Sequences and Summations - Discrete Mathematics and its Applications - Lecture Slides, Slides of Discrete Mathematics

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Lecture 13

Sequences & Summations

Proof by Induction

Sums

geometric progression:

arithmetic progression:

some other useful sums:

0

( 1) (^ 1)

n j

a jd n a d n n

=

1

0

n n j j

ar a if r

ar r

n a if r

=

2 1 3 2 2 1

0 (^12) 1

( 1)(2 1) 6 ( 1) 4 (^1) | | 1 1 (^1) | | 1 (1 )

n k n k k k k k k n n^ n

k n^ n

x (^) x if x

kx (^) x if x

=

= ∞ = ∞ (^) − =

= +^ +

= +

= (^) − <

= (^) − <

derivative

a=1, n  infinity

Cardinality

definition: Two sets have the same cardinality if and only if there is a one-to-one correspondence between them.

This is simple for finite sets, but what if a set has infinite elements?

definition: A set that is finite or has the same cardinality as the set of positive integers (Z+) is called countable.

Example: Consider the sequence {an}, an = n^2, n={1,2,3,4...} Naively speaking, there seem to be much less elements in {an} than in Z+ (since we skip a lot). Infinity is weird! Here is the one-to-one mapping: 1 2 3 4 5 6 7 ... 1 4 9 16 25 36 49 ... infinity^ (intuitively: you can enumerate them)

cardinality

Now what about the positive rational numbers: p/q with p,q integer, q not 0?

1/1 2 /1 3 /1 4 / 1/ 2 2 / 2 3 / 2 4 / 2 1/ 3 2 / 3 3 / 3 4 / 3 1/ 4 2 / 4 3 / 4 4 / 4

→ → → → ↓ ↓ ↓ ↓ 1 2 3 4 5 6 7 8 9 10 1 ½ 2 3 1/3 ¼ 2/3 3/2 4 5

skip

set of positive rational numbers are countable, in fact the set of all rational numbers is countable : 4 countable quadrants.

3.3 Mathematical Induction

If we want to prove propositions P(k) for all positive integers, we may use inductions.

First we prove: P(1) is true. Then we prove P(k)  P(k+1).

So, is P(100) true? yes, use “modus ponens” 99 times.

P(1) P(1)P(2)

---

P(2) P(2)P(3)

---

P(3)

until P(100).

[ P (1) ∧ ∀ k P k ( ( ) → P k ( + 1))] → ∀ nP n ( )

formally:

Examples

prove that the sum of the first n odd positive integers is n^2.

n=1: 1 = 1.

assume it’s true for some k. 1+3+5+...+2k-1 = k^2 is true.

add 2k+1 on each side: 1+3+5+...+2k-1 + 2k+1= k^2 + 2k + 1 1+3+5+...+2(k+1)-1 = (k+1)^



Fun Example

Show that a chessboard with 2^n x 2^n squares where one arbitrary square has been removed can be tiled with L-shapes. (needs drawing)

P(1): All 4 possibilities of removing the square for a 2x2 example are precisely covered with 1 L-shape.

Assume P(k) is true. Now construct a chessboard with that is twice as large in both directions. Equally divide it into 4 pieces. Remove one piece arbitrarily from one of the 4 pieces. Since P(k) is true that piece can be covered with L-shapes. Next place one L-shape in the middle to remove one square from the remaining 3 pieces. Again due to P(k) these can now be covered as well.

Strong Induction

(1) ( ( ) ( 1)) (1) [( (1) (2)) ( (2) (3)) ...( ( ) ( 1))] (1) ( (1) (2)) ( (2) (3)) ...( ( ) ( 1)) (1) (2) ... ( ) ( ( ) ( 1)) (1) [ (1) (2) ... ( )] ( 1)

P k P k P k P k P P P P P k P k P P P P P P k P k arbitrary k P P P k P k P k arbitrary k P k P P P k P k

∧ ∀ → + → ∧ ∀ → ∧ → ∧ → + ∧ → ∧ → ∧ → + ∧ ∧ ∧ → + ∧ ∀ ∧ ∧ ∧ → +

[ P (1) ∧ ∀ k P ( (1) ∧ P (2) ∧ ... ∧ P k ( ) → P k ( + 1))] → ∀ nP n ( )

induction (I):

strong induction (SI):

SII:

P P P k P k

ISI:

they are equivalent

[ P (1) ∧ ∀ k P k ( ( ) → P k ( + 1))] → ∀ nP n ( )

Examples

Proposition: every positive integer n>2 can be written as the product of primes.

P(2): product of itself (it’s prime). Assume k can be written as a product of primes. Can we prove it for k+1? two cases: k+1 = prime (thus it is a product of one number – itself). k+1 = a x b However since both a and b < k+1 and >= 2, we know that a and b can be written as the product of primes  a x b is a product of primes. 

The well ordered property

Every non-empty set of nonnegative integers has a least element.

This is a trivial statement made explicit so we can give it a name in a proof...

Example: Round-Robin tournament: n players play against each other. A cycle is a situation where p1 beats p1, p2 beats p3, pn beats p1. Proposition: If there is a cycle of length greater than 3, then there is also a cycle of 3 among the people in the larger cycle.

Prove: (by contradiction) Assume that there is a cycle of length k, where k is the smallest integer > 3 for which a cycle exist and no cycle of length 3 exists.

Cycle: p1 p2 p3 ... pk k> 3 Look at p1 p2 p3  if p3 beats p1 we have a cycle of length 3 (contradiction)  if p1 beat p3 we can construct the cycle p1 p3 p4 ... pk which also leads to a contradiction. Docsity.com