- #1

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Does it mean that we can simply replace the O part with a function that is a constant times 1/(log N)? What would be the difference between [tex] A + O(\frac{1}{\log N})[/tex] and [tex]O(1)[/tex]?

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- Thread starter Dragonfall
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- #1

- 1,030

- 4

Does it mean that we can simply replace the O part with a function that is a constant times 1/(log N)? What would be the difference between [tex] A + O(\frac{1}{\log N})[/tex] and [tex]O(1)[/tex]?

- #2

shmoe

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Dragonfall said:Does it mean that we can simply replace the O part with a function that is a constant times 1/(log N)?

No it doesn't. If f(n)=g(n)+O(h(n)) then there is a constant C where |f(n)-g(n)|<=C*h(n) in some suitable range of n. It does NOT mean f(n)=g(n)+C*h(n). Consider cos(x)=O(1) but we don't have cos(x)=constant.

Dragonfall said:What would be the difference between [tex] A + O(\frac{1}{\log N})[/tex] and [tex]O(1)[/tex]?

The first gives more information (it implies the second but not vice versa). Even if you don't know the constant A (it can be expressed in terms of an infinite sum over the primes here though) it still says something about the structure of the lower order terms.

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