Some Special Functions

Some Special Functions

Shiju George

Govt. Brennen College, Thalassery, KERALA

18-8-2021

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 1 / 13

 Some Special Functions

POWER SERIES-Review

In the last class, we have studied:

Theorem
∞
X
cn x n converges for
P
Suppose cn converges. Then f (x) =
n=0
∞
X
−1 < x < 1. Further lim f (x) = cn .
x→1
n=0

Using this theorem, we can prove the following result by Abel

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 1 / 13

 Some Special Functions

POWER SERIES-Review

In the last class, we have studied:

Theorem
∞
X
cn x n converges for
P
Suppose cn converges. Then f (x) =
n=0
∞
X
−1 < x < 1. Further lim f (x) = cn .
x→1
n=0

Using this theorem, we can prove the following result by Abel

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 1 / 13

 Some Special Functions

POWER SERIES-Review

In the last class, we have studied:

Theorem
∞
X
cn x n converges for
P
Suppose cn converges. Then f (x) =
n=0
∞
X
−1 < x < 1. Further lim f (x) = cn .
x→1
n=0

Using this theorem, we can prove the following result by Abel

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 1 / 13

 Some Special Functions

Abel’s Theorem

Theorem
P P P
If the series an , bn and cn converge to A, B and C respectively,
where cn = a0 bn + a1 bn−1 , · · · , an b0 , then C = AB.

Proof:
∞
X ∞
X ∞
X
All three series f (x) = an x n , g (x) = bn x n and h(x) = cn x n
n=0 n=0 n=0
converge for for x = 1.

Thus each of them have radius of convergence ≥ 1 and hence converge

absolutely for |x| < 1.

Thus by Merten’s theorem (Theorem 3.50 in our syllabus book),

f (x)g (x) = h(x) for −1 < x < 1.
SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 2 / 13
 Some Special Functions

Abel’s Theorem

Theorem
P P P
If the series an , bn and cn converge to A, B and C respectively,
where cn = a0 bn + a1 bn−1 , · · · , an b0 , then C = AB.

Proof:
∞
X ∞
X ∞
X
All three series f (x) = an x n , g (x) = bn x n and h(x) = cn x n
n=0 n=0 n=0
converge for for x = 1.

Thus each of them have radius of convergence ≥ 1 and hence converge

absolutely for |x| < 1.

Thus by Merten’s theorem (Theorem 3.50 in our syllabus book),

f (x)g (x) = h(x) for −1 < x < 1.
SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 2 / 13
 Some Special Functions

Abel’s Theorem

Theorem
P P P
If the series an , bn and cn converge to A, B and C respectively,
where cn = a0 bn + a1 bn−1 , · · · , an b0 , then C = AB.

Proof:
∞
X ∞
X ∞
X
All three series f (x) = an x n , g (x) = bn x n and h(x) = cn x n
n=0 n=0 n=0
converge for for x = 1.

Thus each of them have radius of convergence ≥ 1 and hence converge

absolutely for |x| < 1.

Thus by Merten’s theorem (Theorem 3.50 in our syllabus book),

f (x)g (x) = h(x) for −1 < x < 1.
SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 2 / 13
 Some Special Functions

Abel’s Theorem

Theorem
P P P
If the series an , bn and cn converge to A, B and C respectively,
where cn = a0 bn + a1 bn−1 , · · · , an b0 , then C = AB.

Proof:
∞
X ∞
X ∞
X
All three series f (x) = an x n , g (x) = bn x n and h(x) = cn x n
n=0 n=0 n=0
converge for for x = 1.

Thus each of them have radius of convergence ≥ 1 and hence converge

absolutely for |x| < 1.

Thus by Merten’s theorem (Theorem 3.50 in our syllabus book),

f (x)g (x) = h(x) for −1 < x < 1.
SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 2 / 13
 Some Special Functions

Abel’s Theorem

We have f (x)g (x) = h(x) for −1 < x < 1.

Hence, lim f (x). lim g (x) = lim h(x)

x→1 x→1 x→1

P P P
Applying the above theorem, we get an bn = cn .

This proves the theorem.

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 3 / 13

 Some Special Functions

Abel’s Theorem

We have f (x)g (x) = h(x) for −1 < x < 1.

Hence, lim f (x). lim g (x) = lim h(x)

x→1 x→1 x→1

P P P
Applying the above theorem, we get an bn = cn .

This proves the theorem.

