Sentence Examples


  • F = aox 2 -alx+a2 =0, �=box2 -blx+b2.
  • Thus to obtain the resultant of aox 3 +a i x 2 +a 2 x+a 3, 4, =box2+bix+b2 we assume the identity (Box+Bi)(aox 3 +aix 2 +a2x+a3) = (Aox 2 +Aix+ A 2) (box2+bix+b2), and derive the linear equations Boa ° - Ac b o = 0, Boa t +B i ao - A 0 b 1 - A 1 bo =0, Boa t +B 1 a 1 - A0b2 - A1b1-A2b° = 0, Boa3+Bla2 - A l b 2 -A 2 b 1 =0, B 1 a 3 - A 2 b 2 =0, = = (y l, y2,...ynl `x1, x2,...xnl for brevity.
  • Thus to obtain the resultant of aox 3 +a i x 2 +a 2 x+a 3, 4, =box2+bix+b2 we assume the identity (Box+Bi)(aox 3 +aix 2 +a2x+a3) = (Aox 2 +Aix+ A 2) (box2+bix+b2), and derive the linear equations Boa ° - Ac b o = 0, Boa t +B i ao - A 0 b 1 - A 1 bo =0, Boa t +B 1 a 1 - A0b2 - A1b1-A2b° = 0, Boa3+Bla2 - A l b 2 -A 2 b 1 =0, B 1 a 3 - A 2 b 2 =0, = = (y l, y2,...ynl `x1, x2,...xnl for brevity.
  • Taking the same example as before the process leads to the system of equations acx 4 +alx 3 +a2x 2 +a3x =0, aox 3 +a1x 2 +a2x+a 3 = 0, box +bix -1-b2x =0, box' +b i x 2 -{-h 2 x = 0, box + b i x + b:: = 0, whence by elimination the resultant a 0 a 1 a 2 a 3 0 0 a 0 a 1 a 2 a3 bo b 1 b 2 0 0 0 bo b 1 b 2000 bo b 1 b2 which reads by columns as the former determinant reads by rows, and is therefore identical with the former.
  • Put (aox 3 -}-a l x 2 +a 2 x +a 3) (box' +b1x'+b2) - (aox'3+aix'2+a2x'+a3) (box' + bix + b2) = 0; after division by x-x the three equations are formed aobcx 2 = aobix+aob2 =0, aobix 2 + (aob2+a1b1-a2bo) x +alb2 -a3bo = 0, aob2x 2 +(a02-a3bo)x+a2b2-a3b1 =0 and thence the resultant aobo ao aob2 aob 1 aob2+a1b1-a2bo alb2-a3b0 aob 2 a1b2 - a 3 bo a2b2 - a3b1 which is a symmetrical determinant.
 

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