Download CALCULUS AND ANALYTIC GEOMETRY UNIVERSITY OF CALICUT 352A

CALCULUS AND
ANALYTIC GEOMETRY
Numerical Skills
STUDY MATERIAL
B.SC. MATHEMATICS
III SEMESTER
CORE COURSE
CU CBCSS
(2014 ADMISSION ONWARDS)
UNIVERSITY OF CALICUT
SCHOOL OF DISTANCE EDUCATION
THENJIPALAM, CALICUT UNIVERSITY P.O. MALAPPURAM, KERALA - 693 635
352A
School of Distance Education
UNIVERSTY OF CALICUT
SCHOOL OF DISTANCE EDUCATION
Study Material
III Sem BSc Mathematics
Core Course
Calculus and Analytic Geometry
2014 Admission
Prepared by:
Module I:
Mohamed Nishad Maniparambath
Assistant Professor,
Department of Mathematics
Farook College, Kozhikkode
Module II: Rashmi. M.P
Assistant Professor,
Department of Mathematics
VTB College, Sreekrishnapurm
Module III: Abdul Rof. V
Assistant Professor,
Department of Mathematics
KAHM Unity Women’s College,
Manjeri
Module IV: Jalsiya. M. P
Assistant Professor
Department of Mathematics
TMG College, Tirur
Scrutinised by:
Dr.D.Jayaprasad,
Principal,
Sreekrishna College, Guruvayur
Type settings and Lay out
Computer Sectio, SDE
Chairman,
Board of Studies in Mathematics (UG)
Calculus and Analytic Geometry
©
Reserved
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MODULE I
MODULE II
MODULES III
MODULE IV
1.
Natural Logarithm
5
2.
The Exponential Function
13
3.
ax and log ax
19
4.
Growth and Decay
25
5.
L’Hospital’s Rule
30
6.
Hyperbolic Functions
38
1.
Sequences of real numbers
52
2.
Infinite series
58
1.
Power Series
73
2
Taylor And Maclaurin Series
81
3.
Convergence of Taylor Series
86
1.
Conic Sections, Parametrized
Curves, And Polar Coordinates
92
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MODULE I
Chapter1: Natural Logarithm
Example 1.
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Example 2:
Example 3:
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Example 4:
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Problem:
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Solution:
We take natural logarithm on both sides
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Chapter 2
The Exponential Function
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Example:1
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Chapter 3
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Chapter 4
Growth and Decay
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Chapter 5
L’Hospital’s Rule
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Chapter 6
Hyperbolic Functions
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Problems: Natural Logarithm
Problem-1
Problem-2
Problem-3
The Exponential Function
Problem-4
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Problem-5
Problem-6.
Problem-7.
Problem-8.
Problem-9.
Problem- 10.
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Problem-11 .
Problem-12.
Problem-13
Problem -14
Problem-15
Problem-16.
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Problem-17.
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Problem-18.
L’Hospital’s Rule
Problem-19.
Problem-20.
Problem- 21.
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Problem-22.
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Problem-23.
Problem-24.
Problem-25
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Problem-26
Hyperbolic Functions
Problem-27.
Problem28.
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Problem-29.
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Problem-30.
Problem-31.
GRAPHS
Calculus and Analytic Geometry
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MODULE II
Chapter 1
Sequences of real numbers
DEFNITION
A Sequece of real numbers is a function defined on the set
= {1,2,3,4, … … … … } of natural numbers whose range is contained in the
set of real numbers .
If : → is a sequence ,the image ( ) of a positive integer n is denoted
by
or ( ) and is known as the nth term of the sequence. we denote this
sequence by the notations { } or { ∶
}.
EXAMPLES
1.
is a sequence even natural
: → ∶ ( )=2
numbers {2,4,6,8……..}
2. The sequence 2,2,2,2………… or eqvilently : → ∶ ( ) =
2
has every term equal to 2.such a sequence in which all terms
are equal is called a constant sequence.
3. The sequence given by the defnition 1 = 1, 2 = 1,
+1=
−
1 + , ℎ
≥ 2 is
known as Fibonacci sequence. Here
= 1, =
1, = +
= 2, =
+
= 1 + 2 = 3, =
+
= 2 + 3 = 5 and
so on.
CONVERGENCE AND DIVERGENCE.
The sequence {an} converges to the number L , if to every positive number ϵ,
however small there corresponds an integer m such that │an-L│<ϵ for all n>m. Here
we call L , the limit of the sequence and we write
→∞
= or
→
→ ∞. If no such number L exist , we say that the sequence { } diverges.
PROBLEM 1
Prove that
Solution
Choose any
≥
→∞
= 0,
= 1/
> 0, we have to find a natural number
Calculus and Analytic Geometry
.
such that │
− 0 │ < for all
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Here │
,
1/
− 0│ <
│1/
< .
,
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− 0│ <
> 1/
Choose m to be a positive integer greater than 1/ . Then we get
so lim
>
= 0.
→
=>
> 1/
=> 1/ <
=> │
− 0│ <
BOUNDED SEQUENCES
DEFNITION
A sequence {
≤
} is said to be bounded from above if there exist a real number
. The number M is an upper bound for { }.
If M is an upper bound for {
is the least upper bound for {
such that
} and no number less than M is an upper bound for{an},then M
}
{ }is said to be bounded from below ,if there exist areal number m such that
n.
The number m is a lower bound for {
for all
≥
}
IF m is a lower bound for {an} and no number greater than m is a lower bound for {
m is the greatest lower bound for { }.
},then
EXAMPLES
1. {1/n} is bounded from below and from above .hence it is a bounded sequence.
Here 1 is the least upper bound and 0 is the greatest lower bound.
2. The Sequence {n} is bounded from below but is not bounded from above
.hence it is not bounded
3. The sequence {(−1) } is bounded above and below. hence it is a bounded
sequence.
THEOREM
If the sequence of real numbers {
Proof.
} is convergent .Then the limit of {
Suppose that L1 and L2 are both limits of {
Corresponding to this
,
Let
=
>
>
,
{ 1,
} and L1≠ L2.Choose = │
> 0 there exist an integer
and there exist an integer
2}.Then for
Calculus and Analytic Geometry
≥
}is unique.
1 such that │
such that │
−
− 1│ <
│.
− 2│ <
we apply the triangle inequality to get
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│
−
│ = │(
=│
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−
−
)−(
│
−
)│ ≤ │
−
│+│
So we get │ − │ < │ − │
This is a contradiction. Hence L1=L2.
THEOREM
Every convergent sequence is bounded.
Proof.
Let { } be a sequence which converges to L. Then take
positive integer such that
−
│<
+
=2
= 1, there exist a
│ − │<
≥
, −1<
< +1
≥
,
( − 1, + 1)
≥
=
{ − 1, , , . . . . . .
}
=
{ − 1, , , . . . . . .
}
ℎ
ℎ
≤
≤
So { } is bounded.
THEOREM
If { }, { } be two sequences of real numbers such that
= and
= , then
a)
(
+ )=
+
= +
(Sum Rule)
b)
c)
d)
e)
(
(
(
(
–
/
)=
) =
) =
)=
= −
(Difference Rule)
.
=
.
(product Rule)
( ) for any number k (Constant multiple rule)
–
(
)/
(
) = / , Provided
THEOREM
Sandwich theorem for sequences
≠ 0 (Quotient Rule)
Let { }, { } and { } be sequences such that
a) for some positive integer
,
≤
≤ for all ≥
b)
=
= , then { } is convergent and
DEFNITION
and
=
A Sequence { } is said to be monotonically increasing or non decreasing if
≤
for all n.
