Section 4'1 Sequences

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Section 4.1 Sequences
A sequence of numbers is an ordered list of
numbers. In this section we mostly consider
infinite sequences of the form
a1, a2, a3, , an,
An explicit formula or general formula for a
sequence is a rule that shows how the value of
ak depend on k.

• Example
• if ak k/(k 1), then the sequence is 1?2 ,
  2?3 , 3?4 , 4?5 , 5?6 ,
• if bk (- ½)k, then the sequence is -1?2 , 1?4
  , -1?8 , 1?16 ,

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Finding an Explicit formula to Fit Given Initial
Terms
Examples 1. 4, 7, 10, 13, 16, 2. 3, 4, 6, 9,
13, 18, 3. 1, 0, -1, 0, 1, 0, -1, 0, 1, 4.
1, 1, 2, 3, 5, 8, 13, 21,
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The Summation Notation
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The Product Notation
Factorial notation n!
n(n 1)(n 2)21 for n gt 0
0! is defined to be 1.
F.Y.I A generalization of the factorial function
is the Gamma function
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The Product Notation
Factorial notation n!
n(n 1)(n 2)21 for n gt 0
0! is defined to be 1.
F.Y.I A generalization of the factorial function
is the Gamma function
and it is not hard to see that G(n) (n 1)!
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Properties of Summation and Products
Theorem If am, am1, am2, and bm, bm1,
bm2, are sequences of real numbers and c is a
constant, then the following equations are true
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Well-ordering principle of whole numbers
For any non-empty (finite or infinite) set of
whole numbers, there must be a smallest element
in that set.
Note this principle is not true for real numbers
or even fractions. Example Consider the set of
fractions 1/q q is a positive
integer
This set does not have a smallest element.
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Well-ordering principle of whole numbers
An application If a and b are positive integers,
then there are integers s and t such that
as bt GCD(a,b)
To prove this result, we have to construct a
non-empty subset S of the positive integers.
Then according to the well-ordering principle, S
must have a smallest element k. We finally
prove that this smallest non-zero k is the GCD of
a and b.
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Two major types of arguments
Deduction
Induction

• from a systematic list
• of examples, extract a
• pattern and then
• distill it into a theorem.
• example on next page

• apply a general theorem
• to a particular case.
• for example, we know
• that is either a
• whole number or an
• irrational. Since
• is not a whole number, it
• must be an irrational.

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Look for a pattern 1 comic 1.00 2
comics 1.80 3 comics 2.60 4 comics 3.40
. What is the cost for 10 comics? If you guess
8.20, then the argument you used is called
induction.
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Observe another pattern
1 1 1 3 4 1 3 5 9 1 3 5 7
16 1 3 5 7 9 25
A pattern The sum of the first n positive odd
numbers is n2
Mathematical formulation P(n) 1 3 5
(2n-1) n2 We would like to show that P(n) is
true for all positive whole number n.
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Observe another pattern
13 1 1
1 13 23 9
1 2 3 13 23 33 36
1 2 3 6 13 23 33 43 100
1 2 3 4 10 13 23 33 43 53
225 1 2 3 4 5 15
A pattern the sum of the first n cubes is equal
to the square of the sum of
the first n natural numbers.
Mathematical formulation P(n) 13 23 n3
(1 n)2 We would like to show that P(n)
is true for all positive whole number n.
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We would like to prove that P(n) is true for all
whole number n.
By contradiction Suppose to the contrary that
P(n) is not true for all n, then we have a
non-empty subset A of whole numbers,
A n P(n) is false
By the well-ordering principle, A must have a
smallest element. Let this smallest element be
k. Since we know that P(1) is true, k must be gt1.
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The smallest bad number.
The red dots are bad numbers
These are all good numbers
If we can prove that P(k-1) ? P(k), then we have
a contradiction and the proof is complete. In
other words, our assumption that P(n) is not
true for all n is false.
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Mathematical Induction

• Let P(n) be a predicate in terms of the variable
  n.
• To show that P(n) is true for all (positive
  integer) n,
• check that P(1) is true.
• prove that P(k-1) ? P(k)
• where k is treated as a constant gt 1.

• Remarks
• some books use P(k) ? P(k1), which is
  equivalent, and it is a matter of taste on which
  one to use.
• the step assume P(k-1) is true is just for
  your own reference.
• sometimes we have to start from an integer
  greater than 1.

