Brett Nortj 52 Posted August 3 This would be [infinity] * [2n] = [3] * [2] = [6] * [x] Then, [5] * [2] = [10] * [x] Then, [7] * [2] = [14] * [x] So, the answer is double prime times by [x]. Or, [N] * [2] *[X]. 0 Share this post Link to post Share on other sites

dimreepr 855 Posted August 3 1 hour ago, Brett Nortj said: This would be [infinity] * [2n] = [3] * [2] = [6] * [x] Then, [5] * [2] = [10] * [x] Then, [7] * [2] = [14] * [x] So, the answer is double prime times by [x]. Or, [N] * [2] *[X]. OK, I'll bite what is x? 0 Share this post Link to post Share on other sites

Brett Nortj 52 Posted August 3 Just now, dimreepr said: OK, I'll bite what is x? [X] is just an expression for the symbols combined, a new symbol - no rule says we cannot add or use our own symbols. 0 Share this post Link to post Share on other sites

dimreepr 855 Posted August 3 Trivial or non-trivial? 0 Share this post Link to post Share on other sites

Brett Nortj 52 Posted August 3 2 hours ago, dimreepr said: Trivial or non-trivial? Why don't we walk on the wild side? Let's make it a process of elimination? 0 Share this post Link to post Share on other sites

dimreepr 855 Posted August 3 8 minutes ago, Brett Nortj said: Why don't we walk on the wild side? Let's make it a process of elimination? Why not, which version of zero do you prefer? 0 Share this post Link to post Share on other sites

Brett Nortj 52 Posted August 3 3 minutes ago, dimreepr said: Why not, which version of zero do you prefer? I think [X] would be the symbol of the log, of course... 0 Share this post Link to post Share on other sites

dimreepr 855 Posted August 3 3 minutes ago, Brett Nortj said: I think [X] would be the symbol of the log, of course... The wrong version, never mind please try again... 0 Share this post Link to post Share on other sites

Brett Nortj 52 Posted August 3 6 minutes ago, dimreepr said: The wrong version, never mind please try again... [X] would be the indent on the log? Look, I have always struggled with logarithms, but I know I have something going here. Why don't you take a shot? Nothing to lose. 0 Share this post Link to post Share on other sites

wtf 99 Posted August 4 23 hours ago, Brett Nortj said: [X] would be the indent on the log? Look, I have always struggled with logarithms, but I know I have something going here. Why don't you take a shot? Nothing to lose. You struggle with logarithms but want to solve the Riemann hypothesis? 0 Share this post Link to post Share on other sites

taeto 37 Posted August 4 20 minutes ago, wtf said: You struggle with logarithms but want to solve the Riemann hypothesis? He said he wants to talk about it, not necessarily solve it, wtf. 0 Share this post Link to post Share on other sites

dimreepr 855 Posted August 4 2 minutes ago, taeto said: He said he wants to talk about it, not necessarily solve it, wtf. X does neither. 0 Share this post Link to post Share on other sites

taeto 37 Posted August 4 (edited) 24 minutes ago, dimreepr said: X does neither. It depends on the X. Maybe it was a very brilliant X. I admit I did not bother to read far enough into the OP to discover what it means. I trust that you investigated and determined that the X is actually kind of stupid. Note taken. Maybe he just has some questions about logarithms or RH. I will attempt to elucidate. The "natural logarithm" function \( \ln, \) or nowadays mostly written \( \log,\) can be conveniently defined by \[ \ln(x) := \int_1^x \frac{1}{t} dt \] for any positive real number \(x.\) Then \( \ln(1) = 0\) is obvious by definition, and using standard substitution of variables in the Riemann integral, you can easily deduce basic properties, such as \( \ln(ab) = \ln(a) + \ln(b) \) for all \( a,b > 0, \) as well as \( \ln(a^b) = b \ln(a), \) for all \(a > 0\) and all \(b.\) To formulate the Riemann Hypothesis in terms of logarithms it also helps to know about primes. I will wait for the OP to indicate whether this is the intention before I go on to explain about primes. Edited August 4 by taeto 0 Share this post Link to post Share on other sites

