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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

This document was uploaded on 03/23/2013.

Ask a homework question - tutors are online