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x
0
506 A COMPANION TO ANALYSIS

sin »t
(We adopt the convention that, if we write f (t) = , then f (0) = ».)
t
(i) Show, by using the formula for the sum of a geometric series, or
otherwise, that
n
sin((n + 1 )x)
2
1+2 cos rx = x
sin 2
r=1

for all |x| < π and deduce that

sin((n + 1 )x)
π
2
2π = dx.
x
sin 2
’π

(ii) If > 0 show that

sin »x sin x
dx ’ dx,
x x
’ ’∞

as » ’ ∞.
(iii) If π ≥ > 0 show, by using the estimates from the alternating series
test, or otherwise, that
1 1
π
sin((n + 2 )x) sin((n + 2 )x)
dx ’ dx = 2π
sin x sin x
’ ’π
2 2

as n ’ ∞.
(iv) Show that

2 1
’ ’0
1
x sin 2 x

as x ’ 0.
(v) Combine the results above to show that

sin x π
dx = .
x 2
0

Exercise K.144. (Big Oh and little oh.) [9.2, T] The following nota-
tions are much used in branches of mathematics like number theory, algo-
rithmics and combinatorics. Consider functions f, g : N ’ R.
We say f (n) = O(g(n)) as n ’ ∞ if there exists a constant A > 0 and
an integer n0 such that |f (n)| ¤ A|g(n)| for n ≥ n0 .
We say f (n) = „¦(g(n)) as n ’ ∞ if there exists a constant A > 0 and
an integer n0 such that |f (n)| ≥ A|g(n)| for n ≥ n0 .
507
Please send corrections however trivial to twk@dpmms.cam.ac.uk

We say f (n) = o(g(n)) if, given any > 0 we can ¬nd an integer n1 ( )
such that |f (n)| ¤ |g(n)| for n ≥ n1 ( ).
We say f (n) ∼ g(n) as n ’ ∞ if, f (n)/g(n) ’ 1 as n ’ ∞.
(i) If f (n) = O(g(n)) does it follow that f (n) = „¦(g(n))? Give a proof
or counterexample. If f (n) = O(g(n)) must it necessarily be false that
f (n) = „¦(g(n))? Give a proof or counterexample.
(ii) Formulate the 11 other possible questions along the lines of (i) and
resolve them. (This is not as tedious as it might seem, they all have two
sentence answers.)
(iii) In more traditional terms, what does it mean to say that f (n) = o(1)
as n ’ ∞? What does it mean to say that f (n) = O(1) as n ’ ∞?
(iv) Give an example of a pair of functions f, g : N ’ R such that
f (n), g(n) > 0 for all n and neither of the two relations f (n) = O(g(n)),
g(n) = O(f (n)) hold.
(v) Consider f, g : R ’ R. Give de¬nitions along the lines given above
for the statement ˜f (x) = O(g(x)) as x ’ ∞™ and for the statement ˜f (x) =
o(g(x)) as x ’ 0™.
(vi) Recall or do exercise K.86 and write the result in the notation of this
question.
Although the notations introduced in this question are very useful for
stating results, they are sharp edged tools which can do as much harm as
good in the hands of the inexperienced. I strongly recommend translating
any statements involving such notations back into classical form before trying
to work with them.

Exercise K.145. [9.2, P, ‘ ] Are the following statements true or false?
Give reasons.
(i) If fj , gj : N ’ R are positive functions such that f1 (n) = O(g1 (n))
and f2 (n) = O(g2 (n)), then f1 (n) + g1 (n) = O(f2 (n) + g2 (n)).
(ii) If fj , gj : N ’ R are functions such that f1 (n) = O(g1 (n)) and
f2 (n) = O(g2 (n)), then f1 (n) + g1 (n) = O(f2 (n) + g2 (n)).
(iii) If fj , gj : N ’ R are positive functions such that f1 (n) = O(g1 (n))
and f2 (n) = O(g2 (n)), then f1 (n) + g1 (n) = O(max(f2 (n), g2 (n)).
2
(iv) n! = O(2n ) as n ’ ∞.
2
(v) cos x ’ 1 ’ (sin x) = o(x4 ) as x ’ 0.
2!
(sin x)2 4
(vi) cos x ’ 1 ’ 2! ’ (sin x) = o(x6 ) as x ’ 0.
4!

Exercise K.146. [9.2, P, ‘ ] (i) If ± > ’1, show that
n
n±+1
±
r∼ .
±+1
r=1
508 A COMPANION TO ANALYSIS

(ii) State and prove a result similar to (i) for ± = ’1.
(iii) If ± < ’1, show that

n±+1
±
r∼ .
(’±) ’ 1
r=n

(iv) Does there exist a modi¬cation of (iii) similar to (ii) for ± = ’1?
Give brief reasons.

Exercise K.147. [9.2, P, ‘ ] Suppose that g : R ’ R is twice di¬erentiable
and

g (x) = O(x’» ) as x ’ ∞

for some » > 0. Prove that
n+1
(g(x) ’ g(n + 1 )) dx = O(n’» )
2
n

X
as n ’ ∞. Deduce that, if » > 1 and | g(x) dx| ’ ∞ as n ’ ∞, then
1

n n
1

g(r + ) g(x) dx.
2
1
r=1

Deduce the result of Exercise K.141 as a special case.

