UNIT - I
SEQUENCES OF FUNCTIONS
Introduction
First we shall discuss a few motivating examples on the behaviour of sequences of functions in
relation to basic concepts like continuity, differentiability and integrability. Then we introduce the notion
of uniform convergence and discuss its usefulness in preserving the basic properties.
1.1
Sequences of Functions
1.
Convergence of sequence of real numbers.
Let { an} be a sequence of real numbers. The sequence { an} is said to converge to l if for
any ∈> 0 there exists a positive integer N such that
an − l <∈ for all n ≥ N .
2.
Definition
Let
x∈ S,
{ f n}
{ fn ( x )}
be a sequence of functions defined on a set S of real numbers. For each
is a sequence of real numbers. As x varies over S we get different sequences
{ fn ( x)} of real numbers. For each
x ∈ S define a function f by
f ( x ) = lim f n ( x )
n →∞
f is called the limit function of the sequence { f n} and we say that { f n} converges pointwise
{ f n}
converges pointwise to f on S if for given ∈> 0 and for each x ∈ S there exists a
positive integer N such that
to f. Thus
f n ( x ) − f ( x) <∈ , for all n ≥ N .
3.
1.
Example
Let { f n} be a sequence of functions where f n ( x) =
x 2n
1 + x 2n
, x∈ ¡.
Show that f n → f where each f n is continuous, but f is not continuous.
Solution :
f n ( x) =
x 2n
1 + x 2n
, x ∈ ¡ , n = 1, 2, 3, .......
1
Since numerator and denominator are polynomial functions and denominator is not zero for
any x ∈ ¡ , the functions f n ’s are continuous for all n. And for x ∈ ¡ ,
lim f n ( x ) = lim
n→∞
x 2n
n →∞ 1 +
x 2n
0, for x < 1
= 1 , for x = 1
2
1, for x > 1
Hence { f n} converges for all x ∈ ¡ and the limit function f is given by,,
0, for x < 1
f ( x ) = 1 , for x = 1
2
1, for x > 1
Clearly f is not continuous at x = ± 1.
Thus { f n} is a sequence of continuous functions which converges to a function f which is not
continuous.
4.
Note
Above example shows that sequence of continuous functions need not converge to a continuous
function.
5.
Example
If { f n} is a sequence of functions defined by
f n ( x) = n2 x (1 − x ) n , 0 < x < 1 , n = 1, 2, 3, ....
Show that f n → f on (0,1) but
1
1
0
0
∫ fn dx → ∫ f
dx .
Solution : For x ∈ (0,1) consider
lim f n ( x ) = lim n x (1 − x ) = 0
2
n→∞
n
n →∞
since (1 − x) < 1
i.e. f n ( x ) → f ( x) = 0 on (0, 1).
1
But f ( x ) = 0 on (0,1) ⇒ ∫ f ( x) dx = 0 .
0
2
On the other hand,
1
∫
0
1
f ( x) dx = ∫ n x (1 − x ) dx
2
n
0
1
=n
2
a
∫ (1 − x) ⋅ x dx
n
(By property
0
1
=n
2
∫ (x
n
n +1
−x
∫
0
a
f ( x) dx = ∫ f ( a − x )dx )
0
) dx
0
1
x n+1 x n+ 2
=n
−
n +1 n + 2 0
2
=
n2
( n + 1)( n + 2)
n2
1
Therefore,
lim ∫ f n ( x ) dx = lim
n→∞
n →∞
0
1
1
0
0
=1
lim ∫ f n ( x ) dx ≠ ∫ f (x ) dx
Thus,
n→∞
1
i.e. f n → f but
∫
1
f n ( x) dx →
0
6.
(n + 1)( n + 2)
∫ f (x )dx .
0
Example
Let { f n} be a sequence of differentiable functions where
f n ( x) =
sin nx
n
, x ∈ ¡ , n = 1, 2, 3, .....
{ }
Show that { f n} converges to a function f but the sequence of the derivatives f n′ need not
converge to f '.
Solution :
Consider,
lim f n ( x ) = lim
n→∞
n →∞
sin nx
n
3
= limsin nx × lim
n →∞
1
n→∞
n
= 0 ,
since −1 ≤ sin nx ≤ 1 for all n.
Thus f n ( x ) → f ( x) = 0 on ¡ . Hence f ′( x) = 0 on ¡ .
Now,
sin nx ′ cos nx
f n′ ( x) =
⋅ n = n cos nx
=
n
n
Therefore, lim f n′ ( x ) = lim n cos nx
n→∞
n →∞
n = ∞ and lim cos nx does not exists.
But nlim
n→∞
→∞
{ }
Hence, lim f n′ ( x) does not exists i.e. f n′ → f ′ on ¡ .
n→∞
{ }
Thus { f n} converges to f but f n′ is not convergetnt on ¡ .
7.
Note
Above examples shows that the pointwise convergence of sequences of functions need not
preserve continuity, integrability or differentiability. Hence we introduce more stronger convergence
concept.
8.
Definition : Uniform Convergence
A sequence of functions { f n} is said to converge uniformly to f on S if for every ∈> 0 there
exists an integer N (∈) , depending only on ∈ , such that
f n ( x ) − f ( x) <∈ for all x ∈ S and for all n ≥ N .
We write f n → f uniformly on S.
9.
Note
(1)
In the case of pointwise convergence for every ∈> 0 we get different values N for different
x ∈ S . For uniform convergence, for every ∈> 0 we have only one N which is independent
of x ∈ S .
(2)
Uniform convergence implies pointwie convergence.
(3)
f n → f uniformly on S.
⇒ f n ( x ) − f ( x ) <∈ for all n ≥ N and for all x ∈ S .
4
⇒ − ∈ < f n ( x ) − f ( x) <∈
⇒ f ( x )−∈< f n ( x ) < f ( x) + ∈, ∀x ∈ S , ∀n ≥ N .
This shows that for all n ≥ N the functions f n lies within a range of 2 ∈ situated symmetrically
about f.
10.
Example
Let { f n} be a sequence of functions where f n ( x) = xn , 0 ≤ x ≤ 1 . Show that { f n} converges
to a function f on [0, 1] pointwise but not uniformly.
Soluton : Consider,
0, if 0 ≤ x < 1
lim f n ( x ) = lim x n =
n→∞
n →∞
1, if x = 1
i.e.
0, if 0 ≤ x < 1
f (x) =
1, if x = 1
Then for any ∈> 0 and for x = 0 or 1 we have,
f n ( x ) − f ( x) = 0 <∈ for all n ≥ 1 and x = 0 or 1.
For 0 < x < 1 we have
f n ( x ) − f ( x ) = x n − 0 = x n <∈ for all n ≥ N where N depends on ∈ and x.
n
1
1
For example if ∈= and x = then < holds for all n ≥ N = 2 .
2
2
2
2
1
1
n
3 3
1
For ∈= and x = , < holds for all n ≥ N = 3 .
2
4 4
2
1
n
9 1
For ∈= and x = , < holds for all n ≥ N = 7 .
2
10 10
2
1
9
Thus N depends also on the point x.
Hence f n ( x ) → f ( x) on [0, 1] pointwise.
The sequence { f n} does not converge to f uniformly. On the contrary if there exists an integer
N such that for ∈=
1
2
, f n ( x ) − f ( x) <
1
2
, ∀n ≥ N and ∀x ∈ (0,1) .
5
1
⇒ xn − 0 <
⇒ xN <
∀n ≤ N
2
1
∀x ∈ (0,1)
2
Taking limit on both sides as x → 1 . Then we get
lim x N <
n→1−
1
2
⇒ 1≤
1
2
which is a contradiction. Hence there is a no N such that
f n ( x ) − f ( x) <
1
2
, ∀n ≥ N and ∀x ∈ [0,1] .
and that is not uniform.
11.
Example
Show that the sequence
{ gn }
of functions converges uniformly where g n ( x) =
0≤ x< ∞.
Solution : g n ( x) =
x
1 + nx
, 0≤ x< ∞
For x = 0, g n ( x) = 0 , ∀n = 1, 2, ....
And for all x > 0, lim gn ( x ) = lim
n→∞
x
n →∞ 1 + nx
=0
gn ( x ) = 0 = g ( x) , for all x ≥ 0 .
Thus nlim
→∞
Now for given ∈> 0 , assume that there is a positive integer N such that,
gn ( x ) − g ( x) <∈ , ∀n ≥ N and ∀x ≥ 0
x
⇒
⇒
x
But
1 + nx
=
1 + nx
x
1 + nx
1
1
x
+n
<
− 0 <∈ , ∀n ≥ N and ∀x ≥ 0
<∈, ∀n ≥ N and ∀x ≥ 0
1
and ∀x ≥ 0
n ∀n
6
x
1 + nx
,
Therefore choose the integer N such that
Then
x
1 + nx
<
1
n
<
1
N
1
N
<∈ or N >
1
∈
.
<∈ , ∀n ≥ N and ∀x ≥ 0 .
⇒ g n ( x) − g ( x ) <∈, ∀n ≥ N and ∀x ≥ 0 .
Hence the sequence { gn } converges to g uniformly on [ 0, ∞) .
12.
Example
Show that the sequence { f n} of functions converges pointwise but not uniformly where
f n (x) =
1
nx + 1
, 0 < x <1
Solution : Consider, lim f n ( x ) = lim
n→∞
n →∞
1
nx + 1
= 0 , ∀x ∈ (0,1)
Thus f n ( x ) → f ( x) = 0 , ∀x ∈ (0,1)
Let ∈> 0 be arbitrary. Then,
f n ( x ) − f ( x ) <∈⇒
⇒
1
nx + 1
1
nx + 1
⇒ n>
− 0 <∈, x ∈ (0,1)
− 0 <∈
11
− 1
x ∈
Therefore, for each x ∈ (0,1) choose an integer N such that N >
11
−1 . And for this
x ∈
integer N we have f n ( x ) − f ( x) <∈ holds for all n ≥ N .
Since N >
11
−1 , it depends on both ∈ and the point x. Hence the convergence is
x ∈
pointwise. Also as x → 0 , N → ∞ and hence no integer N exists for all x ∈ (0,1) . Thus the convergence
is not uniform.
7
Exercise I :
1.
Let { f n} be a sequence of functions defined by
1
f n ( x) = x 1 + , x ∈ ¡ and n = 1, 2, 3, .....
n
Show that { f n} converges uniformly on every bounded interval.
2.
Show that the sequence { f n} of functions defined by f n ( x) =
nx
1+ n2 x2
, n = 1, 2, 3, .... is not
uniformly convergent in any interval containing 0.
{ fn} of functions defined by
f n ( x) = tan −1 nx , x ≥ 0 is uniformly
convergent in any interval [ a, b], a > 0 but only pointwise in [0, b].
3.
Show that the sequence
1.2
1.
Uniform Convergence and Continuity
Theorem
Let { f n} be a sequence of functions and let f n → f uniformly on S. If each f n is continuous
at c ∈ S then the limit function f is also continuous at c.
Proof : Since f n → f uniformly on S, for given ∈> 0 there is a positive integer N such that,
f n ( x ) − f ( x) <
∈
, ∀n ≥ N and ∀x ∈ S
3
Now c ∈ S , and for n = N
⇒ f N ( x) − f ( x) <
and
f N ( c) − f (c ) <
∈
3
, ∀x ∈ S
.... (1)
∈
.... (2)
3
Now each f n is continuous at c ∈ S .Hence f N is also continuous at c.
Therefore for given ∈> 0 there is δ > 0 such that.
f N ( x) − f N (c ) <
∈
3
whenever x − c < δ
Therefore if x − c < δ then we get,
8
.... (3)
f ( x) − f (c ) = f ( x ) −f N ( x) + f N ( x) − fN (c ) + f N (c ) − f ( c)
≤ f ( x ) − f N ( x) + f N ( x) − f N (c ) + f N (c ) − f ( c)
Using (1), (2) and (3) we obtain,
f ( x) − f ( c) <∈ where x − c < δ
This shows that f is continuous at x = c.
2.
Note
The uniform convergence is sufficient condition for preserving continuity but not necessary.
i.e. A sequence { f n} of continuous functions which converges to f (not uniformly) may have a
continuous limit function.
For example, let f n ( x ) = n 2 x n (1 − x) , 0 ≤ x ≤ 1 .
Then each f n is continuous in [0, 1],
lim f n ( x ) = 0 = f ( x) ∀x ∈ [ 0,1]
and
n→∞
Since, f ( x ) = 0 on [0, 1] it is continuous even if the sequence
{ f n}
does not converge
uniformly to f on [0, 1].
Exercise II : Let f n ( x ) =
1
1 + n2 x
, x > 0 and n = 1, 2, 3, ....
Show that each f n is continuous, f n → f where f is also continuous but
{ f n}
does not
converge to f uniformly.
3.
Theorem : (Cauchy condition for uniform convergence)
Let { f n} be a sequence of functions defined on S. Then there exists a function f such that
f n → f uniformly on S if and only if for every ∈> 0 there exists an integer N such that
f m ( x) − fn ( x) <∈, ∀m , n ≥ N and ∀x ∈ S
Proof :
Let f n → f uniformly on S. Then for given ∈> 0 there exists a positive integer N such that
f n ( x ) − f ( x) <
∈
2
, for all n ≥ N and ∀x ∈ S .
9
Taking another integer m ≥ N we get
f m ( x) − f ( x) <
∈
∀x ∈ S
2
Hence, f m ( x) − f n ( x) = f m ( x) − f ( x ) + f ( x ) − fn ( x)
≤ f m ( x) − f ( x) + fn ( x ) − f ( x)
<
∈ ∈
+ =∈
2 2
∀m , n ≥ N and ∀x ∈ S .
Thus for given ∈> 0 there exists a positive integer N such that,
f m ( x) − fn ( x) <∈, ∀m , n ≥ N and ∀x ∈ S . i.e. Cauchy condition holds.
Conversely, suppose that the Cauchy condition is satisfied. Then for given ∈> 0 there exists
an integer N such that
f m ( x) − f n ( x) <
∈
4
, ∀m , n ≥ N and ∀x ∈ S .
But this is a Cauchy condition for a sequence { f n ( x)} of real numbers for each x ∈ S . Hence
it is convergent.
f n ( x ) = f ( x ) , ∀x ∈ S i.e. f ( x ) → f ( x)
Let nlim
n
→∞
We show that f n → f uniformly on S. Let ∈> 0 be given By Cauchy condition, there exists
an integer N such that,
f m ( x) → f n ( x ) <
∈
2
, ∀m , n ≥ N and ∀x ∈ S .
⇒ f n ( x ) → f n+k (x ) <
∈
, ∀n ≥ N and ∀x ∈ S .
2
f n+ k (x ) = f ( x) for all x ∈ S .
Taking limit as k → ∞ we obtain, klim
→∞
Therefore,
fn (x ) − f ( x ) ≤
∈
2
<∈ , ∀n ≥ N and ∀x ∈ S .
Which shows that f n → f uniformly on S.
10
1.3
Uniform Convergence and Riemann Integration
Let R [a, b] denote the set of all Riemann integrable functions on [a, b].
1.
Theorem
Let { f n} be a sequence of functions in R [a, b] such that f n → f uniformly on [a, b].
b
b
a
a
f n ( x ) = ∫ f (x ) dx .
Then f ∈ R[a , b] and nlim
∫ fn (x )dx = ∫ nlim
→∞
→∞
Proof : Given that f n ∈ R[ a, b] , ∀n and f n → f uniformly on [a, b]. First we show that f ∈ R[a , b].
Now f n → f uniformly on [a, b]. Therefore for ∈= 1 there is an integer N such that
f n ( x ) − f ( x ) <∈= 1 , ∀n ≥ N and ∀x ∈ [a , b ] .
⇒ f N ( x ) − f ( x ) < 1 , ∀x ∈ [a , b ]
Since f N ∈ R[ a, b] there exists a positive real number M such that f N ( x ) < M , ∀x ∈ [a , b ].
f ( x) = f ( x) − f N ( x ) + f N ( x )
Therefore,
≤ f ( x ) − f N ( x) + f N ( x)
< 1 + M for all x ∈ [a , b]
Thus f is bounded on [a, b]. Now we prove that f is continuous a.e on [a, b].
Now for each n ∈ N , let En be the set of points of [a, b] at which f n is not continuous. Since
f n ∈ R[ a, b] for all n, En is a set of measure zero for all n = 1, 2, 3, ......
∞
Let E = U En . Then E is also a set of measure zero. Let x ∈ [a , b] − E . Then x ∉ En for all
n =1
n and hence each f n is continuous at x. Since f n → f unformly f is also continuous at x. Thus f is
continuous alomost everywhere on [a, b]. Hence f ∈ R[a , b] .
