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June 3

Function question

Hello, this should be a relatively simple question. Recently I've learned about multi-variable functions, e.g. $f\colon \mathbb {R} ^{n}\to \mathbb {R}$; and of course I am familiar with single-variable functions, e.g. $f\colon \mathbb {R} \to \mathbb {R}$. I realized that in both cases, no matter how many real numbers are input, the function only outputs a single real number. This got me wondering, can the following functions (or if not, perhaps equations instead) exist?

• $f\colon \mathbb {R} ^{n}\to \mathbb {R} ^{n}$
• $f\colon \mathbb {R} ^{n}\to \mathbb {R} ^{n+1}$
• $f\colon \mathbb {R} ^{n}\to \mathbb {R} ^{n-1}$

Thanks! 74.15.5.167 (talk) 02:29, 3 June 2016 (UTC)

Sure. You can have a function from any set to any set, except that the set on the right can't be the empty set, unless the set on the left is also the empty set. (If you allow partial functions, you don't even have that restriction.) --Trovatore (talk) 02:36, 3 June 2016 (UTC)

For an example of a function from Rⁿ to Rⁿ, see Matrix difference equation#Non-homogeneous first-order matrix difference equations and the steady state. For an example from R^n×n to R^n×n, see Matrix difference equation#Nonlinear matrix difference equations: Riccati equations. Loraof (talk) 03:05, 3 June 2016 (UTC)

Vector fields in Euclidean space are a common example of functions from $\mathbb {R} ^{n}$ to itself, for instance. I'd think that you'll study the calculus of such functions once you study the calculus of real-valued functions on multivariate domains.--Jasper Deng (talk) 09:24, 3 June 2016 (UTC)

Point to clarify, regarding "single number". For a function F:R^n->R^n, the input and the output are both single elements of their respective sets. So if F(a,b,c)=(x,y,z), it is still sending a single input to a single output, even though we can characterize the output as an ordered n-tuple that sort of looks like a list of many numbers. For functions that sort of drop dimensions look at Projection_(set_theory) and projection_(mathematics). For A:R^n->R^m, consider $A\in \mathbb {R} ^{m\cdot n},\;x\in \mathbb {R} ^{n}$. Then using matrix multiplication, Ax=b is a mapping that can increase or decrease the dimension. SemanticMantis (talk) 14:42, 3 June 2016 (UTC)

A paper map of the Earth surface is a function from (some subset of) $\mathbb {R} ^{2}$, representing some range of (φ,λ), into (some other subset of) $\mathbb {R} ^{2}$, corresponding to (x,y) coordinates on a sheet of paper.
You can paint a picture of a landscape, which is a kind of a projection from $\mathbb {R} ^{3}$ to $\mathbb {R} ^{2}$.
You can make a 3D model of the ocean bottom's depth, thus creating a real representation of $\mathbb {R} ^{2}\to \mathbb {R}$ function.
You can describe your bike ride in time as $\mathbb {R} \ni t\mapsto (x,y)\in \mathbb {R} ^{2}$ or a projectile trajectory in time as $\mathbb {R} \ni t\mapsto (x,y,z)\in \mathbb {R} ^{3}$
Etc, etc, etc... --CiaPan (talk) 16:48, 3 June 2016 (UTC)

Symbol: X-like, but with a bar on top?

I believe once to have seen a math symbol looking like an X, the upper part being closed with a bar. Is there such a symbol, and if yes, what is it used for? --KnightMove (talk) 06:28, 3 June 2016 (UTC)

In probability theory ${\overline {X}}$ usually means a mean value of a random variable $X$. --CiaPan (talk) 06:42, 3 June 2016 (UTC)

But that bar isn't closing the top of the X.

Looking over the "Mathematical Symbols" sections of the Unicode code charts, I can find symbols looking like an x with a bar closing any of the other three sides, but not the top! The three I found are charted on these two pages and identified as:

• 22C9 (⋉) left normal factor semidirect product
• 22CA (⋊) right normal factor semidirect product
• 2A32 (⨲) semidirect product with bottom closed

Note that the words "with bottom closed" seem to be describing the symbol rather than explaining its meaning. --69.159.60.83 (talk) 06:50, 3 June 2016 (UTC)

According to The Comprehensive LaTeX Symbol List, three different LaTeX symbol packages have the top-closed product as \utimes, suggesting that someone has used it for something, but I don't know who or what. -- BenRG (talk) 08:36, 3 June 2016 (UTC)

(ec) In a set theory it is a complement of a set, and in Boolean algebra it is sometimes used for negation. --CiaPan (talk) 06:42, 3 June 2016 (UTC)

See also

• List of mathematical symbols, section 'Letter modifiers' and
• Overline, section 'Math and science'.