A Sequence { } is said to be monotonically decreasing or non increasing if
≥
for all n
A Sequence is said to be monotone if it is either a non-decreasing sequence
or a non-increasing sequence.
EXAMPLE
1. The Sequence {1,2,3,4 … … … . . }is a non decreasing sequence
2. The sequence{1, , , … … … … } is a nonincreasing sequence
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THEOREM
A nondecreasing sequence of real numbers converges if and only if it is bounded
from above.If a nondecreasing sequence converges ,it converges to its least upper
bound.
PROBLEM
Evaluate lim
→
Solution
We have lim(
→ )
lim
→
PROBLEM
= 0. Using limit rules we can write
−1
1
1
= lim (1 − ) = lim 1 − lim = 1 − 0 = 1
→
→
→
Evaluate lim
→
Solution
lim
→
= lim
PROBLEM
(By limit Rules)
=
→
=
Evaluate lim
Solution
→
lim
= lim
→
(
→
PROBLEM
)(
)
= lim
→
=0
Evaluate lim
Solution
→
We have −1 ≤
So − ≤
We know that
≤
≤1
lim −
→
= lim
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→
= 0
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o by sandwich theorem ,
lim
→
THEOREM
= 0.
Let { } be a sequence of real numbers. If
→ and if f is a function that is
continuous at L and defined at all , then ( ) → ( )
PROBLEM
Evaluate lim
→
Solution
= lim
lim
→
=
→
=2
Let ( ) = √ then f is continuous on the set of all non negative real numbers
and hence continuous at 2. So by previous theorem we get
= lim
lim
→
RESULTS
1.
2.
3.
4.
5.
lim
lim
lim
→
PROBLEM
Evaluate lim
Solution
lim
=1
→
→
→
1+
) = (lim
→
2 / +1 )=
(2) = √2
=0
→
If │ │ < 1,
lim
(
→
= 1,for any
1+
lim
→
=
>0
=0
1+
→
=
(by Result 5)
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PROBLEM
Evaluate
lim
Solution
(4 n)
→
lim (4 n) = lim 4n = 4. lim n = 4.1 = 4
→
→
→
(by Result 2)
EXERCISES
1.Use Definition of limits to show that lim
→
2.Write out the first five terms of the sequence
3.Prove that lim
→
=0
=
= 1,
=
.
4.Find the limits of the following sequences
a)
b)
c)
d)
=
=8
.
=
−√
=
Answers
2) 1, , ,
−
,
4)
(a) converges ,-1. (b) converges ,1
converges,1/2.
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(c )converges ,e2 (d)
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Chapter 2
Infinite series
DEFNITION
Let { } be a sequence of real numbers. Then the sum of the infinite number
of terms of this sequence i.e,
+ + +. . . . . . . . . + + ⋯ … is defined as
an infinite series.
An infinite series is generally denoted by ∑
If
or ∑
.
denotes the the sum of the first n terms of this series i.e,
+ +. . . . . . . . . +
=
+
Then { } is called the sequence of partial sums of the given series.
EXAMPLES
1. Consider the series ∑
∑
=∑
=1+ +
+
=∑
+
= 1 + + ⋯……+
2 .Consider the series ∑
Then ∑
, where
=
+ ⋯………..+
.
, where
+ ……
= (−1) .
= −1 + 1 − 1 + 1 − 1 + ⋯ … … … … … + (−1) and
= -1, if n is odd
0, if n is even
CONVERGENCE AND DIVERGENCE
Let ∑
be aseries of real numbers with partial sums
+. . . . . . . . . +
If the sequence { } of partial sums converges to a limit
series converges and the sum is .
=
+
+
, we say that the
If the sequence of partial sums does not converge ,then the series diverges.
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PROBLEM
Show that the series 1 + + 1/2 + ⋯ … ….converges
Solution
Here the sequence of partial sums is given by
=
lim
=1+
2
1
1
1
+ ⋯……+
.
2
2
→0
→∞
(
1
lim 2 −
, 1/2 < 1)
= 2
2
ℎere the sequence of partial sums converges and so the given series converges and
its sum is 2.
PROBLEM
→
=
1
2
=
1
1−
2
1
=2−
2
1−
→
Discuss the convergence of the series
Solution
Here
= nth term of the series =
Hence
.
+
)
.(
=
.
+
−
.
+ ⋯..+
1
1
−
1
2
1 1
= –
2 3
1 1
= −
3 4
……………………
…………………….
.(
)
+ ⋯……
=
=
Adding we get
Since
→0
=
→∞,
+
+
+ ⋯……+
lim
→
1
−
=1
1
+1
= 1−
here the sequence of partial sums converges .So the given series is converget and
its sum is 1.
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GEOMETRIC SERIES
Geometric series are series of the form
+
+
+
+ ⋯…….= ∑
+ ⋯……..
Where and are fixed real numbers and
ratio of the geometric series.
,
≠ 0.The number r is known as the
RESULT
If │ │ < 1 ,then the geometric series converges and
n-1
∑
=
EXAMPLE
The geometric series with
+
=
( )n-1=
+ ⋯……..= ∑
+
is
=
=1/6
THEOREMS ON CONVERGENCE
THEOREM
If ∑
a)
n=
b)
and ∑
n=
sum rule : ∑(
c)
are convergent series, then
n+bn)
= ∑
n
Difference Rule: ∑( n-bn) =
+ ∑
Constant multiple Rule : = ∑
∑
THEOREM
converges , then the nth term
If ∑
RESULT
If
lim
→
PROBLEM
n=
n
n
A+ B
- ∑
= k ∑
n=
A–B
n
= kA
→ 0.
fails to exist or is different from zero, then ∑
Examine the convergence of the series ∑
Solution
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n
diverges.
( + 1)/
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Here the nth term
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=
lim
→
then by previous result ∑
n
= lim
→
diverges.
+1
= 1 ≠ 0
PROBLEM
Find the sum of the series ∑
(
Solution
∑
∑
(
+
(
)
)=∑
+
(
+∑
is a geometric series with
∑
= /1 −
(
)
=
=
)
)
)
(using sum rule)
= and
converges
∑
(
.here │ │ < 1 so the series
=
=1
is also a geometric series, with
= − and
here │ │ < 1 so the series converges and the sum is
(−1)
=
5
1−
=
= -1/7
Therefore
∑
(
=
−
1
5
1− −
= −
1
5
=
+
(
)
) )=∑
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(
)
= 1 − 1/7 = 6/7
PROBLEM
Find the sum of the series ∑
+∑
(
+
)
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Solution
∑
(
+
)
=∑
+∑
(using sum rule)
Here both series on the right side ,are geometric series . for the first series
= 5 and = and for the second series
= 1 and =
For both the series , the common ratio is less than 1,so both series converges
and their sums are
∑
(
+
)
=∑
=
=
+∑
+
+
= 10 +
=
PROBLEM
Examine the convergence of the series ∑
Solution
The given series ∑
=1+
geometric series with a= 1 and
=
.Here │ │ = │
+
+ ⋯……..+
+ ⋯. Is a
│<1
so the given series converges and its sum is,
∑
=
=
=
=
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PROBLEM
Examine the convergence of the series ∑
cos
Solution
∑
cos
=
0 +
+
2 + ⋯….+
+ ⋯..
