MA1001E MATHEMATICS -I

Dr. Athira T M

Module III

Department of Mathematics
NIT Calicut
November 29, 2023

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 1 / 81

Sequences

Sequence
A sequence is a function whose domain is a set of integers.

Example: 1, 2, 3, · · ·
1, 1/2, 1/4, · · ·
0, 1, 1, 2, 3, · · ·
1, −1, 1, −1, · · ·

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A sequence is convereges to the limit L means,

lim an = L
n→+∞

A sequence that does not converge to some finite limit is said to diverge.

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Suppose that the sequences {an } and {bn } converge to limits L1 and L2 ,
respectively, and c is a constant. Then:
(a) limn→+∞ c = c
(b) limn→+∞ can = c limn→+∞ an = cL1
(c) limn→+∞ (an + bn ) = limn→+∞ an + limn→+∞ bn = L1 + L2
(d) limn→+∞ (an − bn ) = limn→+∞ an − limn→+∞ bn = L1 − L2
(e) limn→+∞  (an bn ) = limn→+∞ an · limn→+∞ bn = L1 L2
limn→+∞ an
(f) limn→+∞ bann = lim n→+∞ bn
= LL12 ( if L2 ̸= 0)

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In each part, determine whether the sequence converges or diverges by
examining the limit as n → +∞.
n o+∞
1 n
2n+1 n=1
n o+∞
n
2 (−1)n+1 2n+1
n=1
+∞
(−1)n+1 n1 n=1

3

4 {8 − 2n}+∞
n=1

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A sequence converges to a limit L if and only if the sequences of
even-numbered terms and odd-numbered terms both converge to L.
Example: The sequence
1 1 1 1 1 1
, , , , , ,...
2 3 22 32 23 33
converges to 0 , since the even-numbered terms and the odd-numbered
terms both converge to 0 , and the sequence
1 1 1
1, , 1, , 1, , . . .
2 3 4
diverges, since the odd-numbered terms converge to 1 and the
even-numbered terms converge to 0 .

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The Squeezing Theorem for Sequences
Let {an } , {bn }, and {cn } be sequences such that

an ≤ bn ≤ cn ( for all values of n beyond some index N)

If the sequences {an } and {cn } have a common limit L as n → +∞, then
{bn } also has the limit L as n → +∞.

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  n! +∞
Find the limit of the sequence nn n=1

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Monotone sequence
A sequence {an }+∞
n=1 is called

strictly increasing if a1 < a2 < a3 < · · · < an < · · ·

increasing if a1 ≤ a2 ≤ a3 ≤ · · · ≤ an ≤ · · ·
strictly decreasing if a1 > a2 > a3 > · · · > an > · · ·
decreasing if a1 ≥ a2 ≥ a3 ≥ · · · ≥ an ≥ · · ·

A sequence that is either increasing or decreasing is said to be monotone,

and a sequence that is either strictly increasing or strictly decreasing is said
to be strictly monotone.

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Monotone-convergence theorem
A bounded monotone sequence is always convergent.

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Infinite series
An infinite series is an expression that can be written in the form
∞
∑ uk = u1 + u2 + u3 + · · · + uk + · · ·
k=1

The numbers u1 , u2 , u3 , . . . are called the terms of the series.

Consider the decimal

0.3333 . . .
This can be viewed as the infinite series

0.3 + 0.03 + 0.003 + 0.0003 + · · ·

or, equivalently,
3 3 3 3
+ 2 + 3 + 4 +···
10 10 10 10

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 s 1 = u1
s2 = u1 + u2
s3 = u1 + u2 + u3
..
.
n
sn = u1 + u2 + u3 + · · · + un = ∑ uk
k=1

The number sn is called the nth partial sum of the series and the
sequence {sn }+∞
n=1 is called the sequence of partial sums.

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Convergence of the series
Let {sn } be the sequence of partial sums of the series

u1 + u2 + u3 + · · · + uk + · · ·

If the sequence {sn } converges to a limit S, then the series is said to

converge to S, and S is called the sum of the series. We denote this by
writing
∞
S= ∑ uk
k=1

If the sequence of partial sums diverges, then the series is said to diverge.
A divergent series has no sum.

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Geometric series
If |r | < 1, the geometric series a + ar + ar 2 + · · · + ar n−1 + · · · converges to
a/(1 − r ) :
∞
a
∑ ar n−1 = 1 − r , |r | < 1.
n=1

If |r | ≥ 1, the series diverges.

