1. The unreasonable eectiveness of algebra (but dont take it for granted) James DavenportUniversity of Bath (visiting Waterloo) 11 June 2009

2. What is the derivative of sin x? 3. What is the derivative of sin x? sin(x + ) sin(x) f (x) = lim0 4. What is the derivative of sin x? sin(x + ) sin(x) f (x) = lim0 sin(x + ) sin(x) > 0D : < D f (x) < 5. What is the derivative of sin x?sin(x + ) sin(x)f (x) = lim 0 sin(x + ) sin(x) > 0D : < D f (x) 0D : < D f (x) 0D : < D f (x) 0D : < D f (x) < sin(x + ) sin(x) = cos(x) sin() sin(x)(1 cos()) so we canmake the choice f (x) = cos(x). We can then take1 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. What is the derivative of sin x?sin(x + ) sin(x) f (x) = lim0 sin(x + ) sin(x) > 0D : < D f (x) < sin(x + ) sin(x) = cos(x) sin() sin(x)(1 cos()) so we canmake the choice f (x) = cos(x). We can then take1 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. /2 What is the integral: 0cos x? 11. /2 What is the integral: 0cos x?I = lim S [cos(x)] = lim S [cos(x)] ||0||0 ( ranges over all dissections of [0, /2]) 12. /2 What is the integral: 0cos x?I = lim S [cos(x)] = lim S [cos(x)] ||0||0 ( ranges over all dissections of [0, /2])Does anyone actually do this? 13. /2 What is the integral: 0cos x?I = lim S [cos(x)] = lim S [cos(x)] ||0||0 ( ranges over all dissections of [0, /2])Does anyone actually do this?Can anyone actually do this? 14. /2 What is the integral:0cos x? I = lim S [cos(x)] = lim S [cos(x)]||0 ||0 ( ranges over all dissections of [0, /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 sin 0 = 1?2 15. /2 What is the integral:0cos x? I = lim S [cos(x)] = lim S [cos(x)]||0||0 ( ranges over all dissections of [0, /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 sin 0 = 1?2We might say therefore, by the Fundamental Theorem ofCalculus, . . . 16. What is the Fundamental Theorem of Calculus? 17. What is the Fundamental Theorem of Calculus? Indeed, what is calculus? 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. 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,dierent subjects. 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,dierent subjects. What you learned in calculus, which I shall write as D : the dierentiation of analysis. Also d d x , and its inverse . 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,dierent subjects. What you learned in calculus, which I shall write as D : the dierentiation of analysis. Also d d x , and its inverse . What is taught in dierential algebra, which I shall write as d DDA : the dierentiation of dierential algebra. Also dDA x , and its inverse DA . 22. D (for functions R R) 23. D (for functions R R) Dene CL(f , x0 ) (the Cauchy Limit) as f (x0 + h) f (x0 ) CL(f , x0 ) = limh0hand D (f ) = x. CL(f , x). 24. D (for functions R R) Dene CL(f , x0 ) (the Cauchy Limit) asf (x0 + h) f (x0 )CL(f , x0 ) = lim h0hand D (f ) = x. CL(f , x). Then the following are theorems. 25. D (for functions R R) Dene CL(f , x0 ) (the Cauchy Limit) asf (x0 + h) f (x0 )CL(f , x0 ) = lim h0hand D (f ) = x. CL(f , x). Then the following are theorems.D (c) = 0 c constants. 26. D (for functions R R) Dene CL(f , x0 ) (the Cauchy Limit) asf (x0 + h) f (x0 )CL(f , x0 ) = lim h0hand D (f ) = x. CL(f , x). Then the following are theorems.D (c) = 0 c constants.CL(f + g , x0 ) = CL(f , x0 ) + CL(g , x0 ), soD (f + g ) = D (f ) + D (g ). 27. D (for functions R R) Dene CL(f , x0 ) (the Cauchy Limit) asf (x0 + h) f (x0 )CL(f , x0 ) = lim h0hand D (f ) = x. CL(f , x). Then the following are theorems.D (c) = 0 c constants.CL(f + g , x0 ) = CL(f , x0 ) + CL(g , x0 ), soD (f + g ) = D (f ) + D (g ).CL(fg , x0 ) = CL(f , x0 )g (x0 ) + f (x0 ) CL(g , x0 ), soD (fg ) = D (f )g + fD (g ). 28. D (for functions R R) Dene CL(f , x0 ) (the Cauchy Limit) asf (x0 + h) f (x0 )CL(f , x0 ) = lim h0hand D (f ) = x. CL(f , x). Then the following are theorems.D (c) = 0 c constants.CL(f + g , x0 ) = CL(f , x0 ) + CL(g , x0 ), soD (f + g ) = D (f ) + D (g ).CL(fg , x0 ) = CL(f , x0 )g (x0 ) + f (x0 ) CL(g , x0 ), soD (fg ) = D (f )g + fD (g ).D (x.f (g (x))) = D (g )x.D (f )(g (x)). (Chain rule) 29. DDA (for any ring R of characteristic 0)Denition: DDA : R R is a derivation on R if it satises: 30. DDA (for any ring R of characteristic 0)Denition: DDA : R R is a derivation on R if it satises:DDA (f + g ) = DDA (f ) + DDA (g ) 31. DDA (for any ring R of characteristic 0)Denition: DDA : R R is a derivation on R if it satises:DDA (f + g ) = DDA (f ) + DDA (g )DDA (fg ) = DDA (f )g + fDDA (g ) 32. DDA (for any ring R of characteristic 0)Denition: DDA : R R is a derivation