Dr. Katz DEq Homework Solutions 52

Dr. Katz DEq Homework Solutions 52 - (3.5.3.7) we must show...

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52 N.M. Katz: A final substitution gives Lie (P)P-1 (f(1) a (2) - f(2) a (1)) - (h (1) a (2) - h (2) a (1)) (3.5.3.3) = g(1)Pa(2)+g(2)Pa(1)modulo dF(V,(_%). Availing ourselves of (3.5.2.3), (3.5.3.3) becomes Lie(P) ~- t(f(l) a (2)-f(2) a(t)) = W-t(f(t)} a(2)- p~-i (f(2)) a(l) (3.5.3.4) modulo dF(V, (9~). Now the right hand member of(3.5.3.4) is the "first term" in the expansion of the left hand member by Leibniz's rule, so that, expanding, (3.5.3.4) becomes (3.5.3.5) Z (Pkl)[pk(f(1))Lie(P)e(a(2)) k+~=p-l,8~:O _ pk (f(2)) Lie (p)e (or (1))] ~ dr(E r By (3.5.1.2), we triay substitute Lie (F) e (a (2)) = Lie (p)e- 1 d f(2) = dP t-~(f(2)) (3.5.3.6) Lie (p)e (a (1)) = Lie (p)e_ ~ df(1)=dpe_l(f(1)).
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Unformatted text preview: (3.5.3.7) we must show (Pk 1) ---(- 1)k mod p for Ogk<_p-1 ~, ( - t)k [p~f(1) dP e-: f(2) (3.5.3.8) k+e=p-l,e*0 -- Pk (f(2)) de e-' f(1)] e dr(V, (gv) Re-indexing the summation by k and re=Y-1, (3.5.3.8) becomes (re- membering that (- 1) k+l -(- !) m mod p) (- 1) k nkf(1) dn"f(2) (3.5.3.9) k+m=p-Z + ~ (- 1)"1Pk(f(2)) de'f(l)edF(V, (gv). k+m=p--2 This is the case; in fact, the left member of (3.5.3.9) is (3.5.3.10) d( E (_ 1)kpk(f(1)) P'(f(2))). k+m=p-2 This proves (3.5.0.3) in case b = 2, and gives a hint of the combinatorial rearrangements necessary in the general case. (3.5.4) We now turn to the general case. We adopt the convention that a product indexed by a subset of Z is to be taken in increasing order, and...
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This note was uploaded on 12/21/2011 for the course MAP 4341 taught by Professor Normankatz during the Fall '11 term at UNF.

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