1. Chapter 3 Growth of Functions

2. (defined as a set of functions)
Used to denote worst case, as in insertion sort T(n) = O(n2)
Allows for possibility that f(n) = cg(n) for large n
To prove T(n) = O(g(n)), find a combination of c and n0 (may not be unique) for which definition is satisfied.

3. Used to denote best case run time
Allows for possibility that f(n) = cg(n) for large n
To prove T(n) = W(g(n)), find a combination of c and n0 (may not be unique) for which definition is satisfied.

4. Theorem 3.1: T(n) = Q(g(n)) iff
T(n) = O(g(n)) and T(n) = W(g(n))
At large n, f(n) and g(n) are equal within a constant factor
To prove T(n) = Q(g(n)), use Theorem 3.1,

5. Important points
Constants in the definitions of O, W, and Q may depend on the
choice of n0 but for n > n0 they cannot depend on n.
Even though O(g(n)) is formally defined by a set of functions,
we write f(n) = O(g(n)) to mean that f(n) is a member of the
set. Same convention used for W, and Q

6. More important points For all forms of asymptotic notation,
f(n)=O(g(n))
f(n)=W(g(n))
f(n)=Q(g(n))
f(n)=o(g(n)) to be defined
f(n)=w(g(n)) to be defined
the argument of asymptotic notation is the bound and the
quantity it is equal to is the function being bounded.

7. Informal proofs of Q notation
If we have an equation for f(n), and g(n) is the dominant term
in that equation at large values of n, then f(n) = Q(g(n))
Example: Show by informal proof that (n+a)b = Q(nb)
real a & b a not zero b > 0
Binomial expansion: (n+a)b = Cbnb + Cb-1nb-1+…+C0
Since nb is dominant term at large n, (n+a)b = Q(nb)
by informal proof

8.  Assignment 3: due 1/14/19
Use Stirling’s approximation (Eq(3.18) text p57) to give an informal proof that lg(n!)=Q(nlgn). Show all steps.

9. Ex text p52
Formal proof that (n+a)b = Q(nb)
real a and b
a not zero
b > 0
Strategy: formal proof of (n+a) = O(n)
Generalize to (n+a)b = O(nb) by algebra
Similarly for (n+a)b = W(nb)
Use theorem 3.1

10. To prove (n+a)=O(n), find c>0 and n0 such that 0 < (n+a) < cn for all n>n0
Need an inequality with n+a < something
a < |a|
n+a < n+|a| < cn
This string of inequalities says that if we find c>0 and n0 such that n+|a| < cn for all n > n0, then they will also work for n+a and the condition (n+a) < cn in the definition of bigO is satisfied.
Hence we focus on inequality n+|a| < cn in the search c and n0.
|a| < cn-n
|a| < (c-1)n
To use values of n to make this inequality true, c > 1
If we choose c = 2, then |a| < (c-1)n becomes n > |a|
n0 = |a| is large enough to make (n+a) > 0 if a is negative.
Conditions for Big O are satisfied

11. To prove (n+a)=W(n), find c>0 and n0 such that 0 < cn < (n+a) for all n>n0
Need an inequality with n+a > something
a > -|a|
n+a > n-|a| > cn
This string of inequalities says that if we find c>0 and n0 such that n-|a| > cn for all n > n0, then they will also work for n+a and the condition (n+a) > cn in the definition of big W is satisfied.
Hence we focus on inequality n+|a| > cn in the search c and n0.
n-cn > |a|
n(1-c) > |a|
To use values of n to make this inequality true, c < 1
If we choose c = 1/2, then n(1-c) > |a| becomes n > 2|a|
Both c and n are positive, so cn > 0
Conditions for Big W are satisfied.

