Lec001 math1 Fall 2021 .pdf

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Dr. Khaled El Helow Mathematics 1 (BAS 011) Fall 2021 Lec 1
Basic Science Department
Bas 011 Mathematics 1
Teacher :
Dr. Khaled El Sayed El Helow.
khaled_elhelw@cic-cairo.com
Tutorial
Eng. Mohamed El Sayed Eng. Dina AbdEl Hamid
Moha_elsayed@cic-cairo.com dina_abdelhamied@cic-cairo.com
Eng. Amera Okasha
amira_okasha@cic-cairo.com

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3 Credit Hrs
Lecture 1
Program Intended Learning Outcomes (By Code)
Knowledge &
Understanding
Intellectual
Skills
Professional
Skills
General
Skills
K1, K5 I1, I2, I7 P1,P7 G6

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Contents
Algebra:
Mathematical induction (ILO)[K1, K5, G6]
- Binomial Theorem (ILO) [K1, K5, I1, I2, I7, P1, P7, G6]
- Partial fractions- Matrices (ILO) [K1, K5, I1, I2, I7, P1, P7, G6]
- Solving System of linear equations (ILO)[K1, K5, I1, I2, I7, P1, P7, G6]
- Solving algebraic equations numerically (ILO) [K1, K5, I1, I2, I7,
P1, P7, G6]

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- Eigen values and Eigen vectors. (ILO) [K1, K5, I1, I2, I7, P1, P7, G6]
Calculus:
Functions (ILO)[K1, K5, G6]
-Limits (ILO)[K1, K5, G6]
-Continuity (ILO)[K1, K5, I1, I2, I7, P1, P7, G6]
-Derivatives (ILO)[K1, K5, I1, I2, I7, P1, P7, G6]
-Indefinite forms (ILO)[K1, K5, I1, I2, I7, P1, P7, G6]
-Taylor and Maclaurine Theorems (ILO)[K1, K5, I1, I2, I7, P1, P7, G6]-
Some mathematical applications on Derivatives (ILO)[K1, K5, I1,

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I2, I7, P1, P7, G6]- Introduction to partial differentiation. (ILO)[K1, K5,
I1, I2, I7, P1, P7, G6]
Cohort 2017
100 Marks
50 final
exam
50 Class
work
20
midterm
30 marks
7
Attendanc
11
Assignments
12
Quizzes
Best two quizzes

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Induction has always existed in mathematics, but the
formal concept of mathematical induction did not appear
until it was developed by Maurolycus in 1575 to prove
that the sum of the first n odd numbers is n2
.
Mathematical Induction
Mathematical induction is a mathematical proof technique,
most commonly used to establish a given statement for

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all natural numbers. The simplest and most common form of
mathematical induction infers that a statement involving a
natural number n holds for all values of n. The proof consists
of two steps:
1. The basis:
Prove that the statement holds for the first natural number n.
2. The inductive step

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Prove that, if the statement holds for some natural number k,
then the statement holds for k + 1.
Example:
Prove that 1 + 2 + 3 + ⋯ ⋯ + 𝑛𝑛 =
𝑛𝑛(𝑛𝑛+1)
2
by using
mathematical induction.
Solution:

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The basis step (we need to show that the statement is hold for
𝑛𝑛 = 1.)
L.H.S = 1 (first term in the left hand side)
R.H.S =
1(1 + 1)
2
= 1
The both sides are equal so the statement is hold for 𝑛𝑛 = 1.
The inductive step:
Suppose the statement is hold at 𝑛𝑛 = 𝑘𝑘. i.e.:

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1 + 2 + 3 + ⋯ ⋯ + 𝑘𝑘 =
𝑘𝑘(𝑘𝑘+1)
2
(1)
So we need to prove the statement at the next step
(𝑛𝑛 = 𝑘𝑘 + 1). i.e. we need to prove:
1 + 2 + 3 + ⋯ ⋯ + (𝑘𝑘 + 1) =
(𝑘𝑘+1)(𝑘𝑘+2)
2
( )
( )
1
3
2
1
1
3
2
1
L.H.S
+
+
+
+
+
=
+
+
+
+
+
=
k
k
k





