  # wtf

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• ### Carrock

1. Today I learned! Another way is math]\displaystyle e^z = \sum_{n=0}^\infty \frac{z^n}{n!}[/math $\displaystyle e^z = \sum_{n=0}^\infty \frac{z^n}{n!}$
2. Can you help me to understand your point? For example if I write 3 + 4 x 5 = ?, everyone agrees that the anseer is 23; because by convention we multiply first and then add. But that's just a convention. If we all agreed to add first then multiply, the answer would be 35. It's a purely arbitrary convention that could easily be different, as long as everyone agrees on what the convention is. As you know, the Internet is full of order-of-operations puzzles that are the result of poor math education regarding operator precedence. If it makes you happy, 1 isn't a prime because that's the convention. We could make a different convention, regard 1 as prime, and adjust all the theeorems accordingly. It really makes no difference. So what is your core issue or concern? It's just a convention. The reason for the convention can be motivated by higher algebra, but in the end it's still a convention and truly makes no substantive difference. It's similar to why we go on green lights and stop on red. At the dawn of the automobile age we could have adopted the opposite convention. It makes no difference as long as everyone agrees.
4. A few weeks ago I had the problem space mapped into my brain for an evening. Too busy to remap it at the moment, so probably won't be able to get back to this. Off the top of my head I don't think the totient function will always give the same answer as inclusion/exclusion, but I could be wrong and probably am. Sorry I can't offer more assistance.
5. @Tinacity, ps -- Wait DUH! We forgot 11/49. I don't know why we both got confused about this. I solved the problem. The trick is that if n is divisible by one of 2, 3, or 5, so is 60 - n. So the pairs (n, 60-n) where both elements are relatively prime to 2, 3, and 5 are exactly the same as the numbers n with the same property. So the solution is to do inclusion/exclusion on 30 to determine how many numbers are not divisible by 2, 3, or 5; and that's the number of pairs. In the case of 60 there are exactly 8 pairs: 1/59, 7/53, 11/49, 13/47, 17/43, 19/41, 23/37, and 29/31. That's eight. You can now write a program to do inclusion/exclusion on your original number N, or half of 2N if you think of it that way (that is, 2x3x5 = 30, multiply by 2 to get 60, then do inclusion/exclusion on 30). The "sum to 60" is a red herring, an aspect of the problem that adds confusion but doesn't change the problem. The number of pairs that sum to 60 where each element of the pair is not divisible by 2, 3, or 5, is exactly equal to the number of numbers between 1 and 30 not divisible by 2, 3, or 5. And this result generalizes under the conditions of your problem. To do inclusion/exclusion on N = 2*3*5*7*11*13*17*19*23 you take: - The sum of all 8-fold products of N (that is, every combination of 8 factors at a time); - Minus the sum of all 7-fold products; - Plus the sum of all 6-fold products; etc. You then subtract the final sum from N, and that's the number of pairs where both elements are relatively prime. I believe that's it, but if I messed up I hope someone will jump in.
7. If order matters then you have 14 + 16 and 16 + 14 and likewise for all the other pairs except for 15 + 15, which is it's own reverse. Do you mean order doesn't count? Rather than saying it does? What happened to 1/59? I had a couple of mistakes in my own list, I had 11/49 which shouldn't be there, and I forgot 7/53. So there are 7 such pairs, 14 if order matters. I assume by your example that order DOESN'T matter. Still we have 7 pairs 1/59, 7/53, 13/47, 17/43, 19/41, 23/37, and 29/31. That's seven. I think we have them all. The question comes down to taking 60 and asking, out of the first 30 positive integers, how many are divisible by at least one of 2, 3, or 5. Half are divisible by 2, 1/3 are divisible by 3 but we counted the ones divisible by 6 twice; and 1/5 are divisible by 5 but we counted the 10's and the 15's twice. But then we subtracted the ones divisible by 30 once too much so we have to add it back in. I believe you attack this kind of problem with the inclusion/exclusion formula. In fact the first example here shows how to count how many numbers from 1 to 100 are divisible by at least one of 2, 3, or 5, our exact problem here. https://en.wikipedia.org/wiki/Inclusion–exclusion_principle I jotted down some numbers but I was off-by-one somewhere, which doesn't surprise me.
8. Can you show your work in detail? I still don't understand the basic question. 2*3*5 = 30. Multiply that by 2 and you get 60. Please explain the rest because I totally do not understand what we are doing here. Where did the -2's come from, that hasn't been part of your exposition. ps -- Ok I totally don't get this. Pairs that sum to 60 and have no divisors among 2,, 3, 5: I get 1,59 11,49 13,47 17, 43 That's already four, eight if you distinguish order, and there are plenty more. So please explain clearly what you are doing. Others are 19, 41 23, 37 29, 31 That's a total of 7, times 2 to account for order as you said earlier, so there are 14 pairs that satisfy your requirement, not 3. Where do these -2's come from? In one example earlier you had 17-2 as a factor but that's not prime. Maybe it's just me but I do not understand what is being calculated. Can you work out a complete example, a simple one? Apologies if I'm being dense and this is obvious to everyone but me. Those of you who wrote programs to solves this problem, what problem are you solving? Am I just missing something that's obvious to everyone else?
