Unformatted text preview: ions guarantees that f is analytic on 1, 1 . 1, 1 the composition theorem for analytic 6. Suppose that a n x n and b n x n are two power series that converge in an interval
suppose that the set of numbers x
r, r for which the equation
Ý r, r where r 0, and Ý anxn
n0 bnxn
n0 has a limit point in r, r . Prove that a n b n for every n.
The desired result follows at once from the fact that the sum of a power series is analytic and that
two analytic functions on an open interval must be identical if they agree on a set that has a limit
point in that interval.
7. Given that f is analytic and nonconstant on an open interval U and that c
positive integer n such that f n c
0.
Since the equation
Ý fx
n0 f n c
n! x c U, prove that there exists a n must hold for all x sufficiently close to c the desired result follows at once.
8. Given that f is analytic and nonconstant on an open set U and that K is a closed bounded subset of U, prove
that the set
x K fx 0
is finite.
If the set 382 x K fx 0
were infinite then it would have to have a limit point in K and so f would have to be the constant
function 0.
9. Given that f is analytic and nonconstant on an open interval U prove that the set
x U fx 0
is countable.
This solution makes use of the distance function of a set. For each positive integer n we define
1.
K n n, n
x U R U x
n
Each set K n is closed and bounded and
U Ý Kn.
n1 Since
x fx 0 U Ý x Kn fx 0 n1 we deduce that the set
x U fx 0 is countable. 16 Integration of Functions of Two Variables
Some Exercises on Iterated Riemann Integrals
1. Given that
fx Þ 0 exp
x gx t 2 dt Þ0 1 exp x 2 t 2 1
t2 1 dt and given
h x f x 2g x
for every real number x, prove that the function h must be constant. What is the value of this constant?
Deduce that
Ý
Þ 0 exp x 2 dx 2 .
For every number x we see from the theorem on differentiating a partial integral that
h x 2f x f x g x
x
1 2x t 2 1 exp x 2 t 2 1
2 exp x 2 Þ exp t 2 dt Þ
dt
0
0
t2 1
2 exp x 2 Þ 0 exp
x t 2 dt 2 exp x 2 Þ 0 x exp
1 x 2 t 2 dt In the latter integral we make the substitution u tx and we deduce that
h x 2 exp x 2 Þ 0 exp
x t 2 dt 2 exp x 2 Þ 0 exp
x for every number x. Therefore the function h is constant. Now since 383 u 2 du 0 h0 Þ 0 exp
0 2 t 2 dt Þ 1
0 exp 0 2 t 2 1
t2 1 dt
4 We deduce that h x for every number x.
4
Now we use the fact that h n for every positive integer n. From the
4
bounded convergence theorem we see that
1 exp n 2 t 2 1
1
lim Þ
dt Þ 2 0 dt 0
21
nÝ 0
0 t 1
t
and so
Ý Þ 0 exp t 2 dt t 2 dt lim h n
nÝ
4 2
.
2 0 from which we deduce that
Ý Þ 0 exp
2. Given that
gy Þ0 Ý exp x 2 cos 2xydx f y exp y 2 g y and for every real number y, prove that the function f must be constant. What is the value of this constant?
Since
exp x 2 cos 2xy  exp x 2
and
D 2 exp x 2 cos 2xy   2x exp x 2 sin 2xy  2x exp x 2 for all x and y we see that the hypotheses of the theorem on differentiation...
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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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