Wikipedia:Reference desk/Archives/Mathematics/2006 August 18

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A question about the trigonometric functions

Why does $f(x)=\sin {\frac {\pi }{x}}$ oscillate wildly when graphed as x approaches zero? Taking the limit doesn't work. And is $\sin {\frac {\pi }{x}}$ a special case, because $\sin {\frac {2}{x}}$ and $\sin {\frac {1}{x}}$ also seem to exhibit this behavior. Taking the base case of $\sin {\frac {k}{x}}$ , how does the oscillation change in behavior when k is increased or decreased? And are there similar occurrences for $f(x)=\cos {\frac {\pi }{x}}$ , $f(x)=\tan {\frac {\pi }{x}}$ as well? Thank you very much. --Ķĩřβȳ ♥Ťįɱé Ø

As x approaches zero, k/x approaches infinity; thus for any periodic function f, f(k/x) would have an infinite number of oscillations in any neighborhood of zero. —I've taken the liberty of changing your Pi (symbol of a product of enumerated factors) to pi (symbol of the circle constant). —Tamfang 01:55, 18 August 2006 (UTC)[reply]

If you increase k, the horizontal scale of the graph shrinks as it were, so in the landscape around the same value of x (different from 0) the slopes get steeper, and the oscillations get wilder. This should be obvious if you take the derivative with respect to x: you get a factor k, so larger k means larger derivatives (in magnitude), which means steeper slopes. Another way of seeing this is that, taking δ = π/k, if k is large enough for δ to be much smaller than x, you get slightly more than one full oscillation between x–δ and x+δ, so the oscillation takes place on a smaller stretch. The reasoning does not appeal to properties of the sine function except for periodicity. Therefore you get qualitatively the same behaviour for the cosine and the tangent functions. --Lambiam ^Talk 03:57, 18 August 2006 (UTC)[reply]

Graph the function sin 2πz, with z going from 1 to something large, perhaps 101; we see 100 cycles of the basic period. Now let z = 1/x. As x goes from 1 to 1/101 those same oscillations are produced. Then as x decreases again by half that amount, an equal number of oscillations emerges. In other words, x decreasing from 1 to 0 is equivalent to z increasing from 1 to infinity, with more and more oscillations compressed into less and less space.

The constant factor, whether 2π or π or some general k, has only a minor role in the qualitative behavior. Likewise, the specific periodic function has only a minor role. So long as we have a periodic function with a reciprocal argument, we will get oscillation sausage. --KSmrq^T 04:36, 18 August 2006 (UTC)[reply]

Thank you very much for your answers. --Ķĩřβȳ ♥Ťįɱé Ø 05:22, 18 August 2006 (UTC)[reply]

Of course, the function

f(x)=\tan {\frac {\pi }{x}}

will look very different from the other two - it will pass an infinite number of times through infinity, instead of oscillating around 0. The graph will look similar to an infinite series of almost vertical lines, getting denser as we get closer to 0. -- Meni Rosenfeld (talk) 07:21, 18 August 2006 (UTC)[reply]

Ok, so how about the $f(x)=\sec {\frac {\pi }{x}}$ , $f(x)=\csc {\frac {\pi }{x}}$ , and $f(x)=\cot {\frac {\pi }{x}}$ ? And $f(x)=\arcsin {\frac {\pi }{x}}$ (and the related arc functions), which appears to be discontinuous when |x| < π ? Thanks again. --Ķĩřβȳ ♥Ťįɱé Ø 08:30, 18 August 2006 (UTC)[reply]

After the previous answers you should be able to work that out by yourself. Do you have any graphing software? The arcsin case is boring. --Lambiam ^Talk 09:07, 18 August 2006 (UTC)[reply]

Yes I do have a TI-84 Graphing Calculator, but even when I graph it, and then proceed to look at the table, I am still mystified. --Ķĩřβȳ ♥Ťįɱé Ø 09:13, 18 August 2006 (UTC)[reply]

If you're curious about these graphs, I suggest you try this graphing software for the PC. It can handle all sorts of functions, including the ones which have been discussed here. --HappyCamper 16:36, 18 August 2006 (UTC)[reply]

What is this called?

