Abel's theorem in problems and solutions - School of Mathematics

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80 Chapter 2 From the solution of Problem 279 we obtain that also by fixing the image of the initial point of the curve C the continuous image of the curve C under the mapping may be defined non-uniquely. The uniqueness is lost when the curve C passes through the point In fact, for the function the uniqueness is lost only in this case, because only in this case do the two images of the point melt into one point. To avoid the non-uniqueness in the definition of the images of curves under the mapping we may exclude the point and forbid any curve to pass through this point. This restriction, however, does not always allows the continuous single-valued branches of the function to be separated. Indeed, if we fix at a point one of the values and we try to define at a certain point by continuity along two distinct curves joining and we may obtain different values of (for example, see 278). Observe now how we can avoid the non-uniqueness of the obtained value. 280. Suppose the variation of the argument of along a curve C be equal to Find the variation of the argument of along an arbitrary continuous image of the curve C under the mapping 281. Let and choose Define the value of by continuity along: a) the segment joining points and b) the curve with the parametric equation c) the curve with the parametric equation 282. Let and choose at the initial point of a curve C, Define by continuity along the curve C the value of at the end point if the curve C has the equation: a) b) c) 283. Let C be a closed curve on the plane (i.e., Prove that the value of the function at the end point of the curve C, defined by continuity, coincides with the value at the initial point if and only if the curve C wraps around the point an even number of times. In the sequel it will be convenient to use the following notation: DEFINITION. Let C be a continuous curve with a parametric equation We shall denote by the curve geometrically identical to C but oriented in the opposite direction; its equation is (cf., 247)
The complex numbers 81 DEFINITION. Suppose that the initial point of a curve coincides with the final point of a curve We will denote by the curve obtained by joining the end point of to the initial point of (cf., 248). 284. Let and be two curves, joining the points and and let one of the values of be chosen. Prove that the values of defined by continuity along the curves and are equal if and only if curve (Figure 30) turns around the point an even number of times. FIGURE 30 FIGURE 31 From the statement of the last problem it follows that, in particular, if the curve turns zero times around the point then the values of the function at the final points of the curves and will coincide if the values at the initial point coincide. So to separate the single-valued continuous branches of the function it suffices to take the curves and in such a way that the curve does not turn around the point For this it suffices to make a cut from the point to infinity avoiding intersecting the curve. We operated exactly in this way drawing the cut from the point to along the negative side of the real axis. If, after having made the cut, one fixes at a certain point one of the values, and if the value at every point is defined by continuity along an arbitrary curve C joining and and not passing through the cut, then at every point of the plane (except those on the cut) a certain single-valued continuous branch of the function is defined. If at the point one fixes the other value then one defines the other branch of the function 285. Prove that for every point outside the cut.
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ABEL’S THEOREM IN PROBLEMS AND SO
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Abel’s Theorem in Problems and So
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Contents Preface for the English ed
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A.10.3 The integrable case A.11 Top
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Preface to the English edition by V
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of the differential equations. Unfo
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Preface In high school algebraic eq
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Introduction We begin this book by
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Introduction 3 where by is indicate
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Introduction 5 By Viète’s theore
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Introduction 7 We will be able to p
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Chapter 1 Groups 1.1 Examples In ar
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Groups 11 the triangle ABC into its
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Groups 13 7. Find all symmetries of
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Groups 15 notations (see the soluti
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Groups 17 element, which we will de
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Groups 19 35. Suppose that is the m
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Groups 21 If a group is isomorphic
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Groups 23 tude); (rotation by 180°
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Groups 25 Hence left cosets generat
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Groups 27 as a permutation of the t
Page 45 and 46: Groups 29 102. Let be the order of
Page 47 and 48: Groups 31 111. Find all normal subg
Page 49 and 50: Groups 33 FIGURE 8 127. Prove that
Page 51 and 52: Groups 35 137. Prove that ker is a
Page 53 and 54: Groups 37 We now observe what happe
Page 55 and 56: Groups 39 FIGURE 12 vertices; 4) ro
Page 57 and 58: Groups 41 Every permutation of degr
Page 59 and 60: Groups 43 180. Prove that by multip
Page 61 and 62: Chapter 2 The complex numbers When
Page 63 and 64: The complex numbers 47 195. Prove t
Page 65 and 66: The complex numbers 49 where is the
Page 67 and 68: The complex numbers 51 Consequently
Page 69 and 70: The complex numbers 53 For complex
Page 71 and 72: The complex numbers 55 think that n
Page 73 and 74: The complex numbers 57 214. Let us
Page 75 and 76: The complex numbers 59 219. Prove t
Page 77 and 78: The complex numbers 61 functions (s
Page 79 and 80: The complex numbers 63 of continuit
Page 81 and 82: The complex numbers 65 erations of
Page 83 and 84: The complex numbers 67 a) b) c) d)
Page 85 and 86: The complex numbers 69 REMARK. If a
Page 87 and 88: The complex numbers 71 2.8 Images o
Page 89 and 90: The complex numbers 73 where all th
Page 91 and 92: The complex numbers 75 in such func
Page 93 and 94: The complex numbers 77 FIGURE 26 no
Page 95: The complex numbers 79 C from to th
Page 99 and 100: The complex numbers 83 FIGURE 32 cu
Page 101 and 102: The complex numbers 85 to infinity
Page 103 and 104: The complex numbers 87 property is
