By Anthony Tromba

One of the main straightforward questions in arithmetic is whether or not a space minimizing floor spanning a contour in 3 area is immersed or now not; i.e. does its by-product have maximal rank far and wide.

The function of this monograph is to provide an uncomplicated facts of this very primary and lovely mathematical end result. The exposition follows the unique line of assault initiated by way of Jesse Douglas in his Fields medal paintings in 1931, specifically use Dirichlet's strength in preference to sector. Remarkably, the writer indicates how you can calculate arbitrarily excessive orders of derivatives of Dirichlet's strength outlined at the endless dimensional manifold of all surfaces spanning a contour, breaking new flooring within the Calculus of adaptations, the place commonly in basic terms the second one by-product or edition is calculated.

The monograph starts with effortless examples resulting in an explanation in numerous circumstances that may be provided in a graduate path in both manifolds or advanced research. therefore this monograph calls for basically the main easy wisdom of research, complicated research and topology and will for that reason be learn via nearly a person with a uncomplicated graduate education.

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3 If 2m + 2 = L(n + 1), L odd and ≥ 5, then by the choice τ = β cw −n−1 + cwn+1 , c ∈ C, and by suitable choices of Dt φ(0) for β ≥ 1 we obtain (j ) E (0) = 0 for j = 1, . . , L − 1 and E (L) (0) = 2(L − 1)! M! 17) with κ := i L−1 (m − n)2 (m − 2n − 1)2 (m − 3n − 2)2 . . (m − Mn − (M − 1))2 . Since m + 1 = 12 L(n + 1) = M(n + 1) + 12 (n + 1), we have 1 m − Mn − (M − 1) = m + 1 − M(n + 1) = (n + 1) > 0, 2 and therefore κ = 0. Hence 2 · (L − 1)! M! for a suitable choice of c. 2 in the case 2m + 2 = L(n + 1).

Let us now turn to the investigation of (C5 ) by means of the fifth derivative E (5) (0). 4 If f (w) := w Zˆ tw (0)τ + w Xˆ w φt (0) is holomorphic, then Zˆ ttw (0) = {iw[iw Xˆ w τ ]w τ + iw Xˆ w φt (0)}w , Zˆ ttw (0) · Xˆ w = −Zˆ tw (0) · Zˆ tw (0) = w 2 Xˆ ww · Xˆ ww τ 2 . 51) 28 2 Higher Order Derivatives of Dirichlet’s Energy and therefore Zˆ ttw (0) = {2H [Re(if )]}w = {if }w = {iw Zˆ tw (0)τ + iw Xˆ w φt (0)}w . 35), Zˆ tw (0) = [iw Xˆ w τ ]w , and so Zˆ ttw (0) = {iw[iw Xˆ w τ ]w τ + iw Xˆ w φt (0)}w .

M − Mn − (M − 1))2 . Since m + 1 = 12 L(n + 1) = M(n + 1) + 12 (n + 1), we have 1 m − Mn − (M − 1) = m + 1 − M(n + 1) = (n + 1) > 0, 2 and therefore κ = 0. Hence 2 · (L − 1)! M! for a suitable choice of c. 2 in the case 2m + 2 = L(n + 1). 17) tells us that for the case 2m+2 = L(n+1) there is a Diophantine polynomial pL (x, y) such that κ = i L−1 pL (n, m), and that pL (n, m) = 0. Note that pL only depends on L, m and n and not on the specific minimal surface. 2 When we treated the case n odd, m even in the preceding section, we omitted the special case n = 1.