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Big-Oh Proof

Posted on 2004-09-28
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Last Modified: 2012-06-21
I'm trying to prove, using the definition of big-Oh, that if we have two functions p(n) and q(n), and p(n) is O( q(n) ), then:

[a x p(n)] - [b x q(n)] is O( q(n) )     where a and b are constants.

Can anyone provide any ideas as to how to tackle this?
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Question by:jshantz4
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ozo earned 500 total points
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p(n) <= C*q(n) for n > n0
[a x p(n)] - [b x q(n)] <= (a*C-b)*q(n) for n>n0
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by:Xxavier
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This is obviously homework.
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by:jshantz4
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Which is why I asked for ideas on how to tackle it -- *not* for someone to solve it for me.
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by:Xxavier
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....which is what ozo has done, ie solve it.
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by:NovaDenizen
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Start with the definition of big-Oh equivalence.   f(x) is big-Oh equivalent to g(x) if-and-only-if the limit of f(x)/g(x) as x approaches +infinity is a nonzero constant.  

The standard notation for big-Oh, O(f(x)) = g(x), is kind of confusing because it makes the big-Oh look like a function when the whole concept of big-Oh really is to define equivalence classes for functions.
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by:NovaDenizen
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I think Ozo didn't solve it right, anyway.  I think he solved something for lower bounds (big-Omega?), not for big-Oh.  Big-oh is about asymptotic proportional equality, not <= comparisons.
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by:NovaDenizen
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Ok, sorry, I'm wrong.  According to Knuth vol 1 p. 107, big-Oh is defined as f(x)<=M*g(x) for all x > x0 for some M and x0 iff f(x) == O(g(x)).  So x^2 == O(x^3), but x^3 is not O(x^2).  Ozo had it right.
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by:jshantz4
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Xavier: Can I control what someone posts on my question?  Besides, I'd hardly call that a solution.  The solution -- which I eventually found on my own -- was several more lines long.  However, since no one else provided anything else, I'm going to have to accept it.
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