CK-12-Calculus

http://www.ck12.org Chapter 8. Infinite Series

So if|x|< 1 ,√ 1 +x=∑∞k= 1 (k^21 )xk= 1 +∑∞k= 1 ( 2 − 21 k−)k 1 −k^1 !((^2 k−k− 12 ))!!xk

Example 2Find a power series representation of( 1 −^1 x)mwheremis a positive integer.
Solution. We need to compute the Binomial coefficients forr=−m(and will replacexby−x)

(−m

k

)

=(−m)(−m−^1 )(−mk−!^2 )...(−m−k+^1 )

=(−^1 )

km(m+ 1 )(m+ 2 )...(m+k− 1 )

k! = (−^1 )

k

(m+k− 1

k

)

So( 1 −^1 x)m=

∞

k∑= 0 (−^1 )k

(m+k− 1

k

)

(−x)k

=

∞

k∑= 0 (−^1 )k

(m+k− 1

k

)

(− 1 )kxk

=

∞

k∑= 0 (−^1 )^2 k

(m+k− 1

k

)

xk

=

∞

k∑= 0

(m+k− 1

k

)

xk

Exercises

1. Find a power series representation of√ 11 +xatx=0.


2. Find a power series representation of( 2 −^1 x) 2 atx=0.


3. Notice 1+x+x^2 =x+^34 +(x+^12 )^2. Find a power series representation of

√

( 1 +x+x^2 )atx=−^12. In what

interval is the equality true?

Choosing Centers
Taylor Series (indeed Taylor polynomials of lower degrees) often provide good approximation of functions. How-
ever, the choice of center could determine

1. whether the intended value ofxis inside the interval of convergence


2. rate of convergence, i.e. how many terms to take to achieve prescribed degree of accuracy

For frequently used functions, the first choice may be the standard center (see the list at the end of this section).
Example 1Approximate ln 0 : 99
Solution. Since .99 is close to the centerx=1, we use the standard Taylor series for ln( 1 −x).

ln 0. 99 =ln( 1 − 0. 01 )≈− 0. 01 −(^0.^01 )

2

2 ≈−^0.^01005

Then we may be able to deduce a useful Taylor Series centered close to the givenx.
Example 2Approximate sin( 1. 1 )to 4 decimal places.
Since 1.1 is close toπ 3 , we would try to find a Taylor Series of sinxatx 0 =π 3. Letf(x) =sinx. Then

f(x) =sinx,f′(x) =cosx,f′′(x) =−sinx,f′′′(x) =−cosxandf(π 3 )=

√ 3

2 ,f′

(π

3

)= 1

2 ,f′′

(π

3

)=−√ 3

2 ,f′′′

(π

3

)=

−^12.