Cauchy Limit”) as f (x0 + h) − f (x0 ) CL(f , x0 ) = lim h→0 h and D δ (f ) = λx. CL(f , x). Then the following are theorems. D δ (c) = 0 ∀c constants.
26. 26. D δ (for functions R → R) Deﬁne CL(f , x0 ) (the “Cauchy Limit”) as f (x0 + h) − f (x0 ) CL(f , x0 ) = lim h→0 h and D δ (f ) = λx. CL(f , x). Then the following are theorems. D δ (c) = 0 ∀c constants. CL(f + g , x0 ) = CL(f , x0 ) + CL(g , x0 ), so D δ (f + g ) = D δ (f ) + D δ (g ).
27. 27. D δ (for functions R → R) Deﬁne CL(f , x0 ) (the “Cauchy Limit”) as f (x0 + h) − f (x0 ) CL(f , x0 ) = lim h→0 h and D δ (f ) = λx. CL(f , x). Then the following are theorems. D δ (c) = 0 ∀c constants. CL(f + g , x0 ) = CL(f , x0 ) + CL(g , x0 ), so D δ (f + g ) = D δ (f ) + D δ (g ). CL(fg , x0 ) = CL(f , x0 )g (x0 ) + f (x0 ) CL(g , x0 ), so D δ (fg ) = D δ (f )g + fD δ (g ).
28. 28. D δ (for functions R → R) Deﬁne CL(f , x0 ) (the “Cauchy Limit”) as f (x0 + h) − f (x0 ) CL(f , x0 ) = lim h→0 h and D δ (f ) = λx. CL(f , x). Then the following are theorems. D δ (c) = 0 ∀c constants. CL(f + g , x0 ) = CL(f , x0 ) + CL(g , x0 ), so D δ (f + g ) = D δ (f ) + D δ (g ). CL(fg , x0 ) = CL(f , x0 )g (x0 ) + f (x0 ) CL(g , x0 ), so D δ (fg ) = D δ (f )g + fD δ (g ). D δ (λx.f (g (x))) = D δ (g )λx.D δ (f )(g (x)). (Chain rule)
29. 29. DDA (for any ring R of characteristic 0) Deﬁnition: DDA : R → R is a derivation on R if it satisﬁes:
30. 30. DDA (for any ring R of characteristic 0) Deﬁnition: DDA : R → R is a derivation on R if it satisﬁes: DDA (f + g ) = DDA (f ) + DDA (g )
31. 31. DDA (for any ring R of characteristic 0) Deﬁnition: DDA : R → R is a derivation on R if it satisﬁes: DDA (f + g ) = DDA (f ) + DDA (g ) DDA (fg ) = DDA (f )g + fDDA (g )
32. 32. DDA (for any ring R of characteristic 0) Deﬁnition: DDA : R → R is a derivation on R if it satisﬁes: DDA (f + g ) = DDA (f ) + DDA (g ) DDA (fg ) = DDA (f )g + fDDA (g )
33. 33. DDA (for any ring R of characteristic 0) Deﬁnition: DDA : R → R is a derivation on R if it satisﬁes: DDA (f + g ) = DDA (f ) + DDA (g ) DDA (fg ) = DDA (f )g + fDDA (g ) Corollaries:
34. 34. DDA (for any ring R of characteristic 0) Deﬁnition: DDA : R → R is a derivation on R if it satisﬁes: DDA (f + g ) = DDA (f ) + DDA (g ) DDA (fg ) = DDA (f )g + fDDA (g ) Corollaries: DDA (c) = 0 for all c algebraic over 1 .
35. 35. DDA (for any ring R of characteristic 0) Deﬁnition: DDA : R → R is a derivation on R if it satisﬁes: DDA (f + g ) = DDA (f ) + DDA (g ) DDA (fg ) = DDA (f )g + fDDA (g ) Corollaries: DDA (c) = 0 for all c algebraic over 1 . f DDA (f )g −fDDA (g ) DDA ( g ) = g2 .
36. 36. DDA (for any ring R of characteristic 0) Deﬁnition: DDA : R → R is a derivation on R if it satisﬁes: DDA (f + g ) = DDA (f ) + DDA (g ) DDA (fg ) = DDA (f )g + fDDA (g ) Corollaries: DDA (c) = 0 for all c algebraic over 1 . f DDA (f )g −fDDA (g ) DDA ( g ) = g2 . DDA extends uniquely to algebraic extensions.
