22 



Axel 



S6DERBL0M, 



r 





1 



= 2.0408 1633 





ul 





h 





= 0.0245 0000 



k 





= 0.0085 7500 



k 





= 0.0000 9804 



k 



> u o 



= 0.0000 2808 



h 



Mj° 



= 0.0000 0295 



h 



B? 



= 0.0000 0007 



k s 





= 0.0000 0001 





So 



= 2.0740 2048 



log S 



log s 

 log s 2 



0.3168 1308 

 0.7964 9920 

 2.0000 0218 5 

 4s 3 - s _ 1 

 12s 2 - 1 



R 3 (s ) = 



32.6120 869 



R 3 (s) 



= 



973.4894 91 





3999959.3971 



in oo 



----- 



50.6187 34 





469.0885 



lag '5, (s ) 



= 



1.5133 7856 



log /e 8 (s) = 



2.9883 312 



log# 3 (s 2 ) 



— 



6.6020 5555 



log ' 3 (s ) = 



1.7043 112 



logR' 3 (s) 





2.6712 547 





4.1848 975 



s 2 — s 





93.741 586 



log (s — 8 Q ) = 



0.6216 848 



log (s 2 _ s) 





1.9719 322 





62401.2702 



fjl 3 (s)]/R 3 (s ) 





178.1783 2 





973.4895 



B 3 (s ) 





32.6120 9 



21986.5420 



-R' s (s )(s - s ) 





105.9170 7 



'•' n i = 



85361.3017 



'.' w 2 





316.7074 8 



log w x = 



4.9312 6103 



log n 2 





2.5006 5816 



log 2(s 2 — s) 2 = 



4.2448 9440 



log 2 (s- s ) 2 





1.5443 9959 





0.6863 6663 







0.9562 5857 





4.8569 835 — 







9.0418 765 



2(s 2 -_s) 2 



2(8 -» y 





S — 



6.2589 180 



So 





2.0740 205 



■ . • o = 



11.1159 015 







11.1158 970 



L'accord parfait des valeurs de a et de o 2 est une confirmation 

 numerique de la formule (15). 



Quant k la solution de liquation (15), nous y reviendrons dans 

 le paragraphe 5. 



