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August 30

Graph question

What is the function for a graph that starts at x=a, y=k, which begins to rise slowly then much more steeply until nearly perpendicular to the x axis, then cuts off at x', and resumes in a mirror image at x'', descending until it reaches y=k again? (Apologies for using non-standard notation, I'm afraid I don't know any better. (The shape I'm envisioning is sort of like the elevation of a cooling tower at a power plant.) I suppose it's a hyperbola, but I looked at that article and it's too much information for me to wade through.--82.113.121.52 (talk) 18:39, 30 August 2010 (UTC)[reply]

Just cutting off

y=x^{-2}

at suitable

\pm a,\pm b

(that is, defining it only when

|x|\in [a,b]

) would do something like that — choose a and b to get the right amount of "slowly/nearly vertical" and the right amount of separation between the two branches. --Tardis (talk) 21:43, 30 August 2010 (UTC)[reply]

Taking

a={\frac {1}{2}},b=2

seems to be a nice choice. Cooling tower says that they are hyperboloid, so the cross section is a hyperbola: in this case, it's

y=-{\sqrt {x^{2}-1}}

, which becomes completely vertical at

x=\pm 1

. Replace x and y with

a_{x}x-b_{x}

and

a_{y}y-b_{y}

to scale/shift to taste. --Tardis (talk) 21:56, 30 August 2010 (UTC)[reply]

Dear Dr. Who, thank you! Reason I was asking was, I read an article about a speculative theory that attempts to explain why gravity is so much weaker a force than, say, electromagnetism. According to the theory, if our universe is envisioned as a two-dimensional sheet lying flat then other universes are stacked on both sides like pancakes and gravity dissipates by becoming spread out among universes. I was trying to picture in my mind how universes could interact gravity-wise without them slamming into each other and hit on the idea of a graph that rises from nothingness asymptotically and, after an interval, reverses back to nothing. The graph is made by one function so the two halves are in fact a single entity but there is a break in between them. Hmm. On second thought, maybe part 1 and part 2 should switch places. A few more, if I may: (1) Why are so few of us mathematically talented? (2) I am still heartbroken about Rose Tyler, she should still be your companion! (3) Do you ever tire of stupid questions from earthlings?--82.113.106.31 (talk) 01:34, 31 August 2010 (UTC)[reply]

I'm not he, of course, but the psychic paper tells me that he read this message in 2045:

That's what's so brilliant! You can't become a talented ballet dancer or a talented werewolf, but anyone can learn mathematics because it's all written down!
Aren't we all — but think of the other you over there to whose world she was added.
Apparently not: I keep coming back for more.
...Four things: just "the Doctor."

--Tardis (talk) 02:44, 1 September 2010 (UTC)[reply]

Magic Numbers

Can someone explain why there are "magic numbers", and particularly why they work.

I was thinking of the number: 142857. When multiplied by 7 and multipiles thereof produce unusual results, or indeed of any non-multipiles-of-seven.

When multiplied by 55 and then 7, look at the result! Why? MacOfJesus (talk) 19:21, 30 August 2010 (UTC)[reply]

There's an article on this at Cyclic number. 85.226.205.150 (talk) 20:02, 30 August 2010 (UTC)[reply]

1/7 has the decimal value 0.142857142857142..., so 142857 is just below 1000000/7. Multiplying by 7 gives a whole number just below 1000000. Similarly, 1/13 is 0.076923076..., so 76923 shows the same result when multiplied by 13.→86.132.161.214 (talk) 20:13, 30 August 2010 (UTC)[reply]

Thank you. MacOfJesus (talk) 21:11, 30 August 2010 (UTC)[reply]

The Cyclic number article mentions a number of restrictions:

This restriction also excludes such trivial cases as:

repeated digits, i.e.: 555
repeated cyclic numbers, i.e.: 142857142857
single digits preceded by zeros, i.e.: "005"

I understand the first two, but how is #3 problematic? 005 * 2 = 010, so it doesn't seem to be cyclic, trivially or not. -- ToE^T 00:14, 31 August 2010 (UTC)[reply]

Is it because "Zero" "0" is used by us so differently? "Zero" can mean "Zilch" or that there is no value in that space, such as the difference between "ten" and "one", a zero!? MacOfJesus (talk) 11:37, 31 August 2010 (UTC)[reply]

I don't buy that. The article explicitly addresses the way leading zeros are handled. -- ToE^T 13:05, 31 August 2010 (UTC)[reply]

Anything like "00005" is trivially cyclic because the cyclic permutations are "50000", "05000", "00500", "00050", and "00005". All of these are multiples of 5. (The fact that 005 * 2 = 10 doesn't mean it's non-cyclic. Cyclic means the permutations are all multiples, not that all multiples are permutations.) Staecker (talk) 13:24, 31 August 2010 (UTC)[reply]

But according to the article, cyclic numbers specifically deal with "successive multiples". -- ToE^T 13:30, 31 August 2010 (UTC)[reply]

FWIW, the three restrictions were in place when the article burst forth fully formed from the prolific keyboard of 198.99.123.63. -- ToE^T 13:38, 31 August 2010 (UTC)[reply]

Ah OK- I missed the bit about "successive". Reading more carefully now... In the bit you quoted above, "This restriction" refers to the "successive multiples" condition. So those three categories are examples of numbers which would trivially be cyclic were it not for the "successive multiples" condition. So "0005" trivially satisfies all properties of cyclic numbers except for the "successive multiples" condition, as you said. I'll try to clarify this a bit in the article. Staecker (talk) 11:57, 1 September 2010 (UTC)[reply]

I too noted the "This restriction" part, and was going to address that after figuring out #3. I disagree with your interpretation and your edit. The "consecutive multiples" condition dose not lead to restrictions #1 and #2 -- they still need to be explicitly excluded as trivial cases. I suspect that you were close to the mark when you mentioned that #3 would need to be exclude as trivial cases of numbers whose permutations are (non-consecutive) multiples. Perhaps one of the sources used that definition of cyclic number. Anyone here have a copy of TPDoCaIN to see if they define it differently than the Wolfram page does? -- ToE^T 00:54, 2 September 2010 (UTC)[reply]

Well, repeated digits (#1) do give multiples when you make the permutations: "555" has permutations "555" and "555", each of which is 555 * 1; so, were it not for the "successive" condition, this would be cyclic. For (#2), the example "142857142857" does indeed give multiples when you make permutations, but they are not successive (e.g. 428571428571 = 142857142857 * 4). So this one too would be cyclic were it not for the "successive" multiples. Staecker (talk) 11:22, 3 September 2010 (UTC)[reply]

Staecker meant to type 142857142857 * 3 in the example above. -- ToE^T 16:36, 3 September 2010 (UTC)[reply]

With #2:

142857142857 × 1 = 142857142857

142857142857 × 2 = 285714285714

142857142857 × 3 = 428571428571

142857142857 × 4 = 571428571428

142857142857 × 5 = 714285714285

142857142857 × 6 = 857142857142

which are successive multiples (and are all six distinct cyclic permutations that exist for that twelve digit number). With #1, 555 only has one distinct cyclic permutation and the single multiple 1 is trivially successive. I'm not trying to pick nits here, but this is why such trivial cases are explicitly excluded. -- ToE^T 16:36, 3 September 2010 (UTC)[reply]

I've made the appropriate changeto the article, explained (and with a link back here) at Talk:Cyclic number#Trivial cases. -- ToE^T 23:41, 5 September 2010 (UTC)[reply]

Thanks to all. MacOfJesus (talk) 17:11, 1 September 2010 (UTC)[reply]