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March 25

Conic sections -- visualisation

I have a solid cone with its base flat on the table. I slice it at an angle to the horizontal. The shape of the new cut surface is an ellipse. BUT, since my cut higher up was at a smaller radius on the cone than the cut lower down, how does it form an elipse and not a sort of eggy shape?--178.111.96.35 (talk) 01:47, 25 March 2016 (UTC)[reply]

I assume you're asking for an intuitive (handwaving) argument rather than a mathematical proof. Then just think about one part of the cut, say the lowest point. If you tilt the cutting plane to a higher angle off the horizontal, it meets the edge of the cone at a steeper angle, which sharpens the curve of the cut edge, right? Conversely, if you tilt it to a gentler angle, it widens the curve. Well, now look at the opposite side of the cut. The cutting plane meets that side at a less steep angle than it does the low side, which, as you've just seen, widens the curve of the cut edge. In fact (and this is the part where you really need a mathematical proof) it widens it by just enough to offset the fact that it's cutting a more sharply curving part of the cone, and therefore produces a symmetrical curve.

By the way, I had the same problem myself when I first learned about conic sections. --69.159.61.172 (talk) 02:45, 25 March 2016 (UTC)[reply]

As 69 says, when the plane that defines the conic section is more nearly parallel to a generator of the cone, it tends to cut the cone in a sharper curve, so that when your plane cuts further up the cone, it is cutting at an angle more nearly parallel to the generators, and so the effect of cutting more sharply and the effect of cutting further away tend to balance each other out. You can also prove that the conic section is an ellipse by taking a pair of spheres whose equators are circular parallels of the cone, that are tangent to the plane on opposite sides of the cut. It is then straightforward to prove that the some of the distances from a point on the conic section to the tangent points of the spheres is a constant, equal to the distance between the equators of the two spheres. Sławomir
Biały 11:30, 25 March 2016 (UTC)[reply]

Sławomir is (as always) perfectly correct, but I too didn't find it immediately obvious that the elliptical conic section is symmetrical around the minor axis. Bo Jacoby (talk) 15:50, 25 March 2016 (UTC).[reply]

A rational argument?

It is widely accepted that 0.999... is precisely equal to 1.000... . They are the same number. Now I'm thinking that if you take pi or e, and compare them to the nearest rational number, i.e. a decimal expansion of those numbers, they are just as close, i.e. they differ by one digit at whatever level of accuracy you choose to look.

Therefore, pi is the same number as a rational number. And so either all numbers are rational (and constructible, and algebraic) or they are all both rational and irrational.

I think I may have toked off the wrong bush this time. But how do you dispute this? Wnt (talk) 11:21, 25 March 2016 (UTC)[reply]

@Wnt:: I don't see anyone having made this exact point below, so maybe I can cut the Gordian knot. You ask about taking the "nearest rational number" to π or e. But there is no "nearest" rational number to π or e. Does that help? --Trovatore (talk) 18:29, 25 March 2016 (UTC) Oh, now I see that Sławomir did actually make this point.[reply]

Yes, in that whatever rational number you choose, like 22/7 as an approximation of pi, you can always find one that's closer, by going to larger and larger numerators and denominators. If we restrict ourselves to just powers of 10 as the numerator, we can see this more easily. 3/10 is a terrible approximation of pi, 31/100 is still bad, 314/1000 is a bit better, 3142/10k is not bad, 31416/100k is fairly good, etc. StuRat (talk) 22:29, 26 March 2016 (UTC)[reply]

You are off by a factor of 10 here. You probably meant 3/1, 31/10, 314/100, 3142/1000, and so on... Dhrm77 (talk) 23:06, 27 March 2016 (UTC)[reply]

Yes I am. Good catch. StuRat (talk) 03:07, 28 March 2016 (UTC)[reply]

Irrational numbers have a unique decimal expansion. Non-uniqueness of decimal representations of real numbers is only true for certain kinds of rational numbers (specifically rationals whose denominator is an integral power of 10). The number