Next f n → f uniformly on [a, b]. Therefore for given ∈> 0 there is an integer N such that.
f n ( x ) − f ( x) <
∈
,
and ∀x ∈ [a , b ] .
2(b − a) ∀n ≥ N
Therefore,
b
∫
a
b
f n ( x) dx − ∫ f ( x) dx =
a
b
∫ [ fn (x ) − f ( x)] dx
a
11
b
≤ ∫ f n ( x) − f ( x) dx
a
∈
∈
dx = , ∀n ≥ N
2( b − a )
2
a
b
≤∫
b
⇒
b
∫
f n ( x ) dx − ∫ f ( x )dx <∈ , ∀n ≥ N
b
b
a
a
a
a
⇒ ∫ f n ( x) dx converges uniformly to
b
and we have
∫
a
2.
∫ f (x )dx
b
f ( x) dx = lim ∫ f n ( x ) dx
n →∞
a
Note
We have used following property of Riemann integration.
For a bounded function f, f ∈ R[a , b] if and only if f is continuous a.e on [a, b].
Also continuous function on a compact set is bounded.
Hence, f ∈ R[a , b] ⇒ f is continuous on [a, b]
⇒ f is bounded on [a, b].
3.
Theorem
∞
Assume that a seies of functions
∑ uk
k =1
converges uniformly to a function f on [a, b] where
each uk is continuous on [a, b]. For each x ∈ [a , b] define
n x
x
k =1 a
a
g n ( x) = ∑ ∫ uk (t )dt and g ( x) = ∫ f (t ) dt
Then g n → g uniformly and we get
∞ x
x ∞
k =1 a
a k =1
∑ ∫ u k (t)dt = ∫ ∑ uk (t )dt
12
∞
∑ uk
Proof : Since the series
converges uniformly to f on [a, b], the corresponding sequence of
k =1
partial sums converges uniformly to f. Let
{ f n}
be the sequence of partial sums where
n
f n ( x) = ∑ u k ( x) , x ∈ [a , b] .
k =1
Then f n → f uniformly on [a, b]. Therefore by above theorem we have,
x
x
a
a
lim ∫ f n (t ) dt = ∫ f (t )dt
n→∞
x
Therefore, g ( x) = ∫ f (t ) dt
a
x
= lim ∫ f n ( t ) dt
n →∞
a
x n
= lim ∫ ∑ uk ( t ) dt
n →∞
a k =1
x
x
x
= lim ∫ u1 (t ) dt + ∫ u2 (t ) dt + ... + ∫ un ( t ) dt
n →∞
a
a
a
n x
= lim ∑ ∫ uk ( t ) dt
n →∞ k =1
a
= lim g n ( x) , ∀x ∈ [a , b ]
n →∞
Thus, g n → g on [a, b]. And
n x
g ( x) = lim g n ( x) = lim ∑ ∫ uk (t ) dt
n →∞
n →∞ k =1
a
∞ x
g ( x) = ∑ ∫ uk (t )dt
k =1 a
x
∞ x
a
k =1 a
⇒ ∫ f (t ) dt = ∑ ∫ uk (t )dt
13
x ∞
∞ x
a k =1
k =1 a
⇒ ∫ ∑ u k ( t ) dt = ∑ ∫ uk (t ) dt
4.
Note
The above theorem states that term by term integration is possible for a series which converges
uniformly.
Exercise III :
1.
Let f n ( x) =
nx
1+ n2 x2
, x ∈ [0,1]
Show that f n → f on [0, 1] and the convergence is pointwise, but still
1
1
0
0
∫ f ( x)dx = nlim
∫ fn (x )dx holds.
→∞
5.
Note
The above example in the exercise shows that uniform convergence is sufficient but not necessary
for perserving convergence under integration.
6.
Note
If { f n} is a sequence of function which converges to f, uniformly on [a, b] and if f n ’s exists
for each n, then we expect that f ' exists and f 'n → f ' uniformly on [a, b]. But this is not true.
sin nx
converges uniformly on ¡ . But the sequence { f 'n } does not
n
even converge pointwise on ¡ . Hence for preserving differentiability under convergence we must
have some stronger conditions.
For example f n ( x) =
7.
Theorem
Assume that each term in the sequence { f n} is a real valued function having a finite derivative
at each point of an open interval (a, b). Let the sequence { f ( x0 )} converges for at least one point x0
in the interval (a, b). Further assume that there exists a function g such that f 'n → g uniformly on (a,
b). Then
(a)
There exists a function f such that f n → f uniformly on (a, b).
(b)
For each x ∈ (a , b ) the derivative f '( x ) exists and f '( x ) = g ( x ) , ∀x ∈ (a , b) .
14
Proof : Let c ∈ (a , b) be arbitrary. Define a new sequence { gn } where
f n ( x) − f n ( c)
,x ≠ c
g n ( x) =
x−c
f ' (c )
,x=c
n
Now g n ( c) = f 'n (c ) and
.... (1)
{ f 'n }
converges uniformly on (a, b), therefore { g n ( c)} also
converges. We show that { gn } converges uniformly on (a, b). Now for x ≠ c consider..
g n ( x) − g m (x ) =
=
Let,
f n ( x ) − f n (c )
x −c
−
fm ( x ) − fm ( c)
x −c
[ fn ( x ) − fm ( c) ] − [ fn (c ) − fm ( c) ]
x−c
h ( x ) = f n ( x ) − fm ( x)
Therefore,
g n ( x) − g m (x ) =
Now
h( x ) − h (c )
x −c
f 'n exists for each n. Hence
f 'n and
f 'm exists and therefore
h '( x ) = f 'n ( x) − f 'm ( x) also exists. Further h is continuous in [x, c]. Therefore by mean value theorem
there is x1 ∈ ( x, c) such that
h( x ) − h (c )
x −c
⇒
= h ' ( x1 ) , x1 ∈ ( x, c)
h( x ) − h(c )
x −c
= f 'n ( x1 ) − f 'm (x1 )
⇒ g n ( x) − g m ( x) = f ' n( x1 ) − f ' m (x1 )
.... (3)
But { f 'n } converges uniformly on (a, b),
⇒ for given ∈> 0 there exists an integer N such that
f 'n ( x) − f 'm ( x) <∈
∀m , n > N , ∀x ∈ (a , b)
In particular
f 'n ( x1) − f 'm (x1 ) <∈
∀m , n > N
15
Using equation (3) we get,
⇒ gn ( x ) −g m (x ) <∈
∀m , n > N and ∀x ≠ c
⇒ g n converges uniformly on ( a ,b ) − {c} .
But at x = c, { g n ( c)} converges uniformly..
Hence { gn } converges uniformly on (a, b).
Next we prove that { f n} converges uniformly on (a, b).
Let x 0 be the special point for which the sequence { f n ( x0 )} converges.
Taking c = x0 we form the sequence { gn } , where
f n ( x) − f n ( x0 )
, if x ≠ x0
g n ( x) =
x − x0
f ' (x )
, if x = x0
n 0
If x ≠ x0 then we have,
( x − x0 ) g n (x ) = fn (x ) − fn ( x0 )
⇒ f n ( x) = f n ( x0 ) + ( x − x0 ) g n (x )
This equation also holds when x = x 0. Hence the equation holds for all x ∈ (a , b ) .
For m , n ∈ ¢ + we get
f n ( x) − f m ( x ) = f n ( x0 ) − f m ( x0 ) + ( x − x0 ) [ g n ( x) − g m (x ) ]
... (4)
Since { f m ( x0 )} converges, for given ∈> 0 there is a positive intreger N, such that
f n ( x0 ) − fm ( x0 ) <
∈
2
∀m, n ≥ N1
... (5)
Also { gn } converges uniformly on (a, b).
Therefore for given ∈> 0 there is a positive integer N2 such that
gn ( x ) − g m (x ) <
∈
∀m, n ≥ N2 and ∀x ∈ (a , b)
2(b − a)
Let N = max { N1, N2 } . Then
16
... (6)
f n ( x ) − f m ( x) = fn ( x0 ) − f m ( x0 ) + ( x − x0 )( g n ( x) − g m (x ) )
≤ f n ( x0 ) − f m ( x 0) + x − x 0 g n ( x ) −g m (x )
<
∈
2
+
x − x0 ∈ ∈ ∈
< + =∈
b−a 2 2 2
(from (5) and (6))
⇒ f n ( x) − f m ( x) <∈, ∀m , n ≥ N and ∀x ∈ (a , b)
Which shows that { f n} converges uniformly on (a, b). Let f n → f uniformly on (a, b).
Next we prove part (b) of the theorem.
We know that the sequence { gn } defined by (1) at an arbitrary point c ∈ (a , b) converges
uniformly.
gn ( x ) = G( x)
Let nlim
→∞
Now f 'n exists for all n and for all x ∈ (a , b ) . Hence f 'n exists at c and
f 'n (c ) = lim
f n ( x) − f n ( c)
x−c
x→c
= lim g n ( x )
x→c
But f 'n ( c) = g n ( c) for the special point c.
gn ( x ) = g n (c ) , ∀n
Hence we get lim
x→c
Which shows that each g n is continuous at c. And since g n → G uniformly on (a, b).
We get G is also continuous at c.
G( x ) = G( c)
i.e. lim
x→ c
... (7)
But for x ≠ c we have
G( x) = lim gn ( x ) = lim
n →∞
f n ( x ) − f n (c )
n →∞
=
x −c
f ( x ) − f ( c)
x −c
17
⇒ lim G ( x ) = lim
x→c
f ( x) − f (c )
x→c
x −c
⇒ G ( c) = f '(c )
And at c, g n ( c) = f 'n (c ) . Hence
G( c) = lim g n ( c) = lim f 'n ( c) = g (c ) ,
n →∞
n →∞
since f 'n → g unformly on (a, b).
Which shows that f '(c ) = g ( c ) . Since c is arbitrary point of (a, b).
We get f '( x ) = g ( x ) for all x ∈ (a , b ) .
8.
Theorem
Let each f n be a real valued function defined on (a, b) such that f 'n (x ) exists for each
∞
x ∈ (a , b ) . Let for at least one point x0 in (a, b) the series
∑ f n ( x0 ) converges.
n =1
∞
Further assume that there exists a function g such that
∑ f 'n (x) = g ( x) uniformly on (a, b).
n =1
Then :
∞
∑ f n ( x) = f ( x ) uniformly on (a, b).
(a)
There exists a function f such that
(b)
If x ∈ (a , b ) the derivative f '( x ) exists and equal to
n =1
∞
Proof : For the series
∑ f 'n ( x ) .
n =1
∞
∞
n =1
n =1
∑ fn , let { fn} be the sequence of partial sums where Sn = ∑ fk .
∞
Since f 'n exists on (a, b), S 'n = ∑ f 'k also exists on (a, b). Also for at least one point x 0,
n =1
∞
∑ f n ( x0 ) converges uniformly..
n =1
⇒ {S n ( x0 )} converges uniformly..
Therefore by above theorem { Sn } converges uniformly on (a, b).
Let Sn → f uniformly on (a, b).
18
∞
Further it is given that
∑ f 'n (x) = g ( x)
n =1
uniformly on (a, b). Therefore the corresponding
sequence {S 'n (x )} also converges uniformly to g ( x ) , x ∈ (a , b ) .
Therefore f '( x ) exists and f '( x ) = g ( x ) for all x ∈ (a , b ) . (By above theorem)
Thus {S 'n (x )} converges uniformly to f '( x ) on (a, b) and hence
19
∞
∑ f 'n ( x) = f '( x) .
n =1
UNIT - II
DOUBLE SEQUENCE
2.1
Double Sequence
1.
Definition
A function whose domain is ¢ + × ¢ + is called the double sequence. The co-domain of the
function is either ¡ or £ . i.e. f : ¢+ × ¢+ → ¡ (or £ ) is a double sequence either real valued or
complex valued.
Example : Let f be a double sequence defined by
f (m , n ) =
mn
m + n2
2
, ( m, n) ∈ ¢ + × ¢ +
= f (1,1), f (1,2), f (1,3)....
f (2,1), f (2,2),....
f (3,1)....
M
2.
M
Definition : Convergence of double sequence.
Let f be a double sequence. We say that the double sequence f converges to a point a if for
every ∈> 0 there exists an integer N such that
f ( p, q ) − a <∈ , ∀p, q > N .
lim f ( p, q ) = a
We write
p ,q →∞
This limit is called a double limit of the double sequence.
3.
Note
(
)
lim f ( p, q ) exists then it is called row limit of the double sequence. If
If plim
→∞ q →∞
(
)
(
lim lim f ( p, q ) exists then it is called column limit of the double sequence. If lim lim f ( p, q )
q →∞ p →∞
q →∞ p →∞
)
exists then it is called column limit of the double sequence. The row limit or the column limit of the
double sequence
{ f ( m, n)} is called iterated limit. The iterated limits of the double sequence may
differ from the double limit. The following theorem establishes the relation between these limits.
20
4.
Theorem
f ( p, q ) = a .
Let f be a double sequence. Assume that the double limit exists for f and plim
,q →∞
(
f ( p, q ) exists. Then the iterated limit, lim lim f ( p, q )
For each fixed integer p, assume that qlim
→∞
p →∞ q →∞
)
also exists and has the same value a.
f ( p, q ) exists.
Proof : Given that for each fixed p, qlim
→∞
f ( p, q ) = a .
Let F ( p ) = lim f ( p, q) . Also plim
,q →∞
q →∞
Therefore by definition of convergence of double sequence, for given ∈> 0 there exists an
integer N1 such that
f ( p, q ) − a <
∈
2
, ∀p , q > N1
Also for given ∈> 0 and for a fixed integer p > N1 there exists an integer N2 such that
f ( p, q ) − F ( p ) <
∈
2
, ∀q > N2
Therefore for all p > N1 and q > max {N1 , N 2 } we have,
F ( p) − a = F ( p ) − f ( p , q) + f ( p , q ) − a
≤ F ( p ) − f ( p, q ) + f ( p, q ) − a
<
∈ ∈
+ =∈
2 2
⇒ F ( p) − a <∈ , ∀p > N
(
)
i.e. lim F ( p ) = a and hence lim lim f ( p, q ) = a .
p →∞
p →∞ q →∞
Thus the iterated limit exists and has the same value a as that of double limit.
5.
Note :
(1)
The above theorem says that the existence of
lim f ( p, q ) and lim f ( p, q )
q →∞
p ,q →∞
f ( p , q ) ) implies the existence of iterated limit.
(or plim
→∞
(2)
Converse of the above theorem need not be true i.e. iterated limit exists but double limit need
not exists.
21
Following example shows that converse is not true.
6.
Example
Let f ( p, q) =
pq
p 2 + q2
, ( p, q ) ∈ ¢ + × ¢ + .
Then for some fixed integer p,
lim f ( p, q ) = lim
q →∞
q →∞
= lim
q →∞
pq
p + q2
2
( p / q)
p2
2 + 1
q
=0
(
)
lim f ( p, q ) = 0
Therefore plim
→∞ q →∞
Thus iterated limit exists. On the other hand,
lim f ( p, q) = lim
p ,q →∞
pq
p ,q →∞
= lim
p →∞
=
p + q2
2
p2
( for p = q)
p 2 + p2
1
2
f ( p, q) = lim
Now if p = 2q, then plim
,q →∞
q →∞
2q ⋅ q
4q + q
2
2
=
2
5
f ( p, q ) is not unique and therefore double limit does not exists. Thus iterated
Hence plim
,q →∞
limit exists but double limit does not exists.
7.
Example :
Discuss the existence of two iterated limits and the double limit of the double sequence f given
by
f ( p, q) =
1
p+q
22
Solution : For fixed integer p ∈ ¢+ ,
lim f ( p, q ) = lim
q →∞
q →∞
(
1
p +q
=0
)
(
)
lim f ( p, q ) = 0 . Also lim lim f ( p, q ) = 0 .
Therefore plim
→∞ q →∞
q →∞ p →∞
Thus iterted limit exists and it is zero.
Next the double limit is given by,
lim f ( p, q) = lim
p ,q →∞
p ,q →∞
1
p+q
=0
Thus both iterated limit and double limit exist and are equal.
Exercise I
Discuss the existence of two iterated limits and the double limit of the double sequence f given
by
(1) f ( p, q) =
p
(2) f ( p, q) =
p+q
(7) f ( p , q) =
( −1) p
(6) f ( p, q) =
q
p
q
p+q
1
p+ q 1
(4) f ( p , q) = ( −1) +
p q
(3) f ( p , q) = ( −1) p + q
(5) f ( p, q) =
(−1) p ⋅ p
cos p
q
n
∑ sin p
q2
n =1
[Answer : (1) Both iterated limits exists and are equal to ). But the double limit doesnot exists.