CiaPan (talk) 06:53, 3 June 2016 (UTC)

For clarity, KnightMove is asking about somthing like $\ltimes$ rotated 90° clockwise. Looks a bit like a folding table. —  crh ₂₃  (Talk) 09:54, 3 June 2016 (UTC)

I have to concede that my wording was ambiguous, so thanks for the clarifying. --KnightMove (talk) 05:58, 4 June 2016 (UTC)

June 5

Noetherian ring with non-Artinian total ring of quotients

Is there a commutative Noetherian ring with a non-Artinian total ring of quotients? GeoffreyT2000 (talk) 16:06, 5 June 2016 (UTC)

How about $\mathbb {R} [x,y]/(xy=0)$? It's Noetherian by basic results: fields are Noetherian; if $R$ is Noetherian, then so is $R[x]$; quotients of Noetherian rings are Noetherian. But its ring of quotients is not Artinian, because consider $(x)\supset (x^{2})\supset (x^{3})\supset \dots$.--2406:E006:45D:1:4022:A8FC:6159:3666 (talk) 05:36, 6 June 2016 (UTC)

June 6

Expected distance between point inside a hypercube and a vertex

Let $I_{n}=\int _{0}^{1}\int _{0}^{1}...\int _{0}^{1}{\sqrt {x_{1}^{2}+x_{2}^{2}+...x_{n}^{2}}}d{x_{1}}...d{x_{n}}$. $I_{1}=1/2,I_{2}=({\sqrt {2}}+\sinh ^{-1}{1})/3,I_{3}\approx 0.960591956455053$. What would be the best way to numerically evaluate this integral in higher dimensions? Or, if there's a simple closed form what is it? 24.255.17.182 (talk) 22:01, 6 June 2016 (UTC)

Use generalized spherical coordinates? See n-sphere#Spherical coordinates. Not that this is gonna make things much easier though - I'm sure that a hypercube's boundaries can be expressed as trigonometric functions in spherical coordinates, and integrating trigonometric functions can always be done symbolically. For example, using the notation from that article, we could write $x_{1}=1$ as $r=\sec(\phi _{1})$.--Jasper Deng (talk) 00:18, 7 June 2016 (UTC)

June 7

Algorithm to enumerate (n^2+n) sets of size n from n^2 items

I am seeking an algorithm to enumerate (n^2+n) sets of size n from n^2 items where every pair of items are in common in one and only one set.

Firstly, I believe it is possible. Secondly I have an algorithm that works for prime n: Arrange the items in a square. Read the columns for the "+n" sets. Then (for the n^2 sets) form n bunches of n sets, each bunch starting with an item from the left hand column and taking one item from each subsequent column using a different "drop per column" for each bunch. I'm using 'bunch' as a non-mathsy word instead of the more natural word "group" which might have math meaning.

Example

a b c
d e f
g h i

Columns adg, beh, cfi;
Drop 0 bunch abc, def, ghi;
Drop 1 bunch aei, dhc, gbf;
Drop 2 bunch ahf, dbi, gec.

This algorithm fails when n is non-prime. I can manually enumerate 20 sets for n=4, but can't extend the method to 6, 8 or 9 yet.

Any suggestions please? -- SGBailey (talk) 06:14, 7 June 2016 (UTC)

Would that be the Steiner system $S(2,n,n^{2})$? I think this may be slightly useful to get started. 5.11 seems to imply that the sets you wish to generate might not be possible for non-prime power n. —  crh ₂₃  (Talk) 14:10, 7 June 2016 (UTC)

As you say, it looks as though Steiner is what I want. I (believe I) can do S(2, prime, prime^2) and S(2,4,16). My first stumbling block would be S(2,6,36). Your refs need careful reading. Whether 5.11 says it is impossible I need to study. Thanks - this will keep me going for a week or too. -- SGBailey (talk) 16:27, 7 June 2016 (UTC)

You might try [1], section 4 in particular. It mentions the Thirty-six officers problem in regard to S(2,6,36). --RDBury (talk) 08:18, 8 June 2016 (UTC)