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The geometric series with a = 1/9 and r = 1/3 is
∞  n−1
1 1 1 1 1 1/9 1
+ + +··· = ∑ = = .
9 27 81 n=1 9 3 1 − (1/3) 6

The series ∞
(−1)n 5 5 5 5
∑ n
= 5− + − +···
n=0 4 4 16 64
is a geometric series with a = 5 and r = −1/4. It converges to

a 5
= = 4.
1−r 1 + (1/4)

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 1
Find the sum of the "telescoping" series ∑n=1 n(n+1) .
Solution: We look for a pattern in the sequence of partial sums that might
lead to a formula for sk . The key observation is the partial fraction
decomposition
1 1 1
= −
n(n + 1) n n + 1
so
k k  
1 1 1
∑ = ∑ −
n=1 n(n + 1) n=1 n n+1
and
       
1 1 1 1 1 1 1 1
sk = − + − + − +···+ − .
1 2 2 3 3 4 k k +1

Removing parentheses and canceling adjacent terms of opposite sign

collapses the sum to
1
sk = 1 − .
k +1
We now see that sk → 1 as k → ∞. The series converges, and its sum is 1 :
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Theorem
If ∑∞
n=1 an converges, then an → 0.

Test for Divergence

∑∞
n=1 an diverges if limn→∞ an fails to exist or is different from zero.

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The following are all examples of divergent series.
(a) ∑∞ 2 2
n=1 n diverges because n → ∞.
(b) ∑n=1 n diverges because n+1
∞ n+1
n → 1. limn→∞ an ̸= 0
∞
(c) ∑n=1 (−1) n+1 diverges because limn→∞ (−1)n+1 does not exist.
−n −n
∞
(d) ∑n=1 2n+5 diverges because limn→∞ 2n+5 = − 12 ̸= 0.

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 1 1 1 1
Show that the harmonic series ∑∞
n=1 n = 1 + 2 + 3 + · · · + n + · · · diverges.
Solution:     
1 1 1 1 1 1 1 1 1 1 1
1+ 2 + + + + + + + + + +··· +···
3 4 5 6 7 8 9 10 11 16
| {z } | {z } | {z }
> 24 = 12 > 84 = 12 8
> 16 = 12
The sum of 2n terms ending with 1/2n+1
is greater than 2n /2n+1 = 1/2.
The sequence of partial sums is not bounded from above: If n = 2k , the
partial sum sn is greater than k/2.Hence by monotone convergence
theorem, the harmonic series diverges.

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Theorem
If ∑ an = A and ∑ bn = B are convergent series, then
1. Sum Rule:
Σ (an + bn ) = Σan + Σbn = A + B
2. Difference Rule:

Σ (an − bn ) = Σan − Σbn = A − B

3. Constant Multiple Rule: Σkan = kΣan = kA (any number k ).

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1. Every nonzero constant multiple of a divergent series diverges.
2. If Σan converges and Σbn diverges, then Σ (an + bn ) and Σ (an − bn )
both diverge.

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Find the sums of the following series.
(a)
3n−1 −1 1 1


∑∞ n=1 6n−1 = ∑∞
n=1 2n−1 − 6n−1
1 1
= ∑∞ ∞
n=1 2n−1 − ∑n=1 6n−1
1 1
= 1−(1/2) − 1−(1/6)
a = 1 and r = 1/2, 1/6
(b)
4 1
∑∞
n=0 2n = 4∑ ∞
 n=0 2n 
1
= 4 1−(1/2) a = 1, r = 1/2
=8

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The Comparison Test
Let ∑ an , Σcn , and ∑ dn be series with nonnegative terms. Suppose that for
some integer N
dn ≤ an ≤ cn for all n > N.
(a) If ∑ cn converges, then ∑ an also converges.
(b) If ∑ dn diverges, then ∑ an also diverges.

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Example:
5
(a) The series ∑∞
n=1 5n−1 diverges because its nth term

5 1 1
= 1
>
5n − 1 n − 5 n

is greater than the nth term of the divergent harmonic series.

(b) The series
∞
1 1 1 1
∑ n! = 1 + 1! + 2! + 3! + · · ·
n=0

converges because its terms are all positive and less than or equal to the
1 1 1
corresponding terms of 1 + ∑∞ n=0 2n = 1 + 1 + 2 + 22 + · · · .
The geometric series on the left converges and we have
1 1
1 + ∑∞n=0 2n = 1 + 1−(1/2) = 3. The fact that 3 is an upper bound for the
partial sums of ∑∞n=0 (1/n!) does not mean that the series converges to 3.

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The Limit Comparison Test
Suppose that an > 0 and bn > 0 for all n ≥ N ( N an integer).
1. If limn→∞ bann = c > 0, then ∑ an and ∑ bn both converge or both diverge.
2. If limn→∞ bann = 0 and ∑ bn converges, then ∑ an converges.
3. If limn→∞ bann = ∞ and ∑ bn diverges, then ∑ an diverges.

