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September 18[edit]

2 functions, whose limit at 0 is undefined, but the limit of the product is defined at zero.[edit]

Does such a beast exist?

Function 1 and Function 2 have an undefined limit at 0

but the product of function one and function two does have a limit at 0.

Thanks Duomillia (talk) —Preceding undated comment added 14:28, 18 September 2015 (UTC)[reply]

Sure, take the function

f(x)={\begin{cases}-1&\quad {\text{if }}x<0\\1&\quad {\text{if }}x\geq 0\end{cases}}

then let both of your functions be f. Their product is constant, and so has a limit at 0. —Kusma (t·c) 14:45, 18 September 2015 (UTC)[reply]

But you have defined f(x) at zero, so it isn't undefined. Do you mean 1 if x > 0 rather than >=? -- SGBailey (talk) 13:15, 19 September 2015 (UTC)[reply]

I agree. Also, since my 2 functions need to be distinct, I bet I can make g(x) simply be f(x) times -1.... or can I? --Duomillia (talk) 15:41, 19 September 2015 (UTC)[reply]

I bet I can, because while the resulting constant function x=-1 will be undefined at zero, the limit is! --Duomillia (talk) 15:48, 19 September 2015 (UTC)[reply]

How about

\sin {\frac {1}{x}}

and

{\frac {1}{\sin {\frac {1}{x}}}}

? Gandalf61 (talk) 21:08, 19 September 2015 (UTC)[reply]

Metaquestion: Why did my question receive no answers?[edit]

Please see this question that I asked a couple of weeks ago. It received no answers. I know that the reference desk contributors are eager to help, and will answer questions that are answerable. Therefore, I suspect that there is a problem with the way the question was asked, that made it either difficult to understand, or difficult to answer. There is also the possibility that it was ignored because it was perceived as lazy, but my impression is that laziness on the part of the questioner tends to be commented. My question now is about the question itself. I am not requesting answers to the original question. I'm pretty confident that I have figured out the answers to Q1 and Q2 myself, some doubt about Q3 remains. I may re-ask the question if I learn how to ask it in a way that is more likely to attract answers. --NorwegianBlue^talk 22:39, 18 September 2015 (UTC)[reply]

Hi,I remember your question, and I have some interest in your topic. I thought I knew a bit of the concepts, and I even looked in to it for a few minutes. But I quickly realized that it was fairly technical and specialized question, that general refs wouldn't help you much, and there wasn't much I could to in <10 min to help you. It may have helped if you included more refs and links in your question, so that people who might be able to help but aren't already as familiar as you with the topic could get up to speed quickly. There's a lot of "luck of the draw" in how many people see a question, what they know, how much time they have to spare, how much interest, etc. But there's also a sense I got of this being an area where there might be considerable variation in how terms are used, rigor of proof, ways of justifying conclusions, and all sorts of gritty details of research that just take a lot of time and attention to sort through. You may have better luck at math.stackexchange, quora, or other similar (sub)sites that can sometimes deal a little better with this kind of thing. SemanticMantis (talk) 01:42, 19 September 2015 (UTC)[reply]

My comments are similar to those above, but I would add that, as a visual thinker, it would have helped me if some graphics had been included. Otherwise it looks like a well posed Q. StuRat (talk) 01:45, 19 September 2015 (UTC)[reply]

You might have better luck with stats.stackexchange.com for something like this; add the roc tag to grab the attention of people who specialize in that area. In general, at least as far as I've seen, Stack Exchanges are more geared to professionals helping other professionals, while the WP reference desks are more for general questions. --RDBury (talk) 01:52, 19 September 2015 (UTC)[reply]

Thanks, everyone! --NorwegianBlue^talk 14:50, 20 September 2015 (UTC)[reply]

@NorwegianBlue:, late answer, but anyway...

I've taken a look at it (don't visit RD/MA often ), the question was a bit confusing:

The tangent at a given point (CO), can be estimated as ΔTPR(CO)/ΔFPR(CO),

hence TPR(CO)/FPR(CO) is the derived function of the ROC curve

Disclaimer: not familiar with the topic, just my interpretation, which could be way off..