(2) Double limit doesnot exists. Iterated limits exists but not equal. (3) Double limit and both the
iterated limits doesnot exists. (4) Double limit exists and zero. But both the iterated limits doesnot
exists. (5) Double limit exists and zero. One of the iterated limit exists and zero, other iterated limit
doesnot exists. (6) Double limit exists and zero. One of the iterated limit exists and zero. Other iterated
limit doesnot exists. (7) Both the iterated limits exists and are equal to 1/2 but the double limit doesnot
exists.]
23
2.2
1.
Uniform Convergence and Double Sequence
Theorem
Let f be a double sequence and let ¢ + denote the set of positive integers.
For each n = 1, 2, 3, .... define a function g n on ¢ + by
g n ( m) = f (m , n) , m ∈¢+
Assume that g n → g uniformly on ¢ + i.e.
g ( m) = lim gn (m) = lim f (m , n )
n →∞
n →∞
(
)
f (m , n ) also exists
= lim lim f ( m, n) exists, then the double limit mlim
If the iterated limit
, n→∞
m→∞ n→∞
and has the same value.
Proof : Since g n → g uniformly on ¢ + .
Therefore for given∈> 0∃ an integer N1 such that
gn ( m) − g ( m) <
∈
2
∀n ≥ N1 and ∀m ∈¢+
,
⇒ f (m , n) − g ( m) <
∈
2
∀n ≥ N1 , ∀m ∈¢+
Now it is given that the iterated limit exists.
(
)
lim f( m, n) = lim g( m) .
Let a = mlim
→∞ n →∞
n →∞
Then for given ∈> 0 there is an integer N2 such that
g ( m) − a <
∈
for all m ≥ N2
2
Let N = max { N1, N2 } .
Then
f (m , n) − a = f (m , n) − g ( m ) + g (m ) − a
≤ f (m , n) − g (m ) + g ( m) − a
<
∈ ∈
+ =∈
2 2
f (m , n) = a .
i.e. mlim
, n→∞
i.e. double limit exists and has the same value a.
24
2.3
1.
Mean Convergence
Definition :
Let
{ f n}
be a sequence of Riemann integrable functions defined on [a, b]. Assume that
f ∈ R[a , b] . The sequence
b
{ f n}
is said to converge in the mean to a function f on [a, b] if
lim ∫ f n ( x ) − f ( x) dx = 0 and we write.
n→∞
2
a
l.i.m f n = f on [ a, b]
n →∞
2.
Theorem
If f n → f uniformly on [a, b] then f n → f in the mean on [a, b].
Proof : If { f n} converges to f uniformly on [a, b] then for any ∈> 0 there exists a positive integer N
such that
f n ( x ) − f ( x) <∈ , ∀n ≥ N and ∀x ∈ [a , b ] .
b
Therefore,
∫
a
b
f n ( x) − f ( x ) dx ≤ ∫ ∈ dx =∈ ( b − a)
2
2
2
a
b
⇒ ∫ f n ( x) − f ( x ) ⋅ dx ≤∈ (b − a ) , ∀n ≥ N and ∀x ∈ [a , b ] .
2
2
a
Since ∈> 0 is arbitrary we have,
b
lim ∫ f n ( x ) − f ( x) dx = 0
n→∞
2
a
l.i.m f n = f on [a, b]
n →∞
This shows that { f n} converges to f in the mean. Thus uniform convergence implies convergence
in the mean.
3.
Note
Converse of the above need not be true i.e. mean convergence does not imply uniform
convergence or even pointwise convergence.
For example, for each n ≥ 0 divide the interval [0, 1] into 2 n equal parts.
{
}
Let I1, I 2 ,...., I k ,...., I 2n be the subintervals.
25
k k + 1
Then, I k = n , n ,
2
2
k = 0, 1, 2, .... (2n – 1)
Define function f n on [0, 1] by,,
1 if x ∈ I n
f n ( x) =
0 if x ∈ [0,1] − I n
1
n n +1
where I n = n , n and l ( I n ) = n
2 2
2
Now each f n is a constant function and hence it is Riemann integrable. Therefore,
1
∫
0
1
f n ( x) − 0 dx = ∫ f n ( x) dx = ∫ f n ( x) dx +
2
2
∫
2
0
In
2
f n ( x) dx
[0,1] − In
1
∫ 1dx = l ( I n ) = 2n
=
In
Therefore,
1
lim ∫ f n ( x ) − 0 dx = lim
n→∞
0
1
2
n →∞
2n
= 0 , ∀x ∈ [0,1]
⇒ l.i.m f n = 0 on [0, 1]
n →∞
Thus { f n} converges to zero in the mean.
But for the sequence { f n} ,
limsup f n ( x ) = 1 and liminf f n ( x) = 0 , ∀x ∈ [0,1] .
n→∞
n→∞
f n ( x ) does not exists ∀x ∈ [0,1] .
Hence nlim
→∞
i.e. { f n} is not convergent pointwise on [0, 1].
Thus mean convergence does not imply pointwise convergence.
4.
Definition : Cauchy Schwarz Inequality
If f and g are square integrable functions over an interval I then
∫
I
2
f ( x ) ⋅ g ( x ) dx ≤ ∫ [ f ( x) ] dx
I
1
26
2
[ g ( x) ]2 dx
∫
I
1
2
2
f ( x ) ⋅ g ( x ) dx ≤ [ f ( x) ]2 dx [ g ( x ) ]2 dx
∫
∫
∫
I
I
I
i.e.
5.
Theorem
Let l.i.m f n = f on [a, b]. If g ∈ R[ a, b] then define function
n →∞
x
x
a
a
h ( x ) = ∫ f (t )g (t ) dt and hn ( x) = ∫ f n (t ) g (t ) dt where x ∈ [a , b] .
Then hn → h uniformly on [a, b].
Proof : Since l.i.m f n = f on [a, b], we have,
n →∞
b
lim ∫ f n (t ) − f (t ) dt = 0
n→∞
2
a
Therefore for given ∈> 0 there exists a positive integer N such that
b
∫
2
f n (t ) − f (t ) dt <
a
b
∈2
A , ∀n ≥ N
where A = 1 + ∫ g (t ) dt
2
a
By Cauchy-Schwarz inequality we have
2
x
x
x
2
2
0 ≤ ∫ f n ( t ) − f (t ) g ( t) dt ≤ ∫ f n ( t ) − f (t ) dt × ∫ g (t ) dt
a
a
a
≤
∈2
A
× ( A − 1) <∈2 , ∀n ≥ N
x
⇒
∫
a
f n (t ) − f (t ) g ( t) dt <∈ , ∀n ≥ N and ∀x ∈ [a , b ] .
x
⇒
∫
f n (t ) g (t ) − f (t ) g ( t) dt <∈
a
⇒ hn ( x ) − h ( x ) <∈, ∀n ≥ N and ∀x ∈ [a , b ]
⇒ hn → h uniformly on [a, b].
27
6.
Note
Next we obtain the generalization of the above theorem.
7.
Theorem
Let l.i.m f n = f and l.i.m g n = g on [a, b].
n →∞
n →∞
x
x
a
a
Define h ( x ) = ∫ f (t )g (t ) dt and hn ( x) = ∫ f n (t ) gn (t )dt , x ∈ [a , b] .
Then hn → h uniformly on [a, b].
Proof : Consider
x
x
a
a
hn ( x) − h( x ) = ∫ f n ( t ) gn (t ) dt − ∫ f (t )g (t )dt
x
= ∫ [ f n (t ) g n ( t ) − f (t )g (t ) ] dt
a
x
= ∫ ( f − fn )( g − g n ) + f n g − fg + f ⋅ g n − fg dt
a
x
x
x
x
x
a
a
a
a
a
= ∫ ( f − fn )( g − g n ) dt + ∫ f n gdt − ∫ f ⋅ gdt +∫ fg ndt − ∫ fgdt
By above theorem
x
x
a
a
∫ fn gdt → ∫ fgdt
and
x
x
a
a
... (1)
∫ fg ndt → ∫ f ⋅ gdt
⇒ For given ∈> 0∃N1, N 2 , such that
x
∫
a
x
∫
a
x
f n gdt − ∫ f ⋅ gdt <
a
x
fg ndt − ∫ f ⋅ gdt <
a
∈
3 , ∀n ≥ N1 and
... (2)
∈
3 , ∀n ≥ N2 .
...(3)
Also f n → f in the mean and g n → g in the mean.
Therefore for given ∈> 0∃ an integer N3 and N4 such that,
28
x
∫
2
f n (t ) − f (t ) dt <
a
x
∫ gn (t ) − g (t )
2
dt <
a
∈
3 , ∀n ≥ N3
... (4)
∈
3 , ∀n ≥ N4
... (5)
By Cauchy-Schwarz inequality we have,
2
x
x
x
∈ ∈ ∈2
2
2
g − g n dt < ⋅ =
∫ ( f − fn )( g − gn ) dt ≤ ∫ f − f n dt
∫
3 3 9
a
a
a
x
⇒
∫( f
− fn )( g − g n ) dt <
a
∈
... (6)
3
From (1) we obtain for n ≥ N = max { N1, N2 , N3 , N4}
hn ( x ) − h( x) ≤
x
x
x
a
a
a
∫ ( f − fn )( g − g n ) dt + ∫ fn gdt − ∫ fgdt
x
+
∫
a
x
f ⋅ g ndt − ∫ f ⋅ gdt <
a
⇒ hn ( x ) − h ( x ) <∈, ∀n ≥ N , ∀x ∈ [a , b ] .
Which proves that hn → h uniformly on [a, b].
29
∈ ∈ ∈
+ + =∈
3 3 3
UNIT - III
SERIES OF FUNCTIONS
3.1
1.
Series of Functions
Introduction
Convergence of series is associated with the convergence of corresponding sequence of partial
sums. Here we define uniform convergence of series of functions defined on a set S of real numbers.
Some equivalent conditions for uniform convergence are proved. Particularly Weierstrass M-test,
Dirichlets test for uniform convergence are proved and numbers of examples are discussed.
2.
Definition
∞
Let { f n} be a sequence of functions defined on S. The infinite sum ∑ f n ( x) , x ∈ S , is called
n =1
a series of functions.
n
Let Sn ( x ) = ∑ f k ( x) , x ∈ S . Then { Sn } is a sequence of functions defined on S. If the
k =1
sequence { Sn } coverges to f on S, then the infinite series
∞
∑ fn ( x) is said to be convergent and
n =1
∞
∑ f n ( x) = f ( x ) , x ∈ S .
n =1
∞
We say that
3.
∑ fn
converges to f on S.
n =1
Definition : Uniform convergence of infinite series of functions.
Let { f n} be a sequence of functions defined on a set S. For each x ∈ S , let
n
Sn ( x ) = ∑ f k ( x) , n = 1, 2, 3, ....
k =1
If there exists a function f such that { Sn } converges uniformly to f on S. Then we say that the
∞
series
∑
n =1
∞
f n converges uniformly on S to the function f. We write
30
∑ fn =
n =1
f on S uniformly..
4.
Theorem : (Cauchy’s condition for uniform convergence of series of functions)
∞
The infinite series
∑ fn ( x) , x ∈ S
n =1
n+ p
there is an integer N such that,
∞
Proof : Let
∑ fn ( x)
n =1
∑
k = n+1
converges uniformly on S if and only if for every ∈> 0
f k ( x) <∈, ∀n ≥ N , ∀x ∈ S and p = 1, 2, 3, ....
converges uniformly ∀x ∈ S . Let { Sn } be a sequence of functions where
∞
n
Sn ( x ) = ∑ f k ( x) , n = 1, 2, 3, .... and x ∈ S . Since
∑ fn ( x)
k =1
n =1
converges uniformly to f on S the
corresponding sequence { Sn } of partial sums, also converges uniformly to f. Therefore { Sn } is a
Cauchy sequence. For given ∈> 0 there is a positive integer N such that
S m (x ) −S n (x ) <∈ , ∀m , n ≥ N and ∀x ∈ S .
We take integers m and n such that m > n > N.
⇒
⇒
m
n
k =1
k =1
∑ fk (x ) − ∑ f k ( x) <∈, ∀x ∈ S and ∀m > n > N .
m
∑
k = n+1
f k ( x ) <∈, ∀x ∈ S and ∀m > n > N .
Let m – n = p then p is a positive integer and m = n + p.
n+ p
Therefore we get
∑
k = n+1
f k ( x) <∈, ∀x ∈ S and ∀n > N , p = 1, 2, 3, .....
Thus Cauchy condition holds.
Conversely, suppose that for given ∈> 0 there exists an integer N such that n > N implies
n+ p
∑
k = n+1
f k ( x) <∈, ∀x ∈ S and p = 1, 2, 3, .....
⇒ f n+1( x) + fn+ 2 ( x) + .... + f n+ p ( x ) <∈, ∀x ∈ S and p = 1, 2, 3, .....
n
⇒ ( f1 ( x ) + f 2 ( x ) + .... + f n+ p ( x) ) − ∑ fk ( x ) <∈ , ∀x ∈ S and p = 1, 2, 3, .....
k =1
31
n+ p
n
∑
⇒
f k ( x) − ∑ f k ( x) <∈ , ∀x ∈ S and p = 1, 2, 3, .....
k =1
k =1
Let n + p = m then m > n > N and
S m (x ) −S n (x ) <∈ , ∀x ∈ S and ∀m > n > N .
This show that Cauchy’s condition for uniform convergence holds. Hence the sequence
∞
{S n}n=1 of partial sums converges uniformly..
S n ( x) = f ( x) , ∀x ∈ S
Let nlim
→∞
∞
Then Sn → f uniformly on S and hence the given series
∑ fk ( x) converges uniformly to
n =1
f ( x ) on S.
∞
i.e.
∑ fn =
n =1
f uniformly on S.
Next we derive an important test for uniform convergence due to Karl Weierstrass (1815–
1897).
5.
Theorem : (Weierstrass M-Test)
Let {M n } be a sequence of nonnegative real numbers such that
f n ( x ) ≤ M n , ∀ n = 1, 2, 3,.... and ∀x ∈ S .
∞
∞
n =1
n =1
∑ fn ( x) converges uniformly on S if ∑ M n
Then
converges.
∞
Proof : Let
∑ Mn
be convergent. The for given ∈> 0 there exists an integer N such that
n =1
n+ p
∑
k = n+1
Now
M k <∈ , ∀n ≥ N and p = 1, 2, 3, ....
f n ( x ) ≤ M n , ∀x ∈ S and n = 1, 2, 3, ....
⇒
n+ p
∑
k =n +1
⇒
n+ p
∑
k = n +1
f k ( x) <
f k ( x) <
n+ p
∑
k = n+1
Mk
n+ p
∑
k =n+1
M k <∈ , ∀n ≥ N , ∀x ∈ S and p = 1, 2, 3, ....
32
⇒
n+ p
∑
k = n+1
f k ( x ) <∈, ∀n ≥ N , ∀x ∈ S and p = 1, 2, 3, ....
∞
Therefore by Cauchy’s condition for uniform convergence of series,
∑ fn ( x)
n =1
converges
uniformly on S.
6.
Examples :
∞
1.
Show that the
∑ x n cos nx , ( 0 < x < 1 ) converges uniformly..
n =1
Solution : Let f n ( x ) = x n cos nx , 0 < x < 1
Therefore f n ( x ) = x n cos nx = x n ⋅ cos nx ≤ x n
⇒ f n ( x ) ≤ xn , 0 < x < 1 and ∀ n = 1, 2, 3, .....
∞
Consider the series
∑ x n , 0 < x < 1 . This is an infinite geometric series which converges for
n =1
∞
n
all x ∈ (0,1) . Hence by Weierstrass M-test the given series ∑ x cos nx converges uniformly
n =1
for all x ∈ (0,1) .
∞
2.
Prove that ∑ an sin nx and
n =1
∞
∑ an cos nx are uniformly convergent on ¡ if
n =1
∞
∑ an
n =1
converges.
Solution : Since sin nx ≤ 1 and cos nx ≤ 1 for all x ∈ ¡ . We get an sin nx ≤ an and
∞
an cos nx ≤ an , ∀n and ∀x ∈¡ . Now if
∞
the series ∑ an sin nx and
n =1
∞
3.
Show that the series
∑
n =0
∑ an
n =1
converges then by Weierstrass M-test
∞
∑ an cos nx converges uniformly on ¡ .
n =1
( −1) n x2 n+1
(2n + 1)!
converges uniformly in every finite interval (a, b).