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1. Which of the following series converge, and which diverge?
(a) 34 + 59 + 16
7 9
+ 25 + · · · = ∑∞ 2n+1 ∞ 2n+1
n=1 (n+1)2 = ∑n=1 n2 +2n+1
(b) 11 + 13 + 17 + 151
+ · · · = ∑∞ 1
n=1 2n −1
(c) 1+29ln 2 + 1+314
ln 3
+ 1+4 21
ln 4
+ · · · = ∑∞ 1+n ln n
n=2 n2 +5
ln n
2. Does ∑∞ n=1 n3/2 converge?

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The Ratio Test
Let ∑ an be a series with positive terms and suppose that
an+1
lim =ρ
n→∞ an
Then,
(a) the series converges if ρ < 1,
(b) the series diverges if ρ > 1 or ρ is infinite,
(c) the test is inconclusive if ρ = 1.

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Investigate the convergence of the following series.
2n +5
(a) ∑∞n=0 3n
(2n)!
(b) ∑∞n=1 n!n!
∞ 4n n!n!
(c) ∑n=1 (2n)!

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The Root Test
Let ∑ an be a series with an ≥ 0 for n ≥ N, and suppose that
√
lim n an = ρ.
n→∞

Then
(a) the series converges if ρ < 1,
(b) the series diverges if ρ > 1 or ρ is infinite,
(c) the test is inconclusive if ρ = 1.

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 (
n/2n , n odd
Example: Consider again the series with terms an =
1/2n , n even.
Does ∑ an converge?
Solution: We apply the Root Test, finding that
(√
n
√ n/2,n odd
n
an =
1/2,n even.

Therefore, √
1 √ n
n
≤ an ≤
n
.
2 2
√ √
Since n n → 1, we have limn→∞ n an = 1/2 by the Sandwich Theorem. The
limit is less than 1 , so the series converges by the Root Test.

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Which of the following series converge, and which diverge?
n2
(a) ∑∞
n=1 2n
2n
(b) ∑∞
n=1 
n3 n
1
(c) ∑∞
n=1 1+n

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A series in which the terms are alternately positive and negative is an
alternating series. Here are three examples:

1 1 1 1 (−1)n+1
1− + − + −···+ +···
2 3 4 5 n
1 1 1 (−1)n 4
−2 + 1 − + − + · · · + +···
2 4 8 2n
1 − 2 + 3 − 4 + 5 − 6 + · · · + (−1)n+1 n + · · ·

We see from these examples that the nth term of an alternating series is of
the form
an = (−1)n+1 un or an = (−1)n un

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The Alternating Series Test (Leibniz’s Test)
The series ∞
∑ (−1)n+1 un = u1 − u2 + u3 − u4 + · · ·
n=1

converges if all three of the following conditions are satisfied:

1. The un ’s are all positive.
2. The positive un ’s are (eventually) nonincreasing: un ≥ un+1 for all
n ≥ N, for some integer N.
3. un → 0.

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The alternating harmonic series
∞
1 1 1 1
∑ (−1)n+1 n = 1 − 2 + 3 − 4 + · · ·
n=1

clearly satisfies the three requirements of Leibniz’s test with N = 1;

therefore, it converges.

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Absolutely convergent
A series ∑ an converges absolutely (is absolutely convergent) if the
corresponding series of absolute values, Σ |an |, converges.

The geometric series converges absolutely because the corresponding series

of absolute values ∞
1 1 1 1
∑ 2n = 1 + 2 + 4 + 8 + · · ·
n=0

converges. The alternating harmonic series does not converge absolutely

because the corresponding series of absolute values is the (divergent)
harmonic series.
Conditionally convergent
A series that converges but does not converge absolutely converges
conditionally.

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The Absolute Convergence Test
∑∞ ∞ ∞
n=1 an converges. If ∑n=1 |an | converges, then ∑n=1 an converges.

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This example gives two series that converge absolutely. (a) For
n+1 1 = 1 − 1 + 1 − 1 + · · · , the corresponding series of absolute
∑∞n=1 (−1) n2 4 9 16
values is the convergent series
∞
1 1 1 1
∑ n2 = 1 + 4 + 9 + 16 + · · · .
n=1

The original series converges because it converges absolutely. (b) For

sin n sin 1 sin 2 sin 3
∑∞
n=1 n2 = 1 + 4 + 9 + · · · , which contains both positive and
negative terms, the corresponding series of absolute values is
∞
sin n | sin 1| | sin 2|
∑ = + +··· ,
n=1 n2 1 4

2 because | sin n| ≤ 1 for

which converges by comparison with ∑∞ n=1 1/n
every n. The original series converges absolutely; therefore, it converges.

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Power Series and Convergence

Power series about x = 0

A power series about x = 0 is a series of the form
∞
∑ cn x n = c0 + c1 x + c2 x 2 + · · · + cn x n + · · ·
n=0

A power series about x = a is a series of the form

∞
∑ cn (x − a)n = c0 + c1 (x − a) + c2 (x − a)2 + · · · + cn (x − a)n + · · ·
n=0

in which the center a and the coefficients c0 , c1 , c2 , . . . , cn , . . . are constants.

