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polymerize-finale

polymerize-finale


quote

 

Quote

Also, if we are being pedantic, O(50M) != O(1). Both mean that the function in question is bound by a constant, but which constant is very much a big deal 

 

Oh for christ's sake....

 

We say f(x) = O(g(x)) if there exists constants C and x0 such that |f(x)| <= C*g(x) for all x > x0.

 

Let f(x) = O(M) where M is a constant and M > 1. Then |f(x)| <= C*M for x > x0. Then f(x) = O(1) because there exists a constant D such that |f(x)| <= D for all x > x0, namely C*M.

 

Let f(x) = O(1). Then |f(x)| <= C for all x > x0. If M > 1, C < C*M, so |f(x)| <= C*M for all x > x0, so f(x) = O(M).

 

Therefore, f(x) = O(M) if and only if f(x) = O(1).

 

Now that I've supplied a LITERAL FUCKING MATHEMATICAL PROOF of this, can we please get back to discussing linux gaming?

polymerize-finale

polymerize-finale

[quote]

Also, if we are being pedantic, O(50M) != O(1). Both mean that the function in question is bound by a constant, but which constant is very much a big deal 

[/quote]

 

Oh for christ's sake....

 

We say f(x) = O(g(x)) if there exists constants C and x0 such that |f(x)| <= C*g(x) for all x > x0.

 

Let f(x) = O(M) where M is a constant and M > 1. Then |f(x)| <= C*M for x > x0. Then f(x) = O(1) because there exists a constant D such that |f(x)| <= D for all x > x0, namely C*M.

 

Let f(x) = O(1). Then |f(x)| <= C for all x > x0. If M > 1, C < C*M, so |f(x)| <= C*M for all x > x0, so f(x) = O(M).

 

Therefore, f(x) = O(M) if and only if f(x) = O(1).

 

Now that I've supplied a LITERAL FUCKING MATHEMATICAL PROOF of this, can we please get back to discussing linux gaming?

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