12. There exist c=2 such that 0 < (n+a) < cn for all n>n0 = |a|
Since b>0, 0 < (n+a)b < cbnb
There exist d=2b such that 0 < (n+a)b < dnb for all n>n0 = |a|
Therefore (n+a)b = O(nb)
There exist c=1/2 such that 0 < cn < (n+a) for all n>n0 =2|a|
Since b>0, 0 < cbnb < (n+a)b
There exist d=(1/2)b such that 0 < dnb < (n+a)b for all n>n0 = 2|a|
Therefore (n+a)b = W(nb)
Therefore (n+a)b = Q(nb) by Theorem 3.1

13. Assignment 4: due 1/16/19
Ex text p52
If f(n) and g(n) are asymptotically non-negative,
show by the definitions of big O and W and theorem 3.1 that max(f(n),g(n)) = Q(f(n)+g(n)).
Note: at any n, max(f(n),g(n)) is the larger of f(n) and g(n)
Address both inequalities in the definitions of big O and W
Use given information to define n0
Use non-negative property of f(n) and g(n) to find constants in the definitions of big O and W

14. Antonio’s Office Hours
Monday 2-4pm in TFLO 145
TuTh 2-4pm TFLO 252

15. Prove that Sk (O(fk(n))) = O(Sk fk(n)) when fk(n) is asymptotically
non-negative for all k.
{fk(n)} is a set of functions of n that are non-negative for all n>n0
Range of k-values is not specified.
What is the bound?
What is being bounded?
If true, result say “big O passes through sums”
(i.e. a sum of big Os is big O of the sum of the bounds)
Let gk(n)=O(fk(n)) and use definition of Big O(fk(n))

16. Little o notation
f(n) = o(g(n)) means f(n) is a member of a set of function
such that for any c > 0 (no matter how small)
0 < f(n) < cg(n) all n > n0
Can only be true if f(n) is insignificant relative to g(n) at large n
Equivalent to limit f(n)/g(n) = 0
n →∞
g(n) is strictly an asymptotic upper bound on f(n)
Asymptotic equality (within a constant) of f(n) and g(n) is
excluded
2n = o(n2) 2n2 is not o(n2) (see text p 50)

17. Prove that (lg(n))k = o(ne)
What is f(n)?
What is g(n)?
What is limit f(n)/g(n)?
n →∞

19. Prove that nk = o(cn)
What is f(n)?
What is g(n)?
What is limit f(n)/g(n)?
n →∞

20. Polynomials are
exponentially dominated

21. Little omega notation
f(n) = w(g(n)) means f(n) is a member of a set of function
such that for any c > 0 (no matter how large)
0 < cg(n) < f(n) all n > n0
Can only be true if g(n) is insignificant relative to f(n) at large n
Equivalent to limit g(n)/f(n) = 0
n →∞
g(n) is strictly an asymptotic lower bound on f(n)
Asymptotic equality of f(n) and g(n) is excluded
n2/2 = w(n) n2/2 is not w(n2) text p51

22. Prove that 2n = w(2n/2)
What is f(n)?
What is g(n)?
What is limit g(n)/f(n)?
n →∞

24. Homework Assignment 5: due 1/23/19
Use Stirling’ approximation, Eq(3.18) p 57,
to show that n! = o(nn) and n! = w(2n)

25. Prove f(n) = o(g(n)) implies f(n) = O(g(n))
If for any c > 0, 0 < f(n) < cg(n) then
there exist some c > 0 such that 0 < f(n) < cg(n)
Similarly f(n) = w(g(n)) implies f(n) = W(g(n))
What is wrong with these arguments?

26. Little o notation
f(n) = o(g(n)) means f(n) is a member of a set of function
such that for any c > 0 (no matter how small)
0 < f(n) < cg(n) all n > n0
Can only be true if f(n) is insignificant relative to g(n) at large n
Equivalent to limit f(n)/g(n) = 0
n →∞
g(n) is strictly an asymptotic upper bound on f(n)
Asymptotic equality (within a constant) of f(n) and g(n) is
excluded
2n = o(n2) 2n2 is not o(n2) text p 50

27. prob 3-2 p 61: work in class
A->f(n) B->g(n)
Which is true O, o, W, w, Q
Can be more than 1 or none

28. Properties of asymptotic notations (text pp51-52)

29. Important math
from Section 3.2
Text pp53-58

30. See text p54 for useful properties
Floors and Ceilings
Greatest integer less than or equal to x
Least integer greater than or equal to x
See text p54 for useful properties

31. Asymptotic notation in equations
Alone on right side: n = o(n2) set definition
Part of an equation: f(n) = 2n2 + Q(n) = Q(n2)
In recursive relation T(n) = T(n-1) + Q(n) says overhead is linear