37. 37. DDA (for any ring R of characteristic 0) Deﬁnition: DDA : R → R is a derivation on R if it satisﬁes: DDA (f + g ) = DDA (f ) + DDA (g ) DDA (fg ) = DDA (f )g + fDDA (g ) Corollaries: DDA (c) = 0 for all c algebraic over 1 . f DDA (f )g −fDDA (g ) DDA ( g ) = g2 . DDA extends uniquely to algebraic extensions.
38. 38. DDA (for any ring R of characteristic 0) Deﬁnition: DDA : R → R is a derivation on R if it satisﬁes: DDA (f + g ) = DDA (f ) + DDA (g ) DDA (fg ) = DDA (f )g + fDDA (g ) Corollaries: DDA (c) = 0 for all c algebraic over 1 . f DDA (f )g −fDDA (g ) DDA ( g ) = g2 . DDA extends uniquely to algebraic extensions. Note that there is no Chain Rule as such, since composition is not necessarily a deﬁned concept on R.
39. 39. A note on the word “constant”
40. 40. A note on the word “constant” We said DDA (c) = 0 for all c algebraic over 1 .
41. 41. A note on the word “constant” We said DDA (c) = 0 for all c algebraic over 1 . “constants diﬀerentiate to 0”
42. 42. A note on the word “constant” We said DDA (c) = 0 for all c algebraic over 1 . “constants diﬀerentiate to 0”
43. 43. A note on the word “constant” We said DDA (c) = 0 for all c algebraic over 1 . “constants diﬀerentiate to 0” By abuse of language, we say that anything that diﬀerentiates to zero is a “constantDA ”.
44. 44. How are the two subjects related? If we deﬁne DDA (x) = 1, then DDA is deﬁned on Z[x], and extends to Q(x) and indeed Q(x).
45. 45. How are the two subjects related? If we deﬁne DDA (x) = 1, then DDA is deﬁned on Z[x], and extends to Q(x) and indeed Q(x). If we interpret (denoted I) Q(x) as functions R → R, then DDA can be interpreted as D δ , i.e. I(DDA (f )) = D δ (I(f ))
46. 46. How are the two subjects related? If we deﬁne DDA (x) = 1, then DDA is deﬁned on Z[x], and extends to Q(x) and indeed Q(x). If we interpret (denoted I) Q(x) as functions R → R, then DDA can be interpreted as D δ , i.e. I(DDA (f )) = D δ (I(f )) (at least up to removable singularities).
47. 47. An aside on interpretation (x−1)2 x−1 Consider L := x 2 −1 and R := x+1 .
48. 48. An aside on interpretation 2 Consider L := (x−1) and R := x−1 . x 2 −1 x+1 0 As elements of Q(x) they are equal, since L − R = x 2 −1 = 0.
49. 49. An aside on interpretation 2 Consider L := (x−1) and R := x−1 . x 2 −1 x+1 As elements of Q(x) they are equal, since L − R = x 20 = 0. −1 But as formulae (viewed as algorithm speciﬁcations), L(1)=“divide by zero error”, whereas R(1) = 0.
50. 50. An aside on interpretation 2 Consider L := (x−1) and R := x−1 . x 2 −1 x+1 As elements of Q(x) they are equal, since L − R = x 20 = 0. −1 But as formulae (viewed as algorithm speciﬁcations), L(1)=“divide by zero error”, whereas R(1) = 0. Hence the warning about removable singularities!
51. 51. An aside on interpretation 2 Consider L := (x−1) and R := x−1 . x 2 −1 x+1 As elements of Q(x) they are equal, since L − R = x 20 = 0. −1 But as formulae (viewed as algorithm speciﬁcations), L(1)=“divide by zero error”, whereas R(1) = 0. Hence the warning about removable singularities! Also, note that −1 is not a removable singularity!