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The assumption from eq(1)
( )
( )
( )
( )( )
R.H.S
2
2
1
1
2
1
1
2
1
L.H.S
=
+
+
=






+
+
=
+
+
+
=
k
k
k
k
k
k
k

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Example:
Prove that 1 + 3 + 5 ⋯ ⋯ + (2𝑛𝑛 − 1) = 𝑛𝑛2
by using
Solution:
The basis step (we need to show that the statement is hold for
𝑛𝑛 = 1.)
L.H.S = 1 (first term in the left hand side)
R.H.S = (1)2
= 1

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The both sides are equal so the statement is hold for 𝑛𝑛 = 1.
The inductive step:
1 + 3 + 5 ⋯ ⋯ + (2𝑘𝑘 − 1) = 𝑘𝑘2
(1)
So we need to prove the statement at the next step
(𝑛𝑛 = 𝑘𝑘 + 1). i.e. we need to prove:
1 + 3 + 5 ⋯ ⋯ + (2𝑘𝑘 + 1) = (𝑘𝑘 + 1)2

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( )
( ) ( )
1
2
1
2
5
3
1
1
2
5
3
1
L.H.S
+
+
−
+
+
+
=
+
+
+
+
+
=
k
k
k




The assumption from eq(1)
L.H.S = 𝑘𝑘2
+ (2𝑘𝑘 + 1) = (𝑘𝑘 + 1)2
= 𝑅𝑅. 𝐻𝐻. 𝑆𝑆

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Example:
Prove that 8|32𝑛𝑛
− 1 ( 32𝑛𝑛
− 1 is divided by 8) by using
Solution:
The base step (we need to show that the statement is hold for
𝑛𝑛 = 1.)
32(1)
− 1 = 9 − 1 = 8. Therefore the statement is divisible
by 8.

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The inductive step:
8|32𝑘𝑘
− 1 (1)
So we need to prove the statement at the next step (𝑛𝑛 = 𝑘𝑘 +
1). i.e. we need to prove: 8|32(𝑘𝑘+1)
− 1

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( )
( )
[ ]
( )
( ) 8
1
3
9
1
9
1
3
9
1
1
1
3
9
1
3
9
1
3
3
1
3
1
3
2
2
2
2
2
2
2
2
1
2
+
−
=
−
+
−
=
−
+
−
=
−
=
−
=
−
=
− +
+
k
k
k
k
k
k
k

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Clear that 8 is divisible by 8 and 32𝑘𝑘
− 1 is divisible by 8
form the assumption (eq(1))
So, 32(𝑘𝑘+1)
− 1 = 9(32𝑘𝑘
− 1) + 8 is divisible by 8.
Therefore the statement is true at 𝑛𝑛 = 𝑘𝑘 + 1. Thus it is true
for all n.
Example:
Prove that 5|7𝑛𝑛
− 2𝑛𝑛
by using mathematical induction.
Solution:

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The base step (we need to show that the statement is hold for
𝑛𝑛 = 1.)
71
− 21
= 5. Therefore the statement is divisible by 5.
The inductive step:
5|7𝑘𝑘
− 2𝑘𝑘
(1)

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So we need to prove the statement at the next step (𝑛𝑛 = 𝑘𝑘 +
1). i.e. we need to prove: 5|7𝑘𝑘+1
− 2𝑘𝑘+1
( )
[ ]
( )
( ) ( ) k
k
k
k
k
k
k
k
k
k
k
k
k
k
k
2
2
7
2
7
7
2
2
2
7
2
7
7
2
2
2
2
7
7
2
2
7
7
2
7 1
1
−
+
−
=
−
+
−
=
−
+
−
=
−
=
− +
+

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7𝑘𝑘+1
− 2𝑘𝑘+1
= 7(7𝑘𝑘
− 2𝑘𝑘) + 5 .2𝑘𝑘
Clear that 5. 2𝑘𝑘
is divisible by 5 and 7𝑘𝑘
− 2𝑘𝑘
is divisible by 5
form the assumption (eq(1))
So, 7𝑘𝑘+1
− 2𝑘𝑘+1
= 7(7𝑘𝑘
− 2𝑘𝑘) + 5. 2𝑘𝑘
is divisible by 5.
Therefore the statement is true at 𝑛𝑛 = 𝑘𝑘 + 1. Thus it is true
for all n.

