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April 22

Notation in combinatorics

In the article "3-dimensional matching" (version of 02:28, 22 April 2016), there are several instances of an inequality sign preceded by a space but not followed by a space. Is that a typographical error, or is it correct for this particular field of mathematics? If it is correct, what does it mean, besides inequality? I did not find answers to these questions at Wikipedia:Manual of Style/Mathematics.
—Wavelength (talk) 20:01, 22 April 2016 (UTC)[reply]

Typographical error. --JBL (talk) 20:13, 22 April 2016 (UTC)[reply]

Now fixed in all 3 instances, I believe. --JBL (talk) 20:16, 22 April 2016 (UTC)[reply]

Thank you for your reply and your corrections. I just noticed, further down, an equation without spaces around the equals sign, and I inserted two non-breaking spaces.

—Wavelength (talk) 20:43, 22 April 2016 (UTC)[reply]

Heh, I remember when a coworker on a programming project objected to my habit of putting spaces around operators. —Tamfang (talk) 07:33, 24 April 2016 (UTC)[reply]

Bet he cut his teeth as a C programmer.Naraht (talk) 21:02, 25 April 2016 (UTC)[reply]

Dunno about that; he seemed old enough for Fortran. I can say he was a big fan of Petri nets, to the bafflement of the rest of us. —Tamfang (talk) 01:48, 28 April 2016 (UTC)[reply]

indirect derivatives

Given $r=r(u,v)$ and its partial derivatives $r_{u},r_{v},r_{uu},r_{uv},r_{vv}$ ; suppose u,v are themselves linear functions of t, with $u_{t}=a,v_{t}=b$ . Is this right?

$r_{t}=ar_{u}+br_{v}$
$r_{tt}=a^{2}r_{uu}+2abr_{uv}+b^{2}r_{vv}$

(Context: $r(u,v)$ defines a point on a surface which I seek to represent by facets whose size is determined by local curvature and the resolution of a 3d printer; I need to compare the second derivative, in various directions on the surface, with the first.) —Tamfang (talk) 23:13, 22 April 2016 (UTC)[reply]

Yes. The second equation is two applications of the first equation. The first equation follows from the multivariate chain rule. Our article isn't as readable as it could be on this point (imo), but the example gives

{\frac {\partial u}{\partial r}}={\frac {\partial u}{\partial x}}{\frac {\partial x}{\partial r}}+{\frac {\partial u}{\partial y}}{\frac {\partial y}{\partial r}}

which is what you need modulo a few letter changes. Btw, it's interesting that you can apply calculus to 3d printers. This type of practical application usually has a hard time making into the classroom. --RDBury (talk) 05:19, 23 April 2016 (UTC)[reply]

I showed off my creations to the mother of a 4yo who was already showing interest in math. "See? Anton made these with math!" —Tamfang (talk) 08:58, 25 April 2016 (UTC)[reply]

April 24

Card models ( making a Bridge Arch)...

Posting in maths , but it's essentialy a geometric construction problem.

I have the elevation of a bridge. ( for example File:Bridges_27.png)

I'd like to use 2 elevations and a centre piece modelled in card.

How do I construct the flattened centre piece from the eleveation? Sfan00 IMG (talk) 11:19, 24 April 2016 (UTC)[reply]

I.E How do I construct a line that is equal in distance to the length of an arc? ShakespeareFan00 (talk) 08:52, 25 April 2016 (UTC)[reply]

If it's not piecewise straight or circular, you'll need to do some integrating. Have you a mathematical description of the arc? —Tamfang (talk) 09:00, 25 April 2016 (UTC)[reply]

I'm not an expert on bridge construction, but my observations of bridge arches lead me to the conclusion that quite a wide range of curves are possible and used in practice. Perhaps an engineer can advise on the strongest. The oldest bridges seem to be closer to a circular arc. Dbfirs 11:24, 25 April 2016 (UTC)[reply]

I don't have a mathematical description, but I do have the elevation (like in the example image given..)ShakespeareFan00 (talk) 12:20, 25 April 2016 (UTC)[reply]

See Catenary arch. Bo Jacoby (talk) 16:37, 25 April 2016 (UTC).[reply]

At first I thought you were building a bridge out of playing cards, but now I think you meant something else by "card". Care to elaborate ? StuRat (talk) 17:11, 25 April 2016 (UTC)[reply]

paper model ShakespeareFan00 (talk) 18:53, 25 April 2016 (UTC)[reply]

I see, card stock versus playing cards. StuRat (talk) 19:03, 25 April 2016 (UTC)[reply]

I suggest a "practical method". In this case, place a string over your illustration, following the arc, measure it's length, and apply the scale difference between the illustration and your model. StuRat (talk) 19:07, 25 April 2016 (UTC)[reply]

Thanks, that will work but I was sure there was a geometrical method I could use in Inkscape. Bridge arches are just the start of what I'm trying to do. Sfan00 IMG (talk) 19:14, 25 April 2016 (UTC)[reply]

My data does not make sense.