52. 52. An aside on interpretation 2 Consider L := (x−1) and R := x−1 . x 2 −1 x+1 As elements of Q(x) they are equal, since L − R = x 20 = 0. −1 But as formulae (viewed as algorithm speciﬁcations), L(1)=“divide by zero error”, whereas R(1) = 0. Hence the warning about removable singularities! Also, note that −1 is not a removable singularity! In the case of Q(x), every element has a “most continuous formula”, so interpreting this as R → R can’t go wrong.
53. 53. An aside on interpretation 2 Consider L := (x−1) and R := x−1 . x 2 −1 x+1 As elements of Q(x) they are equal, since L − R = x 20 = 0. −1 But as formulae (viewed as algorithm speciﬁcations), L(1)=“divide by zero error”, whereas R(1) = 0. Hence the warning about removable singularities! Also, note that −1 is not a removable singularity! In the case of Q(x), every element has a “most continuous formula”, so interpreting this as R → R can’t go wrong. We use I ∗ to indicate “interpretation with removable singularities removed”, since there isn’t always a formulaic way of doing so.
54. 54. An aside on interpretation 2 Consider L := (x−1) and R := x−1 . x 2 −1 x+1 As elements of Q(x) they are equal, since L − R = x 20 = 0. −1 But as formulae (viewed as algorithm speciﬁcations), L(1)=“divide by zero error”, whereas R(1) = 0. Hence the warning about removable singularities! Also, note that −1 is not a removable singularity! In the case of Q(x), every element has a “most continuous formula”, so interpreting this as R → R can’t go wrong. We use I ∗ to indicate “interpretation with removable singularities removed”, since there isn’t always a formulaic way of doing so. Indeed, there may be no way of interpreting without singularities, √ as in z → z : C → C.
55. 55. δ (for functions R → R)
56. 56. δ (for functions R → R) What is naturally deﬁned is integration over an interval I . We let D stand for sub-divisions d1 = a < d2 < · · · < dn = b of I = [a, b], and |D| for the largest distance between neighbouring points in D, i.e. maxi (di+1 − di ).
57. 57. δ (for functions R → R) What is naturally deﬁned is integration over an interval I . We let D stand for sub-divisions d1 = a < d2 < · · · < dn = b of I = [a, b], and |D| for the largest distance between neighbouring points in D, i.e. maxi (di+1 − di ). Let SD = i (di+1 − di ) maxdi+1 ≥x≥di f (x);
58. 58. δ (for functions R → R) What is naturally deﬁned is integration over an interval I . We let D stand for sub-divisions d1 = a < d2 < · · · < dn = b of I = [a, b], and |D| for the largest distance between neighbouring points in D, i.e. maxi (di+1 − di ). Let SD = i (di+1 − di ) maxdi+1 ≥x≥di f (x); SD = i (di+1 − di ) mindi+1 ≥x≥di f (x);
59. 59. δ (for functions R → R) What is naturally deﬁned is integration over an interval I . We let D stand for sub-divisions d1 = a < d2 < · · · < dn = b of I = [a, b], and |D| for the largest distance between neighbouring points in D, i.e. maxi (di+1 − di ). Let SD = i (di+1 − di ) maxdi+1 ≥x≥di f (x); SD = i (di+1 − di ) mindi+1 ≥x≥di f (x); Then δ I f = lim inf |D|→0 SD = lim sup|D|→0 SD if both exist and are equal.
60. 60. Consequences: Fundamental Theorem of Calculus (FTC δ ) b δ [a,b] f a≤b Deﬁne δ a f = − δ [b,a] f a>b
61. 61. Consequences: Fundamental Theorem of Calculus (FTC δ ) b δ [a,b] f a≤b Deﬁne δ a f = − δ [b,a] f a > b (If we’re not careful, we wind up saying “[2,1] is the same set as [1,2] except that if you integrate over it you have to add a − sign”. What we really have here is the beginnings of contour integration.
62. 62. Consequences: Fundamental Theorem of Calculus (FTC δ ) b δ [a,b] f a≤b Deﬁne δ a f = − δ [b,a] f a > b (If we’re not careful, we wind up saying “[2,1] is the same set as [1,2] except that if you integrate over it you have to add a − sign”. What we really have here is the beginnings of contour integration. Apostol theorem 1.20 states that if g (x) ≤ f (x) for every x in [a, b], then b b g (x)dx ≤ f (x)dx, a a and here a < b is implicit.)
