Sequences and Series of Functions and Uniform Convergence - 1 - CSIR-NET Mathematical Science | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 2


The radius of convergence of the series
P
1
j=0
2
j
2
(x + 2)
j
is
1
lim sup
j!1
2
j
=
1
0
= +1:
As we now show, the radius of convergence tells us exactly where the power series con-
verges.
Theorem 8.5. Let
P
1
j=0
a
j
(xa)
j
be a formal power series, and let R be its radius of
convergence.
(a) (Divergence outside of the radius of convergence) If x2R satisesjxaj>R, then
the series
P
1
j=0
a
j
(xa)
j
is divergent at x.
(b) (Convergence inside the radius of convergence) If x2R satisesjxaj < R, then
the series
P
1
j=0
a
j
(xa)
j
is convergent at x.
 For the following items (c),(d) and (e), we assume that R > 0. Then, let f : (a
R;a +R) be the function f(x) =
P
1
j=0
a
j
(xa)
j
, which exists by part (b).
(c) (Uniform convergence on compact intervals) For any 0 < r < R, we know that the
series
P
1
j=0
a
j
(xa)
j
converges uniformly to f on [ar;a +r]. In particular, f is
continuous on (aR;a +R) (by Theorem 3.1.)
(d) (Dierentiation of power series) The function f is dierentiable on (aR;a +R).
For any 0 < r < R, the series
P
1
j=0
ja
j
(xa)
j1
converges uniformly to f
0
on the
interval [ar;a +r].
(e) (Integration of power series) For any closed interval [y;z] contained in (aR;a+R),
we have
Z
z
y
f =
1
X
j=0
a
j
(za)
j+1
 (ya)
j+1
j + 1
:
Exercise 8.6. Prove Theorem 8.5. (Hints: for parts (a),(b), use the root test. For part (c),
use the Weierstrass M-test. For part (d), use Theorem 6.3. For part (e), use Theorem 5.4.)
Remark 8.7. A power series may converge or diverge whenjxaj =R.
Exercise 8.8. Give examples of formal power series centered at 0 with radius of convergence
R = 1 such that
 The series diverges at x = 1 and at x =1.
 The series diverges at x = 1 and converges at x =1.
 The series converges at x = 1 and diverges at x = 1.
 The series converges at x = 1 and at x =1.
We now discuss functions that are equal to convergent power series.
Denition 8.9. Leta2R and letr> 0. LetE be a subset ofR such that (ar;a+r)E.
Let f : E!R. We say that the function f is real analytic on (ar;a +r) if and only if
there exists a power series
P
1
j=0
a
j
(xa)
j
centered at a with radius of convergence R such
that Rr and such that this power series converges to f on (ar;a +r).
Example 8.10. The functionf : (0; 2)!R dened byf(x) =
P
1
j=0
j(x1)
j
is real analytic
on (0; 2).
Page 3


