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3 1 10. In this exercise we define
p Þ 0 sin p d whenever p 0.
a. Prove that is a decreasing function on the interval 0, Ý .
The fact that is decreasing follows from the fact that whenever 0
0 sin 1 and so sin p decreases as p increases. we have b. Prove that
2p 2
1.
2p
Given any positive number p we integrate by parts and obtain
lim pÝ 2p 2 Þ 0 sin 2p2 d Þ 0 sin 2p1
2p 1 sin 2p1 sin d
sin 2p1 0 cos
Þ 0 sin 2p 1 Þ0 cos 0 2p 1 sin 2p cos sin 2 d 2p 1 2p 2p 1 2p 2 and so
lim pÝ 2p 2
2p pÝ
lim 2p 1
1.
2p 2 c. Combine the first two parts of this question and deduce that
2p 1
1.
lim
pÝ
2p
For each p 0 we have
2p 2
2p 1
2p
and so
2p 2
2p 1
1
2p
2p
and the fact that
2p 1
1
lim
pÝ
2p
follows from the sandwich rule.
d. Prove that
4p
p lim
pÝ 2 p1
2p 1 . This assertion is known as Wallis’s formula
From Exercise 7 we know that whenever p 0,
2p 1 Þ0 Þ0 sin 2 p 1 d 2 p 1 1
2
2 p 1
1
2 and
2p sin 2p d which gives us 391 2 p 1
2
2p
2 1 cos d 2p11
2
2p1
1
2
2p1
2
2p
1
2 2p 2 2p 2p 1
2p 2 p 1
2 p1
2 p 1
2 Now from Exercise 8 we have
1 p p 1
2 and so
2p 1
2p
2 p 1
2 2 4p 2
2 p1
2 p
p1 p 1
2
2 4p 4 p 1
2 2
2 2 1
2 p p 2p 1
2 p1
2 2p 2 2p 1 p p1 p p
p 1 2p 2p
2 2 2 4p 2 2 p1
2 2p 1 and so
lim pÝ 2 4p 4 p 1
p 1 2 2p 1
2 1 from which we deduce that
lim
pÝ 4p
p 2 p1
2p 1 pÝ
lim 4p
p 2
1
2 p1
2p 1 p
p 1
2 e. Assuming that p is restricted to be a positive integer, rewrite Wallis’s formula in terms of factorials.
If p is restricted to be a positive integer then Wallis’s formula becomes
4 p p! 2
lim
pÝ
p 2p !
11. The purpose of this exercise is to encourage you to read a proof of an interesting theorem known as
Stirling’s formula that states that
x1
2 .
lim x x
xÝxe
x
Note that if n is a large natural number then Stirling’s formula suggests that an approximate value for n! is
2n n n
.
en
The proof of Stirling’s formula that is provided by the preceding link is based on a proof that is provided
on page 195 of Walter Rudin’s classic text Principles of Mathematical Analysis. The proof provided here
is actually a little simpler because it makes use of the improper integral form of the bounded convergence
theorem. 17 Sets of Measure Zero
Some Exercises on the Measure Zero Concept
1. Prove that an elementary set E has measure zero if and only if m E 0.
392 Suppose that E is an elementary set. In the event that m E 0, it is clear that E has (nineteenth
century) measure zero. On the other hand, if m E 0 then E includes a closed bounded interval of
positive length and it follows from the discussion on closed bounded sets of measure zero that E
can’t have measure zero.
2. Prove that every countable set has measure zero.
This fact follows at once from the theorem on uni...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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