i cant find the O and the Omega of the function (log n)! in the functions that are in the picture.
i tried to do : logn*(log n - 1)!< (log n)^n to find the O but it is dosent help me
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So, I have
T(n) = 2T(n/2) + n^2
And I found a = 2, b = 2, and f(n) = n^2, and to get n^2 I used case 3 for the master theorem which is
f(n) = big Omega ( n ^ logba + E )
and I found big Omega (n^2)
So, how can I solve
big theta ( big Omega (n^2) ) ?
Also, is my calculations correct?
Im currently having a problem with big O notation. I have the following question which I am trying to figure out.
I currently have the formula: T(n) is O(f(n)) and I must use this to prove directly from the definition of big O that 3n^2+11n+6 is O(n^2).
I was wondering if anybody could possibly help me figure out this problem as I am having trouble working it out.
I think this may help:
For n≥k, there is a constant, let's name it "c" which satisfies 3n^2 + 11n + 6 ≤ c∗n^2.
Let's say we pick k = 1.
We know that n^2 ≥ n^2 ≥ n ≥ 1
So :
3n^2 + 11n + 6 ≤ 3n^2 + 11n^2 + 6n^2 =>3n^2 + 11n + 6 ≤ 20n^2
Now, let c = 20.
=>complexity is O(n2).
whats is the big O notation of this function f(x) = logn + 3n i have ridden big o notation but i am confuse in this function so please help me
It is simply O(n).
When you have a composite of multiple parts in big O notation which are added, you have to choose the biggest one. In this case it is O(3n), but there is no need to include constants inside parentheses, so we are left with O(n).
https://en.wikipedia.org/wiki/Big_O_notation
It's O(n).
Proof:
lim n->inf f(n)/3n = lim n->inf (logn + 3n) / 3n = [0 + 1] = 1
so f(n) and 3n have the same asimptotic behavior
according to the "O" notation constants are dropped so f(n) is O(3n) = O(n)
what is the big O complexity of the function 1005 n^1.75 + 100 n^1.5 + 10nlogn ?
I know that the big O complexity of logn is equal to O(logn)^2) but can't figure out this riddle: 1005 n^1.75 + 100 n^1.5 + 10nlogn
f(n) = 1005 n^1.75 + 100 n^1.5 + 10nlogn
Now, in O notation, we only need to consider the highest order terms, also constants can be ignored.
Therefore,
f(n) = n^1.75
Hence,
O(f(n)) = O(n^1.75) or O(n^7/4)
the complexity of that function is:
O(N^(7/4))
I have this:
a) f(n) = n
b) f(n) = 1045n
c) f(n) = n2 + 70
d) f(n) = 7n + 3
e) f(n) = Cn + D (where C and D are both constants)
f) f(n) = 8
g) f(n) = n3 + n + 1
h) f(n) = 4n + 2log n + 5
I want to check if the Big O notation of them is O(n).
How can I determinate it?
And how to find the Big-O notation for the functions below:
a) f(n) = 3n3 + n
b) f(n) = 3 log n + 5n
c) f(n) = 3n2 + 5n + 4
d) f(n) = 3n3 + n2 + 5n + 99
f(x) is O(g(x)) if there exists a constant c such that f(x) < c*g(x)
You should look at the "biggest" asymptoticall factor in your function (highest exponent or something like that) and that would be your big O
Example: f(n) = 2^n + n^5 + n^2 + n*log(n) + n
This function has 5 different factors that influence big O, sorted from biggest to smallest, so this one would be O(2^n). Drop the 2^n and now f(n) is O(n^5).
Constants are O(1).
Hope I explained it well
Generally the Big O notation of a function is measured by the largest power of n that appears. In your case, this would be n², since the only other factor is the 70 which is constant.
Edit: Original post only contained the function f(n) = n² + 70.
See this answer.
In short, there's no set way to determine Big O results. Strictly speaking any function which eventually (for some n) will always be bigger than your function, is Big O of it. In practice you're looking for as tight a bound as you can get. If the only components of the function are polynomial in n, then the Big O will just be the largest power of n that appears (or rather, n to that power). Another useful thing to know is that log n is of a lower order than n (but higher than constant).