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... �Þ ¼ 1 2 þ �ð4=�Þffiffiffi p � 8��2� � � cot ð�Þ2F1 12 ; 4 � ; 3 2 ;�cot ð�Þ2 � � ; ð5Þ where � is ... Hurst exponent.7 From Eq. ( 3), we now conjecture that � ¼ 8= 3 for � 1 < H < � 3= 4 and � ¼ 12=ð3� 2HÞ for ... conjecture that ¼ 8= 3 for 1 < H < 3= 4 and ¼ 12=ð3 2HÞ for 3= 4 < H < 0. Proving this conjecture remains a ... , J. Phys. C 1, 1532 (1968). 3. A. Weinrib, Phys. Rev. B 26, 1352 (1982). 4. D. Bernard, G. Bo®etta, A ... 1850008 1.. 10 ...
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