CK-12-Calculus

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

8.8 Calculations with Series

Binomial Series

We have learned the Binomial Theorem for positive integer exponents:

(a+b)n=an+nan−^1 b+n(n2!−^1 )an−^2 b^2 +...+nabn−^1 +bn(BE)

=

n

k∑= 0

(n

k

)

an−kbk

where the Binomial coefficients are denoted by
(n
0

)=1 and(n

k

)=n(n− 1 )...(n−k+ 1 )
k! fork≥^1
AS a simple Binomial function, takea=1 andb=x. Then( 1 +x)r=∑∞k= 0 (rk)xk
Letrbe a real number andf(x) = ( 1 +x)r. Isf(x)equal to a series in the form of (BE) except that there may be an
infinite series? The answer is yes.
Theorem(Binomial Series) Letrbe a real number and|x|<1. Then

( 1 +x)r= 1 +rx+r(rk−!^1 )x^2 +r(r−^1 3!)(r−^2 )x^3

=

n

k∑= 0

(r

k

)

xk

where the Binomial coefficients are denoted by
(n
0

)=1 and(n

k

)=n(n− 1 )...(n−k+ 1 )
2! fork≥^1
where( 0 r)=1 and(kr)=r(r−^1 )...k(!r−k+^1 )fork≥ 1
Example 1Find a power series representation of√ 1 +x.
Solution. We need to compute the Binomial coefficients forr=−^12

( 1

2

k

)

=

( 1

2

)(− 1

2

)(− 3

2 )...(−^2 k 2 −^3 )

k! =

(− 1 )k−^11 · 3 · 5 ...( 2 k− 3 )

2 kk!

= (−^1 )

k− (^1) ( 2 k− 2 )!
2. 4. 6 ...( 2 k− 2 )) 2 kk!=
(− 1 )k−^1 ( 2 k− 2 )!
2 k−^1 k!(k− 1 ) 2 kk!=
(− 1 )k−^1 ( 2 k− 2 )!
22 k−^1 k!(k− 1 )!
So if|x|< 1 ,√ 1 +x=∑∞k= 0 (k^21 )xk= 1 +∑∞k= 0 ( 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.