What is this symbol $\oplus$ called? And where is it used most often? --HappyCamper 03:18, 18 August 2006 (UTC)[reply]

I don't know if it has a standard name, but with the dominance of ~~Latex~~ LaTeX among mathematicians most will understand "oplus". I've also heard variously "funny plus", "plus circle(d)" and "circle(d) plus", and even "plusol" (or should I write "plussle"?). It is used whenever mathematicians need something that behaves somewhat like "+" but needs to distinguished. The most common use is for Direct sum, and ⊕ even redirects there. See also Coproduct. --Lambiam ^Talk 03:40, 18 August 2006 (UTC)[reply]

I think I've seen it as a symbol for XOR in the context of image processing, or some sort of boolean operation at least. - Rainwarrior 05:08, 18 August 2006 (UTC)[reply]

Yes, definitely as addition on the ring Z/2Z, which is isomorphic to the Booleans under XOR and AND. See for example MD5. Also for Minkowski addition as e.g. here. There are probably dozens of regular and many more irregular uses. --Lambiam ^Talk 06:07, 18 August 2006 (UTC)[reply]

Ah! you are correct! Dusting off my notes, I have used this as XOR myself! Thanks for the reminder. --HappyCamper 16:29, 18 August 2006 (UTC)[reply]

It's called a Phillips head. ;-) —Bkell (talk) 06:40, 18 August 2006 (UTC)[reply]

Or Earth – we're screwed. --Lambiam ^Talk 06:55, 18 August 2006 (UTC)[reply]

Or in discreet math it can mean sum, wherein the sum is within the set the left and right inputs are part of. e.g. in mod3 2

\oplus

2 would be 1. i kan reed 20:20, 18 August 2006 (UTC)[reply]

Is discreet maths used for secret communications, like in MD5? --Lambiam ^Talk 21:23, 18 August 2006 (UTC)[reply]

In the Unicode character set there are characters called CIRCLED PLUS and N-ARY CIRCLED PLUS OPERATOR that look like the one you give. —Bromskloss 00:26, 21 August 2006 (UTC)[reply]

Infinity in chance

I've been thinking about this little problem that I made up, and can't come up with a solution. Perhaps you can help me.

Here's the theorhetical scenario: You have a bag with two balls in it. One blue, one red. You will stick your hand into the bag, and randomly pull out one ball at a time, and put it back into the bag. You do this for an infinite amount of time. Since this is happening for infinity, it means that all the possible outcomes will play out. For example, it is TRUE that you will eventually pull out the red ball 3,000,000 times in a row. It is TRUE that you will eventually pull out the blue ball 10,000,000 times in a row. But why stop there? You should evetually pull out the red ball infinity times in a row, don't you think so? But how would the blue ball later be pulled out infinity times in a row if the red ball is already going on for ever?--Codell ^{[ Talk ▪ Contrib. ]} 12:08, 18 August 2006 (UTC)[reply]

Infinity is a tricky beast. For a little handle on your problem, try reading the article about the probability term almost surely. Essentially in this case you're looking at the limit of a finite drawing of balls. You might also want to look at the geometric distribution, which describes the number of trials you have to undertake before you get your first success, if success happens with a constant probability in each trial. In this case, you will "almost surely" get a success eventually, meaning that you will "almost never" get an infinite number of failures in a row. Confusing Manifestation 12:41, 18 August 2006 (UTC)[reply]

Given any finite cardinal, no matter how large, we are assured that we will at some point draw that number of consecutive balls of one color, say red. However, that is not the same as drawing an infinite number of consecutive reds.

Think of the reciprocal situation in calculus. We often say, "Let ε be a positive real, however small." But since we then go on to do things like dividing by ε, we are not justified in letting it be zero. It was this kind of subtlety that led Bishop George Berkeley to lampoon Sir Isaac Newton in The Analyst, referring to fluxions as the “ghosts of departed quantities”. Similarly, arbitrarily large is not the same as infinite.