The radius of convergence of the series
P
1
j=0
2
j
2
(x + 2)
j
is
1
lim sup
j!1
2
j
=
1
0
= +1:
As we now show, the radius of convergence tells us exactly where the power series con-
verges.
Theorem 8.5. Let
P
1
j=0
a
j
(xa)
j
be a formal power series, and let R be its radius of
convergence.
(a) (Divergence outside of the radius of convergence) If x2R satisesjxaj>R, then
the series
P
1
j=0
a
j
(xa)
j
is divergent at x.
(b) (Convergence inside the radius of convergence) If x2R satisesjxaj < R, then
the series
P
1
j=0
a
j
(xa)
j
is convergent at x.
 For the following items (c),(d) and (e), we assume that R > 0. Then, let f : (a
R;a +R) be the function f(x) =
P
1
j=0
a
j
(xa)
j
, which exists by part (b).
(c) (Uniform convergence on compact intervals) For any 0 < r < R, we know that the
series
P
1
j=0
a
j
(xa)
j
converges uniformly to f on [ar;a +r]. In particular, f is
continuous on (aR;a +R) (by Theorem 3.1.)
(d) (Dierentiation of power series) The function f is dierentiable on (aR;a +R).
For any 0 < r < R, the series
P
1
j=0
ja
j
(xa)
j1
converges uniformly to f
0
on the
interval [ar;a +r].
(e) (Integration of power series) For any closed interval [y;z] contained in (aR;a+R),
we have
Z
z
y
f =
1
X
j=0
a
j
(za)
j+1
 (ya)
j+1
j + 1
:
Exercise 8.6. Prove Theorem 8.5. (Hints: for parts (a),(b), use the root test. For part (c),
use the Weierstrass M-test. For part (d), use Theorem 6.3. For part (e), use Theorem 5.4.)
Remark 8.7. A power series may converge or diverge whenjxaj =R.
Exercise 8.8. Give examples of formal power series centered at 0 with radius of convergence
R = 1 such that
 The series diverges at x = 1 and at x =1.
 The series diverges at x = 1 and converges at x =1.
 The series converges at x = 1 and diverges at x = 1.
 The series converges at x = 1 and at x =1.
We now discuss functions that are equal to convergent power series.
Denition 8.9. Leta2R and letr> 0. LetE be a subset ofR such that (ar;a+r)E.
Let f : E!R. We say that the function f is real analytic on (ar;a +r) if and only if
there exists a power series
P
1
j=0
a
j
(xa)
j
centered at a with radius of convergence R such
that Rr and such that this power series converges to f on (ar;a +r).
Example 8.10. The functionf : (0; 2)!R dened byf(x) =
P
1
j=0
j(x1)
j
is real analytic
on (0; 2).
From Theorem 8.5, if a function f is real analytic on (ar;a +r), then f is continuous
and dierentiable. In fact, f is can be dierentiated any number of times, as we now show.
Denition 8.11. LetE be a subset ofR. We say that a functionf : E!R is once dier-
entiable on E if and only if f is dierentiable on E. More generally, for any integer k 2,
we say that f : E!R is k times dierentiable on E, or just k times dierentiable, if
and only iff is dierentiable andf
0
isk1 times dierentiable. Iff isk times dierentiable,
we dene thek
th
derivativef
(k)
: E!R by the recursive rulef
(1)
:=f
0
andf
(k)
:= (f
(k1)
)
0
,
for all k 2. We also dene f
(0)
:=f. A function is said to be innitely dierentiable if
and only if f is k times dierentiable for every k 0.
Example 8.12. The function f(x) =jxj
3
is twice dierentiable onR, but not three times
dierentiable onR. Note that f
00
(x) = 6jxj, which is not dierentiable at x = 0.
Proposition 8.13. Let a2R and let r > 0. Let f be a function that is real analytic on
(ar;a +r), with the power series expansion
f(x) =
1
X
j=0
a
j
(xa)
j
; 8x2 (ar;a +r):
Then, for any integer k 0, the function f is k times dierentiable on (ar;a +r), and
the k
th
derivative is given by
f
(k)
(x) =
1
X
j=0
a
j+k
(j + 1)(j + 2)


 (j +k)(xa)
j
; 8x2 (ar;a +r):
Exercise 8.14. Prove Proposition 8.13.
Corollary 8.15 (Taylor's formula). Let a2R and let r > 0. Let f be a function that is
real analytic on (ar;a +r), with the power series expansion
f(x) =
1
X
j=0
a
j
(xa)
j
; 8x2 (ar;a +r):
Then, for any integer k 0, we have
f
(k)
(a) =k!a
k
;
wherek! = 12


k, and we denote 0! := 1. In particular, we have Taylor's formula
f(x) =
1
X
j=0
f
(j)
(a)
j!
(xa)
j
; 8x2 (ar;a +r):
Exercise 8.16. Prove Corollary 8.15 using Proposition 8.13.
Remark 8.17. The series
P
1
j=0
f
(j)
(a)
j!
(xa)
j
is sometimes called the Taylor series of f
arounda. Taylor's formula says that if f is real analytic, thenf is equal to its Taylor series.
In the following exercise, we see that even if f is innitely dierentiable, it may not be equal
to its Taylor series.
Page 4


