Number Sequences

Lecture 7: Sep 28

(chapter 4.1 of the book and chapter 9 of the notes)

?overhang

Interesting Sequences

We have seen how to prove these equalities by induction,

but how do we come up with the right hand side?

Finding General Pattern

a1, a2, a3, …, an, …

1,2,3,4,5,6,7,…

1/2, 2/3, 3/4, 4/5,…

1,-1,1,-1,1,-1,…

1,-1/4,1/9,-1/16,1/25,…

General formula

The first step is to find the pattern in the sequence.

ai = i

ai = i/(i+1)

ai = (-1)i+1

ai = (-1)i+1 / i2

Summation

A Telescoping Sum

When do we have such closed form formulas?

Sum for Children

89 + 102 + 115 + 128 + 141 + 154 + ··· + 193 + ··· + 232 + ··· + 323 + ··· + 414 + ··· + 453 + 466

Nine-year old Gauss saw

30 numbers, each 13 greater than the previous one.

1st + 30th = 89 + 466 = 5552nd + 29th = (1st+13) + (30th13) = 5553rd + 28th = (2nd+13) + (29th13) = 555

So the sum is equal to 15x555 = 8325.

Arithmetic Series

Given n numbers, a1, a2, …, an with common difference d, i.e. ai+1 - ai =d.

What is a simple closed form expression of the sum?

Adding the equations together gives:

Rearranging and remembering that an = a1 + (n − 1)d, we get:

Geometric Series

2 n-1 nnG 1+x +x + +x::= +x

What is the closed form expression of Gn?

2 n-1 nnG 1+x +x + +x::= +x

2 3 n n+1nxG x +x +x + +x +x=

GnxGn= 1 xn+1

n+1

n

1- xG =

1- x

Infinite Geometric Series

n+1

n

1- xG =

1- x

Consider infinite sum (series)

2 n-1 n i

i=0

1+x+x + +x + =x + x

n+1n

nn

1-lim x 1limG

1- x 1-=

x=

for |x| < 1 i

i=0

1x =

1- x

Some Examples

The Value of an Annuity

Would you prefer a million dollars today

or $50,000 a year for the rest of your life?

An annuity is a financial instrument that pays out

a fixed amount of money at the beginning of

every year for some specified number of

years.Examples: lottery payouts, student loans, home mortgages.

A key question is what an annuity is worth.

In order to answer such questions, we need to know

what a dollar paid out in the future is worth

today.

My bank will pay me 3% interest. define bankrate

b ::= 1.03

-- bank increases my $ by this factor in 1 year.

The Future Value of Money

So if I have $X today,

One year later I will have $bX

Therefore, to have $1 after one year,

It is enough to have

bX 1.

X $1/1.03 ≈ $0.9709

• $1 in 1 year is worth $0.9709 now.

• $1/b last year is worth $1 today,

• So $n paid in 2 years is worth

$n/b paid in 1 year, and is worth

$n/b2 today.

The Future Value of Money

$n paid k years from now

is only worth $n/bk today

Someone pays you $100/year for 10 years.

Let r ::= 1/bankrate = 1/1.03

In terms of current value, this is worth:

100r + 100r2 + 100r3 + + 100r10

= 100r(1+ r + + r9)

= 100r(1r10)/(1r) = $853.02

$n paid k years from now

is only worth $n/bk today

Annuities

Annuities

I pay you $100/year for 10 years,

if you will pay me $853.02.

QUICKIE: If bankrates unexpectedly

increase in the next few years,

A. You come out ahead

B. The deal stays fair

C. I come out ahead

Annuities

In terms of current value, this is worth:

50000 + 50000r + 50000r2 +

= 50000(1+ r + )

= 50000/(1r)

Let r = 1/bankrate

If bankrate = 3%, then the sum is $1716666

If bankrate = 8%, then the sum is $675000

Would you prefer a million dollars today

or $50,000 a year for the rest of your life?

Suppose there is an annuity that pays im

dollars at the end of each year i forever.

For example, if m = $50, 000, then the

payouts are $50, 000 and then $100,

000 and then $150, 000 and so on…

Annuities

What is a simple closed form expression of the following sum?

