8.7. Taylor and Maclaurin Series http://www.ck12.org
8.7 Taylor and Maclaurin Series
Taylor and Maclaurin Polynomials
We know the linear approximation functionL 1 (x)to a (smooth) functionf(x)atx=x 0 is given by the tangent line
at the point.L 1 (x 0 ) =f(x 0 )andL′ 1 (x 0 ) =f′(x 0 ). Indeed, this is the only linear function with these 2 properties.
Theorem(nth−degree Taylor polynomial) Given a function f with continuousnthderivative in an open interval
containingx∗ 0. There exists uniquenth−degree polynomialp(x)withp(j)(x 0 ) =f(j)(x 0 ), for 0≤j≤n.
?: the functions in this text have continuous derivatives at the centerx 0 unless otherwise stated.
This polynomial
Tn(x) =f(x 0 )+f′(x 0 )(x−x 0 )+f(2!x^0 )(x−x 0 )^2 +...+f(n)(nx!−^0 )(x−x 0 )n)is called thenth−degree Taylor polynomial of
fatx 0. Ifx 0 =0, it is called thenth−degree Maclaurin polynomial offand denoted byMn(x).Rn(x) =f(x)−Tn(x)
is the remainder of the Taylor polynomial.
Example 1Let f(x) =x 63 ,x 0 =1. Thenf′(x) =x 22 and f(x) =x. So f′( 1 ) =^12 andf( 1 ) =1. HenceT 1 (x) =
(^16) + (^12) (x− 1 ),T 2 (x) = (^16) + (^12) (x− 1 )+(x− 1 ) (^2) , andT 3 (x) =f(x)itself.
Example 2Letf(x) =sinx,x 0 =0 and taken=3. Thenf(x) =cosx, f′′(x) =−sinx,f′′′(x) =−cosx. So
f′( 0 ) = 1 ,f′′( 0 ) = 0 ,f′′′( 0 ) =−1.M 3 (x) =x−3!^1 x^3 =x−^16 x^3 is the third-degree Maclaurin polynomial off.
Example 3Find the second-degree Taylor polynomial off(x) =tanxatx 0 =π 4. Solution. f′(x) =sec^2 xand
f′′(x) =2 secx·secxtanx=2 sec^2 xtanx. Sof′(π 4 )=2 andf′′(π 4 )= 2 · 2 =4. ThenT 2 (x) = 1 + 2 (x−π 4 )+
2!^4 (x−π 4 )^2 =^1 +^2 (x−π 4 )+^2 (x−π 4 )^2.
ExercisesFind the Taylor series of the following functions at the givenx 0 with given degreen.
- f(x) =exatx= 0 ,n= 3