3.8. Approximation Errors http://www.ck12.org
Tn(x) =f(a)+f′(a)(x−a)+f′′(a)
2! (x−a)(^2) +...+fn(a)
n! (x−a)
n.
We call this theTaylor Polynomial offcentered ata.
For our discussion, we will focus on the quadratic case. The Taylor Polynomial corresponding ton=2 is given by
T 2 (x) =f(a)+f′(a)(x−a)+^12 f′′(a)(x−a)^2.
Note that this is just our linear approximation with an added term. Hence we can view it as an approximation off
forxvalues close toa.
Example 3:
Find the quadratic approximation of the functionf(x) =
√
x−2 ata=6 and compare them to the linear approxi-
mations from the first example.
Solution:
Recall thatL(x) =^12 +^14 x.
HenceT 2 (x) =L(x)+^12 f′′( 6 )(x− 6 )^2.
f′′(x) =− 4 √(x^1 − 2 ) 3 ; sof′′( 6 ) =− 4 √( 61 − 2 ) 3 =− 321.
HenceT 2 (x) =L(x)− 641 (x− 6 )^2 =^12 +^14 x− 641 (x− 6 )^2 =− 641 x^2 + 167 x− 161.
SoT 2 (x) =− 641 x^2 + 167 x− 161. If we update our table from the first example we can see how the quadratic approxi-
mation compares with the linear approximation.
f(x) =√x− 2 x L(x) =^12 +^14 x T 2 (x) =− 641 x^2 + 167 x− 161 Actual
√ 3. 95 5. 95 1. 9875 1. 9874 1. 9874
√
√^3.^995.^991.^99751.^99741.^9974
√^46222
√^4.^16.^012.^00252.^00242.^0024
4. 05 6. 05 2. 0125 2. 0124 2. 0124As you can see from the graph below,T(x)is an excellent approximation off(x)nearx= 6.