1=1/10^{0}

is such a number, but so is (for example)

0.2=2/10^{1}

, etc. There is no nearest rational number to e or π. These numbers can be approximated to arbitrarily high degrees of accuracy by a sequence of rational numbers (for example, truncations of their decimal expansion), but there is not single rational that is an approximation to all degrees of precision. The reason is that these numbers are known (by a theorem) to be irrational, and the real number system satisfies the Archimedean property. Sławomir
Biały 11:35, 25 March 2016 (UTC)[reply]

"they differ by one digit.." this is where you're going wrong, I think...0.999... and 1 do't differ by one digit...they are the same number just written a different way....68.48.241.158 (talk) 12:58, 25 March 2016 (UTC)[reply]

so you're stating they're the exact same number (which is correct) and then stating that they're different within your question...this is the confusion...not sure the technical answer above is even relevant (though probably contains accurate information in general)...ie they don't differ AT ALL at ANY LEVEL of accuracy...you're defining (perhaps without realizing it) 0.999...as being very, very close to 1 but not quite 1 and posing a question based on this assumption..but this is simply wrong...so it's kind of a meaningless question...68.48.241.158 (talk) 13:41, 25 March 2016 (UTC)[reply]

@Slawekb: I thought of the "no single rational" argument, but it still isn't working for me. Because if 0.999... is 1.000... then 1.000... also equals 0.999... So 1.000... is approximated by a series of different rational numbers - 0.9, 0.99, 0.999, etc.; but when an unlimited number of digits is allowed, there is one number of that form that is truly equal to it. The non-uniqueness of the truncated forms doesn't enter into it. Likewise pi is approximated by 3, 3.1, 3.14 and so forth. I'm just not seeing the difference. Wnt (talk) 16:33, 25 March 2016 (UTC)[reply]

The fact that a number can be approximated by a sequence of rationals doesn't make it rational. Every number (both rational and irrational) is the limit of a sequence of rationals, but that doesn't imply anything about whether the limit itself is rational. Rationals and irrationals share a lot of characteristics (both can be interpreted as a point on the number line, both can be written as an infinite continued fraction, etc), but that doesn't mean that they share ALL characteristics. CodeTalker (talk) 17:10, 25 March 2016 (UTC)[reply]

(EC)Right, so you know that it's wrong to claim that Pi is rational, but you don't know exactly why your argument fails. Reading the Proof_that_π_is_irrational does lay your claim to rest but it doesn't doesn't necessarily help your understanding of the flaw in the argument. Slawekb as usual has a good answer, but there's also some deeper reasons I think.

I wonder if you've ever had a chance to ~~suffer through~~ enjoy a presentation on the construction of the real numbers. That's what's lurking below all this, because the decimal expressions are not defining anything, they are just notation. Anyway, check out the Dedekind schnitt. Turns out what you thought was a number (pi) is actually just an equivalence class of partitions of the rationals. SemanticMantis (talk) 16:36, 25 March 2016 (UTC)[reply]

Odd... so for a Dedekind cut at 1, we see that 9/10 + 9/100 + 9/1000 ... + 9/10^N is beneath the cut for any N. One would think then that limit (N->infinity) of this would be that it is beneath the cut. But it is above the cut for N = infinity. I'm not sure what making this cut can tell me. I mean, if I can't count on the sum of an infinite series to be the limit of summing the series to infinity, how many of these arguments are even valid? Wnt (talk) 17:06, 25 March 2016 (UTC)[reply]

The sum of the infinite series is indeed the limit of the partial sums. Nothing wrong there.

