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FUNCTIONAL STRENGTHS

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  • cours - matière potentielle : security
- 1 - JAMES MICHAEL WILLIAMSON - 503.784.1031 Jim Williamson leverages his knowledge in business practices with in-depth functional and technical abilities. An expert in Oracle Applications and other relational databases, Jim has been functional lead, technical lead or a combination therein on eight successful ERP/CRM implementations of Oracle Applications. Fifteen years experience with relational databases and ERP software and more than ten years experience with Oracle applications and development toolsets have rendered him a vital asset on Oracle projects of all sizes.
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Week 11 lectures, Math 804
1.More on Weirstrass Elliptic FunctionP(z): Integral formula Lemma 1.1.Introduce Z 31/2 (1.1)z=h(ζ(4) = tg2tg3)dt ζ where the path of integration may be any curve which does not pass 3 through a zero of4tg2tg3, thenh(P(z)) =z. Proof.On differentiating (1.1), we obtain  2 3 (1.2) = 4ζg2ζg3 dz so that (1.3)ζ=P(±z+α) =P(z±α) Now, in (1.1), if we letζ→ ∞, we obtainzmeans that0. This P(z+±α) blows up atzThis means that= 0. αis either 0 or a point congruent to it. Thusζ=P(z) and the Lemma is proved.
Lemma 1.2.The elliptic functionPsatisfies the following addition theorem:   0 P(z)P(z) 1 0   (1.4)P(y)P(y= 0) 1 0 P(z+y)−P(z+y) 1 Proof.Consider the following set of two linear equations for determin-ingAandB: 0 (1.5)P(z) =AP(z) +B
0 (1.6)P(y) =AP(y) +B These determineAandBuniquely in terms ofzandy, unlessP(z) =P(y), i.e. unlessz=±y(mod2ω1,2ω2). ForAandBas determined above, consider 0 (1.7)P(ζ)AP(ζ)B It has a triple pole atζ= 0 and consequently three irreducible zeros (i.e. three zeros within a cell). Two of these zeros are clearlyζ=z andζ=yand the third irreducible zero must be congruent tozy 1