Is it about the continuous probability functions? These are turned into binary tests by integrating them (page 2260), and the ROC curve plots those values. (See note 1)

The tangent at a given point equals the ratio of the density functions (those in fig 1 in the link), and is also (by definition) ΔTPR(CO)/ΔFPR(CO), but not TPR(CO)/FPR(CO). (See note 2)

Each point on the curve gives you the fraction of the sick population and the fraction of the healthy population that would test positive at that CO. So if (TPR=0.9, FPR=0.1) is on the curve, 90% of all sick and 10% of healthy would test positive.

Context would make it easier to understand, preferably a link to the specific material and formulas you're asking about, not many people with a medical background here and most maths techniques have numerous applications, so people won't know the conventions/notations, etc. Most sources I found were like this .

TPR(CO)/FPR(CO) = p(m = CO ± eps|Disease)/p(m = CO ± eps|No disease)

Q1: Am I right in thinking that the probability ratio in the previous line should be the probability of 'm' being close to CO (in my notation ± eps), and not greater than CO?

Haven't seen that formula in links I checked. Seems a casual way of saying that both become equal when eps goes to zero. Seems valid to me, no reason to limit me to one side of CO, it's not like this has influence on anything, or I'm missing context. (See note 3)

Q2: it is valid to draw lines between several points on the ROC curve, say corresponding to "negative", "weak positive" and "strong positive", and that the slope of each line is a valid estimate of the likelihood ratio for test results that fall within the corresponding interval.

It's the definition of ROC curve: Y-coordinate = TPR(CO) x-coordinate=FPR(CO), the slope of a line from origin (0,0) to a point is by definition y/x. For a line between two points on the curve you calculate the slope by subtracting the coordinates. For example: Two points TPR=0.5 FPR=0.3 and TPR=0.8 FPR=0.7, segment between: TPR=0.8-0.5=0.3 and FPR=0.7-0.3=0.4; between those points, the positive likelihood ratio is 0.3/0.4=0.75, so more False positives than true positives. (See note 4)

Q3: I've read many places (including in our article) that the area under the curve corresponds to the probability of a randomly chosen diseased individual getting a higher test result than a randomly chosen individual without the disease. Again, this sounds reasonable, but exactly why is it so?

The X and Y coordinates of a ROC are proportional to the number of healthy and infected people. so an increase of 0.10 in either direction represents 10 % of the related group. Movement at 45° represents the same % of people in both groups.

Suppose the curve starts by going up to 0.4, that means 40% of the infected (lets assume low values indicate infection) will test lower than all the rest. Then 0.2 to the right; 0.3 up; 0.7 right, 0.2 up, finally along the diagonal.

Calculate the odds of infected people scoring better (lower), than healthy: 40% score better than all 0.4*1; 30% score better than 80%; 0.3*0.8; 20% better than 10%: 0.2*0.1 and the last 10% have a chance of one in two to do better than the other 10%, so: 0.1*0.1*0.5. In total: 0.665.

The other group: 0.2 *0.6; 0.7*0.3 and 0.1*0.1*0.5: total: 0.335. You can work it out on paper, when you change the curve, the change in area will match the change in odds. Here, the area is 66.5%. (See note 5)

Oeps, I now see that the link you provided had not just the abstract, but the whole article... (9850136?) Ssscienccce (talk) 02:30, 21 September 2015 (UTC)[reply]

Thank you, @Ssscienccce:, those were very helpful answers and links. I took the liberty of inserting yellow labels in your answer, in order to reference your replies more easily. Yes, I see that the question was confusing, reflecting my own confusion.

Note 1: I was thinking about a smoothened ROC curve, and see that that would imply needing continuous probability distributions/densities.
Note 2: Duh! Confused, wrong thinking on my part here. Of course it isn't. I'm forgetting the Quotient rule, and being unclear about exactly what variable I'm differentiating with respect to.
Note 3: Q1 is very confused. I think I'd rather strike it than try to reformulate it.
Note 4: Thanks! I understand it now.
Note 5: Thanks! I'm still struggling a bit with this one, but it's getting late. I'll re-read your reply carefully in the morning, and I'm optimistic that I'll understand it then.

