Prove/disprove these two claims

  • Thread starter Thread starter Melodia
  • Start date Start date
Melodia
Messages
18
Reaction score
0

Homework Statement


I'm having trouble understanding this pair of claims:
cDMJw.jpg



Homework Equations





The Attempt at a Solution



So g is some function whose result is greater or equal to 0... but I don't really get what the rest of the claim means.
Could someone please explain them to me?
Thanks in advance.
 
Physics news on Phys.org
\mbox{ If } f(n)= O(g(n)), \mbox{ then it means that the function f(n) is of order at most g(n)}.

\mbox{ That is, by def., f(n) is big oh of g(n) if ther exists some positive constant C such that}

|f(n)|\leq C|g(n)|

Considering this, then what the problem is asking is that if f+g is of order at most h, then each of f and g must be of order at most h.

I haven't worked any of these problems out myself, but a proof by contradiction looks promising.
 
Ok thanks for clearing it up.
So basically it's saying:
If f(n) + g(n) \leq c[h(n)]
Then f(n) \leq c[h(n)] and g(n) \leq c[h(n)]
The tricky part is that f(n) could be negative right? Since it says "\forall f". Then f(n) could be larger than (f + g)(n)? Or does "\forall f, g \in F" mean "for all f and g that are in F"?
 
Last edited:
Look at the definition of f:N-->R+. Yes ti means for all f, g that are in F.
 
Oh ok, then the answer would be obvious right? I mean if both f and g has to be positive, then either f or g has to be smaller than f + g; then they have to be in O(h) as well.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top