121616949-math.282

# 121616949-math.282 - F t except the ﬁrst term and the...

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268 Chapter 10 Sequences and Series Here we have replaced a by t in the first N + 1 terms of the Taylor series, and added a carefully chosen term on the end, with B to be determined. Note that we are temporarily keeping x fixed, so the only variable in this equation is t , and we will be interested only in t between a and x . Now substitute t = a : F ( a ) = N X n =0 f ( n ) ( a ) n ! ( x - a ) n + B ( x - a ) N +1 . Set this equal to f ( x ): f ( x ) = N X n =0 f ( n ) ( a ) n ! ( x - a ) n + B ( x - a ) N +1 . Since x = a , we can solve this for B , which is a “constant”—it depends on x and a but those are temporarily fixed. Now we have defined a function F ( t ) with the property that F ( a ) = f ( x ). Consider also F ( x ): all terms with a positive power of ( x - t ) become zero when we substitute x for t , so we are left with F ( x ) = f (0) ( x ) / 0! = f ( x ). So F ( t ) is a function with the same value on the endpoints of the interval [ a, x ]. By Rolle’s theorem ( 6.25 ), we know that there is a value z ( a, x ) such that F ( z ) = 0. Let’s look at
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Unformatted text preview: F ( t ), except the ﬁrst term and the extra term involving B , is a product, so to take the derivative we use the product rule on each of these terms. It will help to write out the ﬁrst few terms of the deﬁnition: F ( t ) = f ( t ) + f (1) ( t ) 1! ( x-t ) 1 + f (2) ( t ) 2! ( x-t ) 2 + f (3) ( t ) 3! ( x-t ) 3 + ··· + f ( N ) ( t ) N ! ( x-t ) N + B ( x-t ) N +1 . Now take the derivative: F ± ( t ) = f ± ( t ) + ± f (1) ( t ) 1! ( x-t ) (-1) + f (2) ( t ) 1! ( x-t ) 1 ¶ + ± f (2) ( t ) 1! ( x-t ) 1 (-1) + f (3) ( t ) 2! ( x-t ) 2 ¶ + ± f (3) ( t ) 2! ( x-t ) 2 (-1) + f (4) ( t ) 3! ( x-t ) 3 ¶ + . . . + + ± f ( N ) ( t ) ( N-1)! ( x-t ) N-1 (-1) + f ( N +1) ( t ) N ! ( x-t ) N ¶ + B ( N + 1)( x-t ) N (-1) ....
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