It must suck for people like this guy to have to live in a world with people like, well, me, who probably think he is just talking nonsense. I understand from your comments that he is brilliant. But it's truly amazing to go through life as a pretty bright guy and yet be so bewildered by another human's intellect.
Additionally, a certain assertion on the Mertens function's rate of growth is exactly equivalent to the Riemann Hypothesis.
Finally, one of my own observations (which is easily derived from Ramanujan's sum for the Mobius function) is that the Mertens function is exactly equal to the sum of the cosine function at points corresponding to fractions of one wavelength that are in the Farey sequence of order n.
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GelandnaleG 1 year ago
Pretty fascinating stuff. You've impressed me. However: why isn't the sum of the Möbius function 0?
R1ckr011 1 year ago
ah apologies. I didn't watch the video pertaining to Merten's Function.
R1ckr011 1 year ago
It must suck for people like this guy to have to live in a world with people like, well, me, who probably think he is just talking nonsense. I understand from your comments that he is brilliant. But it's truly amazing to go through life as a pretty bright guy and yet be so bewildered by another human's intellect.
Congratulations to you all - respect.
Blackmaildesigns 1 year ago
@Blackmaildesigns at least you are honest with yourself. i don't understand any of this.
TheSonjaxfactor 1 year ago
wow . . . what? O_O
zwartkatdre 2 years ago
Additionally, a certain assertion on the Mertens function's rate of growth is exactly equivalent to the Riemann Hypothesis.
Finally, one of my own observations (which is easily derived from Ramanujan's sum for the Mobius function) is that the Mertens function is exactly equal to the sum of the cosine function at points corresponding to fractions of one wavelength that are in the Farey sequence of order n.
Kenyai 3 years ago
This is a great video. Just in case anyone was wondering, the Mobius function mu(n) is defined in this way:
It is equal to 0 if n is divisible by a perfect square greater than 1.
It is equal to 1 if n is the product of an even number of distinct prime numbers.
(Note that mu(1) = 1)
It is equal to -1 if n is the product of an odd number of distinct prime numbers.
Kenyai 3 years ago
this is very interesting, i have a question still, if you use the absolute values of the function, shouldn't the graphic be only above 0?
gniziemazity 4 years ago
cause he's not graphing the absolute value, so if abs(f)<1 then -1<f<1 .
alphachapmtl 4 years ago
i get it now thanks
gniziemazity 4 years ago