= 1 + (−1) + 1 + (−1) + ⋯ … + (−1) + ⋯ …
Which is a geometric series with = 1, and = −1.here │ │ = 1,so the
given series diverges.
OR
Here lim →
= lim → (−1) does not exist ,so the given series diverges.
EXERCISES
1.Which of the following series converge and which diverge ? If a series
converge, find its sum:
a) ∑
b) ∑
c) a) ∑
(−1)
( )
(2 − 1)/3
2.Find the values of for which the geometric series converges. Also find the sum of
the series for those values of :
a) ∑
b) ∑
2
c) ∑ (−1) (1 + )
3. Form a Geometric series that converges to the number 5 if = 2.
Answers
1. (a) converges .8.
(b) converges , 1. (c) converges ,3/2.
2. (a)
≠
(
(c) −2 <
)
, integer,
< 0,
. (b) │ │ < 1/2,
.
3.
= 3/5.
SERIES WITH NON NEGATIVE TERMS
THEOREM
A Series ∑
of nonnegative terms converges if and only if its sequence of
partial sums is bounded from above.
PROBLEM
Show that the series ∑
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1/ = 1 + + + + ⋯ . + + ⋯ ….diverges
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Solution
To show that the given series diverges we need only show that the sequence
{ } of its partial sums is not bounded from above.
Let us group the terms of the series:
1+ + ( + ) + ( + + + ) + ( +
The sum of the first two terms is 1.5
The sum of the next two terms is
+
+ ⋯ … … … . . ) + ⋯ ..
> + = =
The sum of the next 2 terms is
1 1 1 1
1 1 1 1 4 1
+ + + > + + + = =
5 6 7 8
8 8 8 8 8 2
The sum of the next 2 terms is
1 1
1
1
1
1
8
1
+
+ ⋯………..+
>
+
+ ⋯….+
=
=
9 10
16 16 16
16 16 2
In general ,the sum of 2 terms ending with
is
1
1
1
1
1
1
+
+ ………+
>
+
+ ⋯…+
2 +1
2 +2
2
2
2
2
Hence if = 2 ,then
= 1+ +( + )+( + + + )+ (
> + + + ⋯ ….( times) =
Hence the sequence {
series is divergent.
RESULT
The -Series ∑
=
/2.
) + ⋯….+ )
=
2
2
=
1
2
} of partial sums is not bounded from above, so the give
+
+
+ ⋯…+
+ ⋯ … where
is areal constant
,converges if > 1 and diverges if ≤ 1.
COMPARISON TESTS FOR NONNEGATIVE TERMS
The Direct Comparison Test
Let ∑
be a series with nonnegative terms. Then
a) ∑ n converges if there is a convergent series ∑ n with
≤
for all > ,
for some positive integer m
b) ∑ n diverges if there is a divergent series of nonnegative terms ∑ n with an
≥ dn for all n>m, for some positive integer m.
The Limit Comparison Test
Suppose that an>0 and bn>0 for all n ≥ m ,for some positive integer m,
a) If limn→∞ an / bn = c>0 ,then ∑ n and ∑ n both converge or diverge
together.
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b) If limn→∞ an/ bn = 0 and ∑ n converges , then ∑
c) If limn→∞ an/ bn = ∞ and ∑ n diverges , then ∑ n
n
converges.
diverges.
PROBLEM
Test for convergence of the series
Solution
Here the nth term
Take
Then
=
=
(
=
Hence limn→∞
)
(
=
=
(
(
Take
Then
=
=
(
(
+ ⋯…
√ .
)
= 1 ,a finite non zero number
Solution
(
+
/ )
Test for convergence of the series ∑
=
√ .
)
Therefore by limit comparison test ∑
together.
But ∑ n= ∑ 1/ diverges .Hence ∑
PROBLEM
Here
+
√ .
)(
)(
)
n
n
and ∑
both converge or diverge
n
diverges.
(
)(
)
)
)
=
(
)(
)
Hence limn→∞ an/ bn = 10 , a finite non zero number.
Therefore by limit comparison test ∑ n and ∑ n both converge or diverge
together.
But ∑ n= ∑ 1/ 2,the p-Series with p=2>1,converges . Hence ∑ n
converges.
PROBLEM
Test for convergence of the series ∑
Solution
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Here an=
Take bn =1/n
Then
/
=
= .
Hence limn→∞ an/ bn = 1 , a finite non zero number.
Therefore by limit comparison test ∑ n and ∑ n both converge or diverge
together.
But ∑ n= ∑ 1/ diverges .Hence ∑ n diverges.
PROBLEM
Test for convergence of the series whose nth term is √ + 1 - √
Solution
Here an =√ + 1 − √
Take
=
√
Then an/bn =
√
=
√
=
√
√
√
√
√
1
1+
√
1
√
=
[(
√
)
√
]
=
√
√
+1
=
Hence limn→∞ an/ bn =1/2, a finite non zero number.
Therefore by limit comparison test ∑ n and ∑ n both converge or diverge
together.
But ∑ n= ∑ 1/ √ ,the p-Series with p=1/2 <1,diverges. Hence ∑ n
diverges.
EXERCISES
1) Test for convergence of the series + +
2)Discuss the convergence of the series ∑
3) discuss the convergence of the series ∑
Calculus and Analytic Geometry
+ 9/25 + ⋯ ..
sin
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4) Test for convergence of the series ∑
1/(1 + 2 + 3 + ⋯ … … … . + )
5) Test for convergence of the series + + +
+ ⋯ … ..
Answers
1)diverges 2)converges 3)diverges 4)converges 5)converges
THE RATIO TEST
THEOREM
Let ∑
n
be a series of positive terms and suppose that lim
Then (a) the series converges if r <1,
(b) the series diverges if r>1 or r is infinite,
(c) the test is inconclusive if r =1
PROBLEM
n
Examine the convergence of the series : ∑
/n!
Solution
Here
Then
=
!
=
(
(
)
/
)!
=
!
= ( + 1) /
(
)
(
lim
→
= r
)!
= 1+
→
= lim
so by Ratio test the above series diverges.
→
1+
=
>1
PROBLEM
Test the convergence of the series ∑
Solution
Here
Then
=
and
=
=
=
(
(
)
)
=
(
)
/
=
Calculus and Analytic Geometry
Page 69
So lim
→
= lim
→
3
2
School of Distance Education
1
1+
1 3
= >1
So by Ratio test the series diverges.
The nth Root Test
Let ∑ n be a series with
≥ 0 for
and suppose that lim
≥
Then (a) the series converges if < 1
(b)the series diverges if > 1 or is infinite,
(c) the test is inconclusive if = 1.
PROBLEM
Examine the convergence of the series
∑ (2 / )
Solution
Here
√
n
=
=
→
√
=
then
=2/ √
limn→∞ √ n = limn→∞ ( 2/( √ )2 =2/1=2>1.
So by nth root test the series is divergent.
PROBLEM
Test for convergence the series ∑ ( √ -1)n
Solution
Here an =( √
Therefore,
-1)n then √
n
= √
-1
limn→∞ √ n = limn→∞ ( √ - 1) =1 -1=0<1
Therefore by nth –Root test the given series diverges.
EXERCISES
1.Test for convergence:
a) ∑
b) ∑
c) ∑
d) ∑
( + 1)( + 2)/ !
n
/n!