Brett Nortj 52 Posted August 4 42 minutes ago, taeto said: It depends on the X. Maybe it was a very brilliant X. I admit I did not bother to read far enough into the OP to discover what it means. I trust that you investigated and determined that the X is actually kind of stupid. Note taken. Maybe he just has some questions about logarithms or RH. I will attempt to elucidate. The "natural logarithm" function ln, or nowadays mostly written log, can be conveniently defined by ln(x):=∫x11tdt for any positive real number x. Then ln(1)=0 is obvious by definition, and using standard substitution of variables in the Riemann integral, you can easily deduce basic properties, such as ln(ab)=ln(a)+ln(b) for all a,b>0, as well as ln(ab)=bln(a), for all a>0 and all b. To formulate the Riemann Hypothesis in terms of logarithms it also helps to know about primes. I will wait for the OP to indicate whether this is the intention before I go on to explain about primes. Okay, seems to be right to me. 0 Share this post Link to post Share on other sites

taeto 37 Posted August 4 (edited) Okay then: primes. Formally a natural number \(p\) is called "prime" if \(p > 1\) and it is true that if you take any integer numbers \(a\) and \(b,\) and \(p\) divides the product \(ab\), then \(p\) divides \(a\) or \(b.\) The first primes are \(2,3,5,7,11,13,\) and then there are some more larger ones as well. The "primorial" \(p\#\) of a prime \(p\) is the product \(2\times 3 \times 5 \times 7 \times 11 \times \cdots \times p\) of all the primes \(2,3,5,7,11,...,p.\) The "Euler totient function" \(\varphi\) is a more complicated function, but all you have to worry about is its value at primorial \(p\#,\) which is simply the product \( \varphi(p\#) = 1 \times 2 \times 4 \times 6 \times 10 \times \cdots \times (p-1)\) of all numbers less by 1 than a prime up to \(p.\) This is enough that you have to know, before you can meaningfully consider the Riemann Hypothesis. The Riemann Hypothesis is logically equivalent to the following statement: \[ \frac{p\#}{\varphi(p\#) \ln \ln p\#} > e^\gamma, \] for all primes \(p,\) where \(\gamma\) is known as the Euler-Mascheroni constant, with \[ \gamma := \int_1^\infty (\frac{1}{\lfloor x \rfloor} - \frac{1}{x}) dx \approx 0.577\ldots, \] where the value of the floor function \( \lfloor x \rfloor \) is the largest integer that is not greater than \(x.\) Reference: J. Nicolas, Petites valeurs de la fonction d’Euler, J. Number Theory 17, 3, 1983, pp. 375–388. Edited August 4 by taeto 0 Share this post Link to post Share on other sites

Brett Nortj 52 Posted August 4 38 minutes ago, taeto said: Okay then: primes. Formally a natural number p is called "prime" if p>1 and it is true that if you take any integer numbers a and b, and p divides the product ab , then p divides a or b. The first primes are 2,3,5,7,11,13, and then there are some more larger ones as well. The "primorial" p# of a prime p is the product 2×3×5×7×11×⋯×p of all the primes 2,3,5,7,11,...,p. The "Euler totient function" φ is a more complicated function, but all you have to worry about is its value at primorial p#, which is simply the product φ(p#)=1×2×4×6×10×⋯×(p−1) of all numbers less by 1 than a prime up to p. This is enough that you have to know, before you can meaningfully consider the Riemann Hypothesis. The Riemann Hypothesis is logically equivalent to the following statement:p#φ(p#)lnlnp#>eγ, for all primes p, where γ is known as the Euler-Mascheroni constant, with γ:=∫∞1(1⌊x⌋−1x)dx≈0.577…, where the value of the floor function ⌊x⌋ is the largest integer that is not greater than x. Reference: J. Nicolas, Petites valeurs de la fonction d’Euler, J. Number Theory 17, 3, 1983, pp. 375–388. Well, it was I that wrote the script on primes, namely; [n] = {[n] + [n] +1}. 0 Share this post Link to post Share on other sites

taeto 37 Posted August 4 1 minute ago, Brett Nortj said: Well, it was I that wrote the script on primes, namely; [n] = {[n] + [n] +1}. I could not figure out what \( [n] \) means. So maybe you are way ahead of me. 0 Share this post Link to post Share on other sites

John Cuthber 3488 Posted August 4 On 8/3/2018 at 12:24 PM, Brett Nortj said: This would be [infinity] * [2n] = [3] * [2] = [6] * [x] Then, [5] * [2] = [10] * [x] Then, [7] * [2] = [14] * [x] So, the answer is double prime times by [x]. Or, [N] * [2] *[X]. I prefer this version There was a young man from Nepal Who had one triangular ball The square of its weight Times his pecker, times eight Is his number, so give him a call 1 Share this post Link to post Share on other sites

studiot 1478 Posted August 4 (edited) 3 minutes ago, John Cuthber said: I prefer this version You obviously played for a different rugby club from me. Edited August 4 by studiot 0 Share this post Link to post Share on other sites

taeto 37 Posted August 4 27 minutes ago, John Cuthber said: I prefer this version There was a young man from Nepal Who had one triangular ball The square of its weight Times his pecker, times eight Is his number, so give him a call Definite +1 for an on-topic Limerick, good work! 0 Share this post Link to post Share on other sites