Lesson 17.pdf - 17 Taylor’s Theorem August 1 2018 If f(x = a0 a1 x a2 x2 we find Taylor theorem a0 = f(0 a1 = f 0(0 a2 = f 00(0 2 In general for a

Lesson 17.pdf - 17 Taylor’s Theorem August 1 2018 If f(x...

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17. Taylor’s Theorem August 1, 2018 Taylor theorem If f ( x ) = a 0 + a 1 x + a 2 x 2 , we find a 0 = f (0) , a 1 = f 0 (0) , a 2 = f 00 (0) 2! . In general, for a polynomial f ( x ) = a 0 + a 1 x + a 2 x 2 + ... + a n x n , we find a 0 = f (0) , a 1 = f 0 (0) , a 2 = f 00 (0) 2! , ..., a n = f ( n ) (0) n ! . (1) J.L. Lagrange once thought that every continuous function f ( x ) near x = a can be written as a power series n =0 a n ( x - a ) n . We now know that this is true only for functions with good differentiable condition 1 ; in this case, a function f ( x ) equals to its Taylor series : f ( x ) = f ( a ) + f 0 ( a )( x - a ) + f 00 ( a ) 2! ( x - a ) 2 + ... + f ( n ) ( a ) n ! ( x - a ) n + ...... When a = 0, we get f ( x ) = f (0) + f 0 (0) x + f 00 (0) 2! x 2 + ... + f ( n ) (0) n ! x n + ... (2) By comparing (1), these coefficients f ( n ) (0) n ! are expected. Here is one of the theorems in this theory. 1 The condition is: any order of derivative f ( n ) exists, and the corresponding radius of convergence of the Taylor series is positive. 143
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Theorem 0.1 (Taylor theorem) Let f and its first n derivatives f 0 , f 00 , f 000 , ..., f ( n ) be con- tinuous on [ a, b ] and differentiable up to order n +1 on ( a, b ) . Let x 0 ( a, b ) . Then for each x ( a, b ) , x 6 = x 0 , there is a number c between x 0 and x such that f ( x ) = f ( x 0 )+ f 0 ( x 0 )( x - x 0 )+ f 00 ( x 0 ) 2! ( x - x 0 ) 2 + ... + f ( n ) ( x 0 ) n ! ( x - x 0 ) n + f ( n +1) ( c ) ( n + 1)! ( x - x 0 ) n +1 . Proof: Fix x [ a, b ] with x 6 = x 0 . Let M be the unique solution of the equation 2 f ( x ) = f ( x 0 ) + f 0 ( x 0 )( x - x 0 ) + f 00 ( x 0 ) 2! ( x - x 0 ) 2 + ... + f ( n ) ( x 0 ) n ! ( x - x 0 ) n + M ( x - x 0 ) n +1 . We want to show that there exists c between x 0 and x such that M = f ( n +1) ( c ) ( n +1)! . Let us define F ( t ) := f ( t ) + f 0 ( t )( x - t ) + ... + f ( n ) ( t ) n ! ( x - t ) n + M ( x - t ) n +1 . We observe that F ( x ) = f ( x ) and F ( x 0 ) = f ( x ). Also, F is continuous on [ a, b ] and differentiable on ( a, b ). Then by Roll’s theorem, there exists c between x and x 0 such that F 0 ( c ) = 0 . By the definition of F , we calculate F 0 ( t ) = f 0 ( t ) + [ f 00 ( t )( x - t ) - f 0 ( t )] + ... + f ( n +1) ( t ) n ! ( x - t ) n - f ( n ) ( t ) n ! n ( x - t ) n - 1 + M ( n + 1)( x - t ) n , and hence F 0 ( c ) = f ( n +1) ( c ) n ! ( x - c ) n - M ( n + 1)( x - c ) n . Then M = f ( n +1) ( c ) ( n +1)! as desired. Remarks: 1. The polynomial p n ( x ) = f ( x 0 ) + f 0 ( x 0 )( x - x 0 ) + ...... + f ( n ) ( x 0 ) n ! ( x - x 0 ) n is called the Taylor polynomial for f of degree n in powers of x - x 0 . 2 If a = b + cx with c 6 = 0, then x = a - b c is the unique solution. 144
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2. The power series f ( x 0 ) + f 0 ( x 0 )( x - x 0 ) + ...... + f ( n ) ( x 0 ) n ! ( x - x 0 ) n + ... is called the Taylor series for f in powers of x - x 0 .
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