Your reference to Johnson 2005, PMID 15236429 was very helpful indeed! --NorwegianBlue^talk 20:38, 21 September 2015 (UTC)[reply]

Wasn't very happy with that last answer myself. This may be a better way:

For every FPR value there exists a corresponding Cutoff value. Hundred FPR values from 0.01 to 1 (with 0.01 increments), have 100 corresponding cutoff values, CO_0.01, CO_0.02... CO_1.00

This way you could divide the healthy people in 100 groups, based on their test result. Each group would contain 1% of the total healthy population. (follows from the definition of FPR).

The same for the diseased people. They too are split in 100 groups, corresponding to the TPR values 0.01, 0.02, 0.03 and so on... (note: the hundred cutoff values differ from those for the healthy group).

Randomly choose one healthy person. He would belong to one of the hundred groups, and since the groups are of equal size, they are all equally likely to be chosen. The same is true for a randomly selected diseased person.

As an example, we'll say that the healthy person is in healthy group 20 (FPR 0.20 to 0.21) and the other one in diseased group 40 (TPR 0.4 to 0.41). those values define a small square of 0.01*0.01 size on the ROC diagram.

If you draw vertical lines at FPR=0.2 and 0.21, they will cross the ROC curve at two points, (the Cutoff values that defined the group he belongs to) and the persons test value must be between those two.

Do the same for the horizontal lines at TPR =0.4 and 0.41, they cross the ROC curve at 2 points, the test value of the person is between the two.

If the small square where the horizontal and vertical lines cross is below the curve, the vertical lines will cross the curve further from the origin than the horizontal lines (on fig 1b at page 2261, the TPR=0.4 line crosses at about 0.1 FPR, FPR=0.2 line crosses at about TPR=0.57). Values further on the curve are lower, since more people test positive the further you go, so the healthy person has a lower test result than the sick person. If the point (little square) is above the curve, the reverse is true: the vertical FPR line intersects the curve closer to the origin, so the healthy person has a higher test value.

The number of 0.01*0.01 squares below the curve are proportional to the area below the curve. If the area is 90% then 9000 squares are below, 1000 above the curve. Picking one healthy and one sick person at random has a 90% probability of corresponding to a square below the curve, which corresponds to the healthy person having a lower test value than the diseased one. Ssscienccce (talk) 02:55, 22 September 2015 (UTC)[reply]

Thanks, @Ssscienccce: I followed your argument carefully, with the help of a spreadsheet, pen and paper. I understand why the number of healthy people in each of the 100 boxes of healthy people is the same, and ditto for the number of diseased people in each of the 100 boxes of diseased people. I drew a diagram (only 10 categories of each), which turned out to look a lot like the example on page 2261 in the reference. Since any intersection point with the ROC curve, whether by a vertical or horizontal line, corresponds to a cutoff value (or in this argument, measurement), a point above the curve for a random pair of healthy and diseased individuals, implies that the healthy person (vertical line that has crossed the curve) has a measurement that is higher (closer to the origin) than that of the diseased person (horizontal line that will intersect the curve at a point further from the origin, i.e. has a measurement that is lower). I think I got it! If there were any errors in my restatement of your explanation, please warn me about it! Thanks a million. --NorwegianBlue^talk 21:44, 22 September 2015 (UTC)[reply]

That funny looking 'f' thing[edit]

There's a funny looking 'f' sort of thing that goes in calculus equations. I saw it just now in an episode of Torchwood right before an equation with a sin^2 following it. What's it called? Please and thank you. Dismas|^(talk) 23:49, 18 September 2015 (UTC)[reply]

If you mean the large S-shaped symbol in an expression like

\int f(x)\,dx.

it is the symbol for Integral. Here it means the integral with respect to

x

of a real-valued function

f (x)

. Integral and differentiation are basic operations in the branch of mathematics called Calculus. 84.209.89.214 (talk) 00:14, 19 September 2015 (UTC)[reply]

