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Prove/disprove these two claims

  • Thread starter Melodia
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
1,631
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[tex]\mbox{ If } f(n)= O(g(n)), \mbox{ then it means that the function f(n) is of order at most g(n)}.[/tex]

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

[tex]|f(n)|\leq C|g(n)|[/tex]

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.
 
  • #3
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Ok thanks for clearing it up.
So basically it's saying:
If f(n) + g(n) [tex]\leq[/tex] c[h(n)]
Then f(n) [tex]\leq[/tex] c[h(n)] and g(n) [tex]\leq[/tex] c[h(n)]
The tricky part is that f(n) could be negative right? Since it says "[tex]\forall[/tex] f". Then f(n) could be larger than (f + g)(n)? Or does "[tex]\forall[/tex] f, g [tex]\in[/tex] F" mean "for all f and g that are in F"?
 
Last edited:
  • #4
1,631
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Look at the definition of f:N-->R+. Yes ti means for all f, g that are in F.
 
  • #5
18
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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.
 

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