Page 105 and 106: The complex numbers 89 means will b
Page 107 and 108: The complex numbers 91 For example,
Page 109 and 110: The complex numbers 93 of the Riema
Page 111 and 112: The complex numbers 95 in Figure 38
Page 113 and 114: The complex numbers 97 permutation
Page 115 and 116: The complex numbers 99 2.13 Monodro
Page 117 and 118: The complex numbers 101 First we pr
Page 119 and 120: The complex numbers 103 and prove t
Page 121 and 122: Chapter 3 Hints, Solutions, and Ans
Page 123 and 124: Solutions 107 positive integer unde
Page 125 and 126: Solutions 109 on the left and by on
Page 127 and 128: Solutions 111 39. a) See Table 8; b
Page 129 and 130: Solutions 113 of the real numbers i
Page 131 and 132: Solutions 115 to all and therefore
Page 133 and 134: Solutions 117 is a subgroup of the
Page 135 and 136: Solutions 119 hypothesis. The group
Page 137 and 138: Solutions 121 The given group is th
Page 139 and 140: Solutions 123 of symmetries of the
Page 141 and 142: Solutions 125 i.e., there exists an
Page 143 and 144: Solutions 127 Answer. The normal su
Page 145 and 146: Solutions 129 In this way the quoti
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Solutions 131 sends every into itse
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Solutions 133 ement of the commutan
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Solutions 135 143. Let be the three
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Solutions 137 150. See 57. Suppose
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Solutions 139 subgroup. Hence if a
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Solutions 141 is commutative. Since
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Solutions 143 180. Since the lower
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Solutions 145 FIGURE 43 190. The pe
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Solutions 147 permutations of type
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Solutions 149 is a field. 195. We h
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Solutions 151 necessarily that and
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Solutions 153 Let C be the field of
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Solutions 155 in as numerators and
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Solutions 157 c) d) e) where 226. W
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Solutions 159 function of real argu
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Solutions 161 i.e., Consider the re
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Solutions 163 Choose as the smalles
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Solutions 165 Since the function is
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Solutions 167 FIGURE 55 FIGURE 56 F
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Solutions 169 FIGURE 60 257. Suppos
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Solutions 171 265. 266. 267. If the
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Solutions 173 following way. If the
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Solutions 175 for all such that and
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Solutions 177 b) In order for to va
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Solutions 179 to the condition coin
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Solutions 181 are continuous functi
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Solutions 183 turn around the other
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Solutions 185 sought is shown in Fi
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Solutions 187 FIGURE 86 FIGURE 87 p
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Solutions 189 sum of multi-valued f
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Solutions 191 316. a) Let and be th
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Solutions 193 by the formal method,
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Solutions 195 The correct scheme is
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Solutions 197 branch of the functio
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Solutions FIGURE 110 FIGURE 111 329
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Solutions 201 It follows that and b
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Solutions 203 340. To every sheet o
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Solutions 205 Since by hypothesis t
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Solutions 207 moving along a curve
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Drawings of Riemann surfaces 209 Th
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FIGURE 119 211
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FIGURE 121 213
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FIGURE 123 215
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FIGURE 125 217
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FIGURE 127 219
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Appendix by A. Khovanskii: Solvabil
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Solvability of Equations 223 functi
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Solvability of Equations 225 Suppos
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Solvability of Equations 227 where
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Solvability of Equations 229 finds
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Solvability of Equations 231 quadra
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Solvability of Equations 233 More p
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Solvability of Equations 235 group.
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Solvability of Equations 237 Every
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Solvability of Equations 239 The gr
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Solvability of Equations 241 FIG. 1
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Solvability of Equations 243 are ob
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Solvability of Equations 245 corres
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Solvability of Equations 249 up to
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Solvability of Equations 251 fined
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Solvability of Equations 253 the qu
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Solvability of Equations 255 the no
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Solvability of Equations 257 THEORE
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Bibliography [1] [2] [3] [4] [5] [6
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[23] [24] [25] [26] [27] [28] 263 V
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Appendix (V.I. Arnold) The topologi
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Index Abel’s theorem, 6, 103 addi
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pack of sheets, 96 parametric equat
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