Prove or disprove the following claims:
Exist function f(n) so f(n-k) is not equal to Big-theta (f(n)). when k>=1 and is positive constant.
Is there any function which this claim is true?
I thought about f(n)=n! but I'm not sure that's is correct answer.
Moreover, if f(n)=n! is correct, how this claim can be proved?
Exist function so (f(n))^2=Big-O(f(n)) and f(n)=Big-omega (log(log(n))).
I think there is not function which make the claim to be true.
If this is correct - how it could be proved?
Correct for f(n) = n!. It suffices to show that for any fixed k >= 1, (n - k)! is not Omega(n!), as for any constant c > 0, it holds for all n large enough that c * n! > (n - k)!.
There is no f(n) such that both f(n)^2 = O(f(n)) and f(n) = Omega(log log n). The latter implies that for some constant c > 0 and all n large enough, f(n) > c log log n, and in particular f(n) > 1 for all n large enough. If we now assume that f(n)^2 = O(f(n)), then there exists a constant r > 0 so that for all n large enough, f(n)^2 < r * f(n), namely f(n) < r. But this implies that log log n < (r / c) for all n large enough, which is false for all n > e^(e^(r / c)) (where e is the basis of log).
Related
Suppose that f(n)=4^n and g(n)=n^n, will it be right to conclude that f(n)=Θ(g(n)).
In my opinion it's a correct claim but I'm not 100% sure.
It is incorrect. f(n) = Theta(g(n)) if and only if both f(n) = O(g(n)) and g(n) = O(f(n)). It is true that f(n) = O(g(n)). We will show that it is not the case that g(n) = O(f(n)).
Assume g(n) = O(f(n)). Then there exists a positive real constant c and a positive natural number n0 such that for all n > n0, g(n) <= c * f(n). For our functions, this implies n^n <= c * 4^n. If we take the nth root of both sides of this inequality we get n <= 4c^(1/n). We are free to assume c >= 1 and n0 >= since if we found a smaller value that worked a larger value would work too. For all c > 1 and n > 1, 4c^(1/n) is strictly less than 4c. But then if we choose n > 4c, the inequality is false. So, there cannot be an n0 such that for all n at least n0 the condition holds. This is a contradiction; our initial assumption is disproven.
A TA told me that this is true today but I was unable to verify this by googling. This is saying functions like log(n)^2, log(n)^3, ... , log(n)^m are all O(n).
Is this true?
Claim
The function f(n) = log(n)^m, for any natural number m > 2 (m ∈ ℕ+) is in
O(n).
I.e. there exists a set of positive constants c and n0 such that
the following holds:
log(n)^m < c · n, for all n > n0, { m > 2 | m ∈ ℕ+ } (+)
Proof
Assume that (+) does not hold, and denote this assumption as (*).
I.e., given (*), there exists no set of positive constants c and n0 such that (+) holds for any value of m > 2. Under this assumption, the following holds, that for all positive constants c and n0, there exists a n > n0 such that (thanks #Oriol):
log(n)^m ≥ c · n, { m > 2 | m ∈ ℕ+ } (++)
Now, if (++) holds, then the inequality in (++) will hold also after applying any monotonically increasing function to both sides of the inequality. One such function is, conveniently, the log function itself
Hence, under the assumption that (++) holds, then, for all positive constants c and n0, there exists a n > n0 such that the following holds
log(log(n)^m) ≥ log(c · n), { m > 2 | m ∈ ℕ+ }
m · log(log(n)) ≥ log(c · n), { m > 2 | m ∈ ℕ+ } (+++)
However, (+++) is obviously a contradiction: since log(n) dominates (w.r.t. growth) over log(log(n)),
we can—for any given value of m > 2—always find a set of constants c and n0 such that (+++) (and hence (++)) is violated for all n > n0.
Hence, assumption (*) is proven false by contradiction, and hence, (+) holds.
=> for f(n) = log(n)^m, for any finite integer m > 2, it holds that f ∈ O(n).