Solution : Let (a, b) be the given finite interval (a < b). Let M be the positive real number such that
a <M , b<M.
33
Let f n ( x ) =
⇒ f n ( x) ≤
( −1)n x 2n+1
. Then f n ( x ) =
(2n + 1)!
M 2n+1
x 2n+1
(2n + 1)!
, a <x <b.
n = 1, 2, 3, .... (Since a < M , b < M )
(2n + 1)!
Consider the series
∞
M 2n+1
M
∑ (2n + 1)! = 1! +
M3
n =0
=
=
3!
+
M5
5!
+ ....
M M2
1 M M 2
1 +
+
+ .... − 1 − +
+ ....
2
1!
2!
1!
2!
1
2
{e M
− e− M } = sin hM
∞
This shows that the infinite series
M 2n+1
∑ (2n + 1)! converges.
n =0
∞
Therefore by Weierstrass M-test the given series
∑
n =0
( −1) n x2 n+1
(2n + 1)!
converges uniformly for all
x in (a, b).
Exercise I : Discuss the uniform convergence of the following series.
∞
1)
x
∑ (n + x2 )2 on any finite interval (a, b).
n =0
∞
2)
3)
n =1
∞
cos nx
n =1
n2
∑
∞
4)
1
∑ (n + x)β
, x≥0
, x∈ ¡
1
∑ x n , x ∈ [0,1] or (0,1]
n =1
∞
5)
∑ sinn x , x ∈ [0, π ]
n =1
34
∞
6)
∑ e − nx x n , x ∈ [0,2]
n =1
∞
7)
∑
−x
2
e
n =1
n
∞
xn
n
, x∈ ¡
8)
∑ 1 + x 2n , x ∈ [ 0, ∞ )
9)
∑ n (1 + nx 2 ) , x ∈ [ a, b]
7.
Definition : Uniform Bound for Sequence
n =1
∞
x
n =1
A sequence { f n} is said to be uniformly bounded on S if there exists a constant M > 0 such
that f n ( x ) ≤ M for all x ∈ S and for all n.
The number M is called uniform bound for { f n} . If each f n is bounded and f n → f uniformly
on S then
8.
{ f n}
is uniformly bounded on S.
Example
∞
Prove that the series
Solution : Let f n ( x ) =
x
∑ nα (1 + nx 2 )
n =1
converges uniformly on any finite interval if α >
x
nα (1 + nx2 )
To obtain maxima for f n ( x) we differentiate w.r.t. x
⇒ f 'n ( x ) =
=
nα (1 + nx 2 ) − xnα ( 2nx )
nα (1 + nx 2 )
2
nα [1 − nx 2 ]
n 2α (1 + n 2 x )
2
α
2
Therefore f 'n (x ) = 0 ⇒ n (1 − nx ) = 0 ⇒ x = ±
35
1
n
1
2
.
−1
1
, f ''n (x ) =
= α −1 < 0
n
n 2n 2
1
Also when x =
Hence f n ( x) if maximum when x =
1
n
1
Thus f n ( x ) =
x
α
(1 + nx )
1
for all n
n
2
≤
n
1
nα 1 + n ⋅
n
1
=
2n
α+ 1
2
1
⇒ fn ( x ) ≤
α+
n
2
∞
1
n =1
α+
Consider the series ∑
n
1
. This series converges is α +
2
2
Weierstrass M-test the given series converges uniformly when α >
9.
1
1
2
> 1 i.e. α >
1
2
. Therefore by
.
Example
∞
Show that the series
x
∑ ( nx + 1) {(n −1) x + 1}
n =1
is uniformly convergent on any interval [a, b],
0 < a < b but only pointwise on the interval [0, b].
Solution : Let f n ( x) = (
x
,
nx + 1) {( n −1) x + 1} a ≤ x ≤ b
Let { Sn } be a sequence of partial sums of the given series where
n
Sn ( x ) = ∑
k =1 ( kx + 1)
x
( ( k − 1) x + 1)
n 1
1
= ∑
+
k =1 kx + 1 ( k −1) x + 1
n
1
1
= ∑
−
k =1 ( k − 1) x + 1 kx + 1
36
1 1
1
1
1
= 1 −
−
−
+
+ ... +
x + 1 x + 1 2 x + 1
( n − 1) x + 1 nx + 1
= 1−
1
nx + 1
nx
Sn ( x) =
nx + 1
Therefore lim S n ( x) = lim
n→∞
nx
n→∞ nx + 1
=1
This shows that { Sn } of partial sums converges to 1. Hence the gives series also converges to
1. Further, we show that { Sn } converges uniformly to 1. Let ∈> 0 be given. Let N be a positive
integer such that
S n ( x )− 1 <∈, ∀n ≥ N and ∀x ∈ [ a , b ]
⇒
⇒
⇒
nx
nx + 1
−1
nx + 1
1
nx + 1
− 1 <∈ , ∀x ≥ N and ∀x ∈ [ a , b ]
<∈
<∈
11
⇒ −1 < n
x ∈
⇒ n>
11
− 1 ∀n ≥ N and ∀x ∈ [ a , b ]
x ∈
Choose an integer N >
11
− 1 .
a ∈
Then for all n ≥ N ,
S n ( x )− 1 <∈ ∀x ∈ [ a , b ]
This shows that { Sn } converges to 1 uniformly on [a, b].
Therefore, the given series converges uniformly to 1 on [a, b] where 0 < a.
37
But on the interval [0, b],
11
− 1 → ∞ as x → 0
x ∈
Hence, there is no integer N such that N >
11
−1 , ∀x ∈ [ a , b ] . Therefore the series
x ∈
converges only pointwise on [0, b].
Exercise II :
∞
1,
Show that
x
∑ n p + x 2n q
n =1
converges uniformly on any finite interval [a, b] where q > 0 and
p >1.
∞
2.
Show that
x
converges uniformly in the interval [ −k , k ] , ∀k ∈ ¡ .
1
converges uniformly in [ −k , k ] , ∀k ∈ ¡ .
∑ n4 + n 2 x2
n =1
∞
∑ n2 + n 4 x2
3.
Show that
10.
Theorem
n =1
∞
Let
∑ fn
n =1
be the series of functions which converges uniformly to a function f on S. If each
f n is continuous at x0 ∈ S then f is also continous at x0 .
Proof : Let { Sn } be the sequence of partial sums correspoonding to the given series
∞
∑ fn , where
n =1
n
Sn ( x ) = ∑ f k ( x) , x ∈ S , n = 1, 2, 3, ....
k =1
∞
Since
∑ fn
n =1
converges uniformly to f, the sequence { Sn } also converges uniformly to f on S.
S n ( x) = f ( x) , ∀x ∈ S
i.e. nlim
→∞
Now each f n is constinuous at x0 ∈ S . Therefore each Sn is also continuous at x0 ∈ S
(finite sum of continuous functions is continuous). Also Sn → f uniformly on S. Therefore the limit
∞
function f is also continuous at x0 . Thus if ∑ f n converges uniformly to f and if each f n is continuous
n =1
at x0 ∈ S then the limit function f is also continuous at x0 ∈ S .
38
11.
Note
Whenever Weiestrass M-text is not applicable there are other test that may be useful. One
such test is the following Dirichlet’s test for uniform convergence. First we prove Abel’s partial summation
formula for series.
12.
Theorem
∞
Let
∑ fn
n =1
be a series of functions and let Fn = f 1 + f 2 + .... + f n . Let { gn } be any sequence
n
n
k =1
k =1
of functions. If Sn ( x ) = ∑ f k ( x) gk ( x ) then Sn = ∑ Fk ( g k − g k +1 ) + Fn ⋅ g n+1 .
Proof : Consider the expression
n
∑ Fk [ g k − g k +1 ] + Fn ⋅ gn +1
k =1
= F1 [ g1 − g 2 ] + F2 [ g2 − g3] + F3 [ g3 − g4 ] + .... + Fn [ gn − g n+1] + Fn ⋅ g n+1
= f1 [ g1 − g2 ] + ( f1 + f 2 ) [ g 2 − g3 ]+ ( f1+ f 2 + f 3 ) [ g 3 − g4 ] + .... + ( f1 + f 2 + .... + f n ) gn
= f1g1 − f1g 2 + f1g 2 − f1 g3 + f 2 g 2 − f 2 g 3 + f1g3 + f 2 g3 + f 3 g3 − f1g 4 − f 2 g4 − f 3 g 4 + ..
− f 2 g4 − f 3 g4 + .... + f1g n + f 2 g n + .... + f n g n
= f1g1 + f 2 g2 + f 3g 3 + .... + f n g n
∞
= ∑ f k gk
n =1
= Sn
∞
Thus, Sn = ∑ Fk [ g k − g k +1 ] + Fn ⋅ g n+1
n =1
where Fn = f 1 + f 2 + .... + f n , n = 1, 2, 3, .....
13.
Theorem : (Sufficient condition for uniform convergence - Dirichlet’s Test)
∞
Let Fn ( x ) be the
nth
partial sum of a series
∞
∑ fn ( x) where each
n =1
function defined on a set S. Assume that {Fn}n=1 is uniformly bounded on S.
39
f n is a complex valued
Let { gn } be a sequence of real valued functions such that g n+1 (x ) ≤ g n ( x) for all x ∈ S and
∞
∑ fn ( x) gn (x ) converges uniformly on S.
let g n → 0 uniformly on S. Then the series
Proof : Let Sn ( x )
n =1
be the nth partial sum of the series
∞
n
n =1
k =1
∑ fn ( x) gn (x ) . Then Sn (x ) = ∑ fk ( x) gk (x ) .
Then by Abel’s partial summation theorem, ∀x ∈ S ,
n
Sn ( x ) = ∑ Fk ( x) [ gk ( x ) − g k +1 ( x) ] + Fn ( x) gn+1( x)
k =1
Hence for n > m, we can write
n
m
k =1
k =1
Sn ( x ) − S m ( x ) = ∑ f k ( x) gk ( x ) − ∑ f k ( x) gk ( x )
n
= ∑ Fk ( x) [ gk ( x) − gk +1( x) ] + Fn ( x ) g n+1 ( x )
k =1
m
− ∑ Fk ( x) [ gk ( x ) − g k +1( x) ] − Fm ( x) gm +1 ( x )
k =1
=
n
∑
k = m+1
Fk ( x) [ gk ( x ) − g k +1( x) ] − Fn ( x) gn+1( x) − Fm ( x ) g m+1( x)
Now {Fn } is uniformly bounded on S. Therefore there exists positive real number M such
that
Fn ( x ) ≤ M , ∀x ∈ S and ∀ n = 1, 2, 3, .....
Therefore,
S n ( x) − S m ( x) ≤
≤
n
∑
k = m+1
n
∑
k = m+1
≤M
Fk ( x) [ gk ( x ) − g k +1 ( x ) ] + Fn ( x) gn +1 ( x) + Fm ( x) gm +1 ( x )
Fk ( x) gk ( x ) − gk +1 ( x) + Fn ( x) gn+ 1( x) + Fm ( x) gm +1( x )
n
∑
k =m +1
g k ( x ) − g k +1 (x ) + M g n+1 ( x ) − g m+1( x)
40
n
⇒ S n ( x) −S m ( x) ≤ M ∑ ( g k ( x ) − g k +1 ( x) ) + g n+ 1( x) + gm+1 ( x )
k = m+1
≤ M {g m+1( x) − g m+2 ( x ) + g m +2 ( x) − gm +3 ( x) + .... + g n ( x) − gn+1( x) + gn +1( x ) + g m+1( x)}
= 2Mgm +1 ( x )
⇒ S n ( x) −S m ( x ) ≤ 2 Mg m+1( x) , ∀n > M , ∀x ∈ S
Now { gn } converges uniformly to zero. Hence for given ∈> 0 there exists N such that,
∈
⇒ g n ( x) − 0 <
⇒ gm( x) <
2M
∈
, ∀m ≥ N and ∀x ∈ S
2M
⇒ gm +1 ( x ) <
, ∀n ≥ N and ∀x ∈ S
∈
, ∀m ≥ N and ∀x ∈ S
2M
Therefore
S n ( x) − S m ( x) < 2 M ⋅
∈
2M
=∈, ∀n > m ≥ N and ∀x ∈ S .
This shows that Cauchy condition for uniform convergence of the sequence { Sn } holds.
Hence { Sn } converges uniformly. But Sn is the nth partial sum of series
∞
∑ fn ( x) gn (x ) .
n =1
Hence the series converges uniformly on S.
14.
Example
Show that
∞
e inx
n =1
n
∑
converges uniformly. Also obtain the domain of uniform convergence.
Solution : Let f n ( x ) = einx and g n ( x) =
1
n
Then { gn } is a decreasing sequence and g n → 0 uniformly on ¡ .
n
Now,
Fn ( x ) = ∑ fk ( x )
n =1
41
= f1( x) + f 2 ( x) + .... + f n ( x)
= eix +e 2ix + .... + e nix
= ( eix ) + ( eix ) + .... + ( eix )
1
2
n
einx − 1
= e ix
e −1
ix
inx
e
=e ⋅
ix
(e
(e
inx
2
ix
e2
−
2
−e
ix
−
2
−e
inx
e 2
n +1
i
x
2
=e
⋅
ix
e2
=
n +1
i
x
e 2
Therefore, Fn ( x) = e
Bur,
n +1
i
x
e 2
2
ix
2
)
)
−
inx
−
ix
−e
2
2i
−e
2
2i
nx
sin
2
⋅
x
sin
2
n +1
i
x
2
=e
inx
n +1
i
x
2
nx
sin
⋅ 2
x
sin
2
nx
sin
2
x
sin
2
n +1
n +1
= cos
x + i sin
2
2
42
n +1
2 n +1
= cos2
+ sin
=1
2
2
nx
and sin ≤ 1 for all n.
2
Hence Fn ( x) ≤
1
x
sin
2
Let δ > 0 be a real number such that x ∈ [δ , 2π − δ ]
1
Then
x
sin
2
Hence, Fn ( x ) ≤
≤
1
sin
δ , x ∈ [δ , 2π − δ ]
∀
2
1
δ
sin
2
, ∀ x ∈ [δ , 2π − δ ] where 0 < δ < π
Thus {Fn } is uniformly bounded on [ δ , 2π − δ ] for any real number δ , 0 < δ < π .
Therefore by Dirichlet’s test the series
∞
∞
e inx
n =1
n =1
n
∑ fn ( x) gn (x ) = ∑
converges uniformly on the
domain [ δ , 2π − δ ] , 0 < δ < π .
Exercise III :
1.
Let { gn } be a monotonic decreasing sequence of function defined on S and let g n → 0
∞
uniformly on S. Prove that
∑ (−1)n−1 g n ( x) converges uniformly on S.
n =1
( −1)n
x
sin 1 + converges uniformly on every compact subset
Prove that the series ∑
n n
n =1
∞
2.
of ¡ .
43
3.
Let { gn } be a monotonic decreasing sequence which converges to zero. Show that
∞
∞
∑ g n (x )sin nx and
∑ g n ( x)cos nx converges uniformly in any interval [a, b],
n =1
n =1
where 0 < a ≤ x ≤ b ≤ 2π .
∞
(−1) n−1
n =1
nx
∑
converges uniformly in [ δ , ∞ ) ∀δ > 0 .
4.
Show that the series
5.
Show that the following series converges uniformly
∞
(i)
(−1) n−1
∑ n + x2
n =1
, x∈ ¡
∞
(ii)
∑ (−1)n−1 x n , 0 ≤ x < 1
n =1
∞
(iii)
(iv)
(v)
3.2
∑
(−1) n−1 x n
n +1
n =1
∞
∑
sin nx
n =1
np
∞
cos nx
n =1
np
∑
, x ≤ δ ,0 < δ < 1
, p > 0,0 < a ≤ x ≤ b < 2π
, p > 0,0 < a ≤ x ≤ b < 2π
Rearrangement of Series
Let Z+ = {1, 2, 3, ....} be the set of positive integers. A function F : S → ¡ is called one-toone on S if an only if F ( x) = F ( y ) implies x = y, for every x, y ∈ S . Such functions are also called
injective functions. These functions are important because they possess inverse in more general sense.
1.
Definition
Let f : ¢+ → ¢+ be a one-to-one function on ¢ + . Let
such that
bn = a f( n ) for n = 1, 2, 3, ....
Then
∑ bn
is said to be a rearrangement of
44
∑ an .
∑ an
and
∑ bn
be the two series
Infinite Series :
2.
