Series 359Step 3:
f(0)
0!x^0 +
f′(0)
1!x^1 +
f′′(0)
2!x^2 +
f′′′(0)
3!x^3 +
f(4)(0)
4!x^4 =0
1
x^0 +1
1
x^1 +− 1
2
x^2 +2
6
x^3 +− 6
24
x^4 =x−1
2
x^2 +1
3
x^3 −1
4
x^4Example 4
Find the Taylor series for the functionf(x)=e−xabout the pointx=ln 2.Step 1: f(n)(x)=e−xwhennis even andf(n)(x)=−e−xwhennis odd.
Step 2: Evaluatef(n)(ln 2)=e−ln 2=1
2
whennis even andf(n)(ln 2)=− 1
2
whennis odd.Step 3: f(x)=e−x =1 / 2
0!
(x − ln 2)^0 +− 1 / 2
1!
(x − ln 2)^1 +1 / 2
2!
(x −ln 2)^2 + ···=∑∞n= 0(−1)n
2 ·n!
(x −ln 2)nExample 5
Find the MacLaurin series for the functionf(x)=xex.Step 1: Investigating the first few derivatives of f(x)=xexshows thatf(n)(x)=xex+nex.
Step 2: Evaluating f(n)(x)=xex+nexatx=0 givesf(n)(0)=n.Step 3: f(x)=∑∞n= 0f(n)(0)
n!
xn=∑∞n= 0n
n!
xn=∑∞n= 1xn
(n−1)!Common MacLaurin Series
MacLaurin Series for the Functionsex,sinx,cosx,and 11 −x
Familiarity with these common MacLaurin series will simplify many problems.f(x)=ex=∑∞n= 0xn
n!= 1 +x+
x^2
2+
x^3
6+···
f(x)=sinx=∑∞n= 0(−1)nx^2 n+^1
(2n+1)!
=x−
x^3
3!+
x^5
5!−
x^7
7!+···
f(x)=cosx=∑∞n= 0(−1)^2 nx^2 n
(2n)!= 1 −
x^2
2+
x^4
24−
x^6
6!+···
f(x)=1
1 −x=
∑∞n= 0xn= 1 +x+x^2 +x^3 +···14.7 Operations on Series
Main Concepts:Substitution, Differentiation and Integration, Error Bounds
Substitution
New series can be generated by making an appropriate substitution in a known series.