On the other hand, true infinities are notorious for having properties that violate an intuition built on finite numbers. Consider the infinite set of integers, both positive and negative. The negative integers alone are an infinite set. In enumerating the integers can we say, "First count all the negative integers, then zero, then all the positive ones"? How about rational numbers; is there any effective way to count them? (Yes.) --KSmrq^T 13:57, 18 August 2006 (UTC)[reply]

For an interesting and non-technical perspective on countable infinities, you might want to look at the article on Hilbert's paradox of the Grand Hotel. -Maelin 12:53, 20 August 2006 (UTC)[reply]

Help with standard error of estimate

Hi. I'm totally clueless with this sort of thing. I've been asked to work out some stats about a score in an interview questionnaire.

If a candidate scored 85 and the test has a standard error of estimation of 5, it means that 68% of the time, the candidate's true score would be between what and what?

Sorry for posting, but I couldn't make heard or tail of the article I linked to. --Dweller 13:50, 18 August 2006 (UTC)[reply]

Depends on what assumptions you can make about the distribution of scores. But 68% is suspiciously close to the probability that a random variable with a normal distribution takes a value within one standard deviation either side of the mean ... Gandalf61 15:08, 18 August 2006 (UTC)[reply]

Errr... this is really hard for me. Does that mean the answer would be 84-86? --Dweller 15:14, 18 August 2006 (UTC) aka "thicko"[reply]

The question is testing whether you know what standard error means. Try reading the first sentence of the standard error article again. Gandalf61 15:31, 18 August 2006 (UTC)[reply]

I know that. I don't understand the sentence. Probably because I've never studied statistics and I've not sat (slept?) in a maths lesson for about 20 years. This isn't a homework question. --Dweller 15:33, 18 August 2006 (UTC)[reply]

If you have never studied statistics at all then you won't understand what the question is asking and you can't even begin to answer it. I think you need to come clean with whoever has asked you this question, and explain that you don't have the background to give them an answer. If someone gives you an answer but you don't understand how they arrived at it, then it won't help either you or the original questioner. Gandalf61 15:55, 18 August 2006 (UTC)[reply]

"The person" knows I've never studied stats, but doesn't really care. Has your boss never asked you to do something he knows you don't know how to do? You're lucky. In the meantime, all I needed was two figures, but I'm now out of time. --Dweller 16:00, 18 August 2006 (UTC)[reply]

Okay, I'm curious ... were you really going to pass on whatever numbers a complete stranger gave you without having any clue as to whether they were right, wrong or not even close ??? Gandalf61 16:21, 18 August 2006 (UTC)[reply]

Sure, why not? We're generally pretty reliable about that, and besides, if what he says is true, it's possible they wouldn't have been double-checked anyway, so no harm, no foul. I'm not certain, though, that there's actually enough information given to give a definite answer. Black Carrot 22:41, 18 August 2006 (UTC)[reply]

I agree there is insufficient information. Yet, if this was a test, I would have answered 85 ± 5, or, if the teacher does not like that notation, [80,90]. I think IQ can be defined as the ability to guess the answer they will approve (rather than giving the correct answer). --Lambiam ^Talk 08:10, 19 August 2006 (UTC)[reply]

No, IQ can be defined as an imprecise but reliable measure of certain basic mental functions. They've managed to get rid of most of the ones that require that kind of thinking. There have been a lot of school tests, though, where I'd agree completely. Black Carrot 03:25, 21 August 2006 (UTC)[reply]

Back from a wikibreak. I ran out of time, so this, ironically, is now "academic", as no "teacher" was involved... as I mentioned, I've not sat in a maths lesson for 20+ years and it was for my boss. It seems you weren't able to answer the question because of a lack of data. So, out of interest, what further data would you have needed to be able to answer the question? My boss seemed quite certain that was sufficient. --Dweller 15:03, 30 August 2006 (UTC)[reply]