Definition
Let { an} be a sequence of real or complex numbers. Let { Sn } be a new sequence defined as
follows
n
Sn = a1 + a2 +.... +an = ∑ ak , n = 1, 2, 3, ....
k =1
∞
The sum Sn is called nth partial sum of the series ∑ ak . The infinite series
k =1
∞
∑ ak
k =1
is said to
converge if the sequence { Sn } is convergent.
The series diverges if the sequence { Sn } diverges.
∞
Since k is dummy variable we denote the same infinite series by
∑ an
n =1
or
∑ an .
n
Cauchy’s condition for convergence of series.,
The series
∑ an
converges if and only if for every ∈> 0 there exists an integer N such that
an+1 + an+ 2 + .... + an+ k <∈ for all N > N and for all p = 1, 2, ....
3.
Definition
A series
∑ an
is called absolutely convergent if
n
convergent if
∑ an
converges. It is conditionally
n
converges but
n
4.
∑ an
∑ an
diverges.
n
Note
Absolutely convergence implies convergence but the converse is not true.
For example the series
∑
n
(−1) n+1
n
1
is convergent but does not converge absolutely as
∑n
n
is divergent series.
45
5.
Theorem
Let
∑ an
be an absolutely convergent series and converges to S. Then every rearrangement
n
of
∑ an
also converges absolutely and has sum S.
n
Proof : Let ∑ an be the given series. Let f : ¢+ → ¢+ be a one-to-one function. Define a sequence
n
{bn } by bn = a f ( n) , n = 1, 2, 3, ....
∞
b1 + b2 + .... + bn = a f(1) + a f(2) + .... + a f ( n) ≤ ∑ ak
Then
k =1
∞
Since ∑ ak converges absolutely,
k =1
n
∑ bk
Therefore,
k =1
∞
∑
k =1
∞
ak is convergent and let
∑ ak
k =1
=l.
≤l
Since r.h.s. is independent of n, this holds for all n = 1, 2, 3, ....
n
But
∑ bk
k =1
is the nth partial sum of the series
∑ bn
. And this nth partial sum is bounded.
n
Hence the sequence of partial sums is convergent. Hence
∑ bn
is a convergent series and therefore
n
∑ bn converges absolutely..
n
Next we prove that the series ∑ bn converges to the same limit S. Let t n = b1 + b2 + .... + bn
n
be the nth partial sum of the series
∑ bn
n
series
∑ an . Since ∑ an
n
n
and let Sn = a1 + a2 + .... + an be the nth partial sum of the
converges to S, the sequence { Sn } also converges to S. Therefore for
given ∈> 0 there exists a positve integer N1 such that
Sn − S <
∈
2
for all n ≥ N .
46
Since
∑ an
converges absolutely for some given ∈> 0∃ a the integer N2 such that
∞
∑ an
<
∈
n =k
∀ k > N2
2
Define N = max [N1, N2]
Then,
tn − S = t n − S N + S N − S
≤ tn − S N + S N − S
tn − S < tn − S N +
∈
2
Now choose an integer M so that
{1,2,3,....N } ⊆ { f(1) , f (2) ,...., f(n )}
tn − S N = b1 + b2 + .... + bn − ( a1+ a2 + ....+ a N )
= a f(1) + a f (2) + .... + a f ( n) − ( a1+ a2 + .... + a N )
=
n
∑
i =1
f( i ) > N
≤
n
∑
i=1
f( i ) > N
≤
a f( i)
∞
∑
k>N
a f (i )
ak =
∞
∑ ak
<
n= k
∈
∀k> N
2
Thus we get tn − S < t n − S N +
∈ ∈ ∈
< + <∈ , ∀ n ≥ N
2 2 2
The sequence {t n} of partial sums converges to S.
Hence
∑ bn
n
converges to S.
47
Subseries
6.
Definition
∑ an
n
Let f : ¢+ → ¢+ be a one-to-one function whose range is an infinite subset of ¢ + . Let
and
∑ bn
be the two series such that
n
bn = a f( n) , n ∈¢+
∑ bn
Then
7.
n
is called a subseries of
∑ an .
n
Example
Let f : ¢+ → ¢+ be a function defined by f ( n) = 2 n . Then f is one-to-one and range of f
is a subset of ¢ . Let
∞
1
∑ bn = ∑ 2n
n
∑ bn
n
8.
n =1
∑ an
1
be a series where an =
n
n
. Define bn = a f( n) = a2n =
1
2n
. Then
∞
1
is a subseries of ∑ an = ∑ . Note that the series ∑ an is divergent but its subseries
n
n
n =1 n
is convergent.
Example
Let f : ¢+ → ¢+ be a function defined by f ( n ) = 2 n − 1 . Then f is one-to-one and range of
f is a subset of ¢ . Let
∑ an
n
be a series where an =
Define bn = a f( n) = a2n −1 =
Then
∑ bn
n
( −1) 2n
2n −1
∞
=∑
( −1) 2n
n =1
∞
is a subseries of
∑ an = ∑
n
( −1) n+1
2n − 1
n
.
( −1) n+1
n =1
n
.
. Note that the series
1
but its subseries
n
is convergent
1
∑ bn = ∑ 2n − 1 is divergent. In this case ∑ an = ∑ n
n
∑ an
n
n
Following theorem gives a condition for convergent subseries.
48
n
is a divergent series.
9.
Theorem
If
∑ an
converges absolutely, every subseries
n
∑ bn
n
also converges absolutely. And we get
∞
∑ bn ≤ ∑ an
n =1
n
Proof : Let f : ¢+ → ¢+ be the one-to-one function whose range is subset of ¢ . Let bn = a f( n) .
Then
∑ bn
is a subseries of
n
∑ an . Let {tn} be the nth partial sum of the series ∑ bn
n
n
n
n
k =1
k =1
.
Then consider, t n = ∑ bk = ∑ a f ( k )
Let N = max { f (1) , f (2) ,...., f ( n) } . Then we get,
n
N
∞
k =1
k =1
k =1
t n = ∑ b k ≤ ∑ a k ≤ ∑ ak
Since ∑ an converges absolutely, the series
n
∞
∑
k =1
∞
ak converges and let
∑ ak
k =1
=l.
Therefore we have,
t n ≤ l for all n = 1, 2, 3, ....
Thus sequence {t n} is bounded and hence convergent. But {t n} is a sequence of partial sums
of the series
∑ bn
n
. Therefore
∑ bn
n
is convergent and hence
∑ an converges absolutely, every subseries ∑ bn
n
10.
n
∑ bn
n
converges absolutely. Thus if
also converges absolutely..
Theorem
Let { f1, f2, f3,....} be a countable collection of functions defined on ¢ + , which satisfies the
following properties.
(a)
Each f n is one-to-one on ¢ + .
(b)
The range f n ( ¢+ ) = Qn is a subset of ¢ + .
(c)
{Q1, Q2 ,....} is a collection of disjoint sets whose union is ¢ + .
Let
∑ an
n
be an absolutely convergent series.
49
Define bk (n ) = a f k ( n) , n ∈¢+ , k ∈¢ + .
Then,
∞
∑ bk (n ) is an absolutely convergent subseries of ∑n an .
(i)
For each k,
(ii)
If Sk = ∑ bk ( n ) then the series ∑ Sk converges absolutely and has the same sum as ∑ ak .
n =1
∞
∞
∞
n =1
k =1
k =1
Proof :
∞
(i)
Since every subseries of an absolutely convergent series converges absolutely,
converges absolutely for each k.
∞
(ii)
Given that Sk = ∑ bk ( n ) , k = 1, 2, 3, ...
n =1
∞
We prove that
∑ Sk
k =1
∞
converges absolutely. i.e.
∑ Sk
k =1
∞
Consider, kth
partial sum of the series,
∑ Sk
k =1
as
t n = S1 + S2 + S3 + ..... + S k
Then
tn =
∞
∞
∞
∑ b1 (n ) + ∑ b2 (n) + ..... +
n =1
n =1
∑ bk (n)
∞
∞
∞
n =1
n =1
n =1
n =1
≤ ∑ b1( n) + ∑ b2 ( n) + ..... + ∑ bk ( n)
∞
≤ ∑ ( b1( n) + b2 ( n) + .... + bk ( n) )
n =1
∞
(
≤ ∑ a f1( n) + a f 2 (n) + .... + a fk (n )
n =1
∞
≤ ∑ an
n =1
50
)
converges.
∑ bk (n )
n =1
But
∑ an
converges absolutely. Hence
n
∑ an
is a convergent series and converges to a
n
k
finite sum. Thus t k = ∑ Sn is bounded for all k = 1, 2, ....
n =1
Hence the series
∑ Sk
converges absolutely..
k
∑ Sk .
Next we find the sum
k
Let ∈> 0 be arbitrary. Choose a positive integer N such that
∞
∑
n
∈
k =1
2
ak − ∑ ak <
k =1
for all n ≥ N
... (1)
Choose sufficient number of functions f1 , f 2 ,...., f r so that each of the terms a1, a2 ,...., aN
will appear somewhere in the sum.
∞
∞
∞
n =1
n =1
n =1
∑ a f1( n) + ∑ a f2 ( n) + ..... + ∑ a f r (n)
The number r depends on N and hence on ∈ .
Then for n > r and n > N we have
n
S1 + S 2 + .... + S n − ∑ ak ≤ aN +1 + aN + 2 + ....
k =1
≤ a N +1 + a N +2 + ....
≤
∈
(By (1))
2
Also
∞
n
k =1
k =1
∑ ak − ∑ ak
=
∞
∑ ak
k =n +1
≤
∞
∑
k = n+1
ak <
∈
2
Therefore,
∞
n
n
∞
k =1
k =1
k =1
k =1
S1 + S 2 + .... + S n − ∑ ak = S1 + S 2 + .... + S n − ∑ a k + ∑ a k − ∑ ak
51
n
≤ S1 + S2 + .... + S n − ∑ ak +
k =1
∞
⇒ S1 + S2 + .... + S n − ∑ ak <
k =1
n
∞
k =1
k =1
∑ ak − ∑ ak
∈ ∈
+ =∈ for all n > r, n > N.
2 2
Which shows that the sequence of nth partial sums of the series
∑ Sk
converges to the
k
∞
sum
∑ ak .
k =1
i.e.
∑ Sk = ∑ ak .
k
3.3
1.
k
Rearrangement Theorem for Double Series
Definition
Let f be a double sequence and let g be a one-to-one function defined from ¢ + to ¢ + × ¢ + .
Let G be the sequence defined by
G ( n) = f ( g ( n) ) , n ∈¢+
Then g is said to be an arrangement of the double sequence f into the sequence G.
2.
Theorem
Let
∑ f (m, n) be a given double series and let g be an arrangement of the double sequence
f into a sequence G. Then
(a)
∑ G(n) converges absolutely if and only if ∑ f (m, n)
Further if
∑ f (m, n)
converge absolutely with sum S then we have
∞
(b)
∑ G(n) = S
n =1
∞
(c)
∑ f (m, n)
n =1
∞
and
∑ f (m, n)
m=1
converges absolutely..
both converge absolutely..
52
∞
(d)
∞
∑ Am
If Am = ∑ f (m , n) and Bn = ∑ f ( m, n ) , both series
n =1
and
m
m=1
∑ Bn
converges
n
abolutely and both have sum S. That is,
∞
∞
∑ ∑
m=1 n =1
∞
f ( m, n) = ∑
∞
∑ f (m, n) = S
n =1 m =1
Proof : Let {Tk } be the sequence of kth partial sums of the series
p
Let S ( p , q ) = ∑
∑ G ( n) .
n
q
∑
m =1 n=1
f (m , n) be the double sequence of partial sums of the series
∑ f (m , n) . Then for each k, there exists a pair (p, q) such that Tk < S (p , q )
m, n
and conversely for
each pair (p, q) there exists an integer r such that S ( p , q ) ≤ Tr . This implies that ∑ G(n) has bounded
n
partial sum if and only if
if and only if
∑ f (m , n)
m, n
∑ f (m, n)
n
converges absolutely..
m, n
Next assume that
Let
has bounded partial sum . Therefore ∑ G(n) converges absolutely
∑ f (m , n)
m, n
converges and therefore
∑ f (m, n)
m, n
converges.
∑ f (m, n) = S .
m, n
We prove that
∑ G(n) = S . First we show that the sum ∑ G(n)
n
is independent of the
n
function g used to construct G from f.
Let h be another rearrangement of the double sequence f into a sequence H. Then we have
G ( n) = f [ g ( n) ] and H (n ) = f [ h (n) ]
⇒ G = fog and H = foh
⇒ G = fog and f = Hoh−1
⇒ G = Hoh −1og
53
⇒ G = H ( h−1og )
⇒ G ( n) = H ( h −1 ( g ( n) ) ) , n = 1, 2, ....
= H (K (n ) ) where K = h−1 og
Since h–1 and g are one-to-one functions onto ¢ + . K is also one-to-one into ¢ + . Hence
∑ H (n ) is a rearrangement of ∑ G(n) . Suppose ∑ H (n ) has the sum say S'. We will show that
n
S' = S.
p
Let T = lim S (p , q) be the limit of the double sequence, S ( p , q ) = ∑
p ,q →∞
q
∑
f (m , n) .
m =1 n=1
Therefore for given ∈> 0 , choose N such that , 0 ≤ T − S (p , q) <
Next let {t k } be the sequence of partial sums corresponding to the series
∈
2
for p > N and q > N.
∑ G(n) . i.e.
n
k
t k = ∑ G ( n)
n =1
And also assume that,
p
s( p , q ) = ∑
q
∑ f (m, n )
m=1 n=1
be the partial sum of the series
∑ f (m, n)
m, n
choose an integer M so that t M includes all the
terms f (m , n) with 1 ≤ m ≤ N + 1 , 1 ≤ n ≤ N + 1
Then t M − S ( N + 1, N + 1) contains all the terms f (m , n) with m > N or n > N.
Therefore for all n > M we have,
tn − s ( N + 1, N + 1) ≤ T − S ( N + 1, N + 1) <
∈
2
Similarly,
S − s ( N + 1, N + 1) ≤ T − S ( N + 1, N + 1) <
54
∈
2
Thus for given ∈> 0 , we can find an integer M such that
tn − S = tn − s ( N + 1, N + 1) + s ( N + 1, N + 1) − S
≤ t n − s ( N + 1, N + 1) + s ( N + 1, N + 1) − S
≤
∈ ∈
+ =∈ for all n > M
2 2
Hence the sequence {t k } converges to S and therefore
But
∑ G ( n)
converges to S.
n
∑ G(n) = S ' . Hence S = S'.
n
Which proves the rearrangement of
∑ f (m, n) also converges to the same limit.
m, n
Next if Am = ∑ f (m , n) and Bn = ∑ f ( m, n) then Am and Bn are subseries of
n
m
Hence by theorem 3.2 (10)
∑ Am
∑ Bn
and
m
∞
i.e.
3.
∞
∑ f (m, n) .
m, n
also converges to the same limit S.
n
∞
∞
∑ ∑ f (m, n) = ∑ ∑ f (m, n) = S
m=1 n =1
n =1 m =1
Note
∞
The series
∞
∑ ∑
m=1 n=1
∞
f ( m, n ) and
∞
∑ ∑ f (m, n )
n =1 m =1
series need not converge to the same limit.
For example, Let
∑ f (m, n)
m, n
f ( m , n )= 1
be the series where
if m = n + 1, n = 1, 2, .....
= – 1 if m = n – 1, n = 2, 3, ....
=0
otherwise
55
are called iterated series. These iterated
∞
Then
∞
∑ ∑
∞
∑ { f (m,1) + f (m,2) + ....}
f ( m, n) =
m=1 n =1
m =1
=
∞
∞
m =1
m =1
∑ f (m,1) + ∑ f (m,2) + ....
= f (2,1) + [ f (1,2) + f (3,2) ] + ....
= 1+ [− 1 + 1] + [ −1 + 1] + ....
=1
∞
And
∞
∑ ∑
n =1 m=1
∞
f ( m, n) = ∑ { f (1, n) + f (2, n ) + .....}
n=1
∞
∞
n =1
n =1
= ∑ f (1, n) + ∑ f (2, n ) + ....
= f (1,2) + [ f (2,1) + f (2,3) ] + ....
= −1 + [1 − 1] + ....
=–1
∞
Hence
∞
∑ ∑
m=1 n =1
∞
f ( m, n ) ≠ ∑
∞
∑ f (m, n)
n =1 m=1
Sufficient Condition for Equality of Iterated Series :
4.
Theorem
∞
Let f be a double sequence and let
∞
Further assume that
∑ f (m, n)
n =1
converges absolutely for each fixed m.