2.Discuss the convergence of the series +
+
+ ⋯ … ..
3.Test the series 1 + 2 + 3 + ⋯ …. for convergence for all positive values of .
n
4.Examine the convergence of the series ∑
/(2n)2
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5. Examine the convergence of the series
∑
Answers
1. (a)diverges (b)converges (c)divergent (d) converges
2. Converges
3. Converges if x<1 ,diverges if x≥1
4. Diverges
5. Converges.
ALTERNATING SERIES
A series in which the terms are alternately positive and negative is an
alternating series.
EXAMPLE
1.
1 −
+
−
+
− + ⋯……+
(
)
+ ⋯ … ..
2. 1 − 2 + 3 − 4 + ⋯ … … … … . +(−1)
+ ⋯…………….
Leibniz’s Test or Alternating series test
∑ (−1)n+1 un =u1 –u2 + u3 – u4 +……
The series
Converges if all three of the conditions are satisfied :
1.The un᾽s are all positive.
2. un ≥un+1 for all positive integer n.
3.
limn→∞ un =0
PROBLEM
Examine the convergence of the series :
⋯ … … … ….(p>0)
Solution
−
Given series is an alternating series of the form ∑
=
( + 1)
=
1
1+
1
<1
+
−
(−1)
(
+
,
> 0)
=
therefore un+1<un
Also
limn→∞ un = limn→∞ 1/np =0 (since p>0)
So the given series satisfies all the three conditions of the Leibniz’s test and
hence the series converges.
ABSOLUTE CONVERGENCE AND CONDITIONAL CONVEGENCE
DEFNITION
Let ∑
be a series real numbers.then
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(a) If ∑| | converges ,we say that ∑ n converges absolutely.
(b) If ∑ n converges ,but ∑| | diverges, we say that ∑ n converges
conditionally.
THEOREM
If ∑ n converges absolutely, then ∑ n converges i.e, every absolutely convergent series
is convergent.
PROBLEM
((−1)
Show that the series ∑
Solution.
Here an = (−1)
Then │
│=│
We know that
converges absolutely.
│ ≤ 1/ 2
∑ 1/
2
converges
(which is a p-series with p=2>1)
So by direct comparison test ∑ │ n│ converges. so the given series converges absolutely.
EXERCISES
1. Examine the convergence of series ∑
2. Examine the convergence of series 1 –
3. show that the series ∑
4. show that the series
5.
∑
(−1)
(
)
.√
+
.√
/
−
.√
+ ⋯…………
converges absolutely.
converges conditionally.
examine the convergence of the series
1. converges
(−1)
∑
(
!
)
Answers
5.converges absolutely.
2.converges
DEFNITION
∑ n of real numbers is a rearrangement of a series ∑
A Series
is a bijection : → ,
Where N is the set of all positive integers, such that
= ( )
EXAMPLE
The series 1 +
series
1 −
+
− +
+
n
if there
− + ⋯ … … … … …is a rearrangement of the
.
− + ……………
THEOREM
Let ∑ n be an absolutely convergent series and ∑ n be a rearrangement of
∑ n .then ∑ n converges absolutely and ∑
n= ∑
n.
Calculus and Analytic Geometry
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MODULE III
Chapter I
Power Series
A power series about = 0 is a series of the form
∑
= +
+
+ ⋯……+
+ ⋯ ..
A power series about =
is a series of the form
∑
( − ) = + ( − ) + ( − ) + ⋯……+ ( − ) +
⋯ ..,
in which the center a and the coefficients
constants.
,
,
,…………
, . . .. are
Example: Consider the series,
=1 + +
+ ⋯……+
+ ⋯ ..
This is a power series with = 1, for every n. See that, the given series is a
geometric series with first term 1 and common ratio x. So the sum of the series
is,
, provided | | < 1. So,
= 1 + + + ⋯……+
+ ⋯ . . , −1 <
Example: Consider the power series,
< 1.
1
1
1
1 − ( − 2) + ( − 2) + ⋯ … … + −
( − 2) + ⋯ …
2
4
2
Find the sum and the interval in which the power series converges to this sum?
Solution: Given series is a geometric series with first term = 1 and common
ratio
= − ( − 2). So it converges when | | < 1 and the sum is
.
Therefore, the series converges when,
| | < 1 ⟹ − ( − 2) < 1 ⟹ ( − 2) < 1 ⟹ −1 < ( − 2) < 1
⟹ −2 < − 2 < 2 ⟹ 0 < < 4
=
1−
=
1
1
1 − − ( − 2)
2
=
1
1
1 + ( − 2)
2
Hence, = 1 − ( − 2) + ( − 2) + ⋯ … … + −
the series converges to the sum when 0 < < 4
Calculus and Analytic Geometry
=
2+
2
−2
=
2
( − 2) + ⋯ …and
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Theorem: If a power series ∑
=
+
+
+ ⋯ … converges
at = ≠ 0, then it converges absolutely for all with | | < | |. If the series
diverges at = , then it diverges for all with | | > | |.
The Radius of Convergence of a Power Series
Consider the power series
∑
( − ) = + ( − )+
⋯ ..
( − ) + ⋯……+
( − ) +
If the power series converges, when | − | < and diverges when| − | >
, then R is called the radius of convergence of the power series. The power
series converges in the interval, − < < + .
For a geometric power series we can find the radius of convergence from the
condition | | < 1.
For other power series’ we may make use of ratio test to find the radius of
convergence.
We have the series ∑
converges when, lim →
< 1 and diverges
when lim
when, lim
→
So choose
(
→
(
= lim
> 1. So the power series ∑
)
< 1 ⟹ lim
)
→
⟹ | − | < lim
→
(
( − ) converges
)
→
<1
Remarks: 1) If = ∞, the power series converges everywhere.
2) If = 0, the power series converges at = and diverges everywhere else.
Example: For what values of x, the following power series converge?
a) ∑
(−1)
b) ∑
(−1)
c) ∑
d) ∑
1.
−
+
=
−
− ⋯……
+
+ ⋯ ..
= 1 + + + + ⋯ … ….
!
!
!
!
= 1 + 1! + 2! + 3! + ⋯ … …
!
Solution: a)
= lim
=
→
(
)
= lim
(
=
→
Calculus and Analytic Geometry
(
.
)
)
=
(
)
= lim −
→
= lim
→
= lim 1 +
→
=
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So the power series converges when | − 0| < 1.
ie: when | | < 1and diverges when | | > 1. Therefore the power series
converges when −1 < < 1. Again, when = 1, the series
∑ (−1)
= 1 − + − + ⋯ …. converges.
(
When = −1, ∑ (−1)
converges when −1 ≤ < 1.
=
b)
(
)
= lim
= lim
→
(
(
.
=
→
)
)
= 1.
(
(
)(
)
)
)
=∑
=
, diverges. So, the power series
(
)
(−1)
= lim 2 − 1
(−1)
→
2 +1
= lim −
→
2 +1
2 +1
= lim
→ 2 −1
2 −1
As in case a) the power series converges when −1 <
1, the series ∑ (−1)
=∑
test. Again when = −1, the series
∑
(−1)
(
=∑ (−1)
−1 ≤ ≤ 1.
c) Here = .
!
)
=∑
(−1)
(−1)
(
< 1. Again, when
=
, converges by Leibniz
)
is also convergent. So the power series converges when
=
(
)!
1
1
!
!