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Example:
Prove that
1 2
+ 2 2
+ 3 2
+ ... + n 2
= n (n + 1) (2n + 1)/ 6
For all positive integers n.
Solution
Statement P (n) is defined by
1 2
+ 2 2
+ 3 2
+ ... + n 2
= n (n + 1) (2n + 1)/ 2
STEP 1: We first show that p (1) is true.

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Left Side = 1 2
= 1
Right Side = 1 (1 + 1) (2*1 + 1)/ 6 = 1
Both sides of the statement are equal hence p (1) is true.
STEP 2: We now assume that p (k) is true
1 2
+ 2 2
+ 3 2
+ ... + k 2
= k (k + 1) (2k + 1)/ 6
and show that p (k + 1) is true by adding (k + 1) 2
to both
sides of the above statement
1 2
+ 2 2
+ 3 2
+ ... + k 2
+ (k + 1) 2
=[ k (k + 1) (2k + 1)/ 6] + (k + 1) 2

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Set common denominator and factor k + 1 on the right side
= (k + 1) [ k (2k + 1)+ 6 (k + 1) ] /6
Expand k (2k + 1)+ 6 (k + 1)
= (k + 1) [ 2k 2
+ 7k + 6 ] /6
Now factor 2k 2
+ 7k + 6.
= (k + 1) [ (k + 2) (2k + 3) ] /6
We have started from the statement P(k) and have shown that

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1 2
+ 2 2
+ 3 2
+ ... + k 2
+ (k + 1) 2
= (k + 1) [ (k + 2) (2k + 3) ] /6
Which is the statement P(k + 1).
Example:
Use mathematical induction to prove that
1 3
+ 2 3
+ 3 3
+ ... + n 3
= n 2
(n + 1) 2
/ 4
for all positive integers n.
Solution :

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1 3
+ 2 3
+ 3 3
+ ... + n 3
= n 2
(n + 1) 2
/ 4
STEP 1: We first show that p (1) is true.
Left Side = 1 3
= 1
Right Side = 1 2
(1 + 1) 2
/ 4 = 1
hence p (1) is true.
1 3
+ 2 3
+ 3 3
+ ... + k 3
= k 2
(k + 1) 2
/ 4

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add (k + 1) 3
to both sides
1 3
+ 2 3
+ 3 3
+ ... + k 3
+ (k + 1) 3
= [k 2
(k + 1) 2
/ 4 ]+ (k + 1)
3
factor (k + 1) 2
on the right side
= (k + 1) 2
[ k 2
/ 4 + (k + 1) ]
set to common denominator and group
= (k + 1) 2
[ k 2
+ 4 k + 4 ] / 4
= (k + 1) 2
[ (k + 2) 2
] / 4

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We have started from the statement P(k) and have shown that
1 3
+ 2 3
+ 3 3
+ ... + k 3
+ (k + 1) 3
= (k + 1) 2
[ (k + 2) 2
] / 4
Which is the statement P(k + 1).
Example:
Prove that for any positive integer number n , n 3
+ 2 n is
divisible by 3
Solution:

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n 3
+ 2 n is divisible by 3
STEP 1: We first show that p (1) is true. Let n = 1 and
calculate n 3
+ 2n
1 3
+ 2(1) = 3
3 is divisible by 3
hence p (1) is true.

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k 3
+ 2 k is divisible by 3
is equivalent to
k 3
+ 2 k = 3 M , where M is a positive integer.
We now consider the algebraic expression (k + 1) 3
+ 2 (k +
1); expand it and group like terms
(k + 1) 3
+ 2 (k + 1) = k 3
+ 3 k 2
+ 5 k + 3
= [ k 3
+ 2 k] + [3 k 2
+ 3 k + 3]
= 3 M + 3 [ k 2
+ k + 1 ] = 3 [ M + k 2
+ k + 1 ]

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Hence (k + 1) 3
+ 2 (k + 1) is also divisible by 3 and therefore
statement P(k + 1) is true.

Lec001 math1 Fall 2021 .pdf

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