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

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206 Problems of Chapter 2 the mapping Since the curve lies entirely in one of the discs and in this disc there is only one image of the point the curve begins and ends at a unique point. So if C is a closed curve lying entirely inside the disc then the value of the function at the final point of the curve C, defined by continuity, coincides with the value at the initial point. In particular, this holds for all circles with centre at and radius smaller than Consequently the point is not a branch point of the function 346. From the result of Problem 343 it follows that for equation (2.8) has four roots: of which one (for example, ) has order 2, and the others are simple. Suppose that the point is close to the point Thus from the solution of Problem 344, we obtain that near the point there are two images of the point under the mapping and near the point and there is a sole image of the point Let C be a circular curve of small radius, with centre starting and ending at As in the solution of Problem 345 we obtain that the continuous images of the circle C under the mapping which start from the points and end at the initial point, whereas the continuous images which start from one of the images of the point near the point may end on the other image of the point which lies near the point as well. Consequently at the point only two sheets can meet, whilst there are no passages between the other three sheets. 347. Let us draw a continuous curve from the point to the point not crossing the images under of the points and It is possible to draw it, because the points and have a finite number of images. Now let C be the image of the curve under the mapping Since is a continuous function and a continuous curve, C is a continuous curve as well. Since and are related, under the mapping by the relation which coincides with that given by the mapping the curve is itself a continuous image of the curve C under the mapping Since the curve does not pass through the images of the points and the curve C does not pass through the points and The initial and final points of the curve C are and Consequently C is the curve we sought. 348. By virtue of the result of Problem 347 one can move from an arbitrary sheet of the Riemann surface of the function to any other,
Solutions 207 moving along a curve which does not pass through the points and Moreover, every passage from one sheet to another, whilst crossing a cut, coincides with the passage indicated on the scheme at that branch point from which this cut starts (cf., note 1 §2.10). Consequently the connection of the sheets at the branch points must be made so as to obtain a connected scheme. Since points different from and are not branch points (cf., 345), and at each one of the points and only two sheets can join (cf., 346), in order to obtain a connected scheme we must put one arrow in correspondence with each one of the points and i.e., all of these four points are branch points. All the distinct connected schemes are shown in Figure 115. Any connected scheme of the Riemann surface of the function matches one of these three schemes after a permutation of the sheets and of the branch points (here we do not claim that all these three schemes can be realized). FIGURE 115 349. We prove that the permutation group of the scheme for all the three schemes shown in Figure 115 contains all the elementary transpositions (cf., §1.15), i.e., the transposition (1, 2), (2, 3), (3, 4), (4, 5). For the first scheme this is evident because these transpositions correspond to the branch points. In the second and in the third scheme one of the branch points corresponds to the transposition (1, 2). The transpositions (2, 3) and (3, 4) are obtained in both cases as the products (2, 3) = (1, 2) · (1, 3) · (1, 2) and (3, 4) = (1, 3) · (1, 4) · (1, 3). The transposition (4, 5) is obtained for the second scheme as the product (4, 5) = (1, 4) · (1, 5) · (1, 4) and for the third scheme it simply coincides with one of the branch points. Consequently the required group contains, in all cases, all elementary transpositions, and consequently (Theorem 4, §1.15) it coincides with the entire group of the permutations of degree 5,
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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
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Groups 29 102. Let be the order of
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Groups 31 111. Find all normal subg
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Groups 33 FIGURE 8 127. Prove that
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Groups 35 137. Prove that ker is a
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Groups 37 We now observe what happe
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Groups 39 FIGURE 12 vertices; 4) ro
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Groups 41 Every permutation of degr
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Groups 43 180. Prove that by multip
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Chapter 2 The complex numbers When
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The complex numbers 47 195. Prove t
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The complex numbers 49 where is the
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The complex numbers 51 Consequently
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The complex numbers 53 For complex
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The complex numbers 55 think that n
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The complex numbers 57 214. Let us
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The complex numbers 59 219. Prove t
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The complex numbers 61 functions (s
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The complex numbers 63 of continuit
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The complex numbers 65 erations of
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The complex numbers 67 a) b) c) d)
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The complex numbers 69 REMARK. If a
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The complex numbers 71 2.8 Images o
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The complex numbers 73 where all th
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The complex numbers 75 in such func
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The complex numbers 77 FIGURE 26 no
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The complex numbers 79 C from to th
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The complex numbers 81 DEFINITION.
