^^ 



IS 



Schuberfs investigation of Kepler'^ s Problem. 



§ 15. If in the equation (E) (§ IS] we put succes§ively i 

 2=3, and so on, ag far as i=i2f and at the same time n=3, n 



1; 



d 



w==12 



&' 



sm 



i;, we shall ^ 



.sin ^+ ~ siu 2,M.-f- — (S siti S,cc--sin^)+ ^ ^ 



( 



e 



(3*.sin 6;« — £6. sin 4fc+5 sin 2*4) 



+ .- 



+ 



210.3^5 

 2^33.5.7 



8 



(7^. sin 7jn — 5^. sin 5/«-f3''.sin S,«— 5 sin ,«•) 



(2 



■3'^ .sin 6iM.-{-2^.r,sin 4jk. — r sin 2^) 



I • nj 



e 



5 -.2 



-6- -.iJ^.^./ 



9 



— (3i^sin9^-— 7 ^-sin r/tc+22.5^sin 0i 



A^) 



+ 



e 



+ 



10 



+ 



2i».3\52.7 

 2s.3'*.o^.7.U 



-(U^smll,w-32^sin9/«,+5.7^<'.sinriM.-3.5^*.sin5M+2.3l^5.8iIl3^^--2.3.^sin.A^) 



,it-5 



li' 



g 



3b 2r=sm 3^+c(s\n S,M.-sm^') ^e* (sin4iit-sin2^) + ~^(53. sin 5^3' .sin 3^+4 sm ^; 

 ^^ (3^sin6«-2^sln4ii6+r,sin2K)+5^^(''*.sin7|it-35.sIn5iM.+S*.11,sin3M-3.5.sin.,^) 



6 



ir. 



+ 



2'.3*.5 



— ;r(2».sin Sa--— o^.sin 6jtt+2^sin4,M,— 13 sin 2ft.) 



') 



+ 



e 



3 



2^32.5.7 



lOAt^-: 



.2i3.sin4iM.+2.r2.sin2At) 



i 



( 



1- 



a 



.f. 



u 



+ 



e 



10 



2^3-* 5^,7 



^ 



-(2^3^sInI2^^5^^sml0jx+2'-^23•sln8^c-3^5a3•sin6^c4-2^3.5J^.sin4.tt-2.S^ 



J. 



,k 



sin 5s 



+ 



+ 



6* 



^■^s 





3 Se-^ 



?(sin4^« — sm%tt)+ ■5Y-(5sin 5jtc — 2,3.sin 3.a-f sin /«.) 



2 



2 



i- 



I: 



■ h 



2*.5 



6 



-(28. sin8^--S*.5.sin 6,«.4- 2^,5. sin 4f<. — 32.sjn2«.) 



4- _il_(39,sin 9f^~2.T^,sin Tfti-5\%m S^c— S-^.lQ.sin 3a«.-1--S sin.M.) 



2^°. 5 



-f 



e 



Q^. 3.5.7 



( 



012 



6/4—2 



_i- _J (ll^sin^ft-i3.SJ*.sin9^-^2^78.sin7^-58.11.sin5^+2,3^Gl.sin3il4-23.^.sinA*) 



21^.3.5.7^ 



e 



.*6-3 



