The Strange Case of Skewes' Number

[@iwersonsch5131]

So the Riemann hypothesis could be disproven by counting the primes up to 10^10^10^34 and finding that the formula consistently overestimates the number of primes? Not the other way around tho

[@iwersonsch5131]

Well, using only Skewes' results. We now know that the upper bounds for this crossover are lower, but is there still this discrepancy in upper bounds between Riemann's hypothesis being true or not?

[@angelmendez-rivera351]

@@iwersonsch5131 No, no such a discrepancy exists anymore. We have unconditional lower and upper bounds for where these crossovers must happen.

[@BigKingJohn]

Fascinating stuff. Thanks for this!

[@GeoffryGifari]

oh so prime number theorem over/under estimates the number of primes, flipping back and forth? does this mean there is a period? is it a *constant* period? maybe somehow its possible to map this to periodic functions we know. and right at the point between overestimation transitions to underestimation, does that mean we get the number of primes exactly?

[@angelmendez-rivera351]

*The prime number theorem over/under estimates the number of primes, flipping back and forth?* No. What the prime number theorem states is that lim π(n)/li(n) (n -> ∞) = 1. Actually, we have a stronger statement to make. We know that |π(n) - li(n)| =< 0.2795·n/ln(n)^(3/4)·exp(-sqrt(ln(n)/6.455)) is always true for all n >= 229. Thus, we know this is at least how well li approximates the prime counting function. However, this statement is not the prime number theorem, but is instead a corollary of the prime number theorem. As for this video, the claim made in the video is false. It has been proven that π(n) < n/ln(n) for all sufficiently large n. This is not an assumption. The video conflates n/ln(n) with li(n). *does this mean there is a period?* No. Most functions that are equal to 0 infinitely many times are not periodic. *is it a **_comstant_** period?* This is nonsense. By definition, the period of a function is a constant. Otherwise, the function is not periodic. *...does that mean we get the number of primes exactly?* No. The intermediate value theorem does not apply, since the prime counting function is a discrete function. The only way you get the exact number of primes at these crossings is if li(n) is an integer, and since li is a transcendental function, this is impossible.

[@GeoffryGifari]

@@angelmendez-rivera351 i see. one more thing, if i have an oscillating real function around zero whose zeros do not lie on regular intervals (like the bessel function), what do you call that? it is aperiodic, but no specific name? this is what i meant in one of the incorrect statements. the rest is clear

[@angelmendez-rivera351]

@@GeoffryGifari Yeah, such functions are simply called non-periodic functions. There is no special name for them, really.

[@GeoffryGifari]

@@angelmendez-rivera351 hmmm i believe i saw a notification for a reply from you in my other comment, and it's not there. i'm not imagining stuff... am I?

[@angelmendez-rivera351]

@@GeoffryGifari I replied to another comment of yours in a different video

[@angelmendez-rivera351]

3:00 - 3:07 No. This is false. It has been proven that π(n) < n/ln(n) is true for all sufficiently large n. This is not an assumption, it is a theorem. 3:08 - 3:24 Again, this is false. The problem here is that you are presenting the prime number theorem inaccurately. The prime number theorem is stated in terms of the logarithmic integral of function, li. We state the prime number theorem not π(n) ~ n/ln(n), but as π(n) ~ li(n). It is from this that we get π(n) < li(n) for sufficiently small values of n. This inequality was assumed to be true without proof, not π(n) < n/ln(n), which had been proven.

[@socraticmathtutor1869]

Thanks for clarifying. What university are you at?

[@GeorgeSmiley77]

As to the oft-asked question, "Is mathematics invented or discovered?" I say "Discovered, duh!" and the existence of prime numbers is my proof, because prime numbers are prime with or without intelligent beings discussing their nature. Not to mention that pi is irrational in all bases except base pi (and rational multiples of pi), in which it is written as "10" (not 1).

