TaylorSeries

TaylorSeries - MTHSC 360 Taylors Theorem February 27, 2008...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MTHSC 360 Taylors Theorem February 27, 2008 1 Taylors 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. Taylors 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 between a and x such that R n ( x ) = Z x a f ( n +1) ( t ) ( x- t ) n n !...
View Full Document

Page1 / 9

TaylorSeries - MTHSC 360 Taylors Theorem February 27, 2008...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online