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Consider the geometric power series,
∞
∑ xn = 1 + x + x2 + ··· + xn + ···
n=0

This is the geometric series with first term 1 and common ratio x. It
converges to 1/(1 − x) for |x| < 1. We express this fact by writing

1
= 1 + x + x2 + ··· + xn + ··· , −1 < x < 1.
1−x

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The power series
 n
1 1 1
1 − (x − 2) + (x − 2)2 + · · · + − (x − 2)n + · · ·
2 4 2

matches with a = 2, c0 = 1, c1 = −1/2, c2 = 1/4, . . . , cn = (−1/2)n . This is

a geometric series with first term 1 and ratio r = − x−2 2 . The series
x−2
converges for 2 < 1 or 0 < x < 4. The sum is

1 1 2
= x−2
=
1−r 1+ 2 x
2
n
so 2
x = 1 − (x−2)
2 +
(x−2)
4 − · · · + − 12 (x − 2)n + · · · , 0 < x < 4.

P0 (x) = 1
1 x
P1 (x) = 1 − (x − 2) = 2 −
2 2
1 1 3x x 2
P2 (x) = 1 − (x − 2) + (x − 2)2 = 3 − + ,
2 4 2 4
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Theorem
For any power series in x, exactly one of the following is true:
(a) The series converges only for x = 0.
(b) The series converges absolutely (and hence converges) for all real
values of x.
(c) The series converges absolutely (and hence converges) for all x in some
finite open interval (−R, R) and diverges if x < −R or x > R. At either of
the values x = R or x = −R, the series may converge absolutely, converge
conditionally, or diverge, depending on the particular series.

R is called the radius of convergence of the power series, and the interval
of radius R centered at x = a is called the interval of convergence. The
interval of convergence may be open, closed, or half-open, depending on
the particular series. At points x with |x − a| < R, the series converges
absolutely. If the series converges for all values of x, we say its radius of
convergence is infinite. If it converges only at x = a, we say its radius of
convergence is zero.

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Theorem
The radius of convergence of the power series ∑∞ n
n=0 an (x − a) is
1
1/n provided this limit is either a real number or ∞.
limn→∞ |an |

Theorem
The radius of convergence of the power series ∑∞ n
n=0 an (x − a) is given by
an
limn→∞ an+1 , provided that this limit is either a real number or equal to ∞.

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1. Find the radius of convergence of the following power series;
(a) ∑∞ n
n=0 n!x
∞ xn
(b) ∑n=0 n!

n+1
2. For what values of x, do the power series ∑∞ nx
n=0 (−1) n+1 converge?

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Taylor and Maclaurin Series

Definition
Let f be a function with derivatives of all orders throughout some interval
containing a as an interior point. Then the Taylor series generated by f at
x = a is
∞
f (k) (a) f ′′ (a)
∑ (x − a)k = f (a) + f ′ (a)(x − a) + (x − a)2
k=0 k! 2!
f (n) (a)
+···+ (x − a)n + · · ·
n!
The Maclaurin series generated by f is
∞
f (k) (0) k f ′′ (0) 2 f (n) (0) n
∑ x = f (0) + f ′ (0)x + x +···+ x +··· ,
k=0 k! 2! n!

the Taylor series generated by f at x = 0.

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Find the Taylor series and the Taylor polynomials generated by f (x) = e x
at x = 0.
Solution: Since f (n) (x) = e x and f (n) (0) = 1 for every n = 0, 1, 2, . . ., the
Taylor series generated by f at x = 0

f ′′ (0) 2 f (n) (0) n

f (0) + f ′ (0)x + x +···+ x +···
2! n!
x2 xn
= 1+x + +···+ +···
2 n!
∞
xk
= ∑ .
k=0 k!

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Find the Taylor series of f (x) = cos x at x = 0.
Solution: The cosine and its derivatives are f (x) = cos x, f ′ (x) = − sin x,
f ′′ (x) = − cos x, f (3) (x) = sin x,
..
.
f (2n) (x) = (−1)n cos x, f (2n+1) (x) = (−1)n+1 sin x. At x = 0, the cosines are
1 and the sines are 0 , so f (2n) (0) = (−1)n , f (2n+1) (0) = 0.
The Taylor series generated by f at 0 is
f ′′ (0) 2 f ′′′ (0) 3 f (n) (0) n
f (0) + f ′ (0)x + x + x +···+ x +···
2! 3! n!
x2 x4 x 2n
= 1 + 0 · x − + 0 · x 3 + + · · · + (−1)n +···
2! 4! (2n)!
∞
(−1)k x 2k
= ∑ .
k=0 (2k)!

