Chap11_Sec10 - 11 INFINITE SEQUENCES AND SERIES INFINITE...

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11 INFINITE SEQUENCES AND SERIES INFINITE SEQUENCES AND SERIES
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In section 11.9, we were able to find power series representations for a certain restricted class of functions. INFINITE SEQUENCES AND SERIES
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Here, we investigate more general problems. Which functions have power series representations? How can we find such representations? INFINITE SEQUENCES AND SERIES
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11.10 Taylor and Maclaurin Series INFINITE SEQUENCES AND SERIES In this section, we will learn: How to find the Taylor and Maclaurin Series of a function and to multiply and divide a power series.
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TAYLOR & MACLAURIN SERIES We start by supposing that f is any function that can be represented by a power series 2 0 1 2 3 4 3 4 ( ) ( ) ( ) ( ) ( ) ... | | f x c c x a c x a c x a c x a x a R = + - + - + - + - + - < Equation 1
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TAYLOR & MACLAURIN SERIES Let’s try to determine what the coefficients c n must be in terms of f . To begin, notice that, if we put x = a in Equation 1, then all terms after the first one are 0 and we get: f ( a ) = c 0
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TAYLOR & MACLAURIN SERIES By Theorem 2 in Section 11.9, we can differentiate the series in Equation 1 term by term: 2 1 2 3 3 4 '( ) 2 ( ) 3 ( ) 4 ( ) ... | | f x c c x a c x a c x a x a R = + - + - + - + - < Equation 2
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TAYLOR & MACLAURIN SERIES Substitution of x = a in Equation 2 gives: f’ ( a ) = c 1
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TAYLOR & MACLAURIN SERIES Now, we differentiate both sides of Equation 2 and obtain: 2 3 2 4 ''( ) 2 2 3 ( ) 3 4 ( ) ... | | f x c c x a c x a x a R = + × - + × - + - < Equation 3
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TAYLOR & MACLAURIN SERIES Again, we put x = a in Equation 3. The result is: f’’ ( a ) = 2 c 2
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TAYLOR & MACLAURIN SERIES Let’s apply the procedure one more time.
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TAYLOR & MACLAURIN SERIES Differentiation of the series in Equation 3 gives: 3 4 2 5 '''( ) 2 3 2 3 4 ( ) 3 4 5 ( ) ... | | f x c c x a c x a x a R = × + × × - + × × - - < Equation 4
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TAYLOR & MACLAURIN SERIES Then, substitution of x = a in Equation 4 gives: f’’’ ( a ) = 2 · 3 c 3 = 3! c 3
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TAYLOR & MACLAURIN SERIES By now, you can see the pattern. If we continue to differentiate and substitute x = a , we obtain: ( ) ( ) 2 3 4 ! n n n f a nc n c = × × ××××× =
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TAYLOR & MACLAURIN SERIES Solving the equation for the n th coefficient c n, we get: ( ) ( ) ! n n f a c n =
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TAYLOR & MACLAURIN SERIES The formula remains valid even for n = 0 if we adopt the conventions that 0! = 1 and f (0) = ( f ). Thus, we have proved the following theorem.
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TAYLOR & MACLAURIN SERIES If f has a power series representation (expansion) at a , that is, if then its coefficients are given by: Theorem 5 0 ( ) ( ) | | n n n f x c x a x a R = = - - < ( ) ( ) ! n n f a c n =
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TAYLOR & MACLAURIN SERIES Substituting this formula for c n back into the series, we see that if f has a power series expansion at a , then it must be of the following form. Equation 6
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TAYLOR & MACLAURIN SERIES Equation 6 ( ) 0 2 3 ( ) ( ) ( ) ! '( ) ''( ) ( ) ( ) ( ) 1! 2! '''( ) ( ) 3! n n n f a f x x a n f a f a f a x a x a f a x a = = - = + - + - + - +×××
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