63. 63. Consequences: Fundamental Theorem of Calculus (FTC δ ) b δ [a,b] f a≤b Deﬁne δ a f = − δ [b,a] f a > b (If we’re not careful, we wind up saying “[2,1] is the same set as [1,2] except that if you integrate over it you have to add a − sign”. What we really have here is the beginnings of contour integration. Apostol theorem 1.20 states that if g (x) ≤ f (x) for every x in [a, b], then b b g (x)dx ≤ f (x)dx, a a and here a < b is implicit.) FTC δ : δ [a,b] D δ f = f (b) − f (a).
64. 64. Consequences: Fundamental Theorem of Calculus (FTC δ ) b δ [a,b] f a≤b Deﬁne δ a f = − δ [b,a] f a > b (If we’re not careful, we wind up saying “[2,1] is the same set as [1,2] except that if you integrate over it you have to add a − sign”. What we really have here is the beginnings of contour integration. Apostol theorem 1.20 states that if g (x) ≤ f (x) for every x in [a, b], then b b g (x)dx ≤ f (x)dx, a a and here a < b is implicit.) FTC δ : δ [a,b] D δ f = f (b) − f (a). x Or: D δ (λx. δ a f ) = f (if the δ exists).
65. 65. Consequences: Fundamental Theorem of Calculus (FTC δ ) b δ [a,b] f a≤b Deﬁne δ a f = − δ [b,a] f a > b (If we’re not careful, we wind up saying “[2,1] is the same set as [1,2] except that if you integrate over it you have to add a − sign”. What we really have here is the beginnings of contour integration. Apostol theorem 1.20 states that if g (x) ≤ f (x) for every x in [a, b], then b b g (x)dx ≤ f (x)dx, a a and here a < b is implicit.) FTC δ : δ [a,b] D δ f = f (b) − f (a). x Or: D δ (λx. δ a f ) = f (if the δ exists). (Of course, this is normally stated without the λ.)
66. 66. DA : FTC becomes a deﬁnition
67. 67. DA : FTC becomes a deﬁnition FTCDA : deﬁne DA f to be any g such that f = DDA g .
68. 68. DA : FTC becomes a deﬁnition FTCDA : deﬁne DA f to be any g such that f = DDA g . If g and h are two such integrals, then DDA (g − h) = 0, i.e. “g and h diﬀer by a constant”.
69. 69. DA : FTC becomes a deﬁnition FTCDA : deﬁne DA f to be any g such that f = DDA g . If g and h are two such integrals, then DDA (g − h) = 0, i.e. “g and h diﬀer by a constant”. One diﬃculty is that this is really a “constantDA ” (something whose DDA is zero), and, for example, a Heaviside function is a constantDA , though not a constant in the usual sense.
70. 70. Will the real FTC please stand up?
71. 71. Will the real FTC please stand up? FTC (as it should be taught).
72. 72. Will the real FTC please stand up? FTC (as it should be taught). If g = DA f , and I(g ) is continuous on [a, b], then b δ I(f ) = I(g )(b) − I(g )(a). a
73. 73. Will the real FTC please stand up? FTC (as it should be taught). If g = DA f , and I(g ) is continuous on [a, b], then b δ I(f ) = I(g )(b) − I(g )(a). a 1 Note the caveat on continuity: g : x → arctan x is discontinuous at x = 0 (limx→0− arctan x = −π whereas 1 2 limx→0+ arctan x = π ), 1 2
74. 74. Will the real FTC please stand up? FTC (as it should be taught). If g = DA f , and I(g ) is continuous on [a, b], then b δ I(f ) = I(g )(b) − I(g )(a). a 1 Note the caveat on continuity: g : x → arctan x is discontinuous 1 −π at x = 0 (limx→0− arctan x = 2 whereas limx→0+ arctan x = π ), which accounts for the invalidity of 1 2 deducing that the integral of a negative function is positive — 1 −1 π −π π = I(g )(1) − I(g )(−1) = − = > 0. −1 x2+1 4 4 2
75. 75. Rescuing the Fundamental Theorem of Calculus −1 1 Of course, another DA x 2 +1 is h(x) = arctan x + H(x) where 0 x <0 H(x) = is a constantDA . This has a removable −π x > 0 singularity at x = 0.