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Find the Taylor series generated by f (x) = 1/x at a = 2. Where, if
anywhere, does the series converge to 1/x ?
Solution: We need to find f (2), f ′ (2), f ′′ (2), . . ..
Taking derivatives we get f (x) = x −1 , f ′ (x) = −x −2 , f ′′ (x) =
2!x −3 , · · · , f (n) (x) = (−1)n n!x −(n+1) , so that
′′ (n) n
f (2) = 2−1 = 21 , f ′ (2) = − 212 , f 2!(2) = 2−3 = 213 , · · · , f n!(2) = (−1) 2n+1
.
The Taylor series is
f ′′ (2) f (n) (2)
f (2) + f ′ (2)(x − 2) + (x − 2)2 + · · · + (x − 2)n + · · ·
2! n!
1 (x − 2) (x − 2)2 n (x − 2)
n
= − + − · · · + (−1) +···
2 22 23 2n+1
This is a geometric series with first term 1/2 and ratio r = −(x − 2)/2. It
converges absolutely for |x − 2| < 2 and its sum is

1/2 1 1
= = .
1 + (x − 2)/2 2 + (x − 2) x

The Taylor series generated by f (x) = 1/x at a = 2 converges to 1/x for

|x − 2| < 2 or 0 < x < 4.
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Convergence of Taylor series
If f has derivatives of all orders in an open interval I containing a, then for
each positive integer n and for each x in I ,

f ′′ (a)
f (x) = f (a) + f ′ (a)(x − a) + (x − a)2 + · · ·
2!
f (n) (a)
+ (x − a)n + Rn (x),
n!
where

f (n+1) (c)
Rn (x) = (x − a)n+1 for some c between a and x.
(n + 1)!

If Rn (x) → 0 as n → ∞ for all x ∈ I , we say that the Taylor series generated

by f at x = a converges to f on I , and we write
∞
f (k) (a)
f (x) = ∑ (x − a)k .
k=0 k!

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 48 / 81

Show that the Taylor series generated by f (x) = e x at x = 0 converges to
f (x) for every real value of x.
Solution: The function has derivatives of all orders throughout the interval
ec
I = (−∞, ∞). and Rn (x) = (n+1)! x n+1 for some c between 0 and x.
Since e x is an increasing function of x, e c lies between e 0 = 1 and e x . When
x is negative, so is c, and e c < 1. When x is zero, e x = 1 and Rn (x) = 0.
When x is positive, so is c, and e c < e x . Thus, for Rn (x) given as above,

|x|n+1
|Rn (x)| ≤ when x ≤ 0, ec < 1
(n + 1)!
x n+1
and |Rn (x)| < e x (n+1)! when x > 0.
x n+1
Finally, because limn→∞ (n+1)! =0 for every x, limn→∞ Rn (x) = 0, and
the series converges to e x for every x. Thus,
∞
xk x2 xk
ex = ∑ = 1 + x + + · · · + +···
k=0 k! 2! k!

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 49 / 81

Periodic functions and Fourier series expansion

Periodic functions
A function f (x) is called a periodic function if f (x) is defined for all real x,
except possibly at some points, and if there is some positive number p,
called a period of f (x) such that

f (x + p) = f (x) for all x

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 50 / 81

The graph of a periodic function has the characteristic that it can be
obtained by periodic repetition of its graph in any interval of length p
The smallest positive period is often called the fundamental period.
If f (x) has period p, it also has the period 2p implies f (x + 2p) =
f ([x + p] + p) = f (x + p) = f (x), etc.; thus for any integer
n = 1, 2, 3, · · · , f (x + np) = f (x) for all x.
If f (x) and g (x) have period p, then af (x) + bg (x) with any
constants a and b also has the period p.

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 51 / 81

Find the fundamental period of following functions
1 f (x) = sin x
2 f (x) = sin 3x
3 f (x) = cos πx
4 f (x) = k

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 52 / 81

Piecewise continuous functions
A function f (x) is said to be piecewise continuous in an interval if
1 the interval can be divided into a finite number of subintervals in each
of which, f (x) is continuous
2 the limits of f (x) as x approches to the endpoints of each of
subinterval are finite.
i.e., a piecewise continuous is one that has atmost a finite number of jump
discontinuities.

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 53 / 81

Fourier Series

Consider the functions,

1, cos x, sin x, cos 2x, sin 2x, · · · , cos nx, sin nx, · · ·

All these functions have the period 2π. They form the so-called
trigonometric system.
Fourier series are infinite series that represent periodic functions in
terms of cosines and sines

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 54 / 81

Fourier expansion
Let f (x) be a periodic function with period 2π. Then the Fourier series or
Fourier expansion corresponding to f (x) is defined to be
∞
a0
+ ∑ (an cos nx + bn sin nx)
2 n=1

where the coefficients an , and bn called the Fourier coefficients, are

determined by the Euler formulas
1 Rπ
a0 = π −π f (x)dx
1 Rπ
an = π −π f (x) cos nxdx, n = 1, 2, · · ·
1 Rπ
bn = π −π f (x) sin nxdx, n = 1, 2, · · · .

