Mathematics for Computer Science MIT 6.042J/18.062J Sums, Products & Asymptotics Copyright © Albert Meyer, 2002. Prof. Albert Meyer & Dr. Radhika Nagpal

C. F. Gauss Picture source: http://www-groups.dcs.st-and.ac.uk/~history/PictDisplay/Gauss.html

Sum for children 89 + 102 + 115 + 128 + 141 + 154 + ··· 193 + ··· 89 + 102 + 115 + 128 + 141 + 154 + ··· 193 + ··· 232 + ··· 323 + ··· + 401

Sum for children Nine-year old Gauss (so the story goes) saw that each number was 13 greater than the previous one.

Sum for children A ::= 89 + (89+13) + (89+2·13) + … + (89+24·13) + 89+(89+24·13)

first + last A = · #terms 2 Sum for children 2A = [89+ (89+24·13)]·25

A = Average · #terms Sum for children 2A = [89+ (89+24·13)]·25 first last #terms A = Average · #terms

Sum for children Example: 1 + 2 + … + (n-1) + n =

Geometric Series

Geometric Series

Geometric Series G-xG= 1 - xn+1

Geometric Series G-xG= 1- xn+1

Geometric Series G-xG= 1- xn+1 n+1

The future value of $$. Annuities I will promise to pay you $100 in exactly one year, if you will pay me $X now.

1.03 X  100. Annuities My bank will pay me 3% interest. If I deposit your $X for a year, I can’t lose if 1.03 X  100.

Annuities I can’t lose if you pay me: X = $100/1.03 ≈ $97.09

Annuities 97.09¢ today is worth $1.00 in a year $1.00 in a year is worth $1/1.03 today $n in a year is worth $nr today, where r = 1/1.03.

Annuities $n in two years is worth $nr2 today $n in k years is worth $nr k today

Annuities I will pay you $100/year for 10 years If you will pay me $Y now. I can’t lose if you pay me 100r + 100r2 + 100r3 + … + 100r10 =100r(1+ r + … + r9) = 100r(1-r10)/(1-r) = $853.02

In-Class Problem Problems 1 & 2

Book Stacking Rosen Rosen Rosen Rosen Rosen Rosen table

Book Stacking How far out? ?

Book Stacking One book center of mass of book 1 2

Book Stacking One book center of mass of book

Book Stacking One book center of mass of book

n books

n books center of mass

n books Need center of mass over table

n books center of mass of the whole stack overhang

n+1 books  ∆overhang center of mass of all n+1 books at table edge of top n books at edge of book n+1  ∆overhang

 overhang = Horizontal distance from n-book to n+1-book centers-of-mass

Choose origin so center of n-stack at x = 0. Now center of n+1st book is at x = 1/2, and x-coordinate for center of n+1-stack is:

n+1 books  center of mass of all n+1 books at table edge of top n books at edge of book n+1 

Bn ::= overhang of n books B1 = 1/2 Bn+1 = Bn + 1/2(n+1) Book stacking summary Bn ::= overhang of n books B1 = 1/2 Bn+1 = Bn + 1/2(n+1) Bn = 1/2(1 + 1/2 + … + 1/n)

nth Harmonic number Bn = Hn/2

Integral Method Estimate Hn : 1 x+1 0 1 2 3 4 5 6 7 8 1 1 2 1 3 1 2 1 0 1 2 3 4 5 6 7 8

overhang can be any desired size. Book stacking So Hn   as n , and overhang can be any desired size.

Book stacking Overhang 3: need Bn  3 Hn  6 Integral bound: ln (n+1)  6 So can do with n  e6-1 = 403 books Actually calculate Hn : 227 books are enough.

Crossing a Desert How big a desert can the truck cross? Gas depot

Dn ::= max distance on n tank

1 tank 1 tank truck D1= max distance on 1 tank = 1

n+1 tanks x 1-2x truck 1-2x 1-2x 1-x

n+1 tanks x 1-2x 1-2x n 1-2x 1-x

n+1 tanks x (1-2x)n + (1-x)

(1-2x)n + (1-x)

(1-2x)n + (1-x) If (1-2x)n + (1-x) = n,

(1-2x)n + (1-x) If (1-2x)n + (1-x) = n, then use n tank strategy from position x.

(1-2x)n + (1-x) If (1-2x)n + (1-x) = n, then use n tank strategy from position x. Dn+1 = Dn + x

(1-2x)n + (1-x) = n 1 2n+1 x = 1 2n+1 Dn+1 = Dn +

Can cross any desert!

In-Class Problem Problem 3