CSE 20Ñ Discrete Math - University of California, San … · Winter , 200 6 February 21, Day 13...

CSE 20—Discrete Math

Winter, 2006

February 21, Day 13Functions

Summations (portions from Concrete Mathematics by Graham, Knuth,

Patashnik)

Instructor: Neil Rhodes

Powers of Permutations

To calculate the power of a function fi, calculate the power of each cycle! Calculate final result for each element (using i mod cycle-length)! 1-cycles stay 1-cycles! For example, if f = (1, 2, 4), (3, 8, 7, 10), (5), (6, 9)

– f10 = (1, 2, 4), (3, 7), (8, 10), (5), (6), (9)

! If all cycle lengths divide i evenly, fi is the identity permutation! If f is a permutation of A, then f|A|! is the identity permutation

2

Functions

Given f:A->B! Inverse Image of f:

! Coimage of f: partition of A where all elements in a block map to the same element of B

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Functions

Given f : A! B:

Inverse Image of f

f"1(b) = {a : a # A and f (a) = b}

Coimage is the partition of A where all elements in ablock map to the same element of B.

Coimage( f )= { f"1(b) : b # Image( f )}

Summations (portions from Concrete Mathematics by Graham, Knuth, Patashnik) – p.2

Functions

Given f : A! B:

Inverse Image of f

f"1(b) = {a : a # A and f (a) = b}

Coimage is the partition of A where all elements in ablock map to the same element of B.

Coimage( f )= { f"1(b) : b # Image( f )}


Partition

Given a set S = {a1, a2, … an}:! A partition P is a set of subsets of S such that:

– the elements of P cover S

– The elements of P are pairwise disjoint

– No element of P is empty

– The elements of P are called the blocks of the partition

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Number of Partitions

Given S = {a1, a2, …, an}:! How many partitions of S are there?! How many partitions of size k are there?

– If an is in a new block:

– If an is in an existing block

! S(n, k) is the Stirling number of the second kind

– S(n, k) =

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Number of Partitions

Given S= {a1,a2, . . . ,an}:• How many partitions of S are there?

• How many partitions of size k are there?• If an is in a new block:• If an is in an existing block:

• S(n,k) or {nk} is the Stirling number of the secondkind

S(n,k) =


Calculating S(n, k)

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1 2 3 4 5

1

2

3

4

5

Number of functions

Given sets A and B, how many functions are there of the form f:A->B where |Image(f)| = k! How many partitions of A are there of size k?

! How many images are there of size k?

! How many ways to map the blocks of Coimage(f) to Image(f)?

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Summing a sequence

Specific sum! 1 + 2 + 3 + ... + n-1 + n

General summation of a sequence of terms:! a1 + a2 + … + an

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Sigma notation

Ellipsis (…) notation is vague and wordy

Sigma notation is more compact:

Parts of the notation! Summand! Index variable! Lower limit! Upper limit

Sigma notation inline:

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Sigma Notation• Three-dots notation is vague and wordy.

• Sigma notation is more compact:

n

!k=1

k

• Parts of the notation• Summand• Index variable• Lower limit• Upper limit

• Sigma notation inline: !nk=1 k


Sigma Notation• Three-dots notation is vague and wordy.

• Sigma notation is more compact:

n

!k=1

k

• Parts of the notation• Summand• Index variable• Lower limit• Upper limit

• Sigma notation inline: !nk=1 k


Generalized Sigma notation

Specify a condition that the index variable must satisfy:

! Compare:

! To:

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Generalized Sigma Notation• Specify a condition that the index variable mustsatisfy

!1!k!n

k

• Compare:

!1!k!100k odd

k2

to:

49

!k=0

(2k+1)2


Generalized Sigma Notation• Specify a condition that the index variable mustsatisfy

!1!k!n

k

• Compare:

!1!k!100k odd

k2

to:

49

!k=0

(2k+1)2


!k=0

(2k +1)2

Changing the index variable

Can always change from one index variable to another:

What if we want to switch from k to k+1?