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 55 / 81

Convergence of Fourier series
If a periodic function f (x) with period 2π is piecewise continuous in the
interval −π ≤ x ≤ π and has a left and right hand derivatives at each point
of that interval, then the Fourier series of f (x) is convergent. Also,
1 If x is a point of continuity, then
∞
a0
f (x) = + ∑ (an cos nx + bn sin nx)
2 n=1
2 If x is a point of discontinuity, then
∞
f (x−) + f (x+) a0
= + ∑ (an cos nx + bn sin nx)
2 2 n=1

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 56 / 81

Find the Fourier series of the function f (x) given by the following which is
extended to R with periodicity 2π :
(
1 if 0 ≤ x < π
f (x) =
2 if π ≤ x < 2π

Solution:
Due to (periodic extension, we can rewrite the function f (x) on [−π, π) as
2 if − π ≤ x < 0
f (x) =
1 if 0 ≤ x < π
Then
Z 0
1 1 π
Z
a0 = f (x)dx + f (x)dx = 3.
π−π π 0
1 0 1 π
Z Z
an = 2 cos nxdx + cos nxdx = 0.
π −π π 0
1 0 1 π (−1)n − 1
Z Z
bn = 2 sin nxdx + sin nxdx = .
π −π π 0 nπ
Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 57 / 81
Notice that b1 = − π2 , b2 = 0, b3 = − 3π2
, b4 = 0, . . . Therefore, Fourier series
of f (x) is given by
 
3 2 sin 3x sin 5x
− sin x + + +··· .
2 π 3 5

Notice that x = 0 is a point of discontinuity of f (x). The last expression

for f (x) holds for all x ∈ R, where ever f (x) is continuous and left and
right hand derivatives exist; in particular, for x ∈ [−π, π). By the
convergence theorem, the Fourier series at x = 0 sums to f (0+)+f 2
(0−)
,
3
which is equal to 2 .

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 58 / 81

Find the Fourier series of f given by
(
−k when − π < x < 0,
f (x) =
k when 0 < x < π

and f (x + 2π) = f (x) and hence deduce the series π

4 = 1 − 31 + 15 − · · ·
Solution: (
0 when n is even
a0 = 0, an = 0, bn = 4k
nπ when n is odd
Fourier series of f (x) is given by

4k sin 3x sin 5x
[sin x + + + ...]
π 3 5
π 4k sin 3 π2 sin 5 π2
f (x) is continuous at x = 2 . So, f ( π2 ) = k = π
π [sin 2 + 3 + 5 + ...]

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 59 / 81

Fourier series expansions in arbitrary interval
The transition from period 2π to be period ρ = 2L is effected by a suitable
change of scale.
Let f (x) have period p = 2L. Then we can introduce a new variable v such
that
p
x= v,
2π
so that
2π π
v= x= x
p L
then v = ±π corresponds to x = ±L.
This means that f , as a function of v , has period 2π.
Thus Fourier series of the form
  ∞
L
f (x) = f v = a0 + ∑ (an cos nv + bn sin nv )
π n=1

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 60 / 81

Fourier series of a function f (x) of period 2L
∞ 
a0 nπ nπ 
f (x) = + ∑ an cos x + bn sin x
2 n=1 L L
with the Fourier coefficients of f (x) given by the Euler formulas
1 RL nπx
(a) an = L −L f (x) cos L dx n = 0, 1, 2, · · ·
1 RL nπx
(b) bn = L −L f (x) sin L dx n = 1, 2, · · · .

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 61 / 81

Find theFourier series of the function
 0 if −2 < x < −1
f (x) = k if −1 < x < 1 p = 2L = 4, L = 2.
0 if 1<x <2

Solution. a0 = k/2
Z 2 Z 1
1 nπx 1 nπx 2k nπ
an = f (x) cos dx = k cos dx = sin .
2 −2 2 2 −1 2 nπ 2
Thus an = 0 if n is even and

an = 2k/nπ if n = 1, 5, 9, · · · , an = −2k/nπ if n = 3, 7, 11, · · · .

bn = 0 for n = 1, 2, · · · .
Hence the Fourier series is,
 
k 2k π 1 3π 1 5π
f (x) = + cos x − cos x + cos x − + · · · .
2 π 2 3 2 5 2

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 62 / 81

Fourier series of even and odd functions

Definition
A function f is said to be even if f (−x) = f (x) and odd if f (−x) = −f (x)
for all x ∈ R.
If f (x) is an odd function, its Fourier series reduces to a Fourier sine series
∞
nπ
∑ bn sin x (f odd )
n=1 L

with coefficients Z L
2 nπx
bn = f (x) sin dx.
L 0 L

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 63 / 81

If f (x) is an even function, it can be shown that its Fourier series reduces
to a Fourier cosine series
∞
a0 nπ
+ ∑ an cos x (f even )
2 n=1 L

with coefficients (note: integration from 0 to L only!)