Yes. If the function it's f(n), it means m is a parameter and f does not depend on it. In fact, we have a different f_m function for each m.
f_m(n) = log(n)^m
Then it's easy. Given m ∈ ℕ, use L'Hôpital's rule repeatively
f_m(n) log(n)^m m * log(n)^(m-1)
limsup ──────── = limsup ────────── = limsup ────────────────── =
n→∞ n n→∞ n n→∞ n
m*(m-1) * log(n)^(m-2) m!
= limsup ──────────────────────── = … = limsup ──── = 0
n n→∞ n
Therefore, f_m ∈ O(n).
Of course, it would be different if we had f(m,n) = log(n)^m. For example, taking m=n,
f(n,n) log(n)^n
limsup ──────── = limsup ────────── = ∞
n→∞ n n→∞ n
Then f ∉ O(n)
In many ways it is more intuitive that for any positive integer m we have:
x^m = O(e^x)
This says that exponential growth dominates polynomial growth (which is why exponential time algorithms are bad news in computer programming).
Assuming that this is true, simply take x = log(n) and use the fact that then x tends to infinity if and only if n tends to infinity and that e^x and log(x) are inverses:
log(n)^m = O(e^log(n)) = O(n)
Finally, since for any natural number m, the root function n => n^(1/m) is increasing, we can rewrite the result as
log(n) = O(n^(1/m))
This way of writing it says that log(n) grows slower than any root (square, cube, etc.) of n, which more obviously corresponds to e^n growing faster than any power of n.
On Edit: the above showed that log(n)^m = O(n) followed from the more familiar x^m = O(e^x). To convert it to a more self-contained proof, we can show the later somewhat directly.
Start with the Taylor series for e^x:
e^x = 1 + x/1! + x^2/2! + x^3/3! + ... + x^n/n! + ...
This is known to converge for all real numbers x. If a positive integer m is given, let K = (m+1)!. Then, if x > K we have 1/x < 1/(m+1)!, hence
x^m = x^(m+1)/x < x^(m+1)/(m+1)! < e^x
which implies x^m = O(e^x). (The last inequality in the above is true since all terms in the expansion for e^x are strictly positive if x>0 and x^(m+1)/(m+1)! is just one of those terms.)
I am having a hard time comparing these two functions,
(logn)!
and
2^n
Any good mathematical proof?
You cannot compare O((logn)!) and O(2^n) since big O notation represents a set. O(g(n)) is the set of of all function f such that f does not grows faster than g, formally is the same is saying that there exists C and n0 such that we have |f(n)| <= C|g(n)| for every n >= n0. The expression f(n) = O(g(n)) is a shorthand for saying that f(n) is in the set O(g(n)). what we can do is check if 2^n=O((logn)!) or (log n)!=O(2^n) (note that it could be that both are not true). Luckily, if we use the Stirling approximation we get that
log((logn)!) = (logn)*(log (logn)) - logn + O(log(log n)) = O(n*(log 2))
since n * cost grows faster than (logn)*(log (logn)) and (logn)*(log (logn)) is the leading term in (logn)*(log (logn)) - logn + O(log(log n)). So we get that log((logn)!) = O(log(2^n)) which is same as saying that (log n)! = O(2^n)
One can easily show that for sufficiently large n it holds that:
log(n)! <= log(n)^{log(n)} <= n^{log(n)} = 2^{log^2(n)}
We can now only consider exponents of 2 in the 2^n and the expression above - n and log^2(n) respectively (we can do that since we consider only sufficiently large n and 2^x is strictly rising for positive x). It is sufficient to show that the limit below diverges to prove that log(n)! is, in fact, o(2^n):
lim[n -> inf] (n)/(log^2(n))
Now we apply L'Hospital rule:
= lim [n -> inf] `n/(2log(n))`
And again:
= lim [n -> inf] `n/(2)`
Which diverges to infinity.
I came across two asymptotic function proofs.
f(n) = O(g(n)) implies 2^f(n) = O(2^g(n))
Given: f(n) ≤ C1 g(n)
So, 2^f(n) ≤ 2^C1 g(n) --(i)
Now, 2^f(n) = O(2^g(n)) → 2^f(n) ≤ C2 2^g(n) --(ii)
From,(i) we find that (ii) will be true.
Hence 2^f(n) = O(2^g(n)) is TRUE.
