So I have to figure out if n^(1/2) is Big Omega of log(n)^3. I am pretty sure that it is not, since n^(1/2) is not even in the bounds of log(n)^3; but I do not know how to prove it without limits. I know the definition without limits is
g(n) is big Omega of f(n) iff there is a constant c > 0 and an
integer constant n0 => 1 such that f(n) => cg(n) for n => n0
But can I really always find a constant c that will satisfy this?
for instance for log(n)^3=>c*n^(1/2) if c = 0.1 and n = 10 then we get 1=>0.316.
When comparing sqrt(n) with ln(n)^3 what happens is that
ln(n)^3 <= sqrt(n) ; for all n >= N0
How do I know? Because I printed out sufficient samples of both expressions as to convince myself which dominated the other.
To see this more formally, let's first assume that we have already found N0 (we will do that later) and let's prove by induction that if the inequality holds for n >= N0, it will also hold for n+1.
Note that I'm using ln in base e for the sake of simplicity.
Equivalently, we have to show that
ln(n + 1) <= (n + 1)^(1/6)
Now
ln(n + 1) = ln(n + 1) - ln(n) + ln(n)
= ln(1 + 1/n) + ln(n)
<= ln(1 + 1/n) + n^(1/6) ; inductive hypethesis
From the definition of e we know
e = limit (1 + 1/n)^n
taking logarithms
1 = limit n*ln(1 + 1/n)
Therefore, there exits N0 such that
n*ln(1 + 1/n) <= 2 ; for all n >= N0
so
ln(1 + 1/n) <= 2/n
<= 1
Using this above, we get
ln(n + 1) <= 1 + n^(1/6)
<= (n+1)^(1/6)
as we wanted.
We are now left with the task of finding some N0 such that
ln(N0) <= N0^(1/6)
let's take N0 = e^(6k) for some value of k that we will are about to find. We get
ln(N0) = 6k
N0^(1/6) = e^k
so, we only need to pick k such that 6k < e^k, which is possible because the right hand side grows much faster than the left.
Related
I am trying to learn how to prove Big O correctly.
what i am trying to do is find some C and N0 for a given function.
the definition given for Big-O is
Let f(n) and g(n) be functions mapping nonnegative integers to real numbers.
We say that f(n) is O(g(n)) if there is a real constant c > 0 and an integer
constant n0 ≥ 1 such that for all n ≥ n0, f(n) ≤ c g(n).
Given the polynomial (n+1)^5 i need to show that it has a runtime of O(n^5).
my question is, how do i find such c and N0 from the definition above, and how do i continue my algebra to see if it runs n^5?
So far by trying induction i have,
(n+1)^5 = n^5 + 5n^4 + n^3 + 10n^2 + 5n^1 + n^0
find the n+1 element so
n^5 + 5n^4 + n^3 + 10n^2 + 5n^1 + n^0 <= n^5 + 5n^5 + n^5 + 10n^5 + 5n^5 + n^5
n^5 + 5n^4 + 10n^2 + 5n + 1 <= 22n^5
You want a constant c such that (n + 1) 5 ≤ c n 5. For that, you do not need induction, only a bit of algebra and it turns out you actually already found such a c, but missed the n0 in the process. So let's start from the beginning.
Note that c does not need to be tight, it can be way bigger than necessary and will still prove time-complexity. We will use that to our advantage.
We can first develop the left side as you did.
(n + 1) 5 = n5 + 5n4 + 10n3 + 10 n2 + 5n + 1
For n ≥ 1, we have that n, n2, n3, n4 ≤ n5, an thus.
(n + 1) 5 ≤ (1 + 5 + 10 + 10 + 5 + 1) n5 = 22n5
And there you got a c such that (n + 1) 5 ≤ c n5. That c is 22.
And since we stated above that this holds if n ≥ 1, then we have that n0 = 1.
Generalization
This generalizes for any degree. In general given the polynomial f(n) = (n + a)b, then you know that there exists a number c that is found by summing all the coefficients of the polynomial after development. It turns out the exact value of c does not matter so you do not need to compute it, all that matter is that we proved its existence and thus (n + a)b is O(nb).
I have to prove that f(n)= 5n+2=O(n^2) and I know that it is true for O(n) so obviously, it will be true for a higher degree of n but how to prove this?
OK. Here is a simple way of proving this. I'm including it here in the hope that it could be useful for others with similar problems
5n + 2 <= 5n + 2n ; n >= 1
= 7n ; always
<= n*n ; n >= 7
= n^2 ; always
Therefore, there exists a constant c, in this case c=1, and an integer N, in this case N=7, such that
5n + 2 <= c*n^2 for all n >= N
Then, by definition
5n + 2 = O(n^2).
Note also that the first two lines
5n + 2 <= 5n + 2n ; n >= 1
= 7n ; always
are enough to show that 5n + 2 = O(n). In this case, c=7 and N=1.
Prove 5n^2+ 2n -1 = O(n^2).
This is what I have attempted so far:
5n^2 + 2n -1 <= 5n^2 + 2n^2 -n2 for n>1
5n^2+ 2n -1 <= 6n^2 for n>1
5n^2+ 2n -1 = O(n^2) [ c = 6n^2, n0 = 1 ]
Is this the right way of proving Big O notation?