I have an Excel spreadsheet. The data is located here: User:Joseph A. Spadaro/Sandbox/Page79. (That is one column of data from my Excel spreadsheet. I just did a copy-and-paste.) Excel tells me that I have 66 data points. The sum of all of the data points is negative 100.12. And the average of all of the data points is negative 0.23. This does not seem to make any sense to me. Can this possibly be accurate? If not, what's the problem in Excel that I am not seeing? Thanks. Joseph A. Spadaro (talk) 18:46, 24 April 2016 (UTC)[reply]

If there is a blank cell in the column, does Excel treat that cell as a value of zero? But, even if that were the case (which I do not think that it is), the average still does not seem correct. What could be going on? Joseph A. Spadaro (talk) 18:53, 24 April 2016 (UTC)[reply]

Excel treats blank entries as if they're not there. I pasted the data into my own Excel spreadsheet and it tells me that average is -1.517 which I assume is correct. But there may be some setting somewhere that tells Excel to treat blanks as 0, which might explain what you're getting. Another likely explanation is that the range of cells you're averaging over doesn't include all the data; this can easily happen if you're adding/removing the numbers from month to month. --RDBury (talk) 19:20, 24 April 2016 (UTC)[reply]

Thanks. What is have is a bunch of rows. Let's say 75. And I fill them up one-by-one (each month). So, apparently, I got down to Row 66. In this column, there are 9 blank rows under the data. I take the sum -- and also the average -- of the entire column (from Row 1 all the way down to Row 75). This way, the sum and average will change month-by-month as each Row value is added. That should work correctly. Also, even if Excel puts "zero" in each blank cell, that average does not come out right. Is there a way to attach/include the actual spreadsheet somehow? Joseph A. Spadaro (talk) 19:43, 24 April 2016 (UTC)[reply]

@Joseph A. Spadaro: You could upload it to google drive (or another cloud platform) and link it from there — crh ₂₃ (Talk) 19:45, 24 April 2016 (UTC)[reply]

Thanks. How do I do all that? I have no idea what that even means. Joseph A. Spadaro (talk) 20:20, 24 April 2016 (UTC)[reply]

Just go to the Google Drive website and either log in (if you already have a Google account such as Gmail), or create a new account. My Excel agrees with RDBury's average of negative 1.517. What formulas appear in your total and average cells? We might be able to diagnose the problem from those. Dbfirs 11:36, 25 April 2016 (UTC)[reply]

These are the three formulas ... =COUNT(H3:H82) ... =SUM(H3:H82) ... and, lastly, ... =AVERAGE(H43:H82) ... Thanks. Joseph A. Spadaro (talk) 13:10, 25 April 2016 (UTC)[reply]

Unless H43 is a typo for H3, that could be where your problem is. Gandalf61 (talk) 13:44, 25 April 2016 (UTC)[reply]

That may be it! I will double check! Thanks. Joseph A. Spadaro (talk) 19:17, 25 April 2016 (UTC)[reply]

O.M.G. --JBL (talk) 14:46, 25 April 2016 (UTC)[reply]

One potential problem is if the range of cells which is averaged is finite, and you keep adding items, then you eventually exceed that range. From your description above it sounds like you have 80 cells in the range, and currently have 66 entries, so you won't hit this limit for another 14 months. If there's a way to define the cell range to be infinite, or at least large enough that you will never hit the end, like 1000, then that will fix the problem. StuRat (talk) 15:52, 25 April 2016 (UTC)[reply]

That's a good point. And I always consider that. When I have to add in a new row, I always add it into the middle of the range somewhere. I never add it right at the end (i.e., the last row). This helps to preclude the common error of having to modify the formula (when you extend the range by adding new rows) yet forgetting to do so. So I can just keep the old formula. And the old formula simply adjusts itself when new rows are added in to the middle of the range. Joseph A. Spadaro (talk) 19:15, 25 April 2016 (UTC)[reply]

Follow up question

Referring to the above problem, it seems that the error came about as a typo in a cell formula. The formula was =AVERAGE(H43:H82), which contained a typo. It should have been =AVERAGE(H3:H82). My question is: does Excel have any type of way to "catch" an error like this? Thanks. Joseph A. Spadaro (talk) 03:44, 26 April 2016 (UTC)[reply]