Exercise K.148. [9.2, P] The function f : [0, 1] ’ R ∪ {∞} is de¬ned as
follows. Each 0 < x < 1 is expressed as a decimal x = .a1 a2 . . . ak . . . with
aj an integer 0 ¤ aj ¤ 9. In ambiguous cases we use the non-terminating
form. We put f (x) = k if the kth digit ak is the ¬rst digit 7 to occur in the
expansion, if there is any digit 7. If there is none (and also at x = 0 and
x = 1) we put f (x) = ∞. Prove that the function fX : [0, 1] ’ R de¬ned by
fX (x) = min(X, f (x)) is integrable for all X ≥ 0 and that
1
fX (x) dx ’ 10
0

as X ’ ∞.
Does it make any di¬erence if we rede¬ne f so at those points x where
we previously had f (x) = ∞ we now have f (x) = 0?

Exercise K.149. [9.2, H] This question investigates another way of de¬n-
ing improper integrals.
509
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(i) If f : [a, b] ’ R and R, S ≥ 0, we write
±
R if f (x) ≥ R,

fR,S (x) = f (x) if R > f (x) > ’S,


’S if ’S ≥ f (x).

If f |R,S ∈ R[a, b] for all R, S ≥ 0 and there exists an L such that, given any
> 0, there exists an R0 > 0 with
b
fR,S (x) dx ’ L <
a

b
for all R, S ≥ R0 , then we say that LV a f (x) dx exists with value L.

Show that, if f |R,S ∈ R[a, b] for all R, S ≥ 0, then LV a |f (x)| dx exists

if and only if LV a |f (x)| dx exists.
(ii) Suppose that f : [a, b] ’ R is continuous except at b and f (x) ’ ∞
as x ’ b. Show that the following two statements are equivalent.
(A) There exists an L such that, given any > 0, there exists an R0 > 0
with
b
fR,S (x) dx ’ L <
a

for all R, S > R0 .
(B) There exists an L such that, given any > 0, there exists a δ0 > 0
with
b’·
f (x) dx ’ L <
a

for all 0 < · < δ0 .
Show further that, if (A) and (B) hold, then L = L .
(iii) Suppose that f : [a, b] ’ R is continuous except at possibly at b.
Show that statement (A) implies statement (B) but that the converse is
false. (Think about Question K.142 and change of variable if you need a
hint.)
[Warning: If you ever need to take the contents of this question seriously
you should switch to the Lebesgue integral.]

Exercise K.150. [9.3, P] By considering both orders of integration in
X b
e’tx dt dx,
0 a
510 A COMPANION TO ANALYSIS

where b > a > 0 and X > 0, show that
X b
e’ax ’ e’bx 1 ’ e’tX
dx = dt.
x t
0 a

Show that
b
e’tX b
dt ¤ e’aX log ,
t a
a

and hence ¬nd

e’ax ’ e’bx
dx.
x
0

Exercise K.151. [9.3, T] Suppose that f : R2 ’ R is continuous. By using
results on the di¬erentiation of integrals, which should be quoted exactly,
show that
t d d t
d d
f (u, v) dv du = f (u, v) du dv.
dt dt
a c c a

Deduce that
b d d b
f (u, v) dv du = f (u, v) du dv.
a c c a

In what way is this result weaker than that of Theorem 9.3.2?
Exercise K.152. [9.3, H] (i) Reread Exercise 5.3.10.
(ii) Let us de¬ne f : [0, 1] — [0, 1] ’ R by

if 2’n < x < 2’n’1 , 2’n < y < 2’n+1 and n ≥ 1, n ∈ Z,
f (x, y) = 22n
if 2’n < x < 2’n’1 , 2’n’1 < y < 2’n and n ≥ 1, n ∈ Z,
f (x, y) = ’22n+1
f (x, y) = 0 otherwise.

Sketch the sets where f (x, y) take various values. Show that the integral
1
f (x, y) dx is well de¬ned and calculate it for all values of y. Show that the
0
1
integral 0 f (x, y) dy is well de¬ned and calculate it for all values of x. Show
that the two integrals
1 1 1 1
f (x, y) dx dy and f (x, y) dy dx
0 0 0 0

are well de¬ned and calculate them. Show that the two integrals are not
equal.
511
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(iii) Suppose that u : R ’ R is continuous function with u(t) = 0 for
1
t < 1/4 and t > 3/4, u(t) ≥ 0 for all t and 0 u(t) dt = 1. Show that the
function g : [0, 1]2 ’ R