The radius of convergence of the series
P
1
j=0
2
j
2
(x + 2)
j
is
1
lim sup
j!1
2
j
=
1
0
= +1:
As we now show, the radius of convergence tells us exactly where the power series con-
verges.
Theorem 8.5. Let
P
1
j=0
a
j
(xa)
j
be a formal power series, and let R be its radius of
convergence.
(a) (Divergence outside of the radius of convergence) If x2R satisesjxaj>R, then
the series
P
1
j=0
a
j
(xa)
j
is divergent at x.
(b) (Convergence inside the radius of convergence) If x2R satisesjxaj < R, then
the series
P
1
j=0
a
j
(xa)
j
is convergent at x.
 For the following items (c),(d) and (e), we assume that R > 0. Then, let f : (a
R;a +R) be the function f(x) =
P
1
j=0
a
j
(xa)
j
, which exists by part (b).
(c) (Uniform convergence on compact intervals) For any 0 < r < R, we know that the
series
P
1
j=0
a
j
(xa)
j
converges uniformly to f on [ar;a +r]. In particular, f is
continuous on (aR;a +R) (by Theorem 3.1.)
(d) (Dierentiation of power series) The function f is dierentiable on (aR;a +R).
For any 0 < r < R, the series
P
1
j=0
ja
j
(xa)
j1
converges uniformly to f
0
on the
interval [ar;a +r].
(e) (Integration of power series) For any closed interval [y;z] contained in (aR;a+R),
we have
Z
z
y
f =
1
X
j=0
a
j
(za)
j+1
 (ya)
j+1
j + 1
:
Exercise 8.6. Prove Theorem 8.5. (Hints: for parts (a),(b), use the root test. For part (c),
use the Weierstrass M-test. For part (d), use Theorem 6.3. For part (e), use Theorem 5.4.)
Remark 8.7. A power series may converge or diverge whenjxaj =R.
Exercise 8.8. Give examples of formal power series centered at 0 with radius of convergence
R = 1 such that
 The series diverges at x = 1 and at x =1.
 The series diverges at x = 1 and converges at x =1.
 The series converges at x = 1 and diverges at x = 1.
 The series converges at x = 1 and at x =1.
We now discuss functions that are equal to convergent power series.
Denition 8.9. Leta2R and letr> 0. LetE be a subset ofR such that (ar;a+r)E.
Let f : E!R. We say that the function f is real analytic on (ar;a +r) if and only if
there exists a power series
P
1
j=0
a
j
(xa)
j
centered at a with radius of convergence R such
that Rr and such that this power series converges to f on (ar;a +r).
Example 8.10. The functionf : (0; 2)!R dened byf(x) =
P
1
j=0
j(x1)
j
is real analytic
on (0; 2).
From Theorem 8.5, if a function f is real analytic on (ar;a +r), then f is continuous
and dierentiable. In fact, f is can be dierentiated any number of times, as we now show.
Denition 8.11. LetE be a subset ofR. We say that a functionf : E!R is once dier-
entiable on E if and only if f is dierentiable on E. More generally, for any integer k 2,
we say that f : E!R is k times dierentiable on E, or just k times dierentiable, if
and only iff is dierentiable andf
0
isk1 times dierentiable. Iff isk times dierentiable,
we dene thek
th
derivativef
(k)
: E!R by the recursive rulef
(1)
:=f
0
andf
(k)
:= (f
(k1)
)
0
,
for all k 2. We also dene f
(0)
:=f. A function is said to be innitely dierentiable if
and only if f is k times dierentiable for every k 0.
Example 8.12. The function f(x) =jxj
3
is twice dierentiable onR, but not three times
dierentiable onR. Note that f
00
(x) = 6jxj, which is not dierentiable at x = 0.
Proposition 8.13. Let a2R and let r > 0. Let f be a function that is real analytic on
(ar;a +r), with the power series expansion
f(x) =
1
X
j=0
a
j
(xa)
j
; 8x2 (ar;a +r):
Then, for any integer k 0, the function f is k times dierentiable on (ar;a +r), and
the k
th
derivative is given by
f
(k)
(x) =
1
X
j=0
a
j+k
(j + 1)(j + 2)


 (j +k)(xa)
j
; 8x2 (ar;a +r):
Exercise 8.14. Prove Proposition 8.13.
Corollary 8.15 (Taylor's formula). Let a2R and let r > 0. Let f be a function that is
real analytic on (ar;a +r), with the power series expansion
f(x) =
1
X
j=0
a
j
(xa)
j
; 8x2 (ar;a +r):
Then, for any integer k 0, we have
f
(k)
(a) =k!a
k
;
wherek! = 12