But that doesn't mean that everything that's true about all the partial sums also has to be true of the limit. All the partial sums are indeed in the lower part of the cut, but the limit is in the upper part. The real numbers are a connected set, so you can't divide them into two pieces without doing something discontinuous. When you have a discontinuous function, the function applied to the limit does not have to equal the limit of the function values. --Trovatore (talk) 18:00, 25 March 2016 (UTC)[reply]

Also remember that "0.999..." is itself just a notational convention. What does that ellispsis really mean? Ellipsis#In_mathematical_notation mentions with reference that some authors call for abandoning it. I don't want to take it that far, but if we're going to use it at all, we should be willing and able to describe what it means in any given case, and to do so with some rigor when called for. In this case, I think

0.999\ldots =_{\mathrm {def} }\sum _{n=1}^{\infty }{\frac {9}{10^{n}}}=\lim _{N\to \infty }\sum _{n=1}^{N}{\frac {9}{10^{n}}}

, and we can further define that limit of a finite partial sum via an epsilon-delta definition if needed. Is that what you thought it meant? If so, maybe we're done. If not, then maybe the way you thought of the notation was the source of your confusion. SemanticMantis (talk) 18:59, 25 March 2016 (UTC)[reply]

My input might have seemed trivial and nontechnical but I think it's the fundamental answer...OP wants to take two different numbers and argue they're actually the same number based on the logic that 0.999.. and 1 are seemingly two different numbers that are actually the same number...but they are NOT different numbers in any sense at all...they are both X and only X but just two different ways of writing X...(and the notion of 'differing by one digit' is literally nonsense regarding this single number)...am I missing the point?? 68.48.241.158 (talk) 19:26, 25 March 2016 (UTC)[reply]

I think that you are exactly right. Bubba73 ^{You talkin' to me?} 00:12, 27 March 2016 (UTC)[reply]

It would be a big number!

anyone want to give this a shot, arose out of previous post......assuming Wikipedia has exactly 5,000,000 articles: what would be the probability of (or odds against) someone hitting the 'random article' button exactly 5,000,000 times and touching every article once with no repeats??? 68.48.241.158 (talk) 15:00, 25 March 2016 (UTC)[reply]

This was already answered by BenRG. --JBL (talk) 15:09, 25 March 2016 (UTC)[reply]

he seemed sort of unsure about it...what would the answer be in scientific notation then, based on 5,000,000 articles?? or how many digits would the number have?? (curious if someone would write the specific number based on 5,000,000 if it's straightforward to do..as don't know much about notation of very big numbers)68.48.241.158 (talk) 15:31, 25 March 2016 (UTC)[reply]

On the first click any article will do, so the probability of getting a valid response is n/n. On the second click, you want any article except the one you already chose on the first click, so the probability is (n-1)/n. On the third click, it's (n-2)/n, and so on. Repeating n times and multiplying them all together, you get a total probability of

n!/n^{n}

. By Stirling's approximation,

n!\sim {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}

, so substituting for n! the

n^{n}

factor cancels and you get

{\sqrt {2\pi n}}\cdot e^{-n}

. For n=5000000, this is about

10^{-2171469}

. CodeTalker (talk) 15:47, 25 March 2016 (UTC)[reply]

okay, cool..meaning the number would have about 2.2 million digits if written out long hand?? that's not so bad...wouldn't even fill a nyc phone book...(note: I actually was able to achieve it last night, but obviously PURE luck!!)68.48.241.158 (talk) 15:54, 25 March 2016 (UTC)[reply]

The number of times you'd have to press the button to have a good chance of having seen all the articles is given by the coupon collectors problem, so unless you go for a more organised way or don't depend on being supremely lucky as in the original question this article gives how long you'd have to wait. 17:52, 25 March 2016 (UTC)

What I was unsure about is the average number of times you'd need to click the "random article" button before getting n consecutive articles with no duplicates, which is a different question, and possibly not a question you were asking anyway. -- BenRG (talk) 18:41, 25 March 2016 (UTC)[reply]

I think that's the same question??? would that answer still be about 10^(2171469) times before it was likely to happen?? maybe not...as in likely that many tries at it before it was likely to happen...68.48.241.158 (talk) 19:31, 25 March 2016 (UTC)[reply]

maybe I'm confused: my notion of it was a long line of people lined up to try to complete the feat...first one to do so wins...some would get closer than others...but only one would eventually complete the feat and win....how many people would have to be lined up for it to be probable that one of them completes the feat?????68.48.241.158 (talk) 20:33, 25 March 2016 (UTC)[reply]

Consider the question of flipping a coin 3 times and getting all heads. The probability of succeeding is 1/8. That means if a large number of people each flip 3 times, about 1/8 of them will succeed in getting all heads. Similarly, if a (very) large number of people select a random article 5000000 times each, about 10^-2171469 of them will succeed in getting all unique articles.