76. 76. Rescuing the Fundamental Theorem of Calculus −1 1 Of course, another DA x 2 +1 is h(x) = arctan x + H(x) where 0 x <0 H(x) = is a constantDA . This has a removable −π x > 0 singularity at x = 0. I ∗ (h) is continuous, so FTC is appropriate and 1 −1 π −π π = I(h)(1) − I(h)(−1) = −π − = − < 0. −1 x2+1 4 4 2
77. 77. “Most continuous expressions” revisited 1 π arctan x + H(x) can also be rewritten as 2 − arctan(x).
78. 78. “Most continuous expressions” revisited arctan x + H(x) can also be rewritten as π − arctan(x). 1 2 In this case π − arctan(x) is a “most continuous formula”. 2
79. 79. “Most continuous expressions” revisited arctan x + H(x) can also be rewritten as π − arctan(x). 1 2 In this case π − arctan(x) is a “most continuous formula”. 2 1 (except at ∞, where the original arctan x is continuous!).
80. 80. “Most continuous expressions” revisited arctan x + H(x) can also be rewritten as π − arctan(x). 1 2 In this case π − arctan(x) is a “most continuous formula”. 2 1 (except at ∞, where the original arctan x is continuous!). z−1 What about arctan z+1 ?
81. 81. “Most continuous expressions” revisited arctan x + H(x) can also be rewritten as π − arctan(x). 1 2 In this case π − arctan(x) is a “most continuous formula”. 2 1 (except at ∞, where the original arctan x is continuous!). z−1 What about arctan z+1 ? An equivalent is √ √ z − 1 + z2 + 6 z − 7 z − 1 + z2 + 6 z − 7 arctan −arctan +1 2(z − 1) 2(z − 1)
82. 82. “Most continuous expressions” revisited arctan x + H(x) can also be rewritten as π − arctan(x). 1 2 In this case π − arctan(x) is a “most continuous formula”. 2 1 (except at ∞, where the original arctan x is continuous!). z−1 What about arctan z+1 ? An equivalent is √ √ z − 1 + z2 + 6 z − 7 z − 1 + z2 + 6 z − 7 arctan −arctan +1 2(z − 1) 2(z − 1) 2 or − arctan −2z−1+z 2 − arctan(z), where the singularities are at 2 z−1+z √ −1 ± 2 (and ∞), or . . . .
83. 83. “Most continuous expressions” revisited arctan x + H(x) can also be rewritten as π − arctan(x). 1 2 In this case π − arctan(x) is a “most continuous formula”. 2 1 (except at ∞, where the original arctan x is continuous!). z−1 What about arctan z+1 ? An equivalent is √ √ z − 1 + z2 + 6 z − 7 z − 1 + z2 + 6 z − 7 arctan −arctan +1 2(z − 1) 2(z − 1) 2 or − arctan −2z−1+z 2 − arctan(z), where the singularities are at 2 z−1+z √ −1 ± 2 (and ∞), or . . . . Still an unsolved problem (Rioboo, . . . ).
84. 84. But I’m not interested in all this DA stuﬀ
85. 85. But I’m not interested in all this DA stuﬀ Surely most people (even engineers) do δ etc.
86. 86. But I’m not interested in all this DA stuﬀ Surely most people (even engineers) do δ etc. They may think they do, but in practice they do DA even when intending to do δ .
87. 87. But I’m not interested in all this DA stuﬀ Surely most people (even engineers) do δ etc. They may think they do, but in practice they do DA even when intending to do δ . 1 What is limh→0 h (1 + h) cos2 (1 + h) − cos2 1 ?
88. 88. But I’m not interested in all this DA stuﬀ Surely most people (even engineers) do δ etc. They may think they do, but in practice they do DA even when intending to do δ . 1 What is limh→0 h (1 + h) cos2 (1 + h) − cos2 1 ? If you immediately said cos2 (1) − 2 sin(1) cos(1)
89. 89. But I’m not interested in all this DA stuﬀ Surely most people (even engineers) do δ etc. They may think they do, but in practice they do DA even when intending to do δ . 1 What is limh→0 h (1 + h) cos2 (1 + h) − cos2 1 ? If you immediately said cos2 (1) − 2 sin(1) cos(1) I bet you actually did DDA (x cos2 x)|x=1 .
