Taylor and Maclaurin Series

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Presentation transcript:

Taylor and Maclaurin Series CHAPTER 2 2.4 Continuity Theorem: If f has a power series representation (expansion) at a, that is, if f(x) =  n=0 cn (x – a)n |x – a| < R Then its coefficients are given by the formula cn = ( f (n)(a) ) / n!. f(x) =  n=0 [ ( f (n)(a) ) / n! ] (x – a)n = f (a) + [ f’(a) / 1!] (x – a) + [ f’’(a) / 2!] (x – a)2 + … f(x) =  n=0 [ ( f (n)) (0) ) / n! ] (x – a)n = f (0) + [ f’(0) / 1!] x + [ f’’(0) / 2!] x2 + …

CHAPTER 2 Theorem: If f (x) = Tn (x) + Rn (x), where Tn is the nth-degree Taylor polynomial of f at a and lim n= Rn (x) = 0 for | x – a | < R, then f is equal to the sum of its Taylor series on the interval | x – a | < R. 2.4 Continuity Taylor’s Inequality: If | f (n+1)(x)|  M for | x – a | < R, then the remainder Rn (x) of the Taylor series satisfies the inequality | Rn (x) |  M | x – a | n+ 1 / (n + 1)! for | x – a | < R.

lim n-> x n / n ! = 0 for every real number x. CHAPTER 2 lim n-> x n / n ! = 0 for every real number x. 2.4 Continuity e x =  n=0 x n / n ! for all x. e =  n=0 1/n ! = 1 + 1 / 1! + 1 / 2! + 1 / 3! +… sin x = x – x3 / 3! + x5 / 5! + x7 / 7! + … =  n=0 (-1) n ( x2n+ 1 / (2n + 1)! ) for all x. cos x = 1 – x2 / 2! + x4 / 4! + x6 / 6! + … =  n=0 (-1)n x2n / (2n)! for all x.

CHAPTER 2 1 / (1 – x ) =  n=0 xn = 1 + x + x2 + x3 + … ( -1 ,1 ) e x =  n=0 x n / n ! = 1 + x + x2 / 2! + x3/ 3!+ … (-  , ) sin x =  n=0 (-1) n x 2n+ 1 / (2n + 1)! = x – x3/ 3! + x5 / 5! + x7 / 7! + … (-  , ) cos x =  n=0 (-1) n x 2n / (2n)! = 1 – x2 / 2! + x4 / 4! + x6 / 6! + … (-  , ) tan x =  n=0 (-1) n x2n+ 1 / (2n + 1) = x – x3/ 3! + x5 / 5!+ x7 / 7! + … [- 1 , 1] 2.4 Continuity