Unformatted text preview: * ) agree at α 1 ,...,α n . Indeed, this would say that the diﬀerence of the two sides is a polynomial of degree less than n having n roots, which would force it to be the zero polynomial. By the product rule for derivatives, we have Q ( α k ) = Y j 6 = k ( α k-α j ) and by direct calculation ± Q ( z ) ( z-α k ) ² z = α ` = δ jk . Now we evaluate each side of ( * ) at α j to obtain: RHS : n X k =1 P ( α k ) δ jk = P ( α j ) LHS : P ( α j ) Hence the RHS and LHS of ( * ) agree at each α j and the proof is complete. ± 1...
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This note was uploaded on 10/19/2009 for the course MATH 814 taught by Professor Cong during the Three '09 term at University of Adelaide.
- Three '09