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 3 / 13

 Some Special Functions

Abel’s Theorem

We have f (x)g (x) = h(x) for −1 < x < 1.

Hence, lim f (x). lim g (x) = lim h(x)

x→1 x→1 x→1

P P P
Applying the above theorem, we get an bn = cn .

This proves the theorem.

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 3 / 13

 Some Special Functions

Abel’s Theorem

We have f (x)g (x) = h(x) for −1 < x < 1.

Hence, lim f (x). lim g (x) = lim h(x)

x→1 x→1 x→1

P P P
Applying the above theorem, we get an bn = cn .

This proves the theorem.

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 3 / 13

 Some Special Functions

DOUBLE SERIES

In the first unit, we have studied some properties of double sequence. Let
us now look into double series.
The first question to be asked is that:
∞ X
X ∞ ∞ X
X ∞
Question: Does aij = aij always hold ?
i=1 j=1 j=1 i=1

Answer: No
Counterexample:

 0 if (i < j)
aij = −1 if (i = j)
 j−i
2 if (i > j)

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 4 / 13

 Some Special Functions

DOUBLE SERIES

In the first unit, we have studied some properties of double sequence. Let
us now look into double series.
The first question to be asked is that:
∞ X
X ∞ ∞ X
X ∞
Question: Does aij = aij always hold ?
i=1 j=1 j=1 i=1

Answer: No
Counterexample:

 0 if (i < j)
aij = −1 if (i = j)
 j−i
2 if (i > j)

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 4 / 13

 Some Special Functions

DOUBLE SERIES

In the first unit, we have studied some properties of double sequence. Let
us now look into double series.
The first question to be asked is that:
∞ X
X ∞ ∞ X
X ∞
Question: Does aij = aij always hold ?
i=1 j=1 j=1 i=1

Answer: No
Counterexample:

 0 if (i < j)
aij = −1 if (i = j)
 j−i
2 if (i > j)

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 4 / 13

 Some Special Functions

DOUBLE SERIES

In the first unit, we have studied some properties of double sequence. Let
us now look into double series.
The first question to be asked is that:
∞ X
X ∞ ∞ X
X ∞
Question: Does aij = aij always hold ?
i=1 j=1 j=1 i=1

Answer: No
Counterexample:

 0 if (i < j)
aij = −1 if (i = j)
 j−i
2 if (i > j)

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 4 / 13

 Some Special Functions

DOUBLE SERIES

In the first unit, we have studied some properties of double sequence. Let
us now look into double series.
The first question to be asked is that:
∞ X
X ∞ ∞ X
X ∞
Question: Does aij = aij always hold ?
i=1 j=1 j=1 i=1

Answer: No
Counterexample:

 0 if (i < j)
aij = −1 if (i = j)
 j−i
2 if (i > j)

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 4 / 13

 Some Special Functions

DOUBLE SERIES

∞ X
X ∞ ∞ X
X i−1 ∞
X
aij = ( aij + aii + aij )
i=1 j=1 i=1 j=1 j=i+1
∞ X
X i−1 ∞
X
j−i
= ( 2 + −1 + 0)
i=1 j=1 j=i+1
∞ X
X i−1 ∞
X
j−i
= ( 2 −1+ 0)
i=1 j=1 j=i+1
∞
X i−1
X
= (2−i 2j − 1)
i=1 j=1
X∞
= (2−i 2(2i−1 − 1) − 1)
i=1

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 5 / 13

 Some Special Functions

DOUBLE SERIES

∞
X
= (2−i (2i − 2) − 1)
i=1
∞
X
= (1 − 2−i+1 − 1)
i=1
∞
X
= −2−i+1
i=1
1 1
= −(1 + + 2 + ···)
2 2
= −2

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 6 / 13

 Some Special Functions

DOUBLE SERIES

∞ X
X ∞ ∞ X
X j−1 ∞
X
aij = ( aij + ajj + aij )
j=1 i=1 j=1 i=1 i=j+1
∞ X
X j−1 ∞
X
= ( 0 + −1 + 2j−i )
j=1 i=1 i=j+1
X∞ ∞
X
= (−1 + 2j 2−i )
j=1 i=j+1
∞
X 1 1 1
= (−1 + 2j ( + + + ···
2j+1 2j+2 2j+3
j=1
X∞
= (−1 + 1)
j=1
= 0
SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 7 / 13
 Some Special Functions

DOUBLE SERIES
∞ X
X ∞ ∞ X
X ∞
Thus we have proved that aij = aij may not hold always.
i=1 j=1 j=1 i=1

However sums can be interchanged under certain conditions. The

following theorem explains this.

