TaylorSeries

# TaylorSeries - MTHSC 360 – Taylor’s Theorem 1...

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Unformatted text preview: MTHSC 360 – Taylor’s Theorem February 27, 2008 1 Taylor’s Theorem with Remainder Let f ( x ) be a function defined for x in an interval containing the point a . If the first derivative of f is also defined and continuous on this interval, then the Fundamental Theorem of Calculus tells us that f ( x )- f ( a ) = Z x a f ( t ) dt (1) Recall the formula for Integration by Parts, Z udv = uv- Z vdu which in more detail can be written as Z x a u ( t ) v ( t ) dt = u ( t ) v ( t ) | x a- Z x a u ( t ) v ( t ) dt If we let u ( t ) = f ( t ) and v ( t ) = 1, then u ( t ) = f 00 ( t ) and v ( t ) = t + C . Consequently. if the second derivative of f is also defined and continuous on the interval containing a , then we can use Integration by Parts, cleverly choosing C =- x (remember that the variable of integration is t ), to obtain Z x a f ( t ) dt = f ( t )( t- x ) | x a- Z x a f 00 ( t )( t- x ) dt (2) = f ( x )( x- x )- f ( a )( a- x ) + Z x a f 00 ( t )( x- t ) dt (3) = f ( a )( x- a ) + Z x a f 00 ( t )( x- t ) dt (4) Substituting Equation 4 into Equation 1 we get f ( x ) = f ( a ) + f ( a )( x- a ) + Z x a f 00 ( t )( x- t ) dt (5) This states that the function f ( x ) is equal to the tangent line to f ( x ) at the point x = a plus the remainder term, R 1 ( x ) = Z x a f 00 ( t )( x- t ) dt. It is important to note that this remainder term gives us the exact difference between the value of f ( x ) and the tangent line approximation, f ( a ) + f ( a )( x- a ). For example, if f ( x ) = e x , then by an elementary integration (using Integration by Parts) we can determine that R 1 ( x ) = Z x a f 00 ( t )( x- t ) dt = Z x a e x ( x- t ) dt = e x- [ e a + e a ( x- a )] If the third derivative of f is also defined and continuous on this interval, then we can use Integration by Parts yet again to get Z x a f 00 ( t )( x- t ) dt =- f 00 ( t ) ( x- t ) 2 2 x a + Z x a f 000 ( t ) ( x- t ) 2 2 dt (6) = f 00 ( a ) ( x- a ) 2 2 + Z x a f 000 ( t ) ( x- t ) 2 2 dt (7) Substituting Equation 7 into Equation 5 we get f ( x ) = f ( a ) + f ( a )( x- a ) + f 00 ( a ) ( x- a ) 2 2 + Z x a f 000 ( t ) ( x- t ) 2 2 dt (8) This shows that f ( x ) is equal to the osculating quadratic, which “kisses” f ( x ) at x = a plus the remainder term, R 2 ( x ) = Z x a f 000 ( t ) ( x- t ) 2 2 dt. We can continue this process as far as we want provided that f ( x ) has as many continuous derivatives as needed. Taylor’s Theorem with Remainder. If f ( x ) has a continuous ( n + 1)-st derivative on an interval containing x and a , then f ( x ) = f ( a ) + f ( a )( x- a ) + f 00 ( a ) ( x- a ) 2 2 + ··· + f ( n ) ( a ) ( x- a ) n n ! + Z x a f ( n +1) ( t ) ( x- t ) n n ! dt Since ( x- t ) n is nonnegative for a ≤ t ≤ x we can apply the Integral Mean Value Theorem to conclude that there exists a point ξ between a and x such that R n ( x ) = Z x a f ( n +1) ( t ) ( x- t ) n n !...
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TaylorSeries - MTHSC 360 – Taylor’s Theorem 1...

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