Question

Can someone help me with some Big O notation?

Answer 1

@mathmale

Answer 2

First, Angelina, what is "big O" notation? Completely new to me.

Answer 3

With the first problem, my first instinct is to multiply these two binomials together. Is that not the correct thing to do here?

Answer 4

Big O notation is like and estimate of how long an algorithm will take to run, basically it's just the domination of terms

Answer 5

I see. While I understand your goal, I don't know how to approach this estimation of run time. Have you tried posting this problem in the Computer Science section? Currently there are 38 people checked in there.

Answer 6

Yeah, nobody responded over there

Answer 7

It's kinda new to me too

Answer 8

Angelina, I think your best bet would be to research "Big O" notation on the net. I've done a preliminary search for you and have come up with the following:


https://www.google.com/search?sourceid=chrome-psyapi2&ion=1&espv=2&es_th=1&ie=UTF-8&q=big%20o%20notation&oq=Big%20O%20notation&aqs=chrome.0.0l6.3950j0j7

Answer 9

\[(n-1)(n+2)=n^2+n-2\le n^2+n\le n^2+n^2=2n^2\]

for \(n\ge 1\)

Answer 10

What's that supposed to mean?

Answer 11

\[(n-1)(n+2)=O(n^2)\]

Answer 12

Oh okay

Answer 13

the second one is O(n)?

Answer 14

no

Answer 15

How come?

Answer 16

\[2+4+\cdots+2n=2(1+2+\cdots+n)=2\frac{n(n+1)}{2}=n(n+1)\]

Answer 17

it is \(O(n^2)\)

Answer 18

So for the third one, what would we do?

Answer 19

I assume that the n^3 should be the dominating term?

Answer 20

almost

Answer 21

\[n^3\log(n)\]

Answer 22

is that because 17logn multiplies with n^3?

Answer 23

that is one of them

Answer 24

and n^3 multiplies log n

Answer 25

Okay so if we expand the 4th one out we get n^5 + 2^n n^3 +3^n n^2 +6^n

Answer 26

so should be O(n^5)?

Answer 27

no

Answer 28

This is confusing

Answer 29

exponential grows much faster than polynomial

Answer 30

hmmmm

Answer 31

is it something^n?

Answer 32

yes

Answer 33

6^n?

Answer 34

yes

Answer 35

so O(6^n)?

Answer 36

it is

Answer 37

though i'm sure your instructor will want to see more work...like I did for the first one

Answer 38

I'm out of time here...past my bed time

Answer 39

okay, so for the last one I expanded it out and got \[n ^{n}n!+ 2^{n} n*n! + 5^{n}n! +5^{n}n ^{n} + 10^{n} + 5^{2n}\]

Answer 40

not sure what is the largest

Answer 41

sorry it should be 10^n * n

Answer 42

i'm thinking it would also be O(5^2n)