= lim
= lim
= lim
= lim ( + 1) = ∞
1
1
→
→
→
→
( + 1)!
( + 1) !
So power series converges for every .ie: −∞ < < ∞
d) Here
= !,
= ( + 1)!
!
!
1
= lim
= lim
= lim
= lim
=0
→
→
→
→ ( + 1)
( + 1)!
( + 1) !
So power series converges at = 0only and diverge everywhere else.
Theorem: Multiplication of Power Series
for | | <
then ∑
(∑
If ( ) = ∑
and ( ) = ∑
converge absolutely
and =
+
+
+ ⋯..+
=∑
,
converges absolutely to ( ) ( ) for | | < :
) (∑
)=∑
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Example:
If ( ) =
=1+ +
find a power series for
Here
=
times]
=
(
)
+ ⋯..= ∑
+ ⋯……+
using power series multiplication.
1
1
1
=
.
(1 − )
1− 1−
= 1, for every . So,
+
+
+ ⋯..+
Therefore,
1
1
1
=
.
(1 − )
1− 1−
=1+2 +3
=(
, −1 <
=
)(
+ ⋯ . . +(( + 1)
=(
+1
)(
< 1,
)
= 1 + 1 + ⋯ … + 1 [(n+1 )
)=
+ ⋯ …., −1 <
=
<1
( + 1)
Theorem: If ∑
converges absolutely for | | < , then
∑
( ( )) converges absolutely for any continuous function on | ( )| <
Example: We have
=1+ +
∑
,converges for | | < 1.
So the series
= 1 + 4 + (4
∑
(4
)
converges for |4
+ ⋯……+
+ ⋯..=
) + ⋯ … … + (4
) + ⋯.=
| < 1 ⟹ |2 | < 1 ⟹ | | <
Theorem: Term-by-Term Differentiation
If ( ) = ∑
( − ) = + ( − ) + ( − ) + ⋯……+
( − ) + ⋯ ..defines a function on − < < + , then ( ) has
derivatives of all orders in the same interval.
( ) = + 2 ( − ) + 3 ( − ) + ⋯……+
( − )
+ ⋯ ..
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:
( )=
Similarly,
( )=∑
Each of the series converge for, − <
Example: Find the series for
( )=
1
1−
Solution:
( )
=
′′( ) =
=1+
+
( ) and
+ ⋯……+
1
=1+2 +3
(1 − )
=
,
( − 1) ( − )
< +
( ) if,
+ ⋯..=
+4
−1 <
2
= 2 + 6 + 12
(1 − )
=
( − )
( − 1)
,
(
)
Solution:
+
= x−
,
−1 <
+ ⋯……+
+ ⋯ … … + ( − 1)
−1 <
+ ⋯..
<1
+ ( − )+
< < + .
−
converges for
<1
+ ⋯..
<1
Theorem: Term-by-Term Integration
Suppose ( ) = ∑
( − ) =
⋯ … … + ( − ) + ⋯ ..converges for −
Then ∫ ( )
=∑
Example: Identify the function,
(−1) x
( )=
2n + 1
,
<
( − ) +
<
x
x
+ + ⋯ … , −1 ≤ x ≤ 1
3
5
+ .
Differentiating the original series, we have,
( )=1−
+
+ ⋯ = 1 − x + x + ⋯….=
(
)
=
1
(Since this is a geometric series with first term 1 and common
ratio− , provided, |− | < 1, : | | < 1)
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, −1 <
<
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1
1+x
Integrating, ( ) = ∫
:
( )=
= tan
+ .
From the given series, (0) = 0 − + ⋯ . . = 0
Therefore, (0) = tan 0 +
: = 0.
Hence, ( ) = tan
, −1 < < 1
Again, the series converges for
(
So, ( ) = ∑
)
= ±1, by Leibniz test.
= x−
+
+ ⋯… =
, −1 ≤ x ≤ 1
Example: Use the series 1 − + − + ⋯ … … , −1 < < 1, to find a
series for ln (1 + )
Solution:
We have,
= 1 − + − + ⋯ … … , −1 < < 1.
Integrating over (0, ), we get,
1+
=
−
2
: [ln (1 + )] =
: ln(1 + ) =
+
−
3
−
2
−
2
+
4
+
3
+ ⋯……
3
−
−
4
4
+ ⋯ ..
+ ⋯..=
(−1)
, −1 <
<1
Remark: Convergence of a power series in the radius of convergence is
absolute convergence.
Additional Problems
1. Find the radius of convergence and interval of convergence of the series,
∑ (−1) (4 + 1) . Also find the sum.
Solution:
∑ (−1) (4 + 1) = 1 − (4 + 1) + (4 + 1) − (4 + 1) + ⋯ .. is a
geometric series with first term = 1
= −(4 + 1). So
it converges when | | < 1 ⟹ |−(4 + 1)| < 1 ⟹ |4 + 1| < 1
So radius of convergence is 1.
1
⟹ −1 < 4 + 1 < 1 ⟹ −2 < 4 < 0 ⟹ − < < 0
2
and diverges otherwise. Again the power series converges to the sum
1
1
1
=
=
=
1−
1— [−(4 + 1)] 1 + 4 + 1 4 + 2
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2. Find the radius of convergence and interval of convergence of the series,
(−1) ( + 2)
= lim
When
→
= −3,
=
= lim
(
)
→
=
and
(
(
)
)
(
)
= lim
= lim 1 +
→
→
The power series converges when | − | <
: | + 2| < 1 ⟹ −1 <
(−1) ( + 2)
(
) (
)
+ 2 < 1 ⟹ −3 <
(−1) (−3 + 2)
=
=∑
=∑
which is divergent. When = −1,
(−1) ( + 2)
=
=
< −1 .
=1.
(−1) (−1 + 2)
(−1) (1)
(−1)
=
which is convergent. So the interval of convergence is, −3 <
3. Find the radius of convergence the power series,
≤ −1.
+1
Solution: The radius of convergence can be evaluated using root test. Consider
lim |
→
| = lim
= lim
→
= lim
→
Calculus and Analytic Geometry
→
+1
1
+1
(1 + )
| |
| |
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= lim
→
1
1
(1 + )
1
| |
| |
So the power series converges when,
| | < 1 or
− < < . So the radius of convergence is .
| |<
Exercises:
1. Find the radius and interval of convergence of
(
)
a) ∑
b) ∑
c) ∑
d) ∑
e) ∑
( + 5)
(
1+
)
(
(−1)
)
2. Find the radius of convergence of the power series
!
a) ∑
b) ∑
. . ……
( !)
(
)!
Answers: 1.a) 10, −8 < < 12 b) 1, −4 < < 6 c) 1/3, ≤
d)1, −1 < < 1 e) 2, −4 ≤ < 0 2. a)3 b)8
Calculus and Analytic Geometry
<1
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Chapter 2
Taylor And Maclaurin Series
Consider the power series,
( )=∑
( − ) = + ( − ) + ( − ) + ⋯……+
( − ) + ⋯.
Substituting = in the equation, we get,
( ) = + ( − ) + ( − ) + ⋯……+ ( − ) + ⋯.=
Again,
( )=
So,
( )=
+ 2 ( − ) + 3 ( − ) + ⋯……+
( )=
+ 2 ( − ) + 3 ( − ) + ⋯……+
( )=2
( )=2
( − )
+ ⋯ ..
+ ⋯ ..