The symbol was originally the long s which was in common use until the 1800's. Presumably it stands for summation. --RDBury (talk) 01:12, 19 September 2015 (UTC)[reply]

Don't you mean "stood" for summation ? Sigma (Σ) is used for that now. StuRat (talk) 01:39, 19 September 2015 (UTC)[reply]

The integral symbol is the symbol for a continuous infinitesimal summation, an integral. The Σ is the symbol for a discrete summation. Robert McClenon (talk) 01:48, 19 September 2015 (UTC)[reply]

Yes! The $\int$ symbol! But what is it called besides 'the integral symbol'? Dismas|^(talk) 01:43, 19 September 2015 (UTC)[reply]

I believe that answer was given above, "the long s". Of course, you would only call it that when talking about the symbol, as opposed to it's usage for integration. When reading it in a math context, you would just say "integral" (or "double integral" or "triple integral" when you have 2 or 3 in a row). If there are numbers or symbols at the top and bottom those are the "from" and "to" numbers. So, you might get something like "the integral from a to b of the function 2x squared". StuRat (talk) 01:52, 19 September 2015 (UTC)[reply]

English grammar note: the Apostrophe punctuation mark is useful to mark an omitted letter as in the contraction of it is to it's. English provides a consistent set of posessive personal pronouns that allow one to write my usage, your usage, his usage, her usage, its usage, etc. in which no apostrophes are required, and to split arbitrarily any of these pronouns by adding an apostrophe serves no good purpose. (In linguistics the singular Grammatical person comprises I, you, he, she and it.) 84.209.89.214 (talk) 11:52, 19 September 2015 (UTC)[reply]

Yes, that rule never made sense to me, so I ignore it, and use "it's" to specify the possessive form, and only use "its" as the plural of "it". (I also use "it's" as the contraction of "it is", but that can be easily distinguished from the possessive form by the context.) StuRat (talk) 14:23, 19 September 2015 (UTC) [reply]

Can I assume you also say Is, yous, hes, shes as the plural form of I, you, he, she? And that you also say I's, you's, he's and she's as the possessive forms? Because if you're going to deliberately use broken English and obfuscate your writings, you may as well be consistent about it. -- Meni Rosenfeld (talk) 17:45, 19 September 2015 (UTC)[reply]

The character in the show said something like "cause sin squared..." but I couldn't quite catch it because they cut to the character reading it just as they finished that first word. I may be a bit off on the pronunciation because the character is Welsh and I'm American. So, does that seem familiar to anyone? Dismas|^(talk) 02:08, 19 September 2015 (UTC)[reply]

Well, if the equation began with ∫sin² then there's no way it would be pronounced anything like "cause sin squared". However, cos may sound like "cause" (depending on your accent) and often appears in equations together with sin. So maybe it was that, and you misheard or the person misspoke or there was a mixup and the wrong equation was written. Or else "cause" was short for "because" (you didn't quote the whole sentence). --174.88.134.156 (talk) 05:01, 19 September 2015 (UTC)[reply]

Would anyone actually say the abbreviated forms "cos" and "sin" ? (Good luck saying "csc" !) I would always say cosine, sine, and cosecant. StuRat (talk) 14:21, 19 September 2015 (UTC)[reply]

Yes, people say /coz/ instead of "cosine" and abbreviate cosecant to "cosec". I'm surprised that you don't. (But I suppose I shouldn't be surprised since you have your own idiosyncratic rules of spelling.) I agree that people don't generally abbreviate sine. Dbfirs 16:05, 19 September 2015 (UTC)[reply]

I certainly do. I pronounce the sentence "sin x" means the sine of x exactly the way it looks, with "sin" and "sine". It doesn't save a syllable the way saying "cos" does, but it's consistent with saying "cos" and "tan". Probably that's the way the high-school teacher who I learned the functions from originally said them, but I don't really remember now. And by the way, "cos" rhymes with "dose" in my usage; it doesn't have a Z sound as Dbfirs indicated. As for "csc", it doesn't come up often enough to matter, but I'd just say "cosecant". --174.88.134.156 (talk) 18:32, 19 September 2015 (UTC)[reply]