Theorem
∞ X
X ∞ ∞
X
Let aij be a double sequence. Suppose that |aij | converge for
i=1 j=1 j=1
∞
X
each i = 1, 2, 3, · · · , say to bi and bi converge. Then
i=1
∞ X
X ∞ ∞ X
X ∞
aij = aij
i=1 j=1 j=1 i=1

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 8 / 13

 Some Special Functions

DOUBLE SERIES
∞ X
X ∞ ∞ X
X ∞
Thus we have proved that aij = aij may not hold always.
i=1 j=1 j=1 i=1

However sums can be interchanged under certain conditions. The

following theorem explains this.

Theorem
∞ X
X ∞ ∞
X
Let aij be a double sequence. Suppose that |aij | converge for
i=1 j=1 j=1
∞
X
each i = 1, 2, 3, · · · , say to bi and bi converge. Then
i=1
∞ X
X ∞ ∞ X
X ∞
aij = aij
i=1 j=1 j=1 i=1

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 8 / 13

 Some Special Functions

DOUBLE SERIES
∞ X
X ∞ ∞ X
X ∞
Thus we have proved that aij = aij may not hold always.
i=1 j=1 j=1 i=1

However sums can be interchanged under certain conditions. The

following theorem explains this.

Theorem
∞ X
X ∞ ∞
X
Let aij be a double sequence. Suppose that |aij | converge for
i=1 j=1 j=1
∞
X
each i = 1, 2, 3, · · · , say to bi and bi converge. Then
i=1
∞ X
X ∞ ∞ X
X ∞
aij = aij
i=1 j=1 j=1 i=1

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 8 / 13

 Some Special Functions

DOUBLE SERIES

1 1
Proof: Let E = {0, 1, , , · · · }
2 3
For each i = 1, 2, 3, · · · , define fi : E → R as follows:
∞
X
fi (0) = aij
j=0
n
1 X
fi ( ) = aij
n
j=0

Clearly each fi is continuous on E .

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 9 / 13

 Some Special Functions

DOUBLE SERIES

1 1
Proof: Let E = {0, 1, , , · · · }
2 3
For each i = 1, 2, 3, · · · , define fi : E → R as follows:
∞
X
fi (0) = aij
j=0
n
1 X
fi ( ) = aij
n
j=0

Clearly each fi is continuous on E .

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 9 / 13

 Some Special Functions

DOUBLE SERIES

1 1
Proof: Let E = {0, 1, , , · · · }
2 3
For each i = 1, 2, 3, · · · , define fi : E → R as follows:
∞
X
fi (0) = aij
j=0
n
1 X
fi ( ) = aij
n
j=0

Clearly each fi is continuous on E .

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 9 / 13

 Some Special Functions

DOUBLE SERIES

1 1
Proof: Let E = {0, 1, , , · · · }
2 3
For each i = 1, 2, 3, · · · , define fi : E → R as follows:
∞
X
fi (0) = aij
j=0
n
1 X
fi ( ) = aij
n
j=0

Clearly each fi is continuous on E .

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 9 / 13

 Some Special Functions

DOUBLE SERIES

1 1
Proof: Let E = {0, 1, , , · · · }
2 3
For each i = 1, 2, 3, · · · , define fi : E → R as follows:
∞
X
fi (0) = aij
j=0
n
1 X
fi ( ) = aij
n
j=0

Clearly each fi is continuous on E .

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 9 / 13

 Some Special Functions

DOUBLE SERIES
Proof(This page may be omitted for exam):
We only need to check the continuity at x = 0 as it is the only limit point.
To show the continuity of fi at x = 0:
Let xn → 0 in E . We shall show that fi (xn ) → fi (0).
Let ε > 0 be given.
X∞ ∞
X
Since aij converges absolutely, there exists N such that |aij | < ε
j=0 j=N
Since xn → 0 in E , there exists N1 such that xn < 1/N for all n ≥ N1 .
But then for any n ≥ N1 , we get
(
0P if xn = 0
|fi (xn ) − fi (0)| = ∞
| j= 1 +1 aij | if xn ̸= 0
xn

∞
X
In either case |fi (xn ) − fi (0)| ≤ |aij | < ε
j=N
SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 10 / 13
 Some Special Functions