To prove that your expression is O(n^2), you need to show that it is bounded by M*n^2, for some constant M and some minimum n value. By inspection, we can show that your expression is bounded by 10*n^2, for n=10:
For n = 10:
5n^2 + 2n -1 <= 10*n^2
500 + 20 - 1 <= 1000
519 <= 1000
We can also show that the expression is bounded by 10*n^2 for any value n greater than 10:
For n > 10:
5n^2 + 2n -1 <= 10*n^2
5*(10+i)^2 + 2*(10+i) -1 <= 10*(10+i)^2
5*(i^2 + 20i + 100) + 2i + 19 <= 10*(i^2 + 20i + 100)
2i + 19 <= 5*(i^2 + 20i + 100)
2i + 19 <= 5i^2 + 100i + 500
5i^2 + 98i + 481 >= 0, which is true for `i > 0`
Here is a link to the Wikipedia article on Big-O notation:
https://en.m.wikipedia.org/wiki/Big_O_notation
Update:
Note that in practice in order to label your expression O(n^2) we won't resort to such an ugly written proof. Instead, we will just recognize that the n^2 term will dominate the expression for large n, and that the overall behavior will be O(n^2). And your expression is also O(n^3) (and O(n^4), etc.).
it looks fine , I'm thinking that if you are doing it for your assignment work or other formal work then you can also do it in more formal way like for selecting the value of constant ( c ) such as by f(n)/g(n).Otherwise it looks correct.
We have f(n) = 5*n^2+2*n-1 and g(n) = n^2
In order to prove f(n) = O(g(n)), we just need to find 2 constants, namely c > 0 and n_0, s.t. f(n) <= c.g(n) for all n >= n0.
Let's choose some value of c = 5.5 say. Let's evaluate and plot f(n) and c*g(n). As we can see from the plot and we can also show it theoretically since n^2/2 - 2*n + 1 = (n^2-4*n+2)/2 = ((n-2)^2-2)/2 >= 0 for all n >= 4, it implies that 5*n^2+2*n-1 <= 5.5*n^2 for all n >= n0 = 4. Hence, f(n) = O(g(n)). (proved)
I'm trying to explain to my friend why 7n - 2 = O(N). I want to do so based on the definition of big O.
Based on the definition of big O, f(n) = O(g(n)) if:
We can find a real value C and integer value n0 >= 1 such that:
f(n)<= C . g(n) for all values of n >= n0.
In this case, is the following explanation correct?
7n - 2 <= C . n
-2 <= C . n - 7n
-2 <= n (C - 7)
-2 / (C - 7) <= n
if we consider C = 7, mathematically, -2 / (C - 7) is equal to negative infinity, so
n >= (negative infinity)
It means that for all values of n >= (negative infinity) the following holds:
7n - 2 <= 7n
Now we have to pick n0 such that for all n >= n0 and n0 >= 1 the following holds:
7n - 2 <= 7n
Since for all values of n >= (negative infinity) the inequality holds, we can simply take n0 = 1.
You're on the right track here. Fundamentally, though, the logic you're using doesn't work. If you are trying to prove that there exist an n0 and c such that f(n) ≤ cg(n) for all n ≥ n0, then you can't start off by assuming that f(n) ≤ cg(n) because that's ultimately what you're trying to prove!
Instead, see if you can start with the initial expression (7n - 2) and massage it into something upper-bounded by cn. Here's one way to do this: since 7n - 2 ≤ 7n, we can (by inspection) just pick n0 = 0 and c = 7 to see that 7n - 2 ≤ cn for all n ≥ n0.
For a more interesting case, let's try this with 7n + 2:
7n + 2
≤ 7n + 2n (for all n ≥ 1)
= 9n
So by inspection we can pick c = 9 and n0 = 1 and we have that 7n + 2 ≤ cn for all n ≥ n0, so 7n + 2 = O(n).
Notice that at no point in this math did we assume the ultimate inequality, which means we never had to risk a divide-by-zero error.
I am going over the Big-Oh notation, and I have a problem understanding the solution to this question:
Is 2n + 10 ≡ O(n)?
Can we find c and n0?
2n + 10 <= cn
(c-2)n >= 10
n >= 10/(c-2)
Pick c = 3 and n0 = 10
There is also a graph used in this example:
I am confused as to how c = 3 and how n0 = 10? Can somebody please enlighten me?
I would solve 2n + 10 <= cn differently.
2n + 10 <= cn
2 + 10/n <= c //divide by n
c >= 2 + 10/n
Clearly c has to be bigger than 2. Now take your favorite number bigger than 2. Uhm. c=2.718. This gives
2n + 10 <= 2.718*n
10 <= 0.718*n //subtract 2n
13.93 <= n
Thus, 2n + 10 <= c*n. If we take c=2.718 and n bigger than 15. (15 because it's bigger than the limit, 13.93, and I like 15.)
Any c bigger than 2 works, c=100000000000000000000000 is fine too (but, it costs much ink and paper to write down.)
It might have been easier to take c=3. That would give
2n + 10 <= 3*n
10 <= n //subtract 2n
This makes 3 the most logical / natural solution.
To say a function f(n) is O(n) means you can find c and n0 such that for all n >= n0, f(n) <= cn.
To verify this in your case: if n >= 10, then:
f(n) = 2n + 10
<= 3n // because 10 <= n
= cn
So f(n) <= cn for all n >= 10, so f(n) is an O(n) function.
Note that other values for c and n0 work; you just need to find one pair.
In Big-O notation you drop numbers added and multiplied.
And also use the largest power.
10*N^3+ 23*N^2 + 43*N + 154 = O(N^3)
let f(n) = 2n+10
as per the Big-Oh notation,
f(n)<= c*g(n) ----->eq1(let)
2n+10 <= 2n+n
2n+10 <= 3n for n>=10 //here the constant 10 is replaced by n ---->eq2
comparing eq1 with eq2
c=3 and g(n)=n for value of n=10
f(n)= O(g(n)) = O(n)