Real Analysis 551

P(x)=

1

4

+

1

16

x+

1

64

x^2.

By the residue formula for Taylor series we have

∣

∣∣

∣P(x)+

1

x− 4

∣

∣∣

∣=

x^3

256

+

1

(ξ− 4 )^4

x^5 ,

for someξ∈( 0 ,x). Since|x|≤1 and also|ξ|≤1, we have x

3

256 ≤

1

256 andx

(^4) /(ξ− 4 ) (^5) ≤
1
243. An easy numerical computation shows that
1
256 +
1
243 <
1
100 , and we are done.
(Romanian Team Selection Test for the International Mathematical Olympiad, 1979,
proposed by O. Stan ̆ ̆a ̧sil ̆a)
490.The Taylor series expansion of cos

√

xaround 0 is

cos

√

x= 1 −

x

2!

+

x^2

4!

−

x^3

6!

+

x^4

8!

−···.

Integrating term by term, we obtain

∫ 1

0

cos

√

xdx=

∑∞

n= 1

(− 1 )n−^1 xn

(n+ 1 )( 2 n)!

∣∣

∣∣

∣

1

0

=

∑∞

n= 0

(− 1 )n−^1

(n+ 1 )( 2 n)!

.

Grouping consecutive terms we see that

(

1

5 · 8!

−

1

6 · 10!

)

+

(

1

7 · 12!

−

1

8 · 14!

)

+···<

1

2 · 104

+

1

2 · 105

+

1

2 · 106

+···

<

1

104

.

Also, truncating to the fourth decimal place yields

0. 7638 < 1 −

1

4

+

1

72

−

1

2880

< 0. 7639.

We conclude that

∫ 1

0

cos

√

xdx≈ 0. 763.

491.Consider the Newton binomial expansion

(x+ 1 )−

(^12)

∑∞

k= 0

(

−^12

k

)

xk=

∑∞

k= 0

(

−^12

)(

−^12 − 1

)(

−^12 − 2

)

···

(

−^12 −k+ 1

)

k!

xk