k, and we denote 0! := 1. In particular, we have Taylor's formula
f(x) =
1
X
j=0
f
(j)
(a)
j!
(xa)
j
; 8x2 (ar;a +r):
Exercise 8.16. Prove Corollary 8.15 using Proposition 8.13.
Remark 8.17. The series
P
1
j=0
f
(j)
(a)
j!
(xa)
j
is sometimes called the Taylor series of f
arounda. Taylor's formula says that if f is real analytic, thenf is equal to its Taylor series.
In the following exercise, we see that even if f is innitely dierentiable, it may not be equal
to its Taylor series.
Exercise 8.18. Dene a function f :R!R by f(0) := 0 and f(x) := e
1=x
2
for x6= 0.
Show that f is innitely dierentiable, but f
(k)
(0) = 0 for all k 0. So, being innitely dif-
ferentiable does not imply thatf is equal to its Taylor series. (You may freely use properties
of the exponential function that you have learned before.)
Corollary 8.19 (Uniqueness of power series). Let a2 R and let r > 0. Let f be a
function that is real analytic on (ar;a +r), with two power series expansions
f(x) =
1
X
j=0
a
j
(xa)
j
; 8x2 (ar;a +r):
f(x) =
1
X
j=0
b
j
(xa)
j
; 8x2 (ar;a +r):
Then a
j
=b
j
for all j 0.
Proof. By Corollary 8.15, we have k!a
k
= f
(k)
(a) = k!b
k
for all k 0. Since k!6= 0 for all
k 0, we divide by k! to get a
k
=b
k
for all k 0. 
Remark 8.20. Note however that a power series can have very dierent expansions if we
change the center of the expansion. For example, the function f(x) = 1=(1x) satises
f(x) =
1
X
j=0
x
j
; 8x2 (1; 1):
However, at the point 1=2, we have the dierent expansion
f(x) =
1
1x
=
2
1 2(x 1=2)
=
1
X
j=0
2(2(x 1=2))
j
=
1
X
j=0
2
j+1
(x 1=2)
j
; 8x2 (0; 1):
Note also that the rst series has radius of convergence 1 and the second series has radius
of convergence 1=2.
8.1. Multiplication of Power Series.
Lemma 8.21 (Fubini's Theorem for Series). Let f :NN!R be a function such that
P
(j;k)2NN
f(j;k) is absolutely convergent. (That is, for any bijection g :N!NN, the
sum
P
1
`=0
f(g(`)) is absolutely convergent.) Then
1
X
j=1
(
1
X
k=1
f(j;k)) =
X
(j;k)2NN
f(j;k) =
1
X
k=1
(
1
X
j=1
f(j;k)):
Proof Sketch. We only consider the case f(j;k) 0 for all (j;k)2N. The general case then
follows by writingf = max(f; 0)min(f; 0), and applying this special case to max(f; 0) and
min(f; 0), separately.
Let L :=
P
(j;k)2NN
f(j;k). For any J;K > 0, we have
P
J
j=1
P
K
k=1
f(j;k) L. Letting
J;K!1, we conclude that
P
1
j=1
P
1
k=1
f(j;k) L. Let " > 0. It remains to nd J;K
such that
P
J
j=1
P
K
k=1
> L". Since
P
(j;k)2NN
f(j;k) converges absolutely, there exists
a nite set XNN such that
P
(j;k)2X
f(j;k) > L". But then we can choose J;K
suciently large such thatf(j;k)2 Xgf(j;k): 1 j J; 1 k Kg. Therefore,
P
J
j=1
P
K
k=1
f(j;k)
P
(j;k)2X
f(j;k)>L", as desired. 
Page 5