Can you tell me if this proof is right? Is there any other way to solve this?
2.f(n) = O((f(n))^2)
How to prove the second example? Here I consider two cases one is if f(n)<1 and other is f(n)>1.
Note: None of them are homework questions.
The attempted-proof for example 1 looks well-intentioned but is flawed. First, “2^f(n) ≤ 2^C1 g(n)” means 2^f(n) ≤ (2^C1)*g(n), which in general is false. It should have been written 2^f(n) ≤ 2^(C1*g(n)). In the line beginning with “Now”, you should explicitly say C2 = 2^C1. The claim “(ii) will be true” is vacuous (there is no (ii)).
A function like f(n) = 1/n disproves the claim in example 2 because there are no constants N and C such that for all n > N, f(n) < C*(f(n))². Proof: Let some N and C be given. Choose n>N, n>C. f(n) = 1/n = n*(1/n²) > C*(1/n²) = C*(f(n))². Because N and C were arbitrarily chosen, this shows that there are no fixed values of N and C such that for all n > N, f(n) < C*(f(n))², QED.
Saying that “f(n) ≥ 1” is not enough to allow proving the second claim; but if you write “f(n) ≥ 1 for all n” or “f() ≥ 1” it is provable. For example, if f(n) = 1/n for odd n and 1+n for even n, we have f(n) > 1 for even n > 0, and less than 1 for odd n. To prove that f(n) = O((f(n))²) is false, use the same proof as in the previous paragraph but with the additional provision that n is even.
Actually, “f(n) ≥ 1 for all n” is stronger than necessary to ensure f(n) = O((f(n))²). Let ε be any fixed positive value. No matter how small ε is, “f(n) ≥ ε for all n > N'” ensures f(n) = O((f(n))²). To prove this, take C = max(1, 1/ε) and N=N'.
(log n)^k = O(n)? For k greater or equal to 1.
My professor presented us with this statement in class, however I am not sure what it means for a function to a have a time complexity of O(n). Even stuff like n^2 = O(n^2), how can a function f(x) have a run time complexity?
As for the statement how does it equal O(n) rather than O((logn)^k)?
(log n)^k = O(n)?
Yes. The definition of big-Oh is that a function f is in O(g(n)) if there exist positive constants N and c, such that for all n > N: f(n) <= c*g(n). In this case f(n) is (log n)^k and g(n) is n, so if we insert that into the definition we get: "there exist constants N and c, such that for all n > N: (log n)^k <= c*n". This is true so (log n)^k is in O(n).
how can a function f(x) have a run time complexity
It doesn't. Nothing about big-Oh notation is specific to run-time complexity. Big-Oh is a notation to classify the growth of functions. Often the functions we're talking about measure the run-time of certain algorithms, but we can use big-Oh to talk about arbitrary functions.
f(x) = O(g(x)) means f(x) grows slower or comparably to g(x).
Technically this is interpreted as "We can find an x value, x_0, and a scale factor, M, such that this size of f(x) past x_0 is less than the scaled size of g(x)." Or in math:
|f(x)| < M |g(x)| for all x > x_0.
So for your question:
log(x)^k = O(x)? is asking : is there an x_0 and M such that
log(x)^k < M x for all x>x_0.
The existence of such M and x_0 can be done using various limit results and is relatively simple using L'Hopitals rule .. however it can be done without calculus.
The simplest proof I can come up with that doesn't rely on L'Hopitals rule uses the Taylor series
e^z = 1 + z + z^2/2 + ... = sum z^m / m!
Using z = (N! x)^(1/N) we can see that
e^(x^(1/N)) = 1 + (N! x)^(1/N) + (N! x)^(2/N)/2 + ... (N! x)^(N/N)/N! + ...
For x>0 all terms are positive so, keeping only the Nth term we get that
e^((N! x)^(1/N)) = N! x / N! + (...)
= x + (...)
> x for x > 0
Taking logarithms of both sides (since log is monotonic increasing), then raising to Nth power (also monotonic increasing since N>0)
(N! x)^(1/N) > log x for x > 0
N! x > (log x)^n for x > 0
Which is exactly the result we need, (log x)^N < M x for some M and all x > x_0, with M = N! and x_0=0