Chapter 4(D) - Differentiation Integration If cn x a n has...

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Differentiation & Integration 0 ( ) ( ) , . n n n f x cx a ahxah = = - -<<+ If ( ) has radius of convergence , it defines a function : n n a hf - The differentiated series converges for . ahxah (i) The function has derivatives of all orders in (, ). The derivatives can be obtained by differentiating the power series term-by-term: f a h ah -+ 1 1 ' ( ) () n n n f x n cxa - = =- 2 2 '' ( ) ( 1 ) n n n f x n n cxa - = = -- ( 29 1 ( ) nn d a n dx - -=-
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Example 2 1 ( ) 1 , 11 1 n f x x x xx x = =++++ + -<< - LL Differentiating 2 3 2 '' ( ) 2 6 ( 1 ) 1 1 ( 1) n f x x n n x - = =+++ - + - 21 2 1 ' ( ) 1 2 3 1 ( n f x x x n x - = = + + + + -
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Differentiation & Integration (ii) The function has anti-derivatives in (, ). The anti-derivatives can be obtained by integrating the power series term-by-term: f a h ah -+ 1 0 () ( ) 1 n n n xa fx d x cC n + = - =+ + The integrated series converges for . ahxah -<<+ 1 () 1 n n nn cxa d x n + - - +
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Geometric Series 23 1 1 , 11 1 r r rt r =+++ + -<< - L Put 1 , 1 1() t t tt =- = =-+- + + -- L
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Example 23 11 1 , 1 1() t t tt = =-+- + -<< + -- L 234 1 1 1 1 1 1 ln( 1) t t d t t t t dt t ttt =-+-+ + = -+-+ + + = -+-+ ∫∫ L L L ( 29 1 ln( 1 d t d += + Is the answer correct ?? Is the working correct ??
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Example 23 234 1 1 1 1 1 1 ln( 1) Need to find . t tt t d t t t t dt t ttt t tC C =-+-+ + = -+-+ + + = -+-++ ∫∫ L L L We put 0: ln( 1 0 )0000 0 t C C = + =-+-++ = L ln( 1 ) -+-+ L
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Example 234 ln( 1 ) ln( 1 0) ( 000 0) ln( 1) xxx xx +- + =-+- +- -+-+ + =-+-+ L L L 23 11 1 , 1 1() t t tt = =-+- + -<< + -- L [ ] 00 0 0 1 1 1 ln( x x d t t t t dt t ttt = +  + =-+-+   ∫∫ L L
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Taylor Series
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Taylor Series = - 0 ) ( ) ( ! ) ( k k k a x k a f Taylor (1685-1731) Maclaurin (1698-1746) = 0 ) ( ) ( ! ) 0 ( n n n x n f
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Taylor Series - Definition Let be a function with derivatives of all orders throughout some interval containing as an interior point. f a () 0 The of at is ( ) ' ( ) ( ) ! ( ! k k k n n fa x a f a faxa k xa n = - = + -++ -+ L L Taylor series We use Taylor series to expand a function into power series.
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() 0 The of at is ( ) ' ( ) ( ) ! ( ! k k k n n fa x a f a faxa k xa n = - = + -++ -+ L L Taylor series We use Taylor series to expand a function into power series. For a given function ( ), to find the at , you just need to find the values of: ( ), ' ( ), '' ( ), , ( and then substitute them into the Taylor series formula.
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Chapter 4(D) - Differentiation Integration If cn x a n has...

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