on R if it satises:DDA (f + g ) = DDA (f ) + DDA (g )DDA (fg ) = DDA (f )g + fDDA (g ) 33. DDA (for any ring R of characteristic 0)Denition: DDA : R R is a derivation on R if it satises:DDA (f + g ) = DDA (f ) + DDA (g )DDA (fg ) = DDA (f )g + fDDA (g )Corollaries: 34. DDA (for any ring R of characteristic 0)Denition: DDA : R R is a derivation on R if it satises: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. DDA (for any ring R of characteristic 0)Denition: DDA : R R is a derivation on R if it satises: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. DDA (for any ring R of characteristic 0)Denition: DDA : R R is a derivation on R if it satises: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. DDA (for any ring R of characteristic 0)Denition: DDA : R R is a derivation on R if it satises: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. DDA (for any ring R of characteristic 0)Denition: DDA : R R is a derivation on R if it satises: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 notnecessarily a dened concept on R. 39. A note on the word constant 40. A note on the word constantWe saidDDA (c) = 0 for all c algebraic over 1 . 41. A note on the word constantWe saidDDA (c) = 0 for all c algebraic over 1 .constants dierentiate to 0 42. A note on the word constantWe saidDDA (c) = 0 for all c algebraic over 1 .constants dierentiate to 0 43. A note on the word constantWe saidDDA (c) = 0 for all c algebraic over 1 .constants dierentiate to 0By abuse of language, we say that anything that dierentiates tozero is a constantDA . 44. How are the two subjects related? If we dene DDA (x) = 1, then DDA is dened on Z[x], andextends to Q(x) and indeed Q(x). 45. How are the two subjects related? If we dene DDA (x) = 1, then DDA is dened on Z[x], andextends to Q(x) and indeed Q(x).If we interpret (denoted I) Q(x) as functions R R, then DDAcan be interpreted as D , i.e. I(DDA (f )) = D (I(f )) 46. How are the two subjects related? If we dene DDA (x) = 1, then DDA is dened on Z[x], andextends to Q(x) and indeed Q(x).If we interpret (denoted I) Q(x) as functions R R, then DDAcan be interpreted as D , i.e. I(DDA (f )) = D (I(f )) (at least up to removable singularities). 47. An aside on interpretation (x1)2 x1Consider L :=x 2 1and R := x+1 . 48. An aside on interpretation 2Consider L := (x1) and R := x1 . x 2 1 x+1 0As elements of Q(x) they are equal, since L R = x 2 1 = 0. 49. An aside on interpretation2Consider L := (x1) and R := x1 .x 2 1 x+1As elements of Q(x) they are equal, since L R = x 20 = 0. 1But as formulae (viewed as algorithm specications), L(1)=divideby zero error, whereas R(1) = 0. 50. An aside on interpretation2Consider L := (x1) and R := x1 .x 2 1 x+1As elements of Q(x) they are equal, since L R = x 20 = 0. 1But as formulae (viewed as algorithm specications), L(1)=divideby zero error, whereas R(1) = 0.Hence the warning about removable singularities! 51. An aside on interpretation2Consider L := (x1) and R := x1 .x 2 1 x+1As elements of Q(x) they are equal, since L R = x 20 = 0. 1But as formulae (viewed as algorithm specications), L(1)=divideby zero error, whereas R(1) = 0.Hence the warning about removable singularities!Also, note that 1 is not a removable singularity! 52. An aside on interpretation2Consider L := (x1) and R := x1 .x 2 1 x+1As elements of Q(x) they are equal, since L R = x 20 = 0. 1But as formulae (viewed as algorithm specications), L(1)=divideby 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 continuousformula, so interpreting this as R R cant go wrong. 53. An aside on interpretation2Consider L := (x1) and R := x1 .x 2 1 x+1As elements of Q(x) they are equal, since L R = x 20 = 0.1But as formulae (viewed as algorithm specications), L(1)=divideby 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 continuousformula, so interpreting this as R R cant go wrong.We use I to indicate interpretation with removable singularitiesremoved, since there isnt always a formulaic way of doing so. 54. An aside on interpretation2Consider L := (x1) and R := x1 .x 2 1 x+1As elements of Q(x) they are equal, since L R = x 20 = 0.1But as formulae (viewed as algorithm specications), L(1)=divideby 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 continuousformula, so interpreting this as R R cant go wrong.We use I to indicate interpretation with removable singularitiesremoved, since there isnt always a formulaic way of doing so.Indeed, there may be no way of interpreting without singularities, as in z z : C C. 55. (for functions R R) 56. (for functions R R)What is naturally dened 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. (for functions R R)What is naturally dened 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 xdi f (x); 58. (for functions R R)What is naturally dened 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 xdi f (x); SD = i (di+1 di ) mindi+1 xdi f (x); 59. (for functions R R) What is naturally dened is integration over an interval I . We letD 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 ). LetSD =i (di+1 di ) maxdi+1 xdi f (x);SD =i (di+1 di ) mindi+1 xdi f (x); Then I f = lim inf |D|0 SD = lim sup|D|0 SD if both exist andare equal. 