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Examples
1. Show that 12 22 32 n2 n(n
1)(2n 1)/6
2. Show that 13 23 33 n3 n(n
1)/22
3. Show that 14 24 34 n4 n(n
1)(2n 1)(3n2 3n 1)/30
4. Show that 15 25 35 n5 n2(n
1)2(2n2 2n 1)/12
5. Show that
6. Show that every number n gt 11 can be written
as the from 4a 5b where a, b are
non-negative integers.
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7. Show that n3 2n is divisible by 3 for any
whole number n.
8. Show that 7n 2n is divisible by 5 for any
whole number n.
9. Show that (n 1)! gt 2n for all n gt 1.
10. Show that
whenever x is not divisible by p.
11. Show that
Hint
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An induction paradox
A teacher told her students that there would be a
pop quiz next week (Mon Fri), and nobody could
predict in advance which day it would be on.
A bright student started to analyze this way.
If no one can predict in advance, then the quiz
cannot be on Friday. And if we know right now
that the quiz cannot be on Friday, then we also
know right now that it cannot be on Thursday
either. By induction, the quiz cannot be on any
day of the week at all. He then
discussed his logic with everyone in class, and
they all believed that he was right, hence no one
prepared for the quiz. Nevertheless, the
pop quiz came on Tuesday unexpected to
everybody. Therefore the teacher was also right.
How could this happen?
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A classical mistake in Mathematical Induction
Theorem For any positive integer n, any group
of n people must be of the same gender.
Incorrect proof (by induction on n) When n 1
this statement is clearly true. Assume that this
statement is true for n k ? 1. Suppose that we
are given k1 people s1, s2, , sk, sk1, then
by induction assumption
s1, s2, , sk must be of the same
gender, say female. In particular, s2 is a
female. Also by induction assumption
s2, , sk, sk1 must
also be of the same gender. Since s2 is a female,
s2, , sk, sk1 must all be female. Hence s1,
s2, , sk, sk1 are of the same gender.

Q.E.D.
Where is the mistake?
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Strong induction.

• These inductions are called strong because
• we need to check a few more initial cases
• we need to make a stronger assumption
• we will obtain a stronger result.

• Example
• We construct a sequence an by the following
  rule
• a1 5,
• a2 10,
• an a(n-1) a(n-2) when n ? 3

Our predicate P(n) is an is divisible by
5. Our prediction is P(n) is true for all n.
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Suppose to the contrary that P(n) is not true for
all n, then the set A n
P(n) is false is not empty. By the
well-ordering principle, A must have a smallest
element, say k . And k must be ? 3.
The smallest bad number.
These are all good numbers
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We may try to prove P(k-1) ? P(k) But this
wont work because ak depends
on a(k-1) and a(k-2) Hence we must use a
stronger condition P(k-1) ?
P(k-2) ? P(k) where k is a constant ? 3.

• Another example
• If an a(n-1) a(n-2) a(n-3) , then we need
  to
• check that P(1), P(2), P(3) are all true,
• prove that P(k-1) ? P(k-2) ? P(k-3) ? P(k)
• whenever k is a constant ? 4

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Example Every whole number ? 2 is either a
prime or a product of primes.

• P(n) is the predicate n is a prime or a product
  of primes.
• check that P(2) is true,
• prove that P(2) ? P(3) ? ? P(k-1) ? P(k)

• The strongest type of induction
• check that P(a) is true
• prove that P(a) ? P(a1) ? P(a2) ? ?
  P(k-1) ? P(k)
• whenever k is a constant gt a.
• conclude that P(n) is true for all n ? a

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This type of induction applies when P(n) depends
on some previous terms P(k) with k lt n but not a
fixed number of previous terms.
Example Prove that every whole number greater
than 1 is either prime or is a product of
primes. Proof Since 2 is clearly a prime
number, the initial case is true. Assume
that this statement is true for all positive
integers greater than 1 and less than some
n (n gt2). To show that this is also true for n,
we consider two cases (1) n is prime there
is nothing to prove in this case.
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Example Prove that every whole number greater
than 1 is either prime or is a product of
primes. Proof Since 2 is clearly a prime
number, the initial case is true. Assume
that this statement is true for all positive
integers greater than 1 and less than some n
(ngt2). To show that this is also true for n, we
consider two cases (1) n is prime there is
nothing to prove. (2) n is not prime then n
is composite by definition and there must be
whole numbers a and b such that
n ab and n gt a, b gt 1 By
our strong induction assumption, a and b are both
products of prime numbers, so n is also a product
of prime numbers.
Q.E.D.
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