Only for strictly positive numbers x 0 the expression

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only for strictly positive numbers x > 0. The expression for Taylor’s series given above may be described as the expansion of f ( x + h ) about the point x . It is also common to expand a function f ( x ) about the point x = 0. The resulting series is described as Maclaurin’s series: f ( x ) = f (0) + xf (0) + x 2 2! f (0) + · · · + x n n ! f ( n ) (0) + · · · . We shall give a number of examples of such expansions; all of which may be memorised profitably. D.S.G. Pollock: stephen [email protected]
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TAYLOR’S THEOREM Example 1. log(1 + x ) f ( x ) = log(1 + x ) f ( x ) = 1 1 + x f ( x ) = 1 (1 + x ) 2 f ( x ) = 2 (1 + x ) 3 f (4) ( x ) = 2 . 3 (1 + x ) 4 f (0) = 0 f (0) = 1 f (0) = 1 f (0) = 2 f (4) (0) = 6 It follows that, for | x | < 1—which is the necessary and sufficient condition for the convergence of the series—we have log(1 + x ) = x x 2 2 + x 3 3 x 4 4 + · · · . Example 2. e x f ( x ) = e x f ( n ) ( x ) = e x f (0) = 1 f ( n ) (0) = 1 It follows that, for | x | < 1, we have e x = 1 + x + x 2 2 + x 3 3! + · · · + x n n ! + · · · . Example 3. sin x f ( x ) = sin x f ( x ) = cos x f ( x ) = sin x f ( x ) = cos x f (4) ( x ) = sin x f (5) ( x ) = cos x f (0) = 0 f (0) = 1 f (0) = 0 f (0) = 1 f (4) (0) = 0 f (5) (0) = 1 It follows that, for | x | < 1, we have sin x = x x 3 3! + x 5 5! x 7 7! + · · · . D.S.G. Pollock: stephen [email protected]
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EC3070 FINANCIAL DERIATIVES Example 4. cos x f ( x ) = cos x f ( x ) = sin x f ( x ) = cos x f ( x ) = sin x f (4) ( x ) = cos x f (0) = 1 f (0) = 0 f (0) = 1 f (0) = 0 f (4) (0) = 1 It follows that, for | x | < 1, we have cos x = 1 x 2 2! + x 4 4! x 6 6! + · · · . Example 5. 1 / (1 x ) f ( x ) = (1 x ) 1 f ( n ) ( x ) = n !(1 x ) n f (0) = 1 f ( n ) (0) = n !
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  • Spring '12
  • D.S.G.Pollock
  • Taylor Series, Taylor’s Theorem, D.S.G. POLLOCK, stephen pollock

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