Manipulating Sums

What is a simple closed form expression of ?

(see an inductive proof in tutorial 2)

Manipulating Sums

for x < 1

For example, if m = $50, 000, then the payouts are $50,

000 and then $100, 000 and then $150, 000 and so on…

For example, if p=0.08, then V=8437500.

Still not infinite! Exponential decrease beats additive increase.

Loan

Suppose you were about to enter college today

and a college loan officer offered you the following

deal:

$25,000 at the start of each year for four years to

pay for your college tuition and an option of

choosing one of the following repayment plans:Plan A: Wait four years, then repay $20,000 at the

start of each year for the next ten years.

Plan B: Wait five years, then repay $30,000 at the

start of each year for the next five years.

Assume interest rate 7% Let r = 1/1.07.

Plan A: Wait four years, then repay $20,000 at the

start of each year for the next ten years.

Plan A

Current value for plan A

Plan B

Current value for plan B

Plan B: Wait five years, then repay $30,000 at the

start of each year for the next five years.

Profit

$25,000 at the start of each year for four years

to pay for your college tuition.

Loan office profit = $3233.

How far out?

?overhang

Book Stacking

book centerof mass

One Book

book centerof mass

One Book

12

book centerof mass

One Book

12

n

More Books

How far can we reach?

To infinity??

centerof mass

12

n

More Books

need center of mass

over table

12

n

More Books

center of mass of the whole stack

12

n

More Books

center of mass of all n+1 booksat table edge

center of mass of top n books at edge of book n+1

∆overhang

12

nn+1

Overhang

center of mass of the new book

1

n

1/2

Overhang

center of n-stack at x = 0.center of n+1st book is at x = 1/2,so center of n+1-stack is at

center of mass of all n+1 books

center of mass of top n books

12

nn+1

1/2(n+1)

Overhang

Bn ::= overhang of n books

B1 = 1/2

Bn+1 = Bn +

Bn =

12(n+1)

1 1 1 1

1+ + + +2 2 3 n

n

1 1 1H ::=1+ + + +

2 3 n

nth Harmonic number

Overhang

Bn = Hn/2

Harmonic Number

n

1 1 1H ::=1+ + + +

2 3 nHow large is ?

…

1 number

2 numbers, each <= 1/2 and > 1/4

4 numbers, each <= 1/4 and > 1/8

2k numbers, each <= 1/2k and > 1/2k+1

Row sum is <= 1 and >= 1/2

Row sum is <= 1 and >= 1/2

Row sum is <= 1 and >= 1/2

The sum of each row is <=1 and >= 1/2.

…

Harmonic Number

n

1 1 1H ::=1+ + + +

2 3 nHow large is ?

…

The sum of each row is <=1 and >= 1/2.

…

k rows have 2k-1 numbers.

If n is between 2k-1 and 2k+1-1,

there are >= k rows and <= k+1

rows,

and so the sum is at least k/2

and is at most (k+1).

1x+1

0 1 2 3 4 5 6 7 8

1

1213

12

1 13

Harmonic Number

Estimate Hn:

n

1 1 1H ::=1+ + + +

2 3 n

n

0

1 1 1 1 dx 1 + + +...+

x+1 2 3 n

n+1

n1

1dx H

x

nln(n+1) H

Now Hn as n , so

Harmonic series can go to infinity!

Integral Method (OPTIONAL)

Amazing equality

http://www.answers.com/topic/basel-problem

Proofs from the book, M. Aigner, G.M. Ziegler, Springer

Spine

Shield

Towers

Optimal Overhang?

(slides by Uri Zwick)

Overhang = 4.2390Blocks = 49

Weight = 100

Optimal Overhang?

(slides by Uri Zwick)

Product

Factorial defines a product:

Factorial

How to estimate n!?

Too rough…

Factorial defines a product:

Factorial

How to estimate n!?

Still very rough, but at least show that it is much larger than Cn

Factorial defines a product:

Turn product into a sum taking logs:

ln(n!) = ln(1·2·3 ··· (n – 1)·n)

= ln 1 + ln 2 + ··· + ln(n – 1)

+ ln(n)n

i=1

ln(i)

Factorial

How to estimate n!?