! Compare:

! To:

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Changing the Index Variable

• Can always change from one variable to another

!1!k!n

ak =

• What if we want to switch from k to k+1?Compare:

!1!k!n

ak =

to:

n

!k=1

ak =




!1!k!n

ak =


!1!k!n

ak =

to:

n

!k=1

ak =




!1!k!n

ak =


!1!k!n

ak =

to:

n

!k=1

ak =


Zero terms are OK

Which is better?

or:

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n!1

!k=2

k(k!1)(n! k)

n

!k=0

k(k!1)(n! k)

Using Iverson Notation

Iverson notation. True-or-false statement enclosed in brackets. Value is 0 or 1

! [p odd] =

Example:

0 in Iverson notation is very strongly zero

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• Iverson notation. True-or-false statementenclosed in brackets. Value is 0 or 1

[p odd] =

• Example

!1!k!100k odd

k2 =!k

k2[1! k ! 100][k odd]

• 0 in Iverson notation is very strongly zero

!k

1

k[k > 0]




[p odd] =

• Example

!1!k!100k odd

k2 =!k

k2[1! k ! 100][k odd]


!k

1

k[k > 0]




[p odd] =

• Example

!1!k!100k odd

k2 =!k

k2[1! k ! 100][k odd]


!k

1

k[k > 0]


Products

Pi notation is similar to Sigma notation

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Products

• Pi notation is similar to Sigma notation

n

!k=1

k = 1!2!3! · · ·! (n"1)!n


Working with summations

Distributive law

Associative law

Commutative law

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Working with Summations

Distributive law

!k!K

cak = c!k!K

ak

Associative law

!k!K

ak+bk = !k!K

ak+ !k!K

bk

Commutative law

!k!K

ak = !p(k)!K

ak



Distributive law

!k!K

cak = c!k!K

ak

Associative law

!k!K

ak+bk = !k!K

ak+ !k!K

bk

Commutative law

!k!K

ak = !p(k)!K

ak



Distributive law

!k!K

cak = c!k!K

ak

Associative law

!k!K

ak+bk = !k!K

ak+ !k!K

bk

Commutative law

!k!K

ak = !p(k)!K

ak


Closed forms for summations

Closed form is a formula for a summation with the summation removed.

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Closed Forms for Summations

Closed form is a formula for a summation, with the

summation removed.

!0!k!n

k =n(n+1)

2


Example

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Example

!0!k!n

(a+bk) =


Perturbation method

Given:

Rewrite Sn+1 by splitting off first and last term:

Then, work on last sum and express in terms of Sn. Finally, solve for Sn.

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Perturbation Method

Given Sn = !0!k!n ak, rewrite Sn+1 by splitting offfirst and last term:

Sn+an+1 = a0+ !1!k!n+1

ak

= a0+ !1!k+1!n+1

ak+1

= a0+ !0!k!n

ak+1

Then, work on last sum and express in terms of Sn.

Finally, solve for Sn.


Perturbation Method

Given Sn = !0!k!n ak, rewrite Sn+1 by splitting offfirst and last term:

Sn+an+1 = a0+ !1!k!n+1

ak

= a0+ !1!k+1!n+1

ak+1

= a0+ !0!k!n

ak+1

Then, work on last sum and express in terms of Sn.

Finally, solve for Sn.


Perturbation method example

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Perturbation Method Example

Sn = !0!k!n

xk


Standard closed forms

Arithmetic series

Sums of squares and cubes

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Standard Closed Forms

Arithmetic series

n

!k=0

k =1

2n(n+1)

Sum of squares and cubes

n

!k=0

k2 =

n(n+1)(2n+1)

6

n

!k=0

k3 =

n2(n+1)2

4



Arithmetic series

n

!k=0

k =1

2n(n+1)

Sum of squares and cubes

n

!k=0

k2 =

n(n+1)(2n+1)

6

n

!k=0

k3 =

n2(n+1)2

4


Standard closed forms

Geometric series (x ! 1)

Infinite Geometric series (|x| < 1)

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Geometric series (x != 1)

n

!k=0

xk =

xn+1"1

x"1


"

!k=0

xk =

1

1" x



Geometric series (x != 1)

n

!k=0

xk =

xn+1"1

x"1


"

!k=0

xk =

1

1" x


Standard closed form

Harmonic series

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Harmonic series

Hn =n

!k=0

1

k! lnn

n 0 1 2 3 4 5 6

Hn 0 13

2

11

6

25

12

137

60

49

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Application of harmonic series

Given a stack of 52 playing cards (of length 2 units). Can you stack them overlapping so the top card completely overhangs the table?

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CSE 20Ñ Discrete Math - University of California, San … · Winter , 200 6 February 21, Day 13...

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