60. Consequences: Fundamental Theorem of Calculus (FTC )b [a,b] f abDene af = [b,a] f a>b 61. Consequences: Fundamental Theorem of Calculus (FTC )b [a,b] f abDene af = [b,a] f a > b(If were 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 contourintegration. 62. Consequences: Fundamental Theorem of Calculus (FTC )b [a,b] f abDene af = [b,a] f a > b(If were 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 contourintegration. Apostol theorem 1.20 states that if g (x) f (x) forevery x in [a, b], then b b g (x)dx f (x)dx, a a and here a < b is implicit.) 63. Consequences: Fundamental Theorem of Calculus (FTC )b [a,b] f abDene af = [b,a] f a > b(If were 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 contourintegration. Apostol theorem 1.20 states that if g (x) f (x) forevery x in [a, b], then b b g (x)dx f (x)dx, a a and here a < b is implicit.)FTC : [a,b] D f = f (b) f (a). 64. Consequences: Fundamental Theorem of Calculus (FTC )b [a,b] f abDene af = [b,a] f a > b(If were 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 contourintegration. Apostol theorem 1.20 states that if g (x) f (x) forevery x in [a, b], then b b g (x)dx f (x)dx, a a and here a < b is implicit.)FTC : [a,b] D f = f (b) f (a).xOr: D (x. a f ) = f (if the exists). 65. Consequences: Fundamental Theorem of Calculus (FTC )b [a,b] f abDene af = [b,a] f a > b(If were 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 contourintegration. Apostol theorem 1.20 states that if g (x) f (x) forevery x in [a, b], then b b g (x)dx f (x)dx, a a and here a < b is implicit.)FTC : [a,b] D f = f (b) f (a).xOr: D (x. a f ) = f (if the exists).(Of course, this is normally stated without the .) 66. DA : FTC becomes a denition 67. DA: FTC becomes a denitionFTCDA : dene DA f to be any g such that f = DDA g . 68. DA: FTC becomes a denitionFTCDA : dene 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. gand h dier by a constant. 69. DA: FTC becomes a denitionFTCDA : dene 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. gand h dier by a constant.One diculty is that this is really a constantDA (somethingwhose DDA is zero), and, for example, a Heaviside function is aconstantDA , though not a constant in the usual sense. 70. Will the real FTC please stand up? 71. Will the real FTC please stand up? FTC (as it should be taught). 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 I(f ) = I(g )(b) I(g )(a). a 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 I(f ) = I(g )(b) I(g )(a). a 1Note the caveat on continuity: g : x arctan x is discontinuousat x = 0 (limx0 arctan x = whereas1 2limx0+ arctan x = ),12 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 I(f ) = I(g )(b) I(g )(a). a1Note the caveat on continuity: g : x arctan x is discontinuous 1 at x = 0 (limx0 arctan x = 2 whereaslimx0+ arctan x = ), which accounts for the invalidity of12deducing that the integral of a negative function is positive 11 = I(g )(1) I(g )(1) = = > 0.1 x2+1 442 75. Rescuing the Fundamental Theorem of Calculus 1 1Of course, another DA x 2 +1 is h(x) = arctan x + H(x) where 0 x 0singularity at x = 0. 76. Rescuing the Fundamental Theorem of Calculus 1 1Of course, another DA x 2 +1 is h(x) = arctan x + H(x) where0 x 0singularity at x = 0.I (h) is continuous, so FTC is appropriate and 11 = I(h)(1) I(h)(1) = = < 0.1 x2+14 42 77. Most continuous expressions revisited1 arctan x + H(x) can also be rewritten as 2 arctan(x). 78. Most continuous expressions revisitedarctan x + H(x) can also be rewritten as arctan(x). 12In this case arctan(x) is a most continuous formula. 2 79. Most continuous expressions revisitedarctan x + H(x) can also be rewritten as arctan(x). 12In this case arctan(x) is a most continuous formula. 2 1(except at , where the original arctan x is continuous!). 80. Most continuous expressions revisitedarctan x + H(x) can also be rewritten as arctan(x). 12In this case arctan(x) is a most continuous formula. 2 1(except at , where the original arctan x is continuous!).z1What about arctan z+1 ? 