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Regarding a proof that Merten's function = Big O of k^(.5 + E )...

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  • Jeff Cook
    All, I have discovered a function for calculating E in O(k^(.5+E)) without fail. Putting the results of this function into the equation gives exactly Merten s
    Message 1 of 1 , Jan 3, 2006
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      All,

      I have discovered a function for calculating E in O(k^(.5+E)) without fail. Putting the results of this function into the equation gives exactly Merten's function M(k). In his book, "Prime Obsession," Derbyshire explains that if M(k) = O(k^(.5+E)) can be proven, if would follow that the Riemann Hypothesis is true. "They are exactly equivalent theorems."

      I will be writing a paper on this function, as it is more intricate that can or should be discussed alone through this forum. However, I could use some help in learning what mathematicians would require to constitute that they are equal. What type of proof is sufficient? Simply providing the function? My limited understanding of big O notation could be a problem for me in explaining this function, however. Actually, what I really have is a function that equals M(k) exactly, using k^(.5+E). As I know that E involves a function of its own, that can be precisely defined. How E is defined as "any number no matter how small" is a bit vague for me. Could someone explain?

      Thanks,

      Jeff





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