If f(a)>g(a), then h(x)=f(x)-g(x), otherwise, h(x)=g(x)-f(x). We can see that it is true that h(a)<0 and h(b)>0. Then, we can apply the theorem given here.
]]>Title please.
]]>Yes, I saw that. Why are we doing that limit?
]]>bobbym wrote:Slower than 1 / n that is for sure.
I'm thinking it's O(log log n). Can you do the same limit, but with an 1/n in front of the sum.
Are you sure that you have the right question?
]]>Slower than 1 / n that is for sure.
I'm thinking it's O(log log n). Can you do the same limit, but with an 1/n in front of the sum.
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