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Taylor Series

# 4 6 x 2k 1k 2k k0 1 a similar calculation shows

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Unformatted text preview: =0 Hence ∞ cos(x) ∼ n=0 f (n) (0) n x n! 0x −1x2 0x3 1x4 + + + + ··· 1! 2! 3! 4! x2 x4 x6 = 1− + − + ··· 2! 4! 6! ∞ x 2k (−1)k = (2k )! k=0 = 1+ A similar calculation shows that x3 x5 x7 + − + ··· 3! 5! 7! ∞ x2k+1 = (−1)k (2k + 1)! k=0 sin(x) ∼ x − The major problem that remains is to determine if ∞ (−1)k cos(x) = k=0 and if ∞ (−1)k sin(x) = k=0 4.5.1 x 2k (2k )! x 2k + ? (2k +)! Introduction to Taylor Polynomials and Approximation Recall that if f (x) is diﬀerentiable at x = a, then if x ∼ a = f (x) − f (a) f (a) ∼ = x−a Cross multiplying gives DE Math 128 f (a)(x − a) ∼ f (x) − f (a) = 312 (B. Forrest)2 CHAPTER 4. Sequences and Series 4.5. Introduction to Taylor Series and ﬁnally that f (x) ∼ f (a) + f (a)(x − a) = This leads us to deﬁne the linear approximation to f (x) at x = a to be the function La (x) = f (a) + f (a)(x − a). The geometrical signiﬁcance of the linear approximation is that it graph is the tangent line to the graph of f (x) through the point (a, f (a)). The linear approximation has the following two important properties: 1. La (a) = f (a). 2. La (a) = f (a). In fact, amongst all polynomials of degree at most 1, that is function...
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