The radius of convergence of the series
P
1
j=0
2
j
2
(x + 2)
j
is
1
lim sup
j!1
2
j
=
1
0
= +1:
As we now show, the radius of convergence tells us exactly where the power series con-
verges.
Theorem 8.5. Let
P
1
j=0
a
j
(xa)
j
be a formal power series, and let R be its radius of
convergence.
(a) (Divergence outside of the radius of convergence) If x2R satisesjxaj>R, then
the series
P
1
j=0
a
j
(xa)
j
is divergent at x.
(b) (Convergence inside the radius of convergence) If x2R satisesjxaj < R, then
the series
P
1
j=0
a
j
(xa)
j
is convergent at x.
 For the following items (c),(d) and (e), we assume that R > 0. Then, let f : (a
R;a +R) be the function f(x) =
P
1
j=0
a
j
(xa)
j
, which exists by part (b).
(c) (Uniform convergence on compact intervals) For any 0 < r < R, we know that the
series
P
1
j=0
a
j
(xa)
j
converges uniformly to f on [ar;a +r]. In particular, f is
continuous on (aR;a +R) (by Theorem 3.1.)
(d) (Dierentiation of power series) The function f is dierentiable on (aR;a +R).
For any 0 < r < R, the series
P
1
j=0
ja
j
(xa)
j1
converges uniformly to f
0
on the
interval [ar;a +r].
(e) (Integration of power series) For any closed interval [y;z] contained in (aR;a+R),
we have
Z
z
y
f =
1
X
j=0
a
j
(za)
j+1
 (ya)
j+1
j + 1
:
Exercise 8.6. Prove Theorem 8.5. (Hints: for parts (a),(b), use the root test. For part (c),
use the Weierstrass M-test. For part (d), use Theorem 6.3. For part (e), use Theorem 5.4.)
Remark 8.7. A power series may converge or diverge whenjxaj =R.
Exercise 8.8. Give examples of formal power series centered at 0 with radius of convergence
R = 1 such that
 The series diverges at x = 1 and at x =1.
 The series diverges at x = 1 and converges at x =1.
 The series converges at x = 1 and diverges at x = 1.
 The series converges at x = 1 and at x =1.
We now discuss functions that are equal to convergent power series.
Denition 8.9. Leta2R and letr> 0. LetE be a subset ofR such that (ar;a+r)E.
Let f : E!R. We say that the function f is real analytic on (ar;a +r) if and only if
there exists a power series
P
1
j=0
a
j
(xa)
j
centered at a with radius of convergence R such
that Rr and such that this power series converges to f on (ar;a +r).
Example 8.10. The functionf : (0; 2)!R dened byf(x) =
P
1
j=0
j(x1)
j
is real analytic
on (0; 2).
From Theorem 8.5, if a function f is real analytic on (ar;a +r), then f is continuous
and dierentiable. In fact, f is can be dierentiated any number of times, as we now show.
Denition 8.11. LetE be a subset ofR. We say that a functionf : E!R is once dier-
entiable on E if and only if f is dierentiable on E. More generally, for any integer k 2,
we say that f : E!R is k times dierentiable on E, or just k times dierentiable, if
and only iff is dierentiable andf
0
isk1 times dierentiable. Iff isk times dierentiable,
we dene thek
th
derivativef
(k)
: E!R by the recursive rulef
(1)
:=f
0
andf
(k)
:= (f
(k1)
)
0
,
for all k 2. We also dene f
(0)
:=f. A function is said to be innitely dierentiable if
and only if f is k times dierentiable for every k 0.
Example 8.12. The function f(x) =jxj
3
is twice dierentiable onR, but not three times
dierentiable onR. Note that f
00
(x) = 6jxj, which is not dierentiable at x = 0.
Proposition 8.13. Let a2R and let r > 0. Let f be a function that is real analytic on
(ar;a +r), with the power series expansion
f(x) =
1
X
j=0
a
j
(xa)
j
; 8x2 (ar;a +r):
Then, for any integer k 0, the function f is k times dierentiable on (ar;a +r), and
the k
th
derivative is given by
f
(k)
(x) =
1
X
j=0
a
j+k
(j + 1)(j + 2)


 (j +k)(xa)
j
; 8x2 (ar;a +r):
Exercise 8.14. Prove Proposition 8.13.
Corollary 8.15 (Taylor's formula). Let a2R and let r > 0. Let f be a function that is
real analytic on (ar;a +r), with the power series expansion
f(x) =
1
X
j=0
a
j
(xa)
j
; 8x2 (ar;a +r):
Then, for any integer k 0, we have
f
(k)
(a) =k!a
k
;
wherek! = 12