90. 90. But I’m not interested in all this DA stuﬀ Surely most people (even engineers) do δ etc. They may think they do, but in practice they do DA even when intending to do δ . 1 What is limh→0 h (1 + h) cos2 (1 + h) − cos2 1 ? If you immediately said cos2 (1) − 2 sin(1) cos(1) I bet you actually did DDA (x cos2 x)|x=1 . (If you said −.617370845, you probably had Maple on your Blackberry, and it did that.)
91. 91. Conclusions
92. 92. Conclusions We already know that δ is a powerful world model
93. 93. Conclusions We already know that δ is a powerful world model DA is well-implemented and very powerful
94. 94. Conclusions We already know that δ is a powerful world model DA is well-implemented and very powerful “e−x 2 has no integral” means “has no in terms of DA already known functions”
95. 95. Conclusions We already know that δ is a powerful world model DA is well-implemented and very powerful “e−x 2 has no integral” means “has no in terms of DA already known functions” I does map from one to the other, often very eﬀectively
96. 96. Conclusions We already know that δ is a powerful world model DA is well-implemented and very powerful “e−x 2 has no integral” means “has no in terms of DA already known functions” I does map from one to the other, often very eﬀectively but not always!
97. 97. I crops up elsewhere
98. 98. I crops up elsewhere √ √ √ Claim 1 − z2 = 1 − z 1 + z.
99. 99. I crops up elsewhere √ √ √ Claim 1 − z 2 = 1 − z 1 + z. Squaring both sides gives 1 − z 2 = (1 − z)(1 + z), so there is some interpretation in which it is true.
100. 100. I crops up elsewhere √ √ √ Claim 1 − z 2 = 1 − z 1 + z. Squaring both sides gives 1 − z 2 = (1 − z)(1 + z), so there is some interpretation in which it is true. √ ?√ √ What about z 2 − 1= z − 1 z + 1.
101. 101. I crops up elsewhere √ √ √ Claim 1 − z 2 = 1 − z 1 + z. Squaring both sides gives 1 − z 2 = (1 − z)(1 + z), so there is some interpretation in which it is true. √ ?√ √ What about z 2 − 1= z − 1 z + 1. √ same √ The arguments apply and there is some interpretation of √ z 2 − 1, z − 1 and z + 1 in which it is true, but
102. 102. I crops up elsewhere √ √ √ Claim 1 − z 2 = 1 − z 1 + z. Squaring both sides gives 1 − z 2 = (1 − z)(1 + z), so there is some interpretation in which it is true. √ ?√ √ What about z 2 − 1= z − 1 z + 1. √ same √ The arguments apply and there is some interpretation of √ 2 − 1, z √ z − 1 and z + 1 in which it is true, but when z = −2 √ √ √ we get 3 = −3 −1, so there is no interpretation of as √ √ √ such, consistent with −n = −1 n even for n ∈ N, for which it is true.
103. 103. I crops up elsewhere √ √ √ Claim 1 − z 2 = 1 − z 1 + z. Squaring both sides gives 1 − z 2 = (1 − z)(1 + z), so there is some interpretation in which it is true. √ ?√ √ What about z 2 − 1= z − 1 z + 1. √ same √ The arguments apply and there is some interpretation of √ 2 − 1, z √ z − 1 and z + 1 in which it is true, but when z = −2 √ √ √ we get 3 = −3 −1, so there is no interpretation of as √ √ √ such, consistent with −n = −1 n even for n ∈ N, for which it is true. √ There is no interpretation of as such for which both are true.
104. 104. I crops up elsewhere √ √ √ Claim 1 − z 2 = 1 − z 1 + z. Squaring both sides gives 1 − z 2 = (1 − z)(1 + z), so there is some interpretation in which it is true. √ ?√ √ What about z 2 − 1= z − 1 z + 1. √ same √ The arguments apply and there is some interpretation of √ 2 − 1, z √ z − 1 and z + 1 in which it is true, but when z = −2 √ √ √ we get 3 = −3 −1, so there is no interpretation of as √ √ √ such, consistent with −n = −1 n even for n ∈ N, for which it is true. √ There is no interpretation of as such for which both are true. √ I interprets individual functions, such as 1 + z, and there is no general composition.