2 L
Z
a0 = f (x)dx
L 0
2 L nπx
Z
an = f (x) cos dx, n = 1, 2, · · ·
L 0 L

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 64 / 81

Half range expansion
Consider a function f (x) defined on the interval [0, L]. We need to express
f (x) as a Fourier series. We could expand f (x) as a function of period L,
but this would give us a series with both sines and cosines. Instead, we
think of extending f (x) as either an odd function or an even function. The
two Fourier series you obtain are called the half range expansions.

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 65 / 81

Find the half-range cosine series for f (x) = 2x − 1 for 0 < x < 1 and show
that
1 1 1 π2
2
+ 2 + 2 +··· =
1 3 5 8
Solution: Here, the even periodic extension is,

f (x), if 0 < x < 1
g (x) = and g (x + 2) = g (x)
f (−x), if − 1 < x < 0.

So the half-range cosine series is

∞
a0  nπx 
+ ∑ an cos
2 n=1 1
R1
where a0 = 2( 11 0 (2x − 1)dx) = 0 and
Z 1
(
2  nπx  4 n 0 for n even
an = (2x −1) cos dx = 2 2 [(−1) − 1] = −8
1 0 1 n π n2 π 2
for n odd.

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 66 / 81

Notice that f (x) is continuous at x = 0. Hence,
 
8 cos πx cos 3πx cos 5πx
2x − 1 = − 2 + + + · · · for 0 < x < 1.
π 12 32 52

At x = 0, the Fourier series sums to limx→0+ (2x − 1) = −1.

2
Hence, 112 + 312 + 512 + · · · = π8 .

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 67 / 81

 (
1 if 0 < x < 1
Find the half-range sine series for f (x) =
0 if 1 ≤ x < 2
Solution:
nπx


Here, L = 2. Thus, the Fourier sine series is ∑∞
n=1 bn sin 2 , where
Z 2 Z 1
2  nπt   nπt  2 h  nπ i
bn = f (x) sin dt = sin dt = 1 − cos .
2 0 2 0 2 nπ 2

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 68 / 81

Fourier Integral

If f (x) is piecewise smooth on any interval [−L, L] or [0, L] then it can be

represented by a Fourier series.

If f (x) is not a periodic function, then it cannot be represented by a

Fourier series over the entire real line.

However, we may be able to represent f (x) in an integral form.

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 69 / 81

Fourier Integral theorem
If f (x) is piecewise continuous in every finite interval and has a righthand
derivative and a left-hand derivative at every point and if the integral
Z 0 Z b
lim |f (x)|dx + lim |f (x)|dx.
a→−∞ a b→∞ 0

exists, then f (x) can be represented by a Fourier integral

Z ∞
f (x) = [A(w ) cos wx + B(w ) sin wx]dw .
0

with A and B given by

1 1
Z ∞ Z ∞
A(w ) = f (v ) cos wvdv , B(w ) = f (v ) sin wvdv
π −∞ π −∞

At a point where f (x) is discontinuous the value of the Fourier integral

equals the average of the left and right hand limits of f (x) at that point.

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 70 / 81

Find the Fourier integral representation of the function

1 if |x| < 1
f (x) =
0 if |x| > 1

Solution:
1
1 1 1 sin wv 2 sin w
Z ∞ Z
A(w ) = f (v ) cos wvdv = cos wvdv = =
π −∞ π −1 πw −1 πw
Z 1
1
B(w ) = sin wvdv = 0
π −1

2 cos wx sin w
Z ∞
f (x) = dw .
π 0 w

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 71 / 81

The average of the left- and right-hand limits of f (x) at x = 1 is equal to
(1 + 0)/2, that is, 12 .
Furthermore, we obtain (multiply by π/2 )

Z ∞
cos wx sin w  π/2 if 0 ≦ x < 1,
dw = π/4 if x = 1,
0 w
0 if x > 1.