DOUBLE SERIES
Proof(contd · · · ):
For each i = 1, 2, 3, · · · , |fi (x)| ≤ bi ∀x ∈ E .
X∞ ∞
X
Since bi converges, by Weierstrass M-test, fi converges uniformly
i=1 i=1
on E and hence it is continuous on E . In particular at x = 0
X∞ ∞
X
Thus we have lim fi (1/n) = fi (0)
n→∞
i=1 i=1
Now ∞
X ∞ X
X n
lim fi (1/n) = lim aij
n→∞ n→∞
i=1 i=1 j=1
∞
n X
X
= lim aij
n→∞
j=1 i=1
∞
XX∞
= aij
SHIJU GEORGE (BRENNEN COLLEGE ) j=1 i=1 18-8-2021 11 / 13
 Some Special Functions

DOUBLE SERIES
Proof(contd · · · ):
For each i = 1, 2, 3, · · · , |fi (x)| ≤ bi ∀x ∈ E .
X∞ ∞
X
Since bi converges, by Weierstrass M-test, fi converges uniformly
i=1 i=1
on E and hence it is continuous on E . In particular at x = 0
X∞ ∞
X
Thus we have lim fi (1/n) = fi (0)
n→∞
i=1 i=1
Now ∞
X ∞ X
X n
lim fi (1/n) = lim aij
n→∞ n→∞
i=1 i=1 j=1
∞
n X
X
= lim aij
n→∞
j=1 i=1
∞
XX∞
= aij
SHIJU GEORGE (BRENNEN COLLEGE ) j=1 i=1 18-8-2021 11 / 13
 Some Special Functions

DOUBLE SERIES
Proof(contd · · · ):
For each i = 1, 2, 3, · · · , |fi (x)| ≤ bi ∀x ∈ E .
X∞ ∞
X
Since bi converges, by Weierstrass M-test, fi converges uniformly
i=1 i=1
on E and hence it is continuous on E . In particular at x = 0
X∞ ∞
X
Thus we have lim fi (1/n) = fi (0)
n→∞
i=1 i=1
Now ∞
X ∞ X
X n
lim fi (1/n) = lim aij
n→∞ n→∞
i=1 i=1 j=1
∞
n X
X
= lim aij
n→∞
j=1 i=1
∞
XX∞
= aij
SHIJU GEORGE (BRENNEN COLLEGE ) j=1 i=1 18-8-2021 11 / 13
 Some Special Functions

DOUBLE SERIES
Proof(contd · · · ):
For each i = 1, 2, 3, · · · , |fi (x)| ≤ bi ∀x ∈ E .
X∞ ∞
X
Since bi converges, by Weierstrass M-test, fi converges uniformly
i=1 i=1
on E and hence it is continuous on E . In particular at x = 0
X∞ ∞
X
Thus we have lim fi (1/n) = fi (0)
n→∞
i=1 i=1
Now ∞
X ∞ X
X n
lim fi (1/n) = lim aij
n→∞ n→∞
i=1 i=1 j=1
∞
n X
X
= lim aij
n→∞
j=1 i=1
∞
XX∞
= aij
SHIJU GEORGE (BRENNEN COLLEGE ) j=1 i=1 18-8-2021 11 / 13
 Some Special Functions

DOUBLE SERIES

Proof(contd · · · ): On the other hand

∞
X ∞ X
X ∞
fi (0) = aij
i=1 i=1 j=1

∞ X
X ∞ ∞ X
X ∞
Thus we get aij = aij .
j=1 i=1 i=1 j=1

This proves the theorem.

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 12 / 13

 Some Special Functions

DOUBLE SERIES

Proof(contd · · · ): On the other hand

∞
X ∞ X
X ∞
fi (0) = aij
i=1 i=1 j=1

∞ X
X ∞ ∞ X
X ∞
Thus we get aij = aij .
j=1 i=1 i=1 j=1

This proves the theorem.

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 12 / 13

 Some Special Functions

DOUBLE SERIES

Proof(contd · · · ): On the other hand

∞
X ∞ X
X ∞
fi (0) = aij
i=1 i=1 j=1

∞ X
X ∞ ∞ X
X ∞
Thus we get aij = aij .
j=1 i=1 i=1 j=1

This proves the theorem.

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 12 / 13

 Some Special Functions

THANK YOU...
HAVE A NICE DAY...

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 13 / 13

 Some Special Functions

THANK YOU...
HAVE A NICE DAY...

SHIJU GEORGE (BRENNEN COLLEGE ) 18-8-2021 13 / 13