∞
∑ ∑ f (m , n)
m=1 n=1
converges. Then,
∞
(a)
The series
∑ f (m, n)
m=1
∞
(b)
Both iterated series
∞
∑
∞
converges aboslutely for each n.
∞
∑ ∑
n =1 m =1
∞
f ( m, n ) and
∞
∑ f (m, n ) = ∑
m=1 n =1
∞
∑ ∑ f (m, n ) converges absolutely and
m=1 n=1
∞
∑ f (m , n) = ∑ f (m , n)
n =1 m=1
m, n
56
Proof : Let g be an arrangement of the double sequence f into a sequence G. Let {t n} be the
sequence of nth partial sums of the series
∑ G(n) , where
n
n
tn = ∑ G( k )
k =1
Then
tn =
n
n
k =1
k= 1
p
q
∑ G (k ) ≤ ∑ G(k ) ≤ ∑ ∑
f (m , n)
m =1 n=1
∞
≤∑
∞
∑
f ( m, n )
m=1 n=1
∞
But the series
∞
i.e.
∞
∑ ∑ f (m , n)
m=1 n=1
converges to some finite limit say l .
∞
∑ ∑ f (m , n) = l
m=1 n=1
Hence tn ≤ l for all n = 1, 2, 3, .....
Since the sequence {t n} is bounded it is convergent. Therefore
hence the series
∑ G(k )
is convergent and
n
∑ G( k ) converges absolutely..
n
By theorem ∑ f ( m, n) also converges absolutely. Further
m, n
∞
∑
m=1
f (m, n) = Bn then by theorem 3.3 (2)
∞
∑ Am and
m
∑ Bn
n
n
m
m ,n
As an application of the above theorem we have the following.
57
n =1
and
also converges absolutely and have the
n
∑ ∑ f (m, n) = ∑ ∑ f (m, n ) = ∑ f (m, n)
m
∑ f (m, n) = Am
∞
same limit.
Therefore,
∞
5.
Theorem
Let
∑ am
and
m
∑ bn
be two absolutely convergent series with sums A and B respectively..
n
Let f be the double sequence defined by the equation.
f ( m, n) = ambn , ( m, n) ∈ ¢ + × ¢ +
Then
∑ f (m, n)
converges absolutely and has the sum A•B.
m, n
Proof : Since
∑ am
and
m
∑ bn
converges absolutely
n
∑ am
m
and
∑ bn
also converges.
n
∞
∞
∑ am ∑ bn = ∑ am ∑ bn
m
n
m =1
n=1
And
=
=
∞
∑ am
m=1
∞
∞
n=1
∑ bn
∞
∑ ∑ am
m=1 n =1
=
∞
bn
∞
∑ ∑ ambn
m=1 n =1
If
∑ am
= a and
m
∞
Then
series
=b
n
∞
∑ ∑ am bn = a ⋅ b and hence convergent. Therefore by above theorem, the double
m=1 n=1
∑ ambn
m, n
∑ an
converges absolutely and
∑ ambn =
m,n
∞
∑
∞
∞
∑ am bn = ∑
m =1 n =1
∞
∑ ambn
n =1 m=1
∞
∞
= ∑ am ∑ bn
m=1 n=1
= A • B.
This shows that the double series
∑ ambn
m, n
58
converges to A • B.
6.
Note
The above theorem holds if ∑ am and ∑ bn converges absolutely. However if either
m
or
∑ bn
n
converges conditionally then the product series
n
converges the sum need not be A•B where
∑ ambn
m, n
∑ am
m
need not converge or even if it
∑ am = A and ∑ bn = B .
m
n
In this case the convergence and the sum will depend on the rearrangement of the series.
One such rearrangement of the product is Cauchy Product.
7.
Definition
∞
∞
Let ∑ an and
∑ bn
n =0
n =0
be the two series.
n
Define, Cn = ∑ ak bn −k , n = 0, 1, 2, 3, ....
k=0
∞
Then series
8.
∑ Cn
n =0
is called the Cauchy product of
∑ an
and
n
∑ bn .
n
Note
∞
∞
a
∑ n ∑ bn = ( a0 + a1 + a2 + ....)( b0 + b1 +b2 + ....)
n= 0 n=0
= ( a0b0 + a0b1 + a0b2 + ....) + ( a1b0 + a1b1 + a1b2 + ....)
+ ( a2b0 + a2b1 + a2b2 + ....) + ....
This sum can be written diagramatically as follows,
a0b0
a1b0
a2b0
a3b0
M
a0b1 a0b2
a1b1 a1b2
a2b1 a2b2
a3b1 a3b2
M
M
L
L
L
L
Diagonal sums are C0 = a0b0 , C1 = a1b0 + a0b1 , C2 = a2b0 + a1b1 + a0b2 , .... in which sum of
the coefficients is always constant. Cn ’s are terms in Cauchy product.
59
9.
Mertens Theorem
∞
Let ∑ an converges absolutely to the sum A and let
n =0
∞
∑ bn
n =0
be a convergent series with sum
B. Then Cauchy product of these two series also converges and has the sum AB.
Proof : Let An , Bn and Cn be the nth partial sums of the series ∑ an ,
n
n
where cn = ∑ ak bn −k is the nth term in Cauchy product of
k =0
∑ an
n
∑ bn
n
and
and ∑ cn respectively,,
n
∑ bn .
n
Therefore,
n
n
n
k=0
k =0
k=0
An = ∑ ak , Bn = ∑ bk , Cn = ∑ ck
n
Let d n = B − Bn and en = ∑ ak dn −k
... (1)
k =0
p
Then
p
CP = ∑ cn = ∑
n =0
n
∑ ak bn −k
n= 0 k = 0
If we define
a b , if n ≥ k
f n ( k ) = k n −k
, if n < k
0
p
Then
CP = ∑
p
∑
n =0 k =0
p
=∑
p
f n (k ) = ∑
p
∑ fn ( k )
k =0 n= 0
p
∑ f n (k)
Since f n ( k ) = 0 if n < k
k = 0 n =k
=
p
p
∑ ∑ ak bn −k
k = 0 n= k
p
p
k=0
n= k
p
p− k
p
m =0
k =0
= ∑ ak ∑ bn −k
=
∑ ak
k=0
n − k = m , ∴ n → k to p, ⇒ m → 0 to p – k
∑ bm = ∑ ak ⋅ Bp −k
But d n = B − Bn . Therefore Bn = B − d n and B p −k = B − d p − k .
60
p
CP = ∑ ak ( B − d p−k )
Therefore,
k =0
=
p
p
k=0
k=0
∑ ak B − ∑ ak d p− k
p
= B ∑ ak − e p
.... (2)
k=0
We show that e p → 0 as p → ∞ .
Now d n = B − Bn As n → ∞ , Bn → B . Hence d n → 0 as n → ∞ .
Choose a positive integer M such that d n ≤ M for all n. (Convergent sequences are bounded).
∞
∞
∑ an
Since
converges absolutely,
n =0
∑ an
n =0
∞
converges. Let
∑ an
n =0
=K.
Since { dn } converges to 0, for given ∈> 0 there is an integer N, such that
dn <
∈
2K
for all n ≥ N1
... (3)
∞
Also ∑ an converges to K, its sequence of partial sums { Sn } also converges to K. Therefore
n =1
for given ∈> 0 there is a positive integer N2 such that
Sn − K <
∈
2M
for all n ≥ N 2
... (4)
Take N = max { N1, N2}
⇒ Sn − K <
⇒
⇒
∈
2M
N
∞
∈
n =1
n =1
2M
∑ an − ∑ an <
∞
∑
n = N +1
an <
∈
... (5)
2M
Next for p > 2N we have
61
eP =
p
∑ ak d p − k
k=0
p
≤ ∑ ak dp −k
k =0
=
N
∑ ak d p−k
+
k=0
=
p
∑
N
∑ ak ⋅ d p−k
N
∈
k =0
2M
+
p
∑
+
k=0
≤ ∑ ak
ak d p − k
k = N +1
k = N +1
p
∑
k = N +1
ak ⋅ d p − k
ak ⋅ M
For, d n ≤ M ∀n and p > 2N, implies p – k > N, ∀ k = 0, 1, 2, .... N.
Therefore,
eP ≤
≤
∈
N
∑
2M
k=0
∈
∞
∑
2M
k =0
ak + M
p
∑
k = N +1
ak + M ⋅
ak
∞
∑
k = N +1
ak <
∈ ∈
+ =∈
2 2
Thus for any ∈> 0 there is a positive integer 2N such that
eP <∈ for all p > 2N
This shows that eP → 0 as p → ∞ .
But from (2)
p
C P = B ∑ ak − e p
k =0
p
⇒ lim CP = lim B ∑ ak − e p
p →∞
p →∞
k=0
∞
= B ⋅ ∑ ak − 0
k =0
=B•A
=A•B
62
This shows that the sequence of partial sums {C p } converges to AB.
∞
Hence the series
∑ ck
k =0
converges to AB.
∞
∞
∞
c
=
a
∑ k ∑ bk is the Cauchy product of
But ∑ k
k =0
k = 0 k = 0
∑ ak
and
k
∑ bk .
k
Therefore Cauchy product converges to A • B.
i.e.
10.
∞
∞
∑ ak ∑ bk → A ⋅ B
k =0 k = 0
Note
Let
,if n ≥ k
a b
f n ( k ) = k n −k
,if n < k
0
Then
p
p
p
∑ ∑ fn (k ) = ∑ [ f n (0) + f n (1) + .... + fn ( p )]
n =0 k = 0
n= 0
p
= ∑ f n (0) +
n =0
p
∑
n= 0
p
f n (1) + .... + ∑ f n ( p)
n= 0
= f 0 (0) + f1(0) + .... + f p (0) + f 0 (1) + f1 (1) + .... + f p (1) + ....
+ f 0 ( p) + f1 ( p ) + .... + f p ( p )
= f 0 (0) + f1 (0) + .... + f p (0)
+ f1 (1) + f 2 (1) + .... + f p (1)
+
f 2 (2) + .... + f p (2)
M
+ f p ( p)
+
= a0 b0 + a0 b1 + a0 b2 + .... + a0 bp
+ a1b0 + a1b1 + .... + a1bp −1
63
+ a2 b0 + .... + a2 bp − 2
M
M
+ a p b0
= a0b0 +( a0b1 +a1b0 ) +( a0b2 + a1b1 + a2b0 )
+.... + ( a pb0 + .... + a0b p )
p
= c1 + c2 +..... + c p = ∑ ck = C p
k =1
64
UNIT - IV
POWER SERIES AND ITS PROPERTIES
4.1
Power Series
1.
Introduction
Functions having power series expansion are important because such functions are not only
continuous inside the disk of convergence but they have derivative of every order in the disk. We shall
study convergence, radius of convergence and other properties like multiplication of power series,
Taylor’s series, Binomial series. Finally we prove Abel’s limit theorem and Tauber’s Theorem.
2.
Definition
∞
An infinite series of the form
∑ an ( z − z0 )n
n =0
is called a power series in ( z − z 0 ) where a n , z
and z0 are complex numbers.
With every power series there associated a disk called as disk of convergence D. Such that
the series converges for all z in the interior of D and diverges for all z outside D. The radius of the disk
is called radius of convergence. Thus D = {z | z < r} .
3.
Root Test for Convergence of Series
∞
For a series
∑ an
n =1
of complex numbers, let
ρ = limsup n an .
n →∞
Then,
(a)
The series converges absolutely if ρ < 1.
(b)
The series diverges if ρ > 1 .
(c)
The test is inconclusive if ρ = 1.
4.
Theorem
∞
For a power series
∑ an ( z − z0 )
n =0
n
n a
, let λ = nlimsup
n and r =
→∞
1
λ
. Then the series
converges absolutely if z − z 0 < r and the series diverges if z − z 0 > r . Further the series converges
on every compact subset interior to the disk of convergence.
65
∞
Proof : Given series is
∑ an ( z − z0 )
n
n =0
.
Applying root test we have,
an ( z − z0 )
ρ = limsup
n →∞
n
= limsup n an ⋅ z − z 0
n →∞
= λ z − z0
1
=
r
1
(Since λ = )
r
z − z0
The series converges absolutely if ρ < 1.
1
i.e.
r
z − z 0 < 1 or z − z 0 < r
And the series diverges if ρ >1 or z − z 0 > r .
Next, if T is a compact subset of the disk of convergence then there is a point P ∈ T such that
z − z 0 < p − z0 < r , for all z ∈ T
n
⇒ z − z0 < p − z 0
T
n
⇒ an z − z0 ≤ an p − z 0
∞
P
n
z
n
z0
D
∞
⇒ ∑ an ⋅ ( z − z0 ) ≤ ∑ an ⋅ ( p − z 0 )
n
n =0
n =0
∞
Now,
∑ an ( z − z0 )
n =0
Therefore
n
converges absolutely for all z, such that z − z 0 < r . Now p − z 0 < r .
∑ an ( p − z 0 )
∞
⇒ ∑ an ( p − z 0 )
n= 0
∞
But
n
n
n
converges
∑ an ( z − z0 )
n =0
converges absolutely..
n
∞
≤ ∑ an ( p − z 0 )
n
n =0
66
∞
⇒ ∑ an ( z − z0 ) converges ∀z ∈ T
n
n =0
∞
⇒ ∑ an ⋅ ( z − z 0 ) converges absolutely, ∀z ∈ T .
n
n =0
5.
Note
On the boundary of the disk the series may or may not converge.
6.
Example
∞
(1)
∑ zn , z ∈£
n =0
n a
We have, λ = nlimsup
n
→∞
∞
But
∑z
n
n =0
∞
= ∑ 1( z − 0 ) . Hence an = 1∀n and z 0 = 0 .
n
n= 0
n
1 = limsup {1} = 1
Therefore, λ = nlimsup
→∞
n →∞
But r =
1
λ
⇒ r =1
Therefore the series converges if z < 1 and the series diverges if z > 1 . For z = 1 the series
∞
becomes,
∞
∑
∞
z n = ∑ z = ∑ 1 , which is divergent.
n =0
n
n =0
n =0
∞
Hence
∞
(2)
∑ zn
n =0
converges absolutely if z < 1 and diverges if z ≥ 1 .
zn
∑ n2 , z ∈ £
n =1
∞
Then
zn
∑ n2
n =1
We have,
∞
=∑
1
n =1 n
2
( z − 0 )n . Therefore an =
λ = limsup n an
n →∞
= limsup n
n →∞
1
n2
67
1
n2
, z0 = 0 .
1
= limsup
n →∞
n
( 2 n)
=1
But r =
1
λ
=
1
1
= 1 . Therefore radius of convergence is 1.
And the series converges if z < 1 and diverges if z > 1 . When z = 1 .
∞
∞
zn
z
n
∞
1
∑ n 2 = ∑ n 2 = ∑ n2 , which is a convergent series. Thus the given series converges if
n =1
n =1
n= 0
z ≤ 1 and diverges if z > 1 .
(3)
∞
zn
n =1
n
∑
, z ∈£
∞
∞ 1
1
n
= ∑ ( z − 0 ) . Therefore, an = , z 0 = 0 .
n n=1 n
n
zn
∑
Now,
n =1
λ = limsup n an
We have
n →∞
= limsup n
1
= limsup
1
n →∞
n →∞
But r =
1
λ
=
1
1
n
n
1
=1
n
= 1 . Therefore radius of convergence is 1. And the series converges if z < 1
and divergs if z > 1 . For z = 1 , z = eix = cos x + i sin x , x ∈ ¡ .
Therefore z n = ( eix ) n = einx .
∞
zn
n =1
n
⇒∑
Thus
e inx
n =1
n
=∑
∞
zn
n =1
n
∑
∞
, which is convergent if δ ≤ x ≤ 2π − δ , 0 < δ < π .
converges if z < 1 and diverges if z > 1 . For z = 1 the series converges on
[ δ , 2π − δ ] where 0 < δ < π .
68
7.
Exercise I
Find the radius of convergence of each of the folowing power series.
∞
∞
n!
(i) ∑ z
(ii)
n =0
8.
∑
( −1) n z n(n+1)
∞
(iii)
n
n =0
∑ k n zn
n =0
Theorem
∞
∑ an ( z − z0 )
If the power series
n =0
n
converges for each z ∈ B ( z0 ; r ) to a function f, then f is
continuous on B ( z0 ; r ) .
Proof : Since an ( z − z0 ) is a polynomial for each n, it is continuous on B ( z0 ; r ) . Each point in
n
B ( z0 ; r ) lies in some compact subset of B ( z0 ; r ) . Hence the series of continuous functions converges
uniformly on B ( z0 ; r ) to the function f. Therefore f is also continuous on B ( z0 ; r ) .