Excel sometimes puts a little green triangle at the top left corner to indicate a problem of some kind, such as a formula that isn't consistent with those around it. When the cell is selected you get an icon you can hover over to see a detailed message: with a similar situation to yours I got "The formula in this cell differs from the formulas in this area of the spreadsheet". AndrewWTaylor (talk) 07:47, 26 April 2016 (UTC)[reply]

I must have missed the part of this discussion where you made a general apology for wasting people's time. --JBL (talk) 13:12, 26 April 2016 (UTC)[reply]

To - Joel B. Lewis (JBL). Give me an honest answer to this question. Do you think that this was an honest and sincere mistake? Or do you think this was some big elaborate scheme that I concocted for the devious and nefarious plot to waste people's time? What's your honest answer to that question? Joseph A. Spadaro (talk) 04:48, 28 April 2016 (UTC)[reply]

That's rather unnecessarily harsh. I see it as an example of tunnel vision, where he couldn't see that the formula was wrong even when he cut and pasted it here, because he wasn't looking at the lower limit as a possible problem. StuRat (talk) 16:16, 27 April 2016 (UTC)[reply]

I agree that this is an easy mistake to make and that we shouldn't be harsh, and I had half a mind to make a comment to that effect myself - however, JBL didn't ask the OP to perform Harakiri for the shame of committing unforgivable sins. He merely asked for an apology. Given that the OP's silly, non-mathematical mistake did bring about a wasting of time, an apology isn't too much to expect. -- Meni Rosenfeld (talk) 21:26, 27 April 2016 (UTC)[reply]

Tell me what, pray tell, I should be apologizing for? Please be specific. And if this whole thing is such a waste of your precious time, why are you expending any time here replying? Joseph A. Spadaro (talk) 04:42, 28 April 2016 (UTC)[reply]

Please note that I didn't ask you for an apology, JBL did. I was merely replying to StuRat and explaining JBL's reasoning. You don't need to be defensive and hostile about everything.

It's not my time that was wasted - I didn't participate in the discussion. I am speaking with 0% skin in the game. As for expending time replying - I believe discussing the finer points of this forum's etiquette is a productive use of my time. (Well, probably not, but that's my own problem).

As for what you should be apologizing for - You made a typing/visual error in your spreadsheet. This has caused you to ask a question in this forum, which is meant for helping people with mathematical problems, including also dealing with mathematical mistakes that they make. This has led several people on a wild goose chase, trying to find a mathematical error that is not there. As it turns out, people have been wasting their time debugging a non-mathematical error, which is not what they signed up for when they volunteered to reply here. To summarize - discomfort to others has been brought about by your actions. By definition, when this happens one should apologize - it doesn't matter that this was all a result of an honest mistake, and an easy one to make at that (see also reply to StuRat below). It's not like giving an apology is such a big deal - it can be as easy as adding "Sorry about that!" to the followup question.

Anyway, I apologize for upsetting you with my comment. (That was a sincere apology, though not particularly heartfelt; and also a demonstration of how easy that is.) -- Meni Rosenfeld (talk) 08:41, 28 April 2016 (UTC)[reply]

If he had intentionally wasted our time I might agree, but an accidental oversight doesn't require an apology, IMHO. StuRat (talk) 23:18, 27 April 2016 (UTC)[reply]

We seem to have different definitions of apology. Intention to do harm is definitely not a prerequisite. Apologies can be about regret for bad actions or poor choices but they don't have to be, they are about empathy more than anything else. When one realizes discomfort/harm was caused to others as a result of his actions, he should apologize - even if he couldn't have done anything better! (In our case, I believe the OP's mistake is excusable but he wasn't 100% in the right either, he could have triple-checked for typos before coming here). -- Meni Rosenfeld (talk) 08:41, 28 April 2016 (UTC)[reply]

I am the OP. I assure you that no apology will be forthcoming. Please bank on that. I do, however, say "thanks" to those who helped answer my query. Most of you, that is. Not all. Thanks. Joseph A. Spadaro (talk) 04:35, 28 April 2016 (UTC)[reply]

This fact, however, cracks me up. An individual editor considers this question a "waste of his time". Yet, he goes out of his way to take the time to read, respond, and so forth. Not once, but multiple times. Indicating that he "followed" the conversation. It's a waste of his time. Yet, he finds time to invest in it. You can't make this shit up! Joseph A. Spadaro (talk) 04:39, 28 April 2016 (UTC)[reply]

Instead of laughing, consider trying to understand other people's motivations. Like me, JBL has found it appropriate to spend time refining this forum's etiquette. This has nothing to do with the effectiveness of spending time on the original question. And also, it doesn't take much time at all to scan the thread looking for an apology - much less time than it took to try answering the original question, as some people did.