22n u(2n x ’ 1)u(2n y ’ 1) ’ 22n+1 u(2n x ’ 1)u(2n+1 y ’ 1)
g(x, y) =
n=1

1
is well de¬ned. Show that the integral 0 f (x, y) dx is well de¬ned and calcu-
1
late it for all values of y. Show that the integral 0 f (x, y) dy is well de¬ned
and calculate it for all values of x. Show that the two integrals
1 1 1 1
g(x, y) dx dy and g(x, y) dy dx
0 0 0 0


are well de¬ned but not equal. Show that g is continuous except at (0, 0),
that the function x ’ g(x, y) is an everywhere continuous function for all y
and that the function y ’ g(x, y) is an everywhere continuous function for
all x. Why does Theorem 9.3.2 fail?
(iv) The traditional counterexample is the following. De¬ne h : [0, 1] 2 ’
R by h(0, 0) = 0 and

xy(x2 ’ y 2 )
h(x, y) = 2
(x + y 2 )3 )
1
otherwise. Show that the integral 0 h(x, y) dx is well de¬ned and calculate
it for all values of y (you may ¬nd the substitution w = x2 + y 2 useful). Show
1
that the integral 0 f (x, y) dy is well de¬ned and calculate it for all values of
x. Show that the two integrals
1 1 1 1
h(x, y) dx dy and h(x, y) dy dx
0 0 0 0


are well de¬ned but not equal. Show that g is continuous except at (0, 0),
that the function x ’ g(x, y) is an everywhere continuous function for all y
and that the function y ’ g(x, y) is an everywhere continuous function for
all x.
(v) Which of examples (ii) and (iv) do you consider ˜more natural™. Which
do you feel you ˜understand better™ ? Which is ˜better™. Why? (Of course
you can refuse to answer questions like this on the grounds that they are not
mathematical but I think you will loose something.)
512 A COMPANION TO ANALYSIS

Exercise K.153. (Interchange of in¬nite integrals.) [9.3, T] (i) Sup-
pose that g : [0, ∞) ’ R. is continuous and |g(x)| ¤ A(1 + x2 )’1 for all x.

Show that 0 g(x) dx exists, that

g(x) dx ¤ 2A,
0

and

R
g(x) dx ¤ AR’1 ,
g(x) dx ’
0 0

for R ≥ 1. [If you do not wish to use knowledge of tan’1 , simply observe
that (1 + x2 )’1 ¤ 1 for 0 ¤ x ¤ 1 and (1 + x2 )’1 ¤ x’2 for x ≥ 1.]
(ii) Suppose that f : [0, ∞)2 ’ R is continuous and |f (x, y)| ¤ Ax’2 y ’2
∞∞ ∞∞
for x, y ≥ 1. Show that 0 0 f (x, y) dx dy and 0 0 f (x, y) dy dx exist
and that
∞ ∞ ∞ ∞
f (x, y) dx dy = f (x, y) dy dx.
0 0 0 0

(iii) (Parts (iii) and (iv) run along similar lines to parts (i) and (ii) of
Exercise K.152. You do not need to have done Exercise K.152 to do this ex-
ercise but, if you have, it is worth noting the similarities and the di¬erences.)
Reread Exercise 5.3.10. Let brs be as in part (ii) of that exercise. De¬ne
F : [0, ∞)2 ’ R by
F (x, y) = brs for r ’ 1 ¤ x < r and s ’ 1 ¤ y < s.
∞ ∞ ∞ ∞
Show that F (x, y) dx dy and F (x, y) dy dx exist and that
0 0 0 0
∞ ∞ ∞ ∞
F (x, y) dx dy = F (x, y) dy dx.
0 0 0 0

(iv) By modifying the construction in part (iii), or otherwise, ¬nd a con-
tinuous function G : [0, ∞)2 ’ R such that supx2 +y2 ≥R |G(x, y)| ’ 0 as
R ’ ∞ but
∞ ∞ ∞ ∞
G(x, y) dx dy = G(x, y) dy dx.
0 0 0 0

Exercise K.154. [9.3, H] It is a weakness of the Riemann integral that
there is no Fubini theorem for general Riemann integrable functions.
Let A = [0, 1] — [0, 1] and de¬ne f : A ’ R by f (x, 0) = 1 if x is rational,
f (x, y) = 0 otherwise. By ¬nding appropriate dissections show that f is
1
Riemann integrable. Explain why F2 (y) = 0 f (t, y) dt is not de¬ned for all
y.
[See also the next exercise.]
513
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise K.155. [9.3, H] We say that a sequence e1 , e2 , . . . of points in
A is dense in A = [0, 1] — [0, 1] if, given x ∈ A, we can ¬nd a j ≥ 1 with
x ’ ej < .
(i) By considering Q — Q, or by direct construction, or otherwise, show
that we can ¬nd a sequence e1 , e2 , . . . of points which is dense in A.
(ii) By induction, or otherwise, show that we can ¬nd a sequence w1 =
(u1 , v1 ), w2 = (u2 , v2 ), . . . of points in A such that wk ’ ek < 1/k and
ui = uj , vi = vj whenever i = j. Show that the sequence w1 , w2 , . . . is dense
in A.
(iii) De¬ne f : A ’ R by f (x) = 1 if x = wj for some j, f (x) = 0
otherwise. Show that, if x is ¬xed f (x, y) = 1 for at most one value of
1
y. Conclude that F1 (x) = 0 f (x, s) ds exists and takes the value 0 for all
1 1 1 1

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