9.
Theorem
∞
Let
∑ an ( z − z0 )
n
n =0
be convergent for all z ∈ B ( z0 ; r ) .
∞
Let f ( z ) = ∑ an ( z − z0 ) , ∀z ∈ S ⊆ B ( z 0 ; r ) . Then for each z1 ∈ S there exists a
n
n =0
∞
neighbourhood B ( z1 ; R ) ⊆ S such that f ( z ) = ∑ bk ( z − z1 ) where
k
k=0
∞ n
n− k
bk = ∑ an ( z1 − z0 ) , k = 0, 1, 2, 3, ...
n =k k
∞
Proof : We have f ( z ) = ∑ an ( z − z0 ) , ∀z ∈ S
n
n =0
∞
= ∑ an ( z − z1 ) + ( z1 − z0 )
n
n= 0
∞
n n
k
n −k
= ∑ an ∑ ( z − z1 ) ( z1 − z0 )
n =0
k =0 k
69
∞
=∑
k
n −k
∑ k an ( z − z1 ) ( z1 − z0 )
n
n
n = 0 k =0
∞
=∑
∞
∑ Cn ( k )
n= 0 k = 0
n
k
n− k
where, Cn ( k ) = k an ( z − z1 ) ( z1 − z0 ) , if k ≤ n
= 0, if k >n.
Choose a real number k > 0 such that B ( z1 , R ) ⊆ S . Then for any z ∈ B ( z1 ; R) consider,,
∞
∑
∞
∞
∑ Cn ( k ) = ∑
n =0 k = 0
∞
k
n− k
∑ k an ( z − z1 ) ( z1 − z 0 )
n
n= 0 k = 0
∞
=∑
∞
k
n −k
∑ k an ( z − z1 ) ( z1 − z 0 )
n
n = 0 k =0
∞
= ∑ an
n =0
∞
n
∑ k z − z1
k
k =0
z1 − z 0
∞
= ∑ an z − z1 + z1 − z0
n− k
n
n= 0
Let, z 2 = z0 + z − z1 + z 1 − z 0
⇒ z2 − z 0 = z − z1 + z1 − z0
≤ z − z1 +z 1 −z 0
( z ∈ B ( z1; R ) ⇒ z − z1
< R + z1 − z0
Also B ( z1; R ) ⊆ S ⊆ B ( z 0; r ) . Therefore R + z1 − z 0 < r .
∞
Therefore,
∞
∑ ∑
n =0 k = 0
∞
Cn ( k ) = ∑ an ( z2 − z0 )
n
n= 0
∞
where z2 − z0 < r ⇒ ∑ an ( z2 − z0 )
n =0
n
converges.
70
< R)
∞
And hence the series to the L.H.S.,
∞
∑ ∑
n =0 k = 0
∞
Cn (k ) also converges i.e.
∞
∑ ∑ Cn (k )
n =0 k =0
converges absolutely. Therefore by theorem, we can interchange the order of summation. Hence we
get,
∞
f (z) = ∑
∞
∑ Cn ( k )
n =0 k =0
=
∞
∞
k
n −k
∑ ∑ k an ( z − z1 ) ( z1 − z0 )
n
k = 0 n=0
=
∞
∞
n
n− k
an ( z1 − z0 ) , ( n ≥ k )
n= k k
k
∑ ( z − z1 ) ∑
k=0
=
10.
∞
∞ n
k
n −k
b
z
−
z
b
=
(
)
∑k
1 where k
∑ k an ( z − z0 )
k=0
n =k
(Since n ≥ k )
Note
The above theorem is useful to obtain the power series representation of the function f about
any interior point of the disk of convergence. The differentiability of the function having power series
representation is establihsed in the following theorem.
11.
Theorem
Let
∑ an ( z − z 0 )
n
converges for each z ∈ B ( z0 ; r ) . Then the function f defined by
∞
f ( z ) = ∑ an ( z − z0 ) , ∀z ∈ B ( z 0; r )
n
n =0
∞
has a derivative f '( z ) for each z ∈ B ( z0 ; r ) and f '( z ) = ∑ nan ( z − z0 )
n =1
n −1
and has the
same radius of convergence r.
∞
Proof : Given that, f ( z ) = ∑ an ( z − z0 ) , ∀z ∈ B ( z 0; r )
n
n =0
Then by theorem, for any z1 ∈ B ( z 0 ; r ) the function f has power series expansion about z1
∞
given by f ( z ) = ∑ bk ( z − z1 ) in the disk B ( z1 ; R) .
k
k=0
∞
n
n− k
Where, bk = ∑ k an ( z1 − z0 )
n =k
71
The for all z ∈ B ( z1 ; R) , z ≠ z1 we have
f ( z ) − f ( z1)
z − z1
∞
k
=
∑ bk ( z − z1 ) − b0
z − z1 k = 0
1
=
(Q f (z1 ) = b0 )
1
b0 + b1 ( z − z1 ) + b2 ( z − z1 ) 2 + .... − b0
z − z1
= b1 + b 2 ( z − z1 ) + b3 ( z − z1 ) + ....
2
Taking limit of both sides as z → z1 we obtain,
f ' ( z1 ) = b1
∞ n
n −1
= ∑ an ( z − z0 )
n =1 1
∞
= ∑ nan ( z − z0 )
(k =1)
n −1
n =1
∞
n −1
Since z1 ∈ B ( z 0 ; r ) is arbitrary we have, f ' ( z ) = ∑ nan ( z − z0 ) , z ∈ B ( z0 ; r )
n =1
Next by root test we have
λ = limsup n an
n →∞
= limsup n nan
n →∞
= limsup n an
n →∞
n
n
= limsup n an ⋅ lim sup n n
n →∞
n →∞
1
= (1)
r
Thus λ =
1
r
or r =
1
λ
. Hence the radius of convergence for the series
∞
n −1
f ' ( z ) = ∑ nan ( z − z0 )
is also r. i.e. radius of convergence for f ( z ) and f ' ( z ) is same and
n =1
equal to r.
72
12.
Note
By repeated application of the above theorem we can show that the derivative f (k ) ( z ) exists
in B ( z0 ; r ) , k = 1, 2, 3, .... and it is given by
f (k ) ( z ) =
∞
n −k
∑ n(n − 1)...(n − k + 1) an ( z − z0 )
n= k
∞
=
n!
∑ (n − k )! an ( z − z0 )
n −k
, ∀z ∈ B ( z 0; r )
n= k
At z = z0 we get
∞
n!
n =k
( n − k )!
f (k ) ( z0 ) = ∑
=
⇒ ak =
k!
0!
n− k
an ( z0 − z0 )
ak = k ! ak
f (k ) ( z0 )
k!
, k = 1, 2, 3, ....
Therefore the power series expansion for f about the point z0 can be uniquely determined by
∞
∞
n =0
n =1
n
n
f ( z ) = ∑ an ( z − z0 ) = a0 + ∑ an ( z − z 0 )
∞
f ( n) ( z0 )
n =1
n!
= a0 + ∑
∞
( z − z0 )n
n =1
n!
f ( z ) = a0 + ∑
13.
( z − z0 ) n
f ( n) ( z0 ) , z ∈ B ( z0 ; r )
Theorem (Multiplication of Power Series)
∞
∞
n =0
n= 0
n
n
Let f ( z ) = ∑ an z and g ( z ) = ∑ bn z be the two power series which converges on
B (0, r) and B (0, R) respectively. Then the product f ( z ) g ( z ) is given by the series,
∞
f ( z ) g ( z ) = ∑ Cn z n , ∀z ∈ B ( 0, r ) I B ( 0, R )
n =0
n
where Cn = ∑ ak bn −k , n = 0, 1, 2, 3, ...
n= 0
73
Proof : By Cauchy product of the two series we have,
∞
∞
f ( z) ⋅ g ( z ) = ∑ an z n ∑ bn z n
n= 0
n= 0
∞
n
n =0
n =1
∑ Cn zn where Cn = ∑ ak bn −k
And by Merten’s theorem, if both the series converges then the product series also converges.
Hence ∀z ∈ B ( 0, r ) I B ( 0, R ) we have
∞
n
n =0
k =1
f ( z) ⋅ g ( z ) = ∑ Cn z n where Cn = ∑ ak bn −k
14.
Note
∞
n
If f ( z ) = ∑ an z , z ∈ B ( 0, r ) then we have
n =0
∞
2
f ( z ) = ∑ Cn z n , ∀z ∈ B ( 0, r ) where
n =0
n
Cn = ∑ ak an − k =
k=0
∑
m1+ m2 = n
am1 am2
Similarly for any integer p > 0 we have
∑
∞
p
f ( z ) = ∑ Cn ( p) z n where Cn ( p) =
n= 0
15.
m1+ m2 +.... +m p =n
am1 ....am p
Theorem ( The Substitution Theorem)
∞
∞
n =0
n= 0
n
n
Let f ( z ) = ∑ an z , z ∈ B ( 0; r ) and g ( z ) = ∑ bn z , z ∈ B ( 0; R ) . If for a z ∈ B ( 0; R )
∞
we have
∑
n =0
bn z < r then for this z we can write f ( g ( z ) ) =
n
follows. For each integer n we have
n
∞
∞
g ( z ) = ∑ bk z k = ∑ bk ( n) z k
k=0
k=0
n
∞
Then Ck = ∑ anbk (n ) , k = 1, 2, 3, ....
n =0
74
∞
∑ Cn z k
k=0
where Ck is obtained as
16.
Thoerom (Reciprocal of Power Series)
∞
n
Let p ( z ) = ∑ pn z , z ∈ B ( 0; h ) where p (0) ≠ 0 .
n= 0
Then there exists a neighbourhood B ( 0; δ ) in which the reciprocal of p has a power series
expansion of the form
1
p( z )
∞
1
n =0
p0
= ∑ qn z n where q0 =
Proof : We assume that p0 = 1. Then p (0) = 1.
∞
n
Let p ( z ) = 1 + ∑ pn z if z ∈ B ( 0; h ) .
n =1
∞
n
Then p ( z ) − 1 = ∑ pn z . Therefore there exists a neigbourhood B ( 0; δ ) such that
n =1
p ( z ) − 1 < 1 , ∀z ∈ B ( 0; δ ) .
1
∞
= ∑ zn
Let
f (z) =
and
g ( z ) = 1 − p ( z ) = ∑ pn z n
1− z
n =0
∞
n =1
∞
Then
∑ pn z n
n =1
< 1 , ∀z ∈ B ( 0, δ ) implies
f ( g ( z ) ) = f (1 − p ( z ) ) =
⇒
⇒
1 − (1 − p ( z ) )
=
∞
1
p( z )
= ∑ qn ⋅ z n
n =1
∞
1
p( z )
= ∑ qn z n
1
where
1
p0
n =0
= q0
75
1
p (z )
∞
= ∑ qn z n
n =1
(By substitution Theorem)
4.2
1.
Real Power Series
Definition
∞
If x, x 0 and an are real numbers then the series
∑ an ( x − x0 )
n =0
n
is called a real power series.
Its disk of convergence intersects the real axis in an interval I = (x 0 − r , x0 + r ) . This interval
is called the interval of convergence.
The real power series converges absolutely inside the interval of convergence and converges
uniformly on any compact subset of the interval I. Thus
∞
f ( x ) = ∑ an ( x − x0 ) , ∀x ∈ I ,
n
n= 0
This is series representation of the function f in the interval of convergent and called as power
series expansion of f about x 0.
Further the series converges uniformly on every compact subset of I = (x 0 − r , x0 + r )
therefore we can integrate the series term by term.
Thus for every x ∈ ( x 0 − r , x0 + r ) we have
x
∫
x0
∞
x
n =0
x0
f (t )dt = ∑ an ∫ ( t − x0 ) dt
∞
=∑
n =0
an
n +1
n
( x − x0 ) n+1
The integrated series has the same radius of convergence. Similarly the function
∞
f ( x ) = ∑ an ( x − x0 ) has derivative of every order in the interval of convergence. And by theorem
n
n= 0
(4.1 (11)) we have f (n) ( x0 ) = n ! an , ∀n . Hence the power series for f becomes,
∞
∞
f (n) ( x0 )
n =0
n!
f ( x ) = ∑ an ( x − x0 ) = ∑
n= 0
n
76
( x − x0 )n
Taylor’s Series Generated by a function
2.
Definition
Let f be a real valued function defined on an interval I. If f has the derivative of every order at
each point of I, then f is said to be C ∞ function and we write f ∈ C ∞ on I.
3.
Definition
Let f be a real valued function defined on an interval I on ¡ . If f ∈ C ∞ on some neighbourhood
of a point c then the power series
∞
f ( x) (c )
n =0
n!
∑
( x − c )n is called the Taylor’s series about c generated
by f and we write,
∞
f (n ) ( c )
n =0
n!
f :∑
4.
( x − c ) n i.e. f generates the series.
Theorem (Taylor’s Formula)
If f ∈ C ∞ on a closed [a, b] and if c ∈ [ a ,b ] , then for every x ∈ [a , b] and for every
integer n
n −1
f ( k ) (c )
k=0
k!
f (x ) = ∑
( x − c )k +
f (n) ( x1 )
n!
( x − c )n
where x < x1 < c (or c < x1 < x )
The necessary and sufficient condition that the series converges to f (x) as n → ∞ is that
lim
f (n) ( x1 )
n→∞
The term
n!
f (n ) ( x1 )
n!
Thus En ( x) =
5.
( x − c)n = 0
( x − c ) is called an error term denoted by En .
f ( n) ( x1 )
n!
( x − c )n where c < x1 < x (or x < x1 < c )
Theorem
Let f ∈ C ∞ on [a, b] and let c ∈ [ a ,b ] . Assume that there is a neighbourhood B (c) and a
constant M (may depend on c) such that f (n) (x ) ≤ M n for every ∀x ∈ B(c ) I [a ,b ] and for n = 1,
2, 3, ... Then for any x ∈ B (c ) I [ a , b] we have
77
∞
f ( n) ( c )
n= 0
n!
f (x ) = ∑
( x − c )n
Proof : We show that error term tends to zero as n → ∞ in Taylor’s formula
Since f (n) (x ) ≤ M n for every x ∈ B (c ) I [ a , b] it is also true for c. (Q c ∈ [ a , b] ) Fix a
point x 1 between x and c. Then
f (n) ( x1 )
n!
( x1 − c )
n
≤
Mn
n!
( x1 − c )
n
Mn
=
n!
x1 − c
n
Since c and x 1 are fixed we take A = M ⋅ x1 − c .
f (n) ( x1 )
Therefore
But
An
n!
n!
≤
An
n!
∞
An
n =0
n!
→ 0 as n → ∞ and for all A. Because the series ∑
f (n) ( x1 )
the error term En =
( x1 − c ) n
n!
n −1
f ( k ) (c )
k=0
k!
Hence f ( x ) = ∑
∞
f (k ) ( c )
k =0
k!
⇒ f (x ) = ∑
i.e.
( x1 − c )
n
tends to zero as n → ∞ .
( x − c)k +
f (n) ( x1 )
n!
( x1 − c )n
( x − c)k
∞
f ( n) ( c )
n= 0
n!
f (x ) = ∑
is a convergent series. Thus
(Taking limit as n → ∞ )
( x − c )k , ∀x ∈ B(c ) I [a ,b ] .
Since c ∈ [ a ,b ] is arbitrary the expansion holds for any c ∈ [ a ,b ] .
6.
Note
Another sufficient condition for convergence of the Taylor’s series of f is obtained in Berstein’s
theorem. First we prove that the error term in the Taylor’s formula can be expressed as an integral.
7.
Theorem
Let f be a function having continuous derivatives of order (n + 1) in some open interval I
containing c.
n
f k (c)
k =0
k!
Let, f ( x ) = ∑
( x − c ) k + En ( x )
78
Then En ( x) is also given by the integral
x
1
(x − t )
n! ∫
En ( x) =
n
f
(n +1)
(t ) dt
c
Proof : The proof is by induction on n.
For n = 1, f ( x ) = f ( c) +
f '(c )
1!
( x − c) + E1 ( x )
(By Taylor’s formula)
⇒ E1 ( x ) = f ( x ) − f (c ) − f '( x )( x − c )
x
Also
x
∫ (x − t ) f ''(x )dt = [ ( x − t) f '(t )] − ∫ f '(t )(−1) dt
x
c
c
c
= −( x − c) f '( c) + [ f ( t ) ]c
x
= f ( x) − f (c ) − ( x − c) f '( c)
1
x
Hence E1 ( x ) = 1! ∫ ( x − t ) f ''(t ) dt
c
Therefore the theorem is true for n = 1.
