A Carleman Function and the Cauchy Problem for the Laplace by Yarmukhamedov Sh.

By Yarmukhamedov Sh.

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Observe the vertical symmetry axis in the triangle, which stems from the obvious fact that n n k = n−k . 1 1 1 1 1 1 1 1 7 3 4 5 6 1 3 6 10 15 21 1 2 1 4 10 20 35 1 5 15 35 1 6 21 1 7 1 ... Fig. 9. The Pascal triangle. This yields the coefficients of (X + Y )n as follows: Proposition 255 If n ∈ N, then the polynomial (X + Y )n ∈ Z[X, Y ] has this representation in terms of monomials: n (X + Y )n = k=0 n X n−k · Y k = X n + nX n−1 · Y + . . nX · Y n−1 + Y n . k 26 Limits and Topology Proof One proves the proposition by induction on n using the recursive formula from lemma 254.

V1 , and clockwise walks u0 . . u0 in four points q, r , s, t, all different from u0 , u1 , v0 , v1 . v1 The proof now closes with an analysis of four cases of possible positions of the points q, r , s, t on the cycle Z, and where each case yields a subgraph isomorphic to K5 or K3,3 . This is a contradiction to the assumption that the original graph Γ (of which Φ is a subgraph) does not contain a subgraph isomorphic to K5 or K3,3 . The details of the proof are described in [12]. It goes back to Gabriel Andrew Dirac and Seymour Schuster, A theorem of Kuratowski.

To do so, one first shows that there is a cycle Z in Φ containing the points x, y defined above. One then makes a drawing of Φ such that there is a maximum of faces interior to the drawing of Z. One considers the components of the subgraph of Φ induced on the vertexes outside the drawing of Z and then defines outer pieces as those subgraphs of Φ which are either induced on outer components, plus the points on Z which they are connected to, or else which are outer edges of the drawing of Z connecting two points of Z.

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