81. Most continuous expressions revisitedarctan x + H(x) can also be rewritten as arctan(x). 12In this case arctan(x) is a most continuous formula. 2 1(except at , where the original arctan x is continuous!).z1What about arctan z+1 ? An equivalent is z 1 + z2 + 6 z 7z 1 + z2 + 6 z 7arctanarctan +12(z 1)2(z 1) 82. Most continuous expressions revisitedarctan x + H(x) can also be rewritten as arctan(x). 12In this case arctan(x) is a most continuous formula. 2 1(except at , where the original arctan x is continuous!).z1What about arctan z+1 ? An equivalent is z 1 + z2 + 6 z 7z 1 + z2 + 6 z 7arctanarctan +12(z 1)2(z 1) 2or arctan 2z1+z 2 arctan(z), where the singularities are at 2 z1+z1 2 (and ), or . . . . 83. Most continuous expressions revisitedarctan x + H(x) can also be rewritten as arctan(x). 12In this case arctan(x) is a most continuous formula. 2 1(except at , where the original arctan x is continuous!).z1What about arctan z+1 ? An equivalent is z 1 + z2 + 6 z 7z 1 + z2 + 6 z 7arctanarctan +12(z 1)2(z 1) 2or arctan 2z1+z 2 arctan(z), where the singularities are at 2 z1+z 1 2 (and ), or . . . .Still an unsolved problem (Rioboo, . . . ). 84. But Im not interested in all this DA stu 85. But Im not interested in all this DA stu Surely most people (even engineers) do etc. 86. But Im not interested in all this DA stu Surely most people (even engineers) do etc.They may think they do, but in practice they do DA evenwhen intending to do . 87. But Im not interested in all this DA stu Surely most people (even engineers) do etc.They may think they do, but in practice they doDA evenwhen intending to do . 1What is limh0 h (1 + h) cos2 (1 + h) cos2 1 ? 88. But Im not interested in all this DA stu Surely most people (even engineers) do etc.They may think they do, but in practice they doDA evenwhen intending to do . 1What is limh0 h (1 + h) cos2 (1 + h) cos2 1 ?If you immediately said cos2 (1) 2 sin(1) cos(1) 89. But Im not interested in all this DA stu Surely most people (even engineers) do etc.They may think they do, but in practice they doDA evenwhen intending to do . 1What is limh0 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. But Im not interested in all this DA stu Surely most people (even engineers) do etc.They may think they do, but in practice they doDA evenwhen intending to do . 1What is limh0 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 yourBlackberry, and it did that.) 91. Conclusions 92. Conclusions We already know that is a powerful world model 93. Conclusions We already know that is a powerful world model DA is well-implemented and very powerful 94. Conclusions We already know that is a powerful world model DA is well-implemented and very powerful ex 2 has no integral means has noin terms ofDA already known functions 95. Conclusions We already know that is a powerful world model DA is well-implemented and very powerful ex 2 has no integral means has noin terms ofDA already known functions I does map from one to the other, often very eectively 96. Conclusions We already know that is a powerful world model DA is well-implemented and very powerful ex 2 has no integral means has noin terms ofDA already known functions I does map from one to the other, often very eectively but not always! 97. I crops up elsewhere 98. I crops up elsewhere Claim 1 z2 = 1 z 1 + z. 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 someinterpretation in which it is true. 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 someinterpretation in which it is true. ? What about z 2 1= z 1 z + 1. 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 someinterpretation in which it is true. ? What about z 2 1= z 1 z + 1. same Thearguments apply and there is some interpretation of z 2 1, z 1 and z + 1 in which it is true, but 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 someinterpretation 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 ofas such, consistent with n = 1 n even for n N, for which itis true. 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 someinterpretation 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 ofas such, consistent with n = 1 n even for n N, for which itis true.There is no interpretation of as such for which both are true. 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 someinterpretation 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 ofas such, consistent with n = 1 n even for n N, for which itis true.There is no interpretation of as such for which both are true. I interprets individual functions, such as 1 + z, and there is nogeneral composition.