+ 3.2 ( − ) + 4.3 ( − ) + ⋯ … … + (
− 1) ( − )
+ ⋯ ..
+ 3.2 ( − ) + 4.3 ( − ) + ⋯ … … + (
− 1) ( − )
+ ⋯ ..
So, ( ) = 2 . In general ( ) = ! . So
Hence the power series can be written as,
( )=∑
( − ) =
( − ) + ⋯.
= ( )+
( − )
( )
( − )+
1!
+
( − )+
=
( )
!
.
( − ) + ⋯……+
( )
( − ) + ⋯……+
2!
( )
( − ) + ⋯.
!
Definitions:
Let be a function with derivatives of all orders throughout some interval
containing as an interior point. Then the Taylor Series generated by at
= is,
( )
( )
( − ) = ( )+
( − )+
!
1!
The Maclaurin Series generated by
(0)
!
= (0) +
the Taylor series generated by
Calculus and Analytic Geometry
is,
(0)
+
1!
at
( )
( − ) +⋯……+
2!
( )
( − ) + ⋯.
!
(0)
2!
+ ⋯,
= 0.
+ ⋯……+
(0)
!
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Example:
Find the Taylor series generated by ( ) =
does the series converge to ?
Solution:
The Taylor series at = 2 is,
(2)
(2)
( − 2) = (2) +
( − 2) +
!
1!
(2)
( − 2) + ⋯ … … +
2!
( )=
!
( ) = (−1)
1
2
2!
(2) = −
2
3!
(3) =
2
… … … … … ….
(2) =
(2) = (−1)
So the Taylor series of ( )
(2) +
2
1
(2)
( − 2) + ⋯.
!
2!
=−
6
3!
( )=
=
( )=−
In general,
= 2. Where, if anywhere,
at
2
!
(2)
(2)
(2)
( − 2) +
( − 2) + ⋯ … … +
( − 2) + ⋯ …
1!
2!
!
(−1) ( − 2)
1 ( − 2) ( − 2)
= −
+
+ ⋯……+
+ ⋯…
2
2
2
2
This is a geometric series with first term and common ratio
when | | < 1 ⟹ −
< 1 ⟹ | − 2| < 2 ⟹ −2 <
4, the series converges to the sum
1+
1
2
−2
2
=
2+
1
−2
=
1
=−
. So
−2<2⟹0<
<
Taylor Polynomials
Let ƒ be a function with derivatives of order k, for = 1,2, … . , in some
interval containing a as an interior point. Then for any integer n from 0 through
N, the Taylor polynomial of order n generated by ƒ at = is the polynomial
( )= ( )+
( )
( − )+
1!
Calculus and Analytic Geometry
( )
( − ) + ⋯…+
2!
( )
( − ) +. . … +
!
( )
( − )
!
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Example:
Find the Taylor series and Taylor Polynomial generated by ( ) =
Solution:
,
= 0.
We have ( ) = , ( ) = , ( ) = , … … … . , ( ) = , …., and
therefore, (0) = 1, (0) = 1, (0) = 1, … … … . , (0) = 1, ….
So, the Taylor series generated by ( ) = ,
= 0, is,
(0)
+
1!
(0) +
(0)
2!
2
=1+
+ ⋯……+
(0)
!
+
+ ⋯ … + +. . … … =
!
1! 2!
and the Taylor Polynomial of degree n is,
( )=1+
+
2
= 1 + 0.
(0)
+
1!
(0)
2!
3
2!
!
+ ⋯…+
!
Example:
Find the Taylor series and Taylor Polynomial generated by
( ) = cos ,
=0
Solution:
( ) = cos , ( ) = − sin , ( ) = − cos , f ( ) = sin and so
We have
(0) = 0, ….
on. Therefore, (0) = 1, (0) = 0, (0) = −1,
So, the Taylor series generated by ( ) = cos ,
= 0, is,
(0) +
1!
+ ⋯ ..
+ ⋯……+
4
(0)
!
2
+ ⋯ ..
+ −1
+ 0. + 1. … … + −1
+. . … … =
1!
2!
3!
4!
(2 )!
and the Taylor Polynomial of degree 2n is,
( )= 1−
2!
+
4!
+ ⋯ … + (−1)
Additional Problems
1. Find the Taylor series generated by ( ) = 2 at = 1.
Solution: ( ) = 2 , ′( ) = 2 ln 2, ( ) = 2 ( 2) ,
(−1)
(2 )!
(2 )!
( )(
) = 2 ( 2)
(1) = 2 = 2,
(1) = 2 ln 2 = 2 ln 2 , (1) = 2 ( 2) = 2( 2) ,
( )( )
1 = 2( 2)
Therefore, the Taylor series generated by ( ) = 2 at = 1 is,
(1) +
(1)
( − 1) +
1!
( − 1)
= 2 + 2 ln 2 .
+2
1!
=
2( 2)
2
(1)
( − 1) + ⋯ … … +
2!
2
2
(1)
( − 1) + ⋯ ..
!
( − 1)
( − 1)
+ ⋯ … + 2 ln 2
+. . … …
!
2!
( − 1)
!
2. Find the first three non-zero terms of the Maclaurin series generated by
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( )=
−
and the values of for which the series converges
absolutely?
Solution:
The Maclaurin series generated by cos is,
+
+ ⋯ … + (−1)
+ ⋯ … ..
2! 4!
(2 )!
and the Maclaurin series generated by
is,
1−
1 + + + ⋯….+
So, the Maclaurin series generated by ( ) =
(1 −
!
+
!
+ ⋯ … + (−1)
(
=− −
)!
+ ⋯.
−
is,
+ ⋯ … ..)-(1 + +
−
+ ⋯.)
+ − + ⋯ … ..
!
3
23
=− −
− −
+ ⋯ ..
2
24
The Maclaurin series of
is valid for every ,but the Taylor series of
converges for −1 < < 1. So the resulting series converges absolutely for
−1 < < 1.
3. Find the Maclaurin series of
Solution:
The Maclaurin series of sin is
( ) (0)
(0)
(0)
(0) +
+
+ ⋯……+
+ ⋯ ..
1!
2!
!
(−1)
= − + − ⋯……+
+ ⋯ … ..
(2 + 1)!
3! 5!
So the Maclaurin series of
is,
!
−
+ ⋯….+
− 2 + 2 − ⋯……+
2
3!
5!
(−1)
2
+ ⋯ … ..
(2 + 1)!
(−1)
= −
+
− ⋯……+
+ ⋯…..
(2 + 1)!
2 2 . 3! 2 . 5!
2
∞
(−1)
=
(2 + 1)!
2
=0
Exercise:
1. Find the Taylor Polynomials of order 0,1,2 and 3 for
a) ( ) =
, =0
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b) ( ) = ,
=2
( ) = sin ,
c)
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=
2. Find the Taylor series of
a) ( ) =
,
b) ( ) =
=2
,
=1
3. Find the first three terms of Maclaurin series and the values of x for which
the series converges absolutely if,
( ) = sin ln (1 + )
Answers:
( ) = 1,
1. a)
b)
( )= ,
( ) = 1+2 ,
( ) = − ( − 2),
− ( − 2) + ( − 2) ,
c)
( )=
2. a) ∑
3.
−
√
( ) = 1+2 +2
( )=
,
!
