By Einar Hille

ISBN-10: 0828402698

ISBN-13: 9780828402699

**Read or Download Analytic Function Theory, Volume I - 2nd Edition (AMS Chelsea Publishing) PDF**

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**Example text**

3) has been proved for m = 1 and for a function h, such that h(a) = 0. We must prove it also for a function h, which is not equal to zero at the point a. Letting h = Jf , and w = f (z), we have Z Jf dz Z dw = 0: = ;f fp f1 : : : fn ;w wp w1 : : : wn (See Sec. 4) k=1 k j =1 with n X rk Bk = @a @g : r=1 From here we see that n Z a det A det @ f dg X ip @g = j =1 ;g = n Z X j =1 r (Here we used the equality gj g1 : : : gn ! n Z ajpdg ; X ajpBk fk dg : ;g gj g1 : : : gn k=1 ;g gj g1 : : : gn det @@gf dg1 ^ : : : ^ dgn = Jg dz and the condition that fj (a) = 0.

The resultant of the polynomials P and Q is equal to the determinant a0 0 ::: D = 0b 0 0 ::: 0 a1 a0 ::: 0 b1 b0 ::: 0 9 > : : : an 0 : : : 0 > = : : : an;1 an : : : 0 s lines ::: ::: ::: ::: ::: > > : : : a0 a1 : : : an 9 : : : : bs 0 : : : 0 > = : : : bs;1 bs : : : 0 > : : : : : : : : : : : : : : : >n lines : : : b0 b1 : : : bs > Proof. 2) On the other hand we can calculate the product DM of the determinants D and M by using the formula for the determinant of a product of matrices. Multiplying the matrices corresponding to our determinants, and taking into account that the j are zeros of P and the k are zeros of Q, we get s;1 s;1 s;1 ::: 0 1 P ( 1 ) 2 P ( 2 ) : : : s P ( s) 0 ::: ::: ::: ::: ::: ::: ::: : : : s P ( s) 0 ::: 0 1P ( 1) 2P ( 2) P ( ) P ( ) : : : P ( ) 0 ::: 0 2 s = DM = 0 1 n ; 1 n;1 0 ::: 0 1 Q( 1 ) : : : n Q( n ) n;2 n;2 0 0 ::: 0 1 Q( 1 ) : : : n Q( n ) ::: ::: ::: ::: ::: ::: ::: 0 0 ::: 0 Q( 1) : : : Q( n) s;1 s;1 : : : s;1 n;1 n;1 : : : n;1 1 2 s 1 2 n s n Y Y : : : : : : : : : : : : : : : : : : : : : : : : = = P ( j ) Q( k ) : : : : : : 1 2 s 1 2 n j =1 k=1 1 1 ::: 1 1 1 ::: 1 26 = Ys j =1 P ( j) Yn k=1 Q( k ) Y Y k

8). To this new system we apply the same method and so on. After iterating this procedure s times we are either left with the zero system, or a system of constants which are not all equal to zero. 8) had no common roots. 10) R1 = 0 : : : Rk = 0 where R1 : : : Rk are polynomials in zs : : : zn and the following step (with respect to zs ) produces the zero system. We can then assign to the variables zs+1 : : : zn arbitrary values s+1 : : : n. 8). 10) may contain super uous equations. 8) to triangle form without unnecessary equations.

### Analytic Function Theory, Volume I - 2nd Edition (AMS Chelsea Publishing) by Einar Hille

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