joigus

Yes, please, keep me informed. I'm glad to have you here, proposing interesting problems. Welcome to the forums!

OK. I see what you mean much more clearly now. There's still something you've understood that I haven't. Problem is I don't have much time now, but I promise to get back. This was gonna be my drawing: There's an idea that maybe could work for the general case. I don't know. I'll get back to you ASAP. It's to do with the green and red arrows in the drawing. The co-tangential condition can be written as, \[\boldsymbol{\nabla}f_{1}\wedge\boldsymbol{\nabla}f_{2}=0\] It involves the determinant, \[\left|\begin{array}{cc} \frac{\partial f_{1}}{\partial x} & \frac{\partial f_{1}}{\partial y}\\ \frac{\partial f_{2}}{\partial x} & \frac{\partial f_{2}}{\partial y} \end{array}\right|\] It could work in the general case. I don't know. The problem is \( k \) doesn't appear there. Let's keep talking. No time now.

Mmmm. I think I know what you mean now. When you said you obtained the solution you were looking for, did you mean (in the 1st, simpler, example)?: \[ x = 5 \left( 1 ± \sqrt{1- \frac{3}{125} k} \right) \] \[ y = 6 \left( 1 ± \sqrt{ 1- \frac{1}{45} k} \right) \] \[ K ≤ \frac{125}{3} \] Can you show me in more detail how you got to that by your method of adding a third equation? The equation you propose is, \[\left(-2x+10\right)-\left(-4x+20\right)=0\] That gives \( x = 5 \) as only solution, and I don't see how that relates to what you're looking for. Your 'tangency' idea got me confused, especially in combination with the Jacobian. Maybe I can't help you after all, but I think I'm closer to understand you and hopefully offer you some insight.

Here you have some help: The Jacobian is a matrix, not a simple derivative. I still don't understand why you want to apply derivatives when your equation can be solved exactly. Neither do I understand how an equation derived from the previous one gives you more information. I'm still missing it. I'm sorry. Maybe somebody else can help.

Can you be a bit more explicit about what you're doing there? Are those partial derivatives, implicit (as in \( \frac{d}{dx}=\frac{\partial}{\partial x}+\frac{dy}{dx}\frac{\partial}{\partial y} \). Do they involve K as a function of x and y, or as an independent variable? There are several ways to take the derivative with respect to x, and some of them are sorely wrong. E.g., are you considering, \[x,\:y\left(x\right),\:K\left(x,y\right)\] or, \[K,\:y\left(x,K\right),\:x\left(K\right)\] perhaps? And why would you want to involve a further equation which is implied by the previous two and contains, if anything, less information? I'm missing something in your argument. By the way, I made a mistake. It's, \[K\leq\frac{125}{3}\]

Because, as I told you, the non-linear system, \[− x^2 +10x − \frac{1}{2} y^2 +6y − K=0 \] \[− 2 x^2 +20x+ \frac{3}{2} y^2 − 18y=0\] decouples into, \[5y^{2}\text{−}60y+4K=0\] \[5x^{2}-50x+3K=0\] Which is trivial: \[x=\frac{50\pm\sqrt{2500-60K}}{10}\] \[y=\frac{60\pm\sqrt{3600-80K}}{10}\] Only solvable under the bounds, \[K<\frac{250}{6}=\frac{125}{3}=41.\overline{6}\] \[K<\frac{360}{8}=45\] The lowest upper bound being, \[K<\frac{125}{3}\]

No general rule in non-linear equations. This system is particularly simple because it's a 'scrambled' set of quadratic equations in x and y. You can un-scramble it easily, find the conditions for K, and overlap.

I am. For starters

How significant is the Arizona flip?

I don't know what the status of the question is, but this is exactly the part I didn't understand at all. It seems like Tina is reconsidering and I on my part think everybody is being considerate and trying to understand the problem. Not at all, @wtf. I think you understand it better than anybody else here. At least than me. I applaud your suggestion to start with simpler numbers.

I wouldn't be too surprised if there were something to those claims. During the Cold War there were similar episodes in science. Thank you.

1 is not a prime and those are not the first 23 primes. Don't trust me with numbers. The first 9 primes, I should have said. Thank you @studiot.

I was a bit surprised that such a simple formula would hold for a problem like this. Prime numbers are quite unpredictable. Prime gaps for example cannot be predicted by a general formula. So a formula giving all possible sum decompositions of products of primes... I didn't see the flaw in the argument, because I didn't see the argument. But again, number theory is not within my comfort zone.

I see what you mean. But I think we basically agree about that: So the decomposition of 223'092'870 into a pair of numbers x+y with x, y, not necessarily being prime, but being relative primes of 2, 3, 5, ..., 23, I think is the defining condition. I'm way past my comfort zone here. Fortunately @Sensei did it for all of us, and even though I'm not a great code reader, I think it's what the OP was asking.

You guys are very valuable members of this community. I've been swept away by both of you several times. Yes, you're right, @studiot. But the OP was about the very special, very non-prime number \( 2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17\cdot19\cdot23 \), factor of the first 9 prime numbers.

Why wouldn't rational and irrational numbers be summable?

Sorry, the problem reduces to how many numbers there are between 1 and 223'092'869 that are not divided by 1, 2, 3, 5, ..., 23.

Now that I think about it, the product of the first 23 is, as you say, 223092870. Now, the number of ways in which you can sum two (arbitrary) natural numbers to give N is just 111'546'435, because 223'092'870 is even. So the problem reduces to finding how many numbers there are between 1 and 111'546'435 that are not divided by 1, 2, 3, 5, ..., 23 (the 1st 23 primes). The only thing I can say is that's not an elementary problem. How did you get your conjecture @Tinacity? Counterexample: \( \pi \) is irrational; \( 1-\pi \) is irrational too. but \( \pi + 1- \pi = 1 \).

She didn't specify, but I thought it was kind of implied... For arbitrary products of primes, certainly nothing like that can be proved with the present mathematics. But for 223092870, I just don't know.

The maths for predicting that kind of thing is called number theory. I know very little about number theory. It studies connections between numbers. The result that you propose reminds me of some theorems by Fermat. Have you proved it, or is it just an intuition? Maybe @wtf or @Sensei, or @mathematic, or @taeto can help you. @studiot is encyclopedic. Maybe he can help you too. Number theory is not very interesting for physics, AFAIK. And physics and mathematical physics are my turf.

It's a mark-up language; a subset of XML: https://www.w3.org/Math/whatIsMathML.html Sorry what I posted was inline LateX, but you can read it as MathML by right-clicking.

This could be hard proof of tunneling for macroscopic objects...

Thank you, Zapatos. I've found a little bit more of the geological history on Wikipedia: https://en.wikipedia.org/wiki/Kegon_Falls From https://www.visitnikko.jp/en/spots/kegon-falls/ The place is such a tourist magnet that it's kinda difficult to find something more science-oriented about it: Geology, fauna and flora, etc.

Another real-estate agency's hopes have been shattered, as we speak.