√
( )=
( − 2)
+
+
√2
2
√
+
+ ⋯…
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−
√2
2
b) ∑
,
( )=1+2 +2
( )=
,
( ) = − ( − 2) + ( − 2) −
−
( )=
4
+
√
√2
4
+
√
−
−
4
2
−
+
√2
12
(−1) ( + 1)( − 1)
√
−
−
4
3
+
( − 2)
,
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Chapter 3
Convergence of Taylor Series
Taylor’s Theorem:
If and it’s first derivatives , ,
, … … . . ( ) are continuous on
the closed interval between and , and ( ) is differentiable on the open interval
between and , then there exists a number
and such that,
( )= ( )+
Taylor’s Formula:
( )( − ) +
( )
( − ) + ⋯.+
2!
( )
(
)
( )
( )
( − ) +
( − )
( + 1)!
!
If has derivatives of all orders on an open interval , containing , then
for each positive integer and each in ,
( )= ( )+
( )( − ) +
( )=
(
(
)( )
)!
( )
!
( − )
( − ) +⋯.+
, for
( )( )
!
( − ) +
some c between
( ), where,
and .
This equation is called the Taylor’s Formula, and the function ( ) is called the
reminder of order or the error term for approximation of ( ) by ( )
Remark: We can express ( ) as the sum of Taylor Polynomial of degree ,
( ) and ( ).
: ( )=
( )+
Convergence
If
( ).
( ) → 0 as → ∞, for all ∈ , we say that the Taylor series generated by
= converges to on , and we write
( )=
∞
=0
( )
!
at
( − )
Example: Show that the Taylor series generated by ( ) =
( ), for every real values of .
at
= 0 converges to
Solution:
The function ( ) =
(−∞, ∞).
has derivatives of all orders throuout the interval =
We have ( ) = , ( ) = , ( ) = , … … … . , ( ) ( ) =
therefore, (0) = 1, (0) = 1, (0) = 1, … … … . , ( ) (0) = 1.
and
So, the Taylor formula is,
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( )
( ) = (0 ) +
(
( )=
Where
Where
Since
!
( )=
(
(
( )
+
)!
:
=1+
1!
for some
is finite and lim
converges to
+ ⋯……+
!
)( )
)!
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→
(
)!
for every . Thus,
=
∞
!
=0
+
+
!
+ ⋯…+
!
between 0 and .
1!
+
( )
( ) = 0.nSo the series
→
=1+
( ),
+
= 0 for every , lim
The Reminder Estimation Theorem
between
2!
( )( )
+ ⋯…+
2!
!
+ ⋯.
If there is a positive constant such that ( ) ( ) ≤ for all
and , inclusive, then the reminder term satisfies the inequality
|
( )| ≤
| − |
( + 1 )!
If this inequality holds for every and the other conditions of Taylors’s Theorem are
satisfied by , then the series converges to ( ).
Example:
Show that the Taylor series for
at
Solution:
= 0 converges for all .
The function and its derivatives are
( ) = cos
( ) = sin ,
( ) = −sin ,
( )
So,
(
)
( ) = −cos
( )(
( ) = sin ,
) = cos
………………………………
( ) = (−1) sin ,
(0) = sin 0 = 0,
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( ) = 0,
(
)(
) = (−1) cos
( ) = cos 0 = 1
( ) = −1
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( )
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( )(
( ) = 0,
)=1
…………………………………………………………………………….
( )
(
)( )
(0) = (−1) sin 0 = 0,
0 = (−1) cos 0 = (−1) .
By Taylor formula,
( ) = (0 ) +
Where
: sin
Where
( )=
( )
!
+
(
)( )
(
=0+
( )
+ ⋯……+
!
)!
1
0
+
1!
2!
: sin
( )=
=
−
(
3!
)( )
(
−1
3!
+
+
)!
5!
( )( )
+
!
+ ⋯………+
− ⋯……+
( ),
(−1)
(2 + 1 )!
(−1)
(2 + 1 )!
+
+
( )
To apply reminder estimation theorem we have to find such that, (
. All the derivatives of sin have absolute value less than or equal to 1.
So,
= 1 and
But we have, lim
→
!
|
∞
=
.| |
(
)!
= 0 for every , so that
Hence the Taylor series of
sin
( )| ≤
=0
at
=
−
3!
Example: Show that the Taylor series for cos at
=1+
!
+
ie cos
!
+
=1−
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!
2!
+ ⋯………+
+
4!
(
) ≤
+
( ) → 0, as
→ ∞.
for all . Thus,
5!
−
7!
+ ⋯……
= 0 converges for all .
As in the above example, we have,
cos
)(
.
= 0 converges to
(−1)
(2 + 1 )!
( )
(
)
− ⋯………+
)!
+
(−1)
(2 )!
( )
+
( )
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All the derivatives of cos have absolute value less than or equal to 1.
So,
= 1 and
But we have, lim
→
Using Taylor Series
( )| ≤
at
= 0 converges to
(
= 0 for every , so that
!
Hence the Taylor series of
cos
.| |
|
∞
=
(−1)
(2 )!
=0
=1−
2!
)!
+
.
( ) → 0, as
4!
→ ∞.
for all . Thus,
−
6!
+ ⋯……
Since every Taylor series is a power series, the operations of adding,
subtracting, and multiplying Taylor series are all valid on the intersection of their
interval of convergence.
Example: Find first few terms of the Taylor series of the following functions using
the power series operations.
a) (2 + cos )
cos
b)
Solution:
a) We have the Taylor series generated by cos at = 0 is
∞
(−1)
cos =
= 1 − + − + ⋯……
(2 )!
2! 4! 6!
=0
=
1
2
(2 + cos ) =
+ cos
3
3 3
2
+
1− + − + ⋯……
3 3
2! 4! 6!
=
2
+ − +
− ⋯……
3 3 6 72
b) We have the Taylor series generated by
=
Therefore,
cos
= 1+
1!
+
2!
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∞
=0
+
!
3!
=1+
+
4!
1!
at
+
2!
+ ⋯……
= 0 is
+
3!
+ ⋯……
1−
2!
+
4!
−
6!
+ ⋯……
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=1+
1!
+
−
2!
=1+
2!
−
−
1! 2!
−
3
6
+
3!
+
4!
−
2! 2!
+ ⋯ … … … … … ….
+
4!
…
Remark: If ( ) is a continuous function, we can use the Taylor series of ( ) to find
the Taylor series of ( ( ))
Example: We have the Taylor series generated by cos at
cos
=
∞
=0
(−1)
(2 )!
=1−
2!
+
4!
−
6!
= 0 is
+ ⋯……
So the Taylor series generated by cos 2 at = 0 is
∞
(−1) (2 )
(2 )
(2 )
(2 )
cos 2 =
=1−
+
−
+ ⋯……
(2 )!
2!
4!
6!
=0
=1−
Example: For what values of
2
2!
=
+
∞
=0
2
4!
−
2
6!
(−1) 2
(2 )!
+ ⋯……
can we replace sin by
magnitude no greater than 3 × 10 ?
−
!
with an error of
We have the Taylor series representation,
sin
=
−
3!
This is an alternating series for positive
⋮+
5!
−
values.
7!
+ ⋯……
By Alternating Series Estimation Theorem, the error in truncating after
than,
5!
=
| |
120
Therefore the error will be less than or equal to 3 × 10
or
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!
is no greater
if
| |
< 3 × 10
120
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| |<
Exercises:
360 × 10
≈ 0.514
1. Use the method of substitution to find the Taylor series at
functions.
