A General Type of Singular Point by Hille E. PDF

By Hille E.

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Since the arcs RW and QV are parallel, ∠DF E = ∠DM H and ∠DEF = ∠DHM , so that triangle DM H is similar to triangle DF E. Since 2DH = DE, it follows that arc M N = 2 arc M H = arc F E. Now, arc QV = (3/4) arc RW = 3 arc F E and arc QM = arc N V . Therefore, the arc QV is trisected by M and N , and so the construction is valid. Second proof. Since arc RW = 2 arc P U , arc P D = arc RF . Therefore, F D is parallel to RP , and so arc QM = arc RF . Similarly, arc N V = arc GW = arc RF . Since arc QV = 3 arc RF , QV is trisected by M and N.

Since (a + b + c) 2 ≥ 0, it follows that 1 + 2(ab + bc + ca) ≥ 0, so ab + bc + ca ≥ − 12 . The lower bound should have been further restricted to − 12 . But shouldn't the CauchySchwarz inequality always be right? " What would you tell the student? 14. Surprising symmetry David Wells, in his book You are a mathematician (John Wiley & Sons, 1995, p. 88), makes the interesting observation that the nonsymmetric condition a = b + c leads to the symmetric result a4 + b4 + c4 = 2b2 c2 + 2c2 a2 + 2a2 b2 .

Denoting the right side by f(t), the prover wants the result that, if p(t) = f(t) everywhere, then p(t) and f(t) must have the same degree. This seems to depend on knowing that corresponding coefficients of p(t) and f(t) agree, or that, because p(t) − f(t) vanishes everywhere, its coefficients are all zero. But this is what has to be established. ♣ U: Hmm. You make a telling point. But we should look more closely. Do you agree that we should distinguish two types of equality between polynomials?

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