[@angelmendez-rivera351]

*...and the existence of prime numbers is my proof, because prime numbers are prome with or without intelligent beings discussing their nature.* No, this is false. Prime numbers are only prime because of the axioms and definitions that we have chosen. Had we defined primeness to represent a different concept, they may not be prime. If we choose the axioms differently, that also changes their primeness. For instance, the Gaussian integer 5 is not a prime number, since 5 = (2 + i)·(2 - i), and in the rational numbers, there are no prime numbers. The only reason there are prime numbers in the integers is because of how we choose to axiomatize the integers. If we decided that the integers are something else entirely, then the results would be different. Also, even if we have the same axioms, changing the underlying logic changes which theorems can be proven. This is why constructive mathematics lacks certain theorems: they cannot be proven, since it uses intuitionistic logic, rather than first order logic. *Not to mention that π is irrational in all bases except base π (and rational multiples of π), in which it is written as "10."* This is false. Whether a quantity is rational or not does not depend on how you represent it. What the representation looks like in different bases does not change the properties of what π is as a real number. π is a real number that is not a rational number. This is true in all systems of notation, and thus, in all bases, since π is well-defined. Perhaps what you _meant_ to say is that π cannot be represented using periodic or terminating strings of digits, unless the base is a rational multiple of π. However, this is still false. sqrt(π) is not a rational multiple of π. This is because sqrt(π) = 1/sqrt(π)·π, but 1/sqrt(π) is not a rational number. Yet, in base sqrt(π), π can be represented by the string "100." This is because π = sqrt(π)^2 = 1·sqrt(π)^2 + 0·sqrt(π)^1 + 0·sqrt(π)^0 := 100. In base cbrt(π), π can be represented as "1000." These are trivial examples, but here is a non-trivial example. Consider the principal root of the real polynomial x^7 + x^2 - π. Call said principal root u. Then, in base u, π is represented as 10000100. Yet u is not a rational multiple of π, nor can it be expressed in terms of π using radical symbols either. Anyway, the fact that π has the properties that it has does not mean mathematics are discovered. The truth is, the question itself is a bit stupid. "Discovery" and "invention" are two facets of the exact same thing: theory crafting and development.

[@GeorgeSmiley77]

@@angelmendez-rivera351 Sorry but you lost me at your 2nd sentence by quibbling with the definition of 'prime'. By doing that you effectively remove yourself from the topic by trying to change the topic. It doesn't help that in your last paragraph you suggest that invention and discovery are the same thing, but in doing so you disrespect all the great people who have wrestled with the same question. Dismissiveness invites dismissiveness. I consider the question of invented vs discovered to be difficult yet interesting, not merely trivial.

[@angelmendez-rivera351]

@@GeorgeSmiley77 *Sorry but you lost me at your 2nd sentence by quibbling with the definition of 'prime'.* I never quibbled with the definition of 'prime.' I demonstrated how, even if you keep the definition the same, changing the axioms changes whether something is an a prime number or not, and I provided examples, something you are not able to do, because you lack the intellectual honesty and capacity to do so. *By doing that you effectively remove yourself from the topic by trying to change the topic.* No. I am not removing myself from the topic. I am proving that your understanding of the topic is ill-informed. *It doesn't help that in your last paragraph you suggest that invention and discovery are the same thing, but in doing so you disrespect all the great people who have wrestled with the same question.* Saying that they are wrong is not disrespectful. All this proves is that you find correction to be unwelcome, and that you are not receptive to changing your mind. It demonstrates that you are an unreasonable individual without the capacity to learn and change your mind. *Dismissiveness invites dismissiveness. I consider the question of invented vs discovered to be difficult yet interesting, not merely trivial.* I never said the question is trivial, I am merely implying the question is being presented in a naive, unnuanced fashion, being treated as a black-or-white issue, where it is actually not. But you are right one thing: dismissivness does invite dismissiveness, and since you initiated dismissiveness by actively misunderstanding and misrepresenting my arguments and proving yourself to be dishonest and unwilling to listen, I can dismiss you as a troll. Now your unreasonableness is public for everyone to see, and for everyone to know that no one ever should take you seriously in a conversation.