k, and we denote 0! := 1. In particular, we have Taylor's formula
f(x) =
1
X
j=0
f
(j)
(a)
j!
(xa)
j
; 8x2 (ar;a +r):
Exercise 8.16. Prove Corollary 8.15 using Proposition 8.13.
Remark 8.17. The series
P
1
j=0
f
(j)
(a)
j!
(xa)
j
is sometimes called the Taylor series of f
arounda. Taylor's formula says that if f is real analytic, thenf is equal to its Taylor series.
In the following exercise, we see that even if f is innitely dierentiable, it may not be equal
to its Taylor series.
Exercise 8.18. Dene a function f :R!R by f(0) := 0 and f(x) := e
1=x
2
for x6= 0.
Show that f is innitely dierentiable, but f
(k)
(0) = 0 for all k 0. So, being innitely dif-
ferentiable does not imply thatf is equal to its Taylor series. (You may freely use properties
of the exponential function that you have learned before.)
Corollary 8.19 (Uniqueness of power series). Let a2 R and let r > 0. Let f be a
function that is real analytic on (ar;a +r), with two power series expansions
f(x) =
1
X
j=0
a
j
(xa)
j
; 8x2 (ar;a +r):
f(x) =
1
X
j=0
b
j
(xa)
j
; 8x2 (ar;a +r):
Then a
j
=b
j
for all j 0.
Proof. By Corollary 8.15, we have k!a
k
= f
(k)
(a) = k!b
k
for all k 0. Since k!6= 0 for all
k 0, we divide by k! to get a
k
=b
k
for all k 0. 
Remark 8.20. Note however that a power series can have very dierent expansions if we
change the center of the expansion. For example, the function f(x) = 1=(1x) satises
f(x) =
1
X
j=0
x
j
; 8x2 (1; 1):
However, at the point 1=2, we have the dierent expansion
f(x) =
1
1x
=
2
1 2(x 1=2)
=
1
X
j=0
2(2(x 1=2))
j
=
1
X
j=0
2
j+1
(x 1=2)
j
; 8x2 (0; 1):
Note also that the rst series has radius of convergence 1 and the second series has radius
of convergence 1=2.
8.1. Multiplication of Power Series.
Lemma 8.21 (Fubini's Theorem for Series). Let f :NN!R be a function such that
P
(j;k)2NN
f(j;k) is absolutely convergent. (That is, for any bijection g :N!NN, the
sum
P
1
`=0
f(g(`)) is absolutely convergent.) Then
1
X
j=1
(
1
X
k=1
f(j;k)) =
X
(j;k)2NN
f(j;k) =
1
X
k=1
(
1
X
j=1
f(j;k)):
Proof Sketch. We only consider the case f(j;k) 0 for all (j;k)2N. The general case then
follows by writingf = max(f; 0)min(f; 0), and applying this special case to max(f; 0) and
min(f; 0), separately.
Let L :=
P
(j;k)2NN
f(j;k). For any J;K > 0, we have
P
J
j=1
P
K
k=1
f(j;k) L. Letting
J;K!1, we conclude that
P
1
j=1
P
1
k=1
f(j;k) L. Let " > 0. It remains to nd J;K
such that
P
J
j=1
P
K
k=1
> L". Since
P
(j;k)2NN
f(j;k) converges absolutely, there exists
a nite set XNN such that
P
(j;k)2X
f(j;k) > L". But then we can choose J;K
suciently large such thatf(j;k)2 Xgf(j;k): 1 j J; 1 k Kg. Therefore,
P
J
j=1
P
K
k=1
f(j;k)
P
(j;k)2X
f(j;k)>L", as desired. 
Theorem 8.22. Let a2R and let r> 0. Let f and g be functions that are real analytic on
(ar;a +r), with power series expansions
f(x) =
1
X
j=0
a
j
(xa)
j
; 8x2 (ar;a +r):
g(x) =
1
X
j=0
b
j
(xa)
j
; 8x2 (ar;a +r):
Then the function fg is also real analytic on (ar;a +r). For each j 0, dene c
j
:=
P
j
k=0
a
k
b
jk
. Then fg has the power series expansion
f(x)g(x) =
1
X
j=0
c
j
(xa)
j
; 8x2 (ar;a +r):
Proof. Fix x2 (ar;a +r). By Theorem 8.5, both f and g have radius of convergence
Rr. So, both
P
1
j=0
a
j
(xa)
j
and
P
1
j=0
b
j
(xa)
j
are absolutely convergent. Dene
C :=
1
X
j=0


a
j
(xa)
j


; D :=
1
X
j=0


b
j
(xa)
j


:
Then both C;D are nite.
For any N 0, consider the partial sum
N
X
j=0
N
X
k=0


a
j
(xa)
j
b
k
(xa)
k


:
We can re-write this sum as
N
X
j=0


a
j
(xa)
j


N
X
k=0


b
k
(xa)
k



N
X
j=0


a
j
(xa)
j


DCD:
Since this inequality holds for all N 0, the series
X
(j;k)2NN


a
j
(xa)
j
b
k
(xa)
k


is convergent. That is, the following series is absolutely convergent.
X
(j;k)2NN
a
j
(xa)
j
b
k
(xa)
k
:
Now, using Lemma 8.21,
X
(j;k)2NN
a
j
(xa)
j
b
k
(xa)
k
=
1
X
j=0
a
j
(xa)
j
1
X
k=0
b
k
(xa)
k
=
1
X
j=0
a
j
(xa)
j
g(x) =f(x)g(x):
Rewriting this equality,
f(x)g(x) =
X
(j;k)2NN
a
j
(xa)
j
b
k
(xa)
k
=
X
(j;k)2NN
a
j
b
k
(xa)
j+k
: