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Lecture36-rev

# Lecture36-rev - MAT A26 Lecture 36 1 Lagrange Form of the...

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MAT A26 Lecture 36 1 Lagrange Form of the Remainder We have already established that if f ( x ) is n + 1 times differentiable on an interval ( a - h, a + h ) then for x ( a - h, a + h ) we have f ( x ) = f ( a ) + f ( a )( x - a ) + · · · + f ( n ) ( a ) ( x - a ) n n ! f n,a ( x ) + x a f ( n +1) ( t ) ( x - t ) n n ! dt E n,a ( x ) Theorem 1 (Lagrange form of the remainder) If f ( x ) is n + 1 times differ- entiable on an interval ( c, d ) and f ( n +1) ( x ) is continuous on ( c, d ) , then, for a and x in ( c, d ) we have E n,a ( x ) = f ( n +1) ( ξ ) ( x - a ) n +1 ( n + 1)! for some ξ between a and x so that f ( x ) = f ( a ) + f ( a )( x - a ) + · · · + f ( n ) ( a ) ( x - a ) n n ! + f ( n +1) ( ξ ) ( x - a ) n +1 ( n + 1)! Proof: Let us suppose that a < x . (The case x < a is similar.) Since f ( n +1) is continuous on [ a, x ] we let m denote the minimum and M the max- imum values there. (These exist by EVT) Thus m f ( n +1) ( t ) M for all t [ a, x ] and so x a m ( x - t ) n n ! dt x a f ( n +1) ( t ) ( x - t ) n n ! dt x a M ( x - t ) n n ! dt But x a ( x - t ) n n ! dt = - ( x - t ) n +1 ( n + 1)! x a = ( x - a ) n +1 ( n + 1)! so m ( x - a ) n +1 ( n + 1)! x a f ( n +1) ( t ) ( x - t ) n n ! dt M ( x - a ) n +1 ( n + 1)! 1

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Thus 1 ( x - a ) n +1 ( n +1)! x a f ( n +1) ( t ) ( x - t ) n n ! dt [ m, M ] So by IVT this is f ( n +1) ( ξ ) for some ξ ( a, x ) 2 Properties of Taylor Polynomials Let f and g be functions. Then 1. ( αf + βg ) n,a = αf n,a + βg n,a where α and β are constant (“The Taylor polynomial of a linear combination is the linear combination of the Taylor polynomials.”)
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