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The complex numbers 83 FIGURE 32 cu
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The complex numbers 85 to infinity
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The complex numbers 87 property is
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The complex numbers 89 means will b
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The complex numbers 91 For example,
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The complex numbers 93 of the Riema
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The complex numbers 95 in Figure 38
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The complex numbers 97 permutation
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The complex numbers 99 2.13 Monodro
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The complex numbers 101 First we pr
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The complex numbers 103 and prove t
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Chapter 3 Hints, Solutions, and Ans
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Solutions 107 positive integer unde
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Solutions 109 on the left and by on
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Solutions 111 39. a) See Table 8; b
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Solutions 113 of the real numbers i
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Solutions 115 to all and therefore
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Solutions 117 is a subgroup of the
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Solutions 119 hypothesis. The group
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Solutions 121 The given group is th
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Solutions 123 of symmetries of the
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Solutions 125 i.e., there exists an
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Solutions 127 Answer. The normal su
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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
Page 171 and 172: Solutions 155 in as numerators and
Page 173 and 174: Solutions 157 c) d) e) where 226. W
Page 175 and 176: Solutions 159 function of real argu
Page 177 and 178: Solutions 161 i.e., Consider the re
Page 179 and 180: Solutions 163 Choose as the smalles
Page 181 and 182: Solutions 165 Since the function is
Page 183 and 184: Solutions 167 FIGURE 55 FIGURE 56 F
Page 185 and 186: Solutions 169 FIGURE 60 257. Suppos
Page 187 and 188: Solutions 171 265. 266. 267. If the
Page 189 and 190: Solutions 173 following way. If the
Page 191 and 192: Solutions 175 for all such that and
Page 193 and 194: Solutions 177 b) In order for to va
Page 195 and 196: Solutions 179 to the condition coin
Page 197 and 198: Solutions 181 are continuous functi
Page 199 and 200: Solutions 183 turn around the other
Page 201 and 202: Solutions 185 sought is shown in Fi
Page 203 and 204: Solutions 187 FIGURE 86 FIGURE 87 p
Page 205 and 206: Solutions 189 sum of multi-valued f
Page 207 and 208: Solutions 191 316. a) Let and be th
Page 209 and 210: Solutions 193 by the formal method,
Page 211 and 212: Solutions 195 The correct scheme is
Page 213 and 214: Solutions 197 branch of the functio
Page 215 and 216: Solutions FIGURE 110 FIGURE 111 329
Page 217 and 218: Solutions 201 It follows that and b
Page 219 and 220: Solutions 203 340. To every sheet o
Page 221: Solutions 205 Since by hypothesis t
Page 225 and 226: Drawings of Riemann surfaces 209 Th
Page 227 and 228: FIGURE 119 211
Page 237 and 238: Appendix by A. Khovanskii: Solvabil
Page 239 and 240: Solvability of Equations 223 functi
Page 241 and 242: Solvability of Equations 225 Suppos
Page 243 and 244: Solvability of Equations 227 where
Page 245 and 246: Solvability of Equations 229 finds
Page 247 and 248: Solvability of Equations 231 quadra
Page 249 and 250: Solvability of Equations 233 More p
Page 251 and 252: Solvability of Equations 235 group.
Page 253 and 254: Solvability of Equations 237 Every
Page 255 and 256: Solvability of Equations 239 The gr
Page 257 and 258: Solvability of Equations 241 FIG. 1
Page 259 and 260: Solvability of Equations 243 are ob
Page 261 and 262: Solvability of Equations 245 corres
Page 263 and 264: Solvability of Equations 247 where
Page 265 and 266: Solvability of Equations 249 up to
Page 267 and 268: Solvability of Equations 251 fined
Page 269 and 270: Solvability of Equations 253 the qu
Page 271 and 272: Solvability of Equations 255 the no
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Solvability of Equations 257 THEORE
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Solvability of Equations 259 where
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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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Abel's theorem in problems and solutions - School of Mathematics

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