Let the theorem be true for n. i.e.
1
x
n ! ∫c
En ( x) =
(x − t ) f
n
(n +1)
(t ) dt holds.
We prove that the expression is true for (n + 1)
Now
n +1
f (k ) ( c )
k =0
k!
n
f (k ) ( c )
k =0
k!
En+1 ( x ) = f ( x) − ∑
= f ( x) − ∑
= En ( x ) −
(n + 1)!
( x − c)k −
( x − c )n+1
x
−( x − t ) n+1
( x − c ) n+1
(
x
−
t
)
dt
=
=
∫
n + 1 c
n +1
c
x
Now
f (n+1) ( c )
( x − c)k
n
Therefore,
79
f ( n+1) ( c )
( n + 1)!
( x − c ) n+1
En+1 (x ) = En ( x ) −
f ( n+1) (c )
( n + 1)!
x
1
( x − t)
n! ∫
=
n
f
x
⋅ ( n + 1) ∫ ( x − t) n dt
c
(n +1)
(t ) dt −
c
x
1
f
n! ∫
=
1
x
(x − t )
n! ∫
n
f
(n +1)
(c )dt
c
(n +1)
(t ) − f
( n +1)
n
(c ) ( x − t ) dt
c
Let u (t ) = f
( n +1)
(t ) − f
(n +1)
( c ) and v (t ) = −
( x − t ) n+1
n +1
Therefore dν (t ) = ( x − t ) n dt
⇒ En+1 ( x ) =
1
x
n! ∫c
u (t ) dv(t )
x
1
x
= [ u (t ) ⋅ v (t ) ]c − ∫ v( t) du (t )
n !
c
x
1
( x − t ) n+1 (n+ 2)
= u ( x ) ⋅ v( x) − u (c ) ⋅ v( c) − ∫ −
⋅f
( t ) dt
n !
n +1
c
=
1
1
x
n +1
⋅f
( n +2)
n +1
⋅f
(n +1+1)
( x − t)
n! n + 1 ∫
(t ) dt
c
=
x
1
( x − t)
( n + 1)! ∫
(t ) dt
c
Thus the theorem is true for n + 1. Hence by induction the error term is given by
En ( x) =
1
x
(x − t )
n! ∫
n
f
(n +1)
(t ) dt
c
80
∀ n = 1, 2, 3, ....
8.
Bernstein’s Theorem
Let f and all its derivatives are non-negative on a compact interval [b, b + r ] . Then if
f ( k ) (b)
k=0
k!
f (x ) = ∑
i.e.
f (k ) ( b )
k =0
k!
∑
b ≤ x ≤ b + r then the Taylor’s series
∞
∞
( x − b )k converges to f (x).
( x − b )k ,
∀b ≤ x ≤ b + r
Proof : Without loss of generality we assume that b = 0. Hence 0 ≤ x ≤ r . If x = 0 then the the result
is trivial since the series whose terms are all zero converges to zero. Therefore we assume that
0 < x < r.
By Taylor’s formula with integral form of error term
n
f (x ) = ∑
f ( k ) (0 )
k!
k=0
Now
+ En ( x ) , where En ( x) =
En ( x) =
1
x
(x − t )
n! ∫
n
f
n +1
(t ) dt
(Taking c = b = 0)
0
x
1
(x − t )
n! ∫
n
(n +1)
f
(t ) dt
0
Put t = (1 − u ) x . Then dt = − xdu .
And as t varies from 0 to x, u varies from 1 to 0.
Then
En ( x) =
1
0
[ x − (1 − u ) x]
n! ∫
n
f
( n +1)
( (1 − u ) x ) ( − xdu )
1
=
1
1
( ux )
n! ∫
n
f
(n +1)
( (1 − u ) x ) ⋅ xdu
0
=
x n+1
n!
1
∫u
n
f (n+1) ( (1 − u ) x) du , 0 < x < r
0
Since x > 0 we define Fn ( x ) =
Fn ( x ) =
1
1
u
n! ∫
n
f
( n +1)
E n ( x)
xn +1
. Then
( (1 − u ) x) du , 0 < x < r
0
Since all the derivatives of f are non-negative. f is monotonic increasing function.
81
Therefore,
f (n+1) ( x (1 − u ) ) ≤ f ( n+1) ( r (1 − u ) )
( Since x < r)
Hence for 0 ≤ u ≤ 1 we get,
1
1
u
n! ∫
Fn ( x ) =
n
( n +1)
f
( x(1 − u ) ) du
0
≤
1
1
n ! ∫0
n
u f
(n +1)
( r (1 − u ) ) du
= Fn (r )
i.e
Fn ( x ) ≤ Fn (r )
⇒
En ( r )
x n+1
≤
∀0 < x < r
En ( r )
r n+1
n +1
x
⇒ En ( r ) ≤
r
But
n
f (k ) ( 0)
k =0
k!
En ( r ) = f ( r ) − ∑
≤ f (r)
n +1
f (r) .
n +1
x
Also 0 < x < r ⇒ < 1 . Therefore
r
r
x
But
xk
(since all the terms in the series are positive)
x
Hence En ( r ) ≤
r
Hence
∀0 < x < r
En ( r ) ,
→ 0 as n → ∞
En ( x) → 0 as n → ∞
n
f ( k ) (0 )
k=0
k!
f (x ) = ∑
x k + En ( x )
Taking limit as n → ∞ . We get
∞
f (k ) ( 0 )
k =0
k!
f (x ) = ∑
xk
82
4.3
1.
Binomial Series
Theorem
For any real number a,
∞ a
(1 + x) a = ∑ x n ,
n =0 n
−1 < x < 1
a a( a − 1)(a − 2).....( a − n + 1)
n =
n!
where
Proof :
Let f ( x ) = (1 − x) − c where c > 0 and x < 1
f '( x ) = c(1 − x) − c−1
Then
f ''( x) = c( c + 1)(1 − x) − c −2
........................................
f (n ) ( x ) = c( c +1)( c + 2).........(c + n −1) ⋅ (1 − x) − c− n
Since c > 0 and x < 1, (1 – x) > 0 and hence all the derivatives are positive, f (n ) ( x ) > 0 ,
∀n and ∀x < 1 .
Hence by Berstein’s theorem,
∞
f (k ) ( b )
k =0
k!
f (x ) = ∑
( x − b )k , b ≤ x < b + r
Take b = 0 and r = 1. Then,
∞
f k (0 )
k =0
k!
f (x ) = ∑
=
=
( x − 0) k , 0 ≤ x < 1
∞
c (c + 1)( c + 2).....(c + k − 1)
k=0
k!
∑
xk , 0 ≤ x < 1
∞
( −c )( −c −1)( −c − 2).....(−c − k + 1)( −1) k
k=0
k!
∑
∞ −c
f ( x ) = ∑ ( −1)k x k
k=0 k
0 ≤ x <1
83
xk
⇒
∞ −c
−c
k k
=
(1
−
x
)
=
∑
k (−1) x
c
(1 − x)
k =0
1
This expansion is valid for −1 < x ≤ 0
Replacing – c by a and x by –x. We get
∞ a
1
f ( x ) = ∑ xk =
= (1 − x) − c
c
k
(1 − x )
k=0
∞ a
⇒ (1 − x) a = ∑ x k for all a ∈¡ and –1 < x < 1.
k =0 k
Further if a is positive integer then for a = m,
a
n
=
m m( m − 1).......(m − n + 1)
n =
n!
= 0 if n > m
Hence
∞ m
(1 + x) m = ∑ x k
k=0 k
m
m
m
= x 0 + x 1 + x 2 + ......
0
1
2
= 1 + mx +
m( m − 1)
2!
m
x 2 + ..... + x m
n
Thus expansion terminates finitely if m is a positive integer otherwise the series continues to
infinity.
2.
Abel’s Limit Theorem
For – 1 < x < 1 we have
(1 − x) −1 = 1 + x + x 2 + .....
⇒
1
(1 − x)
∞
= ∑ x n , –1 < x < 1
n =0
84
∞
3.
n
Theorem : Let f ( x ) = ∑ an x , – r < x < r
n =0
∞
If the series
∑ an x n
converges at x = r then lim f ( x) exists and we have
x→ r −
n =0
∞
lim f ( x ) = ∑ an r n .
x→ r −
n= 0
Proof : For simplicity we take r = 1. (This amounts to change of scale).
∞
f ( x ) = ∑ an x n , – 1 < x < 1
Then
n =0
∞
And for x = 1,
∑ an
converges.
n =0
∞
Let
∑ an = f (1)
.... (i)
n =0
We have to prove that lim− f ( x) = f (1) i.e. we have to prove that f is continuous from left.
x→1
Now consider
∞
∞
f ( x) = ∑ x n ∑ an x n
1− x
n=0 n= 0
1
∞
∞
∞
∞
= ∑ an x n ∑ x n = ∑ an x n ∑ bn x n
n=0
n=0 n=0
n= 0
f (x)
Then
1− x
∞
n
n =0
k=0
where bn = 1 ∀n
= ∑ Cn x n where Cn = ∑ ak bn −k
Since bn = 1 ∀n we have
n
Cn = ∑ ak
...... (2)
k =0
Consider
∞
f ( x ) − f (1) = (1 − x) ∑ Cn x n − f (1)
n= 0
∞
1
n= 0
1− x
= (1 − x ) ∑ Cn x n − (1 − x )
85
f (1)
∞
∞
= (1 − x ) ∑ Cn x n − (1 − x ) ∑ x n f (1)
n= 0
n=0
∞
= (1 − x ) ∑ ( Cn − f (1) ) x n
n=0
Now
n
∑ ak
n→∞
lim Cn = lim
n→∞
..... (3)
k=0
∞
= ∑ ak = f (1)
k =0
Therefore given ∈> 0 we can find N such that Cn − f (1) <
∈
2
for all n ≥ N . .... (4)
∞
n
From (3) f ( x ) − f (1) = (1 − x) ∑ ( Cn − f (1) ) x
n= 0
N −1
= (1 − x ) ∑ ( Cn − f (1) )
xn
n= 0
Let,
∞
+ (1 − x) ∑ ( Cn − f (1) ) x n
n= N
M = max { C1 − f (1) , C2 − f (1),..., C N −1 − f (1) }
Then for all x, 0 < x < 1,
N −1
∞
n= 0
n= N
f ( x) − f (1) ≤ (1 − x ) ∑ ( Cn − f (1) ) x n + (1 − x) ∑ ( Cn − f (1) ) x n
N −1
∞
n= 0
n= N
≤ (1 − x ) ∑ Cn − f (1) x n + (1 − x) ∑ Cn − f (1) x n
N −1
∈
n= 0
2 n= N
≤ (1 − x) M ∑ x n + (1 − x)
N −1
≤ (1 − x ) M ∑ 1 + (1 − x )
n =0
∈
∞
∑
∑ xn
2
n= 0
∈ 1
≤ (1 − x) MN + (1 − x) ⋅
2 1− x
= (1 − x ) MN +
∞
∈
2
86
xn
Let δ =
∈
then 0 < (1 − x ) < δ =
2MN
f ( x) − f (1) ≤ MN ⋅
=
∈
2 MN
+
∈
2 MN
gives
∈
2
∈ ∈
+
2 2
⇒ f ( x) − f (1) <∈ when (1 − x) < δ
f ( x) = f (1)
This show that xlim
→1−
∞
∞
n= 0
n= 0
i.e. xlim
∑ an x n = ∑ an
→1−
∞
i.e. the series
4.
∑ an x n
converges for x < 1.
n =0
Example
∞
1
We have
1− x
= ∑ xn
(–1 < x < 1)
n= 0
Integrating both sides w.r.t. x we get
∞
1
x n dx
∫ 1 − x dx = n∑
∫
=0
∞
x n+1
n =0
n +1
∞
xn
n =0
n
⇒ − log(1 − x) = ∑
⇒ log(1 − x ) = − ∑
But
∑
( −1)n
n
(–1 < x < 1)
converges (x = –1)
∞
( −1) n
n =1
n
f ( x) = −∑
Hence, xlim
→−1
∞
(−1) n+1
n =1
n
⇒ lim log(1 − x ) = ∑
x→−1
87
∞
( −1) n+1
n =1
n
⇒ log2 = ∑
5.
Example
Converse of the Abel’s limit theorem is false in general.
∞
i.e.
∑ an x n
f ( x) also exists but the series
converges to f ( x ) , for – r < x < r. The xlim
→r−
n =0
∞
∞
n =0
n =0
Cn = ∑ (−1) n x n converges for
∑ anr n may not converge. e.g. For an = (−1)n , the series nlim
→∞
–1 < x < 1, and
∞
∑ (−1)n x n = 1 − x + x 2 − x3 + .....
n =0
=
1
1+ x
= f ( x)
1
Also lim f ( x) = lim
x →1− 1 +
x→1−
x
=
1
2
∞
But
∑ (−1) n
n =0
does not converge. (Oscillatory series)
Converse of Abel’s theorem can be obtained by putting some restrictions on the
coefficients an
6.
Theorem (Tauber)
∞
n
n ⋅ an = 0 .
Let f ( x ) = ∑ an x , – 1 < x < 1 and suppose that nlim
→∞
n =0
f ( x) = S then
If xlim
→1−
∞
∑ an
n =0
converges and has the sum S.
n ⋅ an = 0
Proof : Given that nlim
→∞
⇒ lim nan = 0
n→∞
⇒ lim n an = 0
n→∞
Let σ n =
a1 + 2 a2 + ..... + n an
n
88
As n an → 0 , σ n → 0 as n → ∞
Therefore for given ∈> 0 there exists a positive integer N1 such that,
σn − 0 <
σn <
i.e.
∈
∀n ≥ N1
3
∈
∀n ≥ N1
3
.... (1)
Also lim n ⋅ an = 0 . Therefore ∃ an integer N2 such that
n→∞
n ⋅ an <
∈
∀n ≥ N2
3
.... (2)
lim f ( x) = S . For xn = 1 − 1 , x → 1 − as n → ∞
n n
Now
x→1−
⇒ lim f ( xn ) = lim f ( xn ) = S
n →∞
(Given)
xn →1−
⇒ For ∈> 0 ∃ N3 such that
f ( xn ) − S <
∈
∀n ≥ N3
3
..... (3)
Let N = max { N1, N2 , N3} . Then for all n ≥ N .
σn <
∈
3
n ⋅ an <
,
∈
3
f ( xn ) − S <
,
∈
... (4)
3
n
Sn = ∑ ak .
Next, let
k =0
For – 1 < x < 1, consider
n
n
k =0
k =0
Sn − S = ∑ ak − S + f ( x ) − ∑ ak x
n
n
k=0
k=0
∞
k
Q f ( x) = ∑ ak x (–1 < x < 1)
k =0
k
= ( f ( x) − S ) + ∑ ak − ∑ ak x k −
n
= f ( x) − S + ∑ ak (1 − xk ) −
k =0
∞
k = n+1
∞
∑
k = n+1
89
∑
ak x k
ak x k
.... (5)
For x ∈ (0,1)
(1 − x k ) = (1 − x ) (1 + x + x2 + ....x k −1 ) ≤ (1 − x ) ⋅ k
where k is +ve integer
... (6)
Thus for all n ≥ N and 0 < x < 1,
∞
n
S n − S ≤ f ( x) − S + ∑ ak 1 − xk +
k=0
∑
ak x k
k = n+1
∞
n
⇒ S n − S ≤ f ( x) − S + (1 − x ) ∑ k ak +
∑
ak x k
n
∈
∞
k =0
3N
k = n+1
n
∈
∞
k =0
⇒ S n − S ≤ f ( x) − S + (1 − x ) ∑ k ak +
⇒ S n − S ≤ f ( x) − S + (1 − x ) ∑ k ak +
k =0
k =n+1
∑
k =0
Since x ∈ (0,1) we take x = xn = 1 −
Sn − S <
∈ 1
+ 1 − 1 − nσ n +
3 n
⇒ Sn − S <
⇒ Sn − S <
∈
3
1
∈ ∈
< , ∀n ≥ N
ak <
3k 3n
xk
∑
3n
k =0
∈
n
⇒ S n − S ≤ f ( x) − S + (1 − x ) ∑ k ak +
xk
3n(1 − x)
, n ≥ N . Therefore,
n
∈
1
3n 1− 1−
n
+σn +
∀n ≥ N
∈
3
∈ ∈ ∈
+ + =∈
3 3 3
∀n ≥ N
n
⇒ lim Sn = S where Sn = ∑ ak
n →∞
k =0
∞
⇒ ∑ an converges to S.
n =0
90
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