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

2. C. F. Gauss
Picture source:

3. Sum for children 89 + 102 + 115 + 128 + 141 + 154 + ··· 193 + ···
···
···
···
···

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

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

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

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

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

9. Geometric Series

10. Geometric Series

11. Geometric Series
G-xG=
xn+1

12. Geometric Series
G-xG=
1- xn+1

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

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

15. 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.

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

17. 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.

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

19. 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 + 100r r3 + … + 100r10
=100r(1+ r + … + r9)
= 100r(1-r10)/(1-r) = $853.02

20. In-Class Problem
Problems 1 & 2

21. Book Stacking
Rosen
Rosen
Rosen
Rosen
Rosen
Rosen
table

22. Book Stacking
How far out? ?

23. Book Stacking
One book
center of mass
of book 1 2

24. Book Stacking
One book
center of mass
of book

25. Book Stacking
One book
center of mass
of book

26. n books

27. n books
center of
mass

28. n books
Need
center of mass
over table

29. n books
center of mass
of the whole stack
overhang

30. 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

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

32. 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:

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

34. 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)

35. nth Harmonic number
Bn = Hn/2

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

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

39. 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.

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

41. Dn ::= max distance on n tank

42. 1 tank
1 tank
truck
D1= max distance on 1 tank = 1

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

44. n+1 tanks x
1-2x
1-2x n
1-2x
1-x

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

46. (1-2x)n + (1-x)

47. (1-2x)n + (1-x)
If (1-2x)n + (1-x) = n,

48. (1-2x)n + (1-x)
If (1-2x)n + (1-x) = n,
then use n tank strategy from position x.

49. (1-2x)n + (1-x)
If (1-2x)n + (1-x) = n,
then use n tank strategy from position x.
Dn+1 = Dn + x

50. (1-2x)n + (1-x) = n 1
2n+1
x = 1
2n+1
Dn+1 = Dn +

51. Can cross any desert!

52. In-Class Problem
Problem 3