= 0 of the following
a)
b) sin( )
c) cos(5
)
2. Using power series operations find the Taylor series at
functions.
sin
a)
b)
cos
c) cos
(Hint: use cos
=
= 0 of the following
)
3. Find the first three terms of the Maclaurin series of (tan
4. For approximately what values of
magnitude no greater than 5 × 10 ?
can we replace sin by
)
−
with an error of
Answers:
1. a) ∑
b) ∑
(−1) 5
!
(−1)2 +1
22 +1 2 +1 !
(−1)2
c) ∑
−
c) 1 + ∑
3.
−
5
4
2
2 !
(−1)2 +1 2 +3
2.a) ∑
b)
2 +1 2 +1
!
+
(
+
2 +1 !
!
−
.(
)!
) (
)
−
!
+ ⋯.= ∑
4. | | < (0.06) < 0.56968
(−1)
(
)!
+ ⋯ ….
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MODULE IV
Chapater 1
Conic Sections, Parametrized Curves,
And Polar Coordinates
Conic Sections and Quadratic Equations
Definition:A circle is the set of points in a plane whose distance from a given fixed point in the
plane is constant. The fixed point is the center of the circle; the constant distance is the
radius.


the standard equation of a circle with center (h, k) and radius a is (x - h)2 + (y - k)2 =
a2.
the standard equation of a circle with center at the origin and radius a is x2 + y2 = a2.
Definition:A set that consists of all the points in a plane equidistant from a given fixed point and a
given fixed line in the plane is a parabola. The fixed point is the focus ofthe parabola. The
fixed line is the directrix.
Standard-form equations for parabolas with vertices at the origin
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The parabola x2 = 4py ,p > 0
The parabola y2 = -4px,p > 0
The parabola x2 = -4py, p > 0
The parabola y2 = -4px, p > 0
 The equation ( − ℎ) = 4 ( − ) represent a parabola having vertex at (h,k) and
axis of symmetry is = ℎ
Definition:An ellipse is the set of points in a plane whose distances from two fixed points
in the plane have a constant sum. The two fixed points are the foci of the ellipse.
Definition:-
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The line through the foci of an ellipse is the ellipse’s focal axis. The point on the axis
halfway between the foci is the center. The points where the focal axis and ellipse cross are
the ellipse’s vertices
The ellipse defined by the equation PF1 + PF2 = 2a is the graph of the equation
+
=1
The major axis of the ellipse
+ = 1 is the line segment of length 2a joining the points
(± , 0) . The minor axis is the line segment of length 2b joining the points (0,± ).The
number a itself is the semimajor axis, the number b the semiminor axis. The number
=√ −
is the center-to-focus distance of the ellipse.
Standard-Form Equations for Ellipses Centered at the Origin
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
(
)
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(
)
the equation
+
= 1 represent an ellipse having center at (h,k)
and axes parallel to the coordinate axes. The length of the semi-major axis and
semi-minor axis are a and b respectively.
Definition:A hyperbola is the set of points in a plane whose distances from two fixed points in
the plane have a constant difference. The two fixed points are the foci of the hyperbola.
Definition:The line through the foci of a hyperbola is the focal axis. The point on the axis
halfway between the foci is the hyperbola’s center. The points where the focal axis and
hyperbola cross are the vertices
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 Hyperbolas have two branches. For points on the right-hand branch of the hyperbola
shown here, PF1 – PF2 =2a . For points on the left-hand branch, PF2 – PF1 =2a,
 the line = ±
are the two asymptotes of the hyperbola
Standard-Form Equations for Hyperbolas Centered at the Origin
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Classifying Conic Sections by Eccentricity
Definition:
The eccentricity of the ellipse


The eccentricity of the hyperbola
The eccentricity of the parabola is
+
= 1 ( > ) is
− = 1 is
=1
=
=
=
√
=
√
Remark: In both ellipse and hyperbola, the eccentricity is the ratio of the distance between the
foci to the distance between the vertices (because c>a = 2c>2a).
Eccentricity =( distance between foci) /(distance between vertices)
 For ellipse and hyperbola , the lines = ± act as directrices
 The “focus–directrix” equation PF=e.PD unites the parabola, ellipse, and hyperbola
in the following way. Suppose that the distance PF of a point P from a fixed point F
(the focus) is a constant multiple of its distance from a fixed line (the directrix). That
is, suppose PF=e.PD where e is the constant of proportionality. Then the path traced
by P is
(a) a parabola if e=1
(b) an ellipse of eccentricity e if e<1 and
(c) a hyperbola of eccentricity e if e>1.
EXAMPLE 1
Find the ellipse’s standard form equation if the foci are(3,0) and (-3,0) and the
eccentricity is 0.5
Solution
Here c=3 and e=c/a=0.5
So a = 6, b= √ −
=
√6 − 3
=5
Therefore the ellipse’s standard form equation is
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+
=1
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Quadratic Equations and Rotations
The cross product term
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for
In practice, this means determining a from one of the two equation
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Possible graphs of quadratic equations
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The discriminant test
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Parametrization of plane curves
Definition:-
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Calculus and Analytic Geometry
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EXAMPLE
A Parametrization of the ellipse
+
=1
Solution
The position P(x, y) of a particle moving in the xy-plane is given by the
equations and parameter interval
x = a cos t , y = b sin t ,
0≤ ≤ 2 .
Solution We find a Cartesian equation for the coordinates of P by eliminating t between
the equations
x = a cos t , y = b sin t
We accomplish this with the identity
+
= 1 which yields + = 1 .
Since the particle’s coordinates (x, y) satisfy the equation the motion takes place
somewhere on this ellipse. When t=0 , th particle’s coordinate are x = a cos(0) = a, y = b
sin(0) = 0 so the motion starts at (a,0). As t increases, the particle rises and moves towards
the left , moving counter clockwise. It traverses the ellipse once, returning to its starting
position (a,0) at time t=2 .
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Calculus with parametrized curves
Formula for finding
The parametric formula for
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Length of parametrized curve
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The area of the surface of revolution
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Polar Coordinates
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To define polar coordinates, we first fix an origin O (called the pole) and an
initial ray from O . Then each point P can be located by assigning to it a polar
coordinate pair(r, ) in which r gives the directed distance from O to P and
gives the directed angle from the initial ray to ray OP.
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Polar equations and graphs
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Equations Relating Polar and Cartesian Coordinates
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Calculus and Analytic Geometry
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Graphing in Polar Coordinates
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Calculus and Analytic Geometry
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Calculus and Analytic Geometry
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Calculus and Analytic Geometry
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Calculus and Analytic Geometry
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Conic Sections in Polar Coordinates
The Standard Polar Equation for Lines
EXAMPLE 1
Write the standard polar and cartesian equation for the line in figure
.
Solution
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Circles
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Ellipses, Parabolas, and Hyperbolas
 Polar Equation for a Conic with Eccentricity e is
vertical directrix.
Calculus and Analytic Geometry
=
where x=k > 0 is the
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 Polar Equation for the Ellipse with Eccentricity e and Semimajor Axis a
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Integration in Polar Coordinates
Area in the Plane
Area of the Fan-Shaped Region Between the Origin and the Curve
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= ( ),
≤
≤
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a.
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Area of the Region 0 ≤
( )≤
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≤
( ),
≤
≤
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Length of a Polar Curve
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Area of a Surface of Revolution of a Polar Curve
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Calculus and Analytic Geometry
Page 135