The case x = 0 is of particular interest. If x = 0, then

sin w
Z ∞
π
dw = .
0 w 2

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 72 / 81

Find the Fourier integral representation of

0, x < 0

f (x) = 1, 0 ≤ x ≤ 1

0, x > 1


Hence, show that

sin(x/2)
Z ∞
π
dx = .
0 x 2
h i1
R1 sin(ωv )
= sinωω
R∞
Solution: A (ω) = −∞ f (v ) cos(ωv )dv =
0 cos(ωv )dv = ω
h i1 0
f (v ) sin(ωv )dv = 01 sin(ωv )dv = − cos(ωv )
= ω1 (1 − cos ω).
R∞ R
B(ω) = −∞ ω 0

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 73 / 81

 1 ∞ 1
Z
f (x) = [sin ω cos(ωx) + (1 − cos ω) sin(ωx)]dω
π 0 ω
1 ∞ 2
Z ω h ω  ω  i
= sin cos cos(ωx) + sin sin(ωx) dω
π 0 ω 2 2 2
 
2 ∞ 1 1
Z ω 
= sin cos ω x − dω
π 0 ω 2 2
Let x = 1/2. Then f (1/2) = 1
1 = π2 0∞ ω1 sin ω2 dω or 0∞ ω1 sin ω2 dω = π2 or 0∞ sin(x/2)
R
R
R
x dx = π2

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 74 / 81

If f has a Fourier integral representation and is even, then B(w ) = 0. This
holds because the integrand of B(w ) is odd. Then the Fourier cosine
integral is, Z ∞
f (x) = A(w ) cos wxdw
0
where
2
Z ∞
A(w ) = f (v ) cos wvdv .
π 0
Note the change in A(w ) : for even f the integrand is even, hence the
integral from −∞ to ∞ equals twice the integral from 0 to ∞.

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 75 / 81

Similarly, if f has a Fourier integral representation and is odd, then
A(w ) = 0. This is true because the integrand of A(w ) is odd. Then we get
a Fourier sine integral
Z ∞
f (x) = B(w ) sin wxdw
0

where
2
Z ∞
B(w ) = f (v ) sin wvdv .
π 0

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 76 / 81

Laplace Integrals:
Consider the function f (x) = e −kx , where x > 0 and k > 0. The result will
be used to evaluate the so-called Laplace integrals.
Solution. R
A(w ) = π2 0∞ e −kv cos wvdv .
R −kv k −kv − w sin wv + cos wv .


e cos wvdv = − k 2 +w 2e k
If v = 0, the expression on the right equals −k/ k 2 + w 2 . If v approaches

infinity, that expression approaches zero because of the exponential factor.

Thus 2/π times the integral from 0 to ∞ gives

2k/π
A(w ) = .
k2 + w 2
Thus the Fourier cosine integral representation is
2k cos wx
Z ∞
−kx
f (x) = e = dw (x > 0, k > 0)
π 0 k2 + w 2
cos wx
Z ∞
π −kx
i.e., dw = e
0 k2 + w 2 2k
Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 77 / 81
 2 R ∞ −kv
Similarly B(w ) =π 0 e sin wvdv .
R −kv w −kv k sin wv + cos wv .


e sin wvdv = − k 2 +w 2e w
This equals −w / k 2 + w 2 if v = 0, and approaches 0 as v → ∞.

Thus
2w /π
B(w ) = 2 .
k +w2
The Fourier sine integral representation is
2 w sin wx
Z ∞
f (x) = e −kx = dw
π 0 k2 + w 2
From this we see that
w sin wx
Z ∞
π −kx
dw = e
0 k2 + w 2 2

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 78 / 81

Show that the integral represents the indicated functions

R ∞ cos xw +w sin xw  0 if x < 0
1
0 1+w 2
dx = π/2 if x = 0

πe −x if x > 0
 π
R ∞ sin πw sin xw
2 sin x if 0 ≦ x ≦ π
2
0 1−w 2
dw =
0 if x >π
 1
R ∞ 1−cos πw
2π if 0 < x < π
3
0 w sin xwdw =
0 if x >π
 1
R ∞ cos 12 πw
2 π cos x if 0 < |x| < 12 π
0 1−w 2 cos xwdw =
4
0 if |x| ≧ 12 π

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 79 / 81

Represent f (x) as Fourier cosine integral

1 f (x) =
1 if 0 < x < 1
0 if x >1
 2
2 f (x) =
x if 0 < x < 1
0 if x >1
Represent f (x) as Fourier sine integral

1 f (x) =
x if 0 < x < a
0 if x >a

2 f (x) =
cos x if 0 < x < π
0 if x >π

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 80 / 81

Fourier Transform

∞ R
Let f (x) be piecewise continuous on (−∞, ∞) and −∞ |f (x)|dt converges.
Then, the Fourier transform of f (t) denoted by F [f (t)] is defined as
Z ∞
F [f (t)] = f (t)e −iωt dt = F (ω).
−∞
∞ R
Assume now that −∞ |F (ω)|dω converges. Then, we define the inverse
Fourier transform of F (ω) as

Dr. Athira T M MA1001E MATHEMATICS -I November 29, 2023 81 / 81