A different question is asking, how many coins do I have to flip before it's "likely" that I'll get 3 heads in a row. It's different because you're not talking about a bunch of independent triples of flips. You've got one long string of flips and you're looking for 3 heads in a row anywhere in that sequence. Say the first 3 flips are THH. In the original question, that's a failed triple so you discard it and try another 3 flips. The probability of succeeding in the next triple remains 1/8. In the second question, you're almost there, you just need one more H, and the probability of success on the next flip is now 1/2. CodeTalker (talk) 20:49, 25 March 2016 (UTC)[reply]

not sure as if a duplicate appears the feat by definition has not been achieved...and you're starting over...so if go Duplicate/Duplicate right away one just starts over, or lets the next person try, instead of seeing if one gets through all the rest of the articles with no more duplicates....ie there's no holdover that the next person can build upon....see how the holdover concept can work with the coin flips but having trouble seeing the analogue with the random article button...ie only one person or one trial will hit the button 5 million times, the one that accomplishes the feat...68.48.241.158 (talk) 01:47, 26 March 2016 (UTC)[reply]

If you start over every time you get a duplicate, then the expected number of restarts before you succeed is

n^{n}/n!\approx e^{n}/{\sqrt {2\pi n}}

. -- BenRG (talk) 07:18, 26 March 2016 (UTC)[reply]

So there's a difference between 1. how many attempts would have to be made/restarts to likely accomplish the feat and 2. how many times the button would have to be hit before getting the requisite streak..??? Guess that's what you're saying. Having a little trouble getting my head around it...would the numbers involved be significantly different?? 68.48.241.158 (talk) 13:41, 26 March 2016 (UTC)[reply]

maybe I get it: the first you're counting simply the number of restarts...the second you're counting up all the clicks (and the clicks that constitute a restart are a subset of these)...so the second would be a larger number.....?? — Preceding unsigned comment added by 68.48.241.158 (talk) 15:02, 26 March 2016 (UTC)[reply]

Sort of. It's a bit more complicated. For example, if there are three articles, and the random article button takes you to 1, 2, 1, 3, ..., then in the restarting case, you'll restart after 121, then you'll just have 3 (and you'll keep clicking). In the setup I was imagining, you "win" at that point, because the last three clicks (213) include all three articles. So the expected number of clicks for that setup is less than if you restarted from scratch. (I'm not sure it matters, though, since that evidently wasn't what you had in mind.) -- BenRG (talk) 20:24, 26 March 2016 (UTC)[reply]

okay thanks...see what you mean with that example of just a few articles....68.48.241.158 (talk) 01:35, 27 March 2016 (UTC)[reply]

It would be a big number? No! Any number is small because most numbers are bigger. Bo Jacoby (talk) 13:36, 27 March 2016 (UTC).[reply]

not if looking at the integer number line with negatives...there every number sits between an infinity in each direction...68.48.241.158 (talk) 13:48, 27 March 2016 (UTC)[reply]

The numbers considered here were nonnegative integers. Bo Jacoby (talk) 18:25, 27 March 2016 (UTC).[reply]

I the real world numbers tend to follow Benford's Law. By that you'll get about as many positive numbers below any particular positive number as above it. Dmcq (talk) 19:23, 27 March 2016 (UTC)[reply]

Benford's law refers to positive real numbers, not to nonnegative integers. Bo Jacoby (talk) 19:59, 27 March 2016 (UTC)[reply]