There are evidently regional differences across the Atlantic. We usually abbreviate cosecant to "cosec" not "csc". Hyperbolic cosine is normally abbreviated to "cosh" here in the UK, and pronounced that way, but there are variations for hyperbolic sine (sinh). I recall many years ago hearing it pronounced as "shine", but I considered that very odd. Dbfirs 20:53, 19 September 2015 (UTC)[reply]

In Hebrew we pronounce the functions "sinus" and "cosinus", so in principal there are more syllables for us to spare by using the short forms. And yet, as far as I recall, we usually say the full form when reading out loud the shortened notation. -- Meni Rosenfeld (talk) 10:43, 20 September 2015 (UTC)[reply]

If you want to get into that level of detail it would help if you could specify which episode you're talking about; it is available online. --RDBury (talk) 12:06, 19 September 2015 (UTC)[reply]

Sure, it's "Captain Jack Harkness". I don't recall exactly where in the episode the scene is but it's at least half way through. When Gwen finds a portion of the formula in an old paint can. Dismas|^(talk) 13:18, 19 September 2015 (UTC)[reply]

The plot revolves around sending some "equations" from Sato's laptop back through a time warp to Torchwood HQ. The first line appears to be a variation on

\int \sin ^{2}{ax}\;\mathrm {d} x={\frac {x}{2}}-{\frac {1}{4a}}\sin 2ax+C

which can be found in List of integrals of trigonometric functions. The second line is just an expression which also involves an integral. Not sure what these would have to do with finding your way through a time warp or anything else in the story, but it's just a show after all. --RDBury (talk) 14:05, 19 September 2015 (UTC)[reply]

By the way, while it's correct that integral sign derives from the long s, also called the "tall s", that used to be used when the letter wasn't at the end of a word, it's specifically the italic version of that letter. In ordinary roman type, the long s was written like a lower-case "f", but with the crossbar only extending to one side, as you can see here. --174.88.134.156 (talk) 18:32, 19 September 2015 (UTC)[reply]

It is not always italic. See integral sign, the Germans and Russians tend to do it different. Dmcq (talk) 11:29, 21 September 2015 (UTC)[reply]

Okay, so I got to where I can watch with subtitles. According to them, she definitely says "Cos sin x squared..." and she is pronouncing it "coz". So yeah, technical gaff since there is no cosine before the sine. So, now I know it's just called a "long s" and the dialogue was a gaff. Thanks, Dismas|^(talk) 20:42, 19 September 2015 (UTC)[reply]

No, in a math context it's just an integral sign (pronounced "integral"). --174.88.134.156 (talk) 00:09, 20 September 2015 (UTC)[reply]

The character Gwen Cooper is a police officer described as a "girl next door" archetype. She doesn't have a scientific background so it's possible the show deliberately made her mispronounce it. I agree with others that anyone who knows the notation would pronounce it "integral". Others have noted the error. [1] says: When Gwen finds the card that Tosh wrote the remainder of the formula on, she reads it as "cos sine squared". Whereas the formula is actually the "integral of sine squared". PrimeHunter (talk) 12:20, 21 September 2015 (UTC)[reply]

I'm getting the feeling that people are reading way more into this than is actually there. The first line of the equations (given above) is the only part that really makes sense. In the course of the episode various people read off different parts of the equations, sometimes they match what's on the screen and sometimes they don't; sometimes what's on the screen changes from scene to scene. The equations are a MacGuffin and what they are exactly does matter to the story, so it appears that no one took much time to make sure they were accurate or consistent. I commend the writers for working some math(s) into the story so perhaps some young people (hopefully not too young as the series is definitely not for little kids) will be inspired to take a bit more interest in their homework. But the show is geared for the general public who won't notice or care whether the mathematical details are correct, so it's hard to blame the people who made the show for not taking time to get them completely correct. --RDBury (talk) 16:29, 21 September 2015 (UTC)[reply]