Abstract convex analysis by Singer I.

By Singer I.

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Example text

Moller - Maersk. 18 In this example we shall derive a formula which approximates the length of a circular arc. This formula is due to Huygens (1629–1695). We shall use the figure and the notation given in π the text. We shall assume that the angle ϕ satisfies ϕ ∈ 0, . 2 ϕ below. The arc is denoted by , Figure 11: Circle of radius r, centre angle ϕ, thus periphery angle 2 and the corresponding cord is denoted by d. Finally, we let s denote the height on the dotted vertical diagonal. The approximating expression ˜ of the length is given in the form ˜ = ad + bs, where a and b are constants which will be found below.

I. When f (−x) = −f (x) is odd we get by differentiation, dn f (−x) = (−1)n f (n) (x) = −f (n) (x). dxn By a rearrangement followed by putting x = 0 we get {(−1)n + 1}f (n) (0) = 0. If n = 2m is even, then (−1)2m + 1 = 2 = 0, and f (2m) (0) = 0. We conclude that the Taylor polynomial only contains powers of x of odd exponents. If instead f (−x) = f (x) is even, then we get by differentiation, dn f (−x) = (−1)n f (n) (−x) = f (n) (x). dxn If we put x = 0, we get by a rearrangement, {(−1)n − 1}f (n) (0) = 0.

Function. I. When f (−x) = −f (x) is odd we get by differentiation, dn f (−x) = (−1)n f (n) (x) = −f (n) (x). dxn By a rearrangement followed by putting x = 0 we get {(−1)n + 1}f (n) (0) = 0. If n = 2m is even, then (−1)2m + 1 = 2 = 0, and f (2m) (0) = 0. We conclude that the Taylor polynomial only contains powers of x of odd exponents. If instead f (−x) = f (x) is even, then we get by differentiation, dn f (−x) = (−1)n f (n) (−x) = f (n) (x). dxn If we put x = 0, we get by a rearrangement, {(−1)n − 1}f (n) (0) = 0.

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