9. Have you tried this out by hand for a simpler case, say N = 2 x 3 x 5 = 30? Then dropping the first and last gives you 3. Can you make your idea come out with that example? Is there something special about the case you're presenting?
10. Haven't followed the thread but this seems a little ambiguous. Let's take a simpler example, N = 2 x 3 x 5 x 7 = 210. Now you want to consider the pairs of numbers that sum to 210, such as (1, 209), (2,208), ..., (209, 1). Do you care about order? Is (1,209) the same as (209,1) or different? Not a big issue, just a factor of 2, but nice to know what is the intended intepretation. Now "... where neither is divisible by any of the primes which make its product?" was confusing to me. What is "its" in this context? Do you mean that since 209 = 11x19, and neither 11 nor 19 is one of 2, 3, 5, or 7, we count (1,209) and (209,1) as satisfying your condition? And you want to count the number of such pairs? Just want to make sure I'm understanding the question. Apologies to all if this has already been covered in the thread.
11. The original interpretation is ordinal exponentiation. Here's an overview. If we "keep counting" after 0, 1, 2, 3, ... we can tack on an element after all those, called $\omega$. So we have the ordered set 0, 1, 2, 3, 4, ..., $\omega$. Now we keep on going: $\omega + 1, \omega + 2, ...$ and when we're done with all those we've reached $\omega + \omega$ or $\omega 2$. Ordinal multiplication is notated backwards, $\omega 2$ means 2 copies of $\omega$, one after another, and NOT $\omega$ copies of 2, which would actually be equal to $\omega$. [Exercise for you: figure out why $2 \omega = \omega$. You have to learn about ordinals. This is not an easy exercise until you have worked through the elementary material on what ordinals are]. This example shows that ordinal addition is not commutative. As a historical note, I tracked this backward notation down once. It turns out that Cantor himself went back and forth on whether to call $\omega + \omega$ $\omega 2$ or $2 \omega$. In the end for whatever reason he went with the notation that I consider backwards. But like everything else in math, you just get used to it. So now we have $\omega 2, \omega 3, \omega 4, ...$, and at the end of that we tack on $\omega \omega$ or $\omega^2$. Then we have $\omega^3, \omega^4$, and we keep on going till we get $\omega^\omega$. If we keep on going we get $\omega^{\omega^\omega}, \omega^{\omega^{\omega^\omega}}$, and so forth. If we keep this process going, at the end we tack on a countably infinite power tower of $\omega$'s. This infinite power tower is given the name $\epsilon_0$, and it is the smallest ordinal $\epsilon$ such that $\omega^\epsilon = \epsilon$. Exercise: Can you see why this equation is true of $\epsilon_0$? This "tacking on at the end" idea can be formalized by letting the limit ordinal, as it's called, simply be the set of everything that's come before. So if the natural numbers are 0, 1, 2, 3, ..., then we define $\omega = \{0, 1, 2, 3, \dots \}$. These are the von Neumann ordinals. https://en.wikipedia.org/wiki/Ordinal_number#Von_Neumann_definition_of_ordinals. By the way I recommend that you read this entire Wiki page to begin to learn about the ordinal numbers. It's important, and mind boggling, to realize that $\epsilon_0$ is a countable ordinal; that is, it's a countable set. The countable ordinals are very strange and hard to get one's mind around. It's an amazing theorem that every countable ordinal can be embedded in some subset of the rationals in their usual order. This is the subject of the theorem you are interested in. This is what it's about, not something else. You can be interested in something else if you like, but if you are interested in this particular idea, this is what it's about. https://en.wikipedia.org/wiki/Epsilon_numbers_(mathematics) Interestingly in physics, the notation $\epsilon_0$ refers to something called the vacuum permitivity, but that has nothing to do with what we're talking about here. I don't expect you to understand all this right now. It's taken me a long time to get even a little bit of understanding of $\epsilon_0$. But if you want to learn this material, you have to make an attempt to learn it, starting with understanding what ordinal numbers are. Otherwise you are having a vacuous conversation with yourself.
12. It's ordinal exponentiation in this case. You need to start with learning about ordinal numbers and then ordinal arithmetic, working up to exponentiation.
13. Best way to start would be to take the time to understand relativization then.
14. This refers to ordinal exponentiation, it's not relativization. https://en.wikipedia.org/wiki/Ordinal_arithmetic#Exponentiation
15. 1 x 1 = 1. That's the multiplication table for the nonzero elements. What's the problem?
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