…ln 2ln 3ln 4

ln 5ln n-1

ln nln 2

ln 3ln 4ln 5

ln n

2 31 4 5 n–2 n–1 n

ln (x+1)ln (x)

Integral Method (OPTIONAL)

ln(x) dx ln(i) ln (x+1)dxi=1

nn n

1 0

x

lnxdx =xlne

Reminder:

n

i=1

1 nln(i) n+ ln

2 eso guess:

n ln(n/e) ln(i) (n+1) ln((n+1)/e)

Analysis (OPTIONAL)

exponentiating:

nn

n! n/ e e

n

i=1

1 nln(i) n+ ln

2 e

nn

n! 2πne

~Stirling’s formula:

Stirling’s Formula

More Integral Method

What is a simple closed form expressions of ?

Idea: use integral method.

So we guess that

Make a hypothesis

Sum of Squares

Make a hypothesis

Plug in a few value of n to determine a,b,c,d.

Solve this linear equations gives a=1/3, b=1/2, c=1/6, d=0.

Go back and check by induction if

Cauchy-Schwarz

(Cauchy-Schwarz inequality) For any a1,…,an, and any b1,…bn

Proof by induction (on n): When n=1, LHS <= RHS.

When n=2, want to show

Consider

Cauchy-Schwarz

(Cauchy-Schwarz inequality) For any a1,…,an, and any b1,…bn

Induction step: assume true for <=n, prove n+1.

induction

by P(2)

Cauchy-Schwarz

(Cauchy-Schwarz inequality) For any a1,…,an, and any b1,…bn

Exercise: prove

Answer: Let bi = 1 for all i, and plug into Cauchy-Schwarz

This has a very nice application in graph theory that hopefully we’ll see.

Geometric Interpretation

(Cauchy-Schwarz inequality) For any a1,…,an, and any b1,…bn

•The left hand side computes the inner

product of the two vectors

• If we rescale the two vectors to be of

length 1, then the left hand side is <= 1

•The right hand side is always 1.

a

b

Arithmetic Mean – Geometric Mean Inequality

(AM-GM inequality) For any a1,…,an,

Interesting induction (on n): • Prove P(2)

• Prove P(n) -> P(2n)

• Prove P(n) -> P(n-1)

Arithmetic Mean – Geometric Mean Inequality

(AM-GM inequality) For any sequence of non-negative numbers a1,…,an,

Interesting induction (on n): • Prove P(2)

Want to show

Consider

Arithmetic Mean – Geometric Mean Inequality

(AM-GM inequality) For any sequence of non-negative numbers a1,…,an,

Interesting induction (on n): • Prove P(n) -> P(2n)

induction

by P(2)

Arithmetic Mean – Geometric Mean Inequality

(AM-GM inequality) For any sequence of non-negative numbers a1,…,an,

Interesting induction (on n): • Prove P(n) -> P(n-1)

Let the average of the first n-1 numbers.

Arithmetic Mean – Geometric Mean Inequality

(AM-GM inequality) For any sequence of non-negative numbers a1,…,an,

Interesting induction (on n): • Prove P(n) -> P(n-1)

Let

Geometric Interpretation

(AM-GM inequality) For any sequence of non-negative numbers a1,…,an,

•Think of a1, a2, …, an are the side lengths of a high-dimensional rectangle.

•Then the right hand side is the volume of this rectangle.

•The left hand side is the volume of the square with the same total side length.

•The inequality says that the volume of the square is always not smaller.

e.g.

Arithmetic Mean – Geometric Mean Inequality

(AM-GM inequality) For any sequence of non-negative numbers a1,…,an,

Exercise: What is an upper bound on ?

•Set a1=n and a2=…=an=1, then the upper bound is 2 – 1/n.

•Set a1=a2=√n and a3=…=an=1, then the upper bound is 1 + 2/√n – 2/n.

•…

•Set a1=…=alogn=2 and ai=1 otherwise, then the upper bound is 1 + log(n)/n