If you'll excuse my armchair psychology, I'd say that both your reluctance to apologize, and the lack of interest in understanding people's motivations, might indicate a problem with empathizing. You might want to work on that. -- Meni Rosenfeld (talk) 08:41, 28 April 2016 (UTC)[reply]

Rings surjecting to localizations

Let R be a commutative ring. When is it true that the natural map from R to S^-1R is surjective for every multiplicative subset S of R? Is this equivalent to every prime ideal of R being maximal? GeoffreyT2000 (talk) 21:38, 24 April 2016 (UTC)[reply]

Do you mean Epimorphism, not Surjection?John Z (talk) 22:21, 24 April 2016 (UTC)[reply]

If R→S^-1R is surjective then S contains only units. Since if s is in S, there must be r in R so r→r/1=1/s, which implies rs=1.
If P is a prime ideal then S=R-P is multiplicative. If R→S^-1R is onto then R-P must contain only units, in other words P is maximal.

I believe that proves one direction, but Z is a counterexample for the other direction. Z-pZ is multiplicative but certainly contains non-units. Actually I'm having a hard time coming up with an example where the first condition holds other than a field. Maybe a local ring?--RDBury (talk) 08:27, 27 April 2016 (UTC)[reply]

April 25

What's the chance of getting AKQJT suited or AKQJT9 suited in a trick taking game?

I just got a "6 card" "super" royal flush (AKQJT9). What's the chance of that? I wonder if this'd still be worth much if poker was 13 cards not 5. It's an app so there's no guarantee there aren't bugs but assume there aren't any. Sagittarian Milky Way (talk) 22:26, 25 April 2016 (UTC)[reply]

There are

{\binom {52}{13}}

possible hands, of which

{\binom {46}{7}}

have the AKQJT9 in a particular suit. So your probability is very slightly less than four times the ratio of these numbers, roughly 1/3000. The probability that any of the four players gets such a hand is somewhat larger. --JBL (talk) 02:44, 26 April 2016 (UTC)[reply]

So not crazy unlikely then (not that I would've guessed that). Sagittarian Milky Way (talk) 05:22, 26 April 2016 (UTC)[reply]

Playing bridge once or twice a week, I've gotten 5-card flushes from time to time and royal flushes once or twice. I got a seven-card straight flush once, I think. Bubba73 ^{You talkin' to me?} 05:30, 26 April 2016 (UTC)[reply]

You did mean 5-card straight flushes, right? --69.159.61.172 (talk) 05:50, 26 April 2016 (UTC)[reply]

I've had 5-card straight flushes out of 13 cards pretty frequently and I think I had a 7-card straight flush once. Bubba73 ^{You talkin' to me?} 22:56, 26 April 2016 (UTC)[reply]

To clarify, Anon's question followed from the fact you wrote "5-card flushes" instead of "5-card straight flushes". Mere flushes are easy, of course. -- Meni Rosenfeld (talk) 16:18, 27 April 2016 (UTC)[reply]

Oh, sorry, I missed that point. I get five-card flushes every time I play! Bubba73 ^{You talkin' to me?} 23:50, 27 April 2016 (UTC)[reply]

April 27

Generalization of perimeter and surface area to higher dimensions

Is there a known name for the generalization of perimeter and surface area to higher dimensions, which produces an (n − 1)-dimensional quantity from an n-dimensional figure? GeoffreyT2000 (talk) 14:43, 27 April 2016 (UTC)[reply]

I would call it the "edge", as in the "edge length" (perimeter) of a circle or the "edge area" (surface area) of a sphere or "edge volume" of a 4D object. However, edge (geometry) defines that term far more restrictively. Boundary (topology) may be the proper term. (For proof that my use of the word "edge" isn't totally off the wall (although it is "outside the box"), see Manifold#Manifold_with_boundary, where they seem to use it as a synonym for "boundary".) StuRat (talk) 14:51, 27 April 2016 (UTC)[reply]

The standard term is "boundary". I've never heard "edge" used in this way. There are various ways of measuring it to get a number. One is the Hausdorff measure, which can be defined for quite general sets. If the boundary is rectifiable, then one can calculate the measure as an integral, in a manner analogous to the calculation of arc length and surface area in basic multivariable calculus. Sławomir
Biały 15:04, 27 April 2016 (UTC)[reply]

From Hypersurface: "Suppose an enveloping manifold M has n dimensions; then any submanifold of M of n − 1 dimensions is a hypersurface." So would the correct term be the hypervolume of the object's hypersurface? Loraof (talk) 16:05, 27 April 2016 (UTC)[reply]

Hypervolume, though the term is rarely used, generally means the n-dimensional volume. I think the standard term is

(n-1)

-dimensional measure. Often this implicitly means the Hausdorff measure, but it can sometimes mean other things like Jordan content. Sławomir
Biały 19:20, 27 April 2016 (UTC)[reply]

April 28