Elegant Solution to Sharp Inequality

How could we prove, without the aid of a calculator, that ${\sqrt[{5}]{12}}-{\sqrt[{12}]{5}}>{\dfrac {1}{2}}$ ? Moving the negative term to the right hand side, and then exponentiating, is —for painfully obvious reasons— unfeasible. Perhaps some clever manipulation might show the way out of this impasse, but I fail to see how... — 82.137.54.252 (talk) 07:51, 7 June 2016 (UTC)

Why do you think there is an easy way? ${\sqrt[{5}]{12}}-{\sqrt[{12}]{5}}\approx 1.64375-1.14353=0.50022$ That is so close to 0.5 that it doesn't leave much room for approximations, and you seem to reject both calculating the root or expanding a long exponential, so I don't see where you have any path forward. Dragons flight (talk) 08:10, 7 June 2016 (UTC)

Given my username and the section title, I feel somewhat obligated to attempt to reply to this, but I don't see a way either. The difference is so close to 0.5 that any approximations that would be simple enough to use without a calculator would not be accurate enough to prove the inequality. Double sharp (talk) 09:01, 7 June 2016 (UTC)

(Huh. So the word "inequation" is apparently a thing. I confess that I do not remember having heard it.) Double sharp (talk) 13:38, 7 June 2016 (UTC)

Hm. inequation says that it is a statement of inequality (which is a relation). But equation says that one of those is an equality. This is a troublesome state of affairs... SemanticMantis (talk) 14:27, 7 June 2016 (UTC)

I've tweaked the opening sentence of equation accordingly. Loraof (talk) 15:17, 7 June 2016 (UTC)

The usual term for these sorts of sentences is "inequality", not "inequation". I smell the odor of MathWorld here, which is one of the two references at inequation, the other being a book called "The A to Z of Mathematics" or some such, another name that does not inspire high confidence (sounds like a tertiary ref).

Wikipedia should not be used for these sorts of language-reform efforts. My opinion is that we should rework the articles to refer to the sentences primarily as "inequalities", while mentioning somewhere that the term "inequation" also exists. --Trovatore (talk) 16:01, 7 June 2016 (UTC)

I have a hazy recollection from long ago, an undergrad math professor using these words a bit differently to distinguish assertions from assignations, different from the notion in our articles of distinguishing statements from relations. E.g. 4=2+2 is an assertion (and a statement of equality) but 4=x is more like an assignment; it is an equation by fiat only. Does that ring a bell with anyone else? My first thought when seeing 'inequation' was that maybe somebody was taking this distinction to inequality as well, e.g. maybe 4>2 is considered an inequation while 4>x is considered an inequality. I cannot support this idea with any references at present though, so maybe my hazy recollections are playing tricks on me :) Finally, I'll add that cursory googling seems to indicate that 'inequation' has more use in BrEng (e.g. BBC [2]), so there may be an WP:ENGVAR issue conflated with a semantic one. SemanticMantis (talk) 18:29, 7 June 2016 (UTC)

I think you need to look up assignation :-) I doubt that's the issue. 4=x is a fine equation in general, albeit one containing a free variable. Assignments are generally distinguished by context (for example, "let x=4"), and there's no obvious corresponding concept in inequalities (you can say "assume x<4", but that's an assumption, which is different from an assignment, though you might be able to dispense with assignments by replacing them with assumptions). However the ENGVAR thing is a possibility; I would be interested to hear more evidence on that. --Trovatore (talk) 18:40, 7 June 2016 (UTC)

It came out as slip of the tongue, but then I did look it up, and decided it was fine, as the second definition in my OSX copy of NOAD is pretty much what we mean when we talk about assignments for variables, and this is also supported by sense 2 of your your wiktionary link. In case you're curious, I haveAssignation: "the allocation or attribution of someone or something as belonging to something." Pretty much a synonym to the second sense of Assignment: "the attribution of someone or something as belonging." So maybe you could benefit by looking in to Muphry's_law ;)

Anyway, I do know how to talk about free variables and assumptions and assignments, I just thought I'd share a thought I had about a possible distinction. SemanticMantis (talk) 19:01, 7 June 2016 (UTC)

• Does this work?

It can be shown by multiplication and division that 263/160 is an underestimate of ${\sqrt[{5}]{12}}$

It can be shown by multiplication and division that 183/160 is an overestimate of ${\sqrt[{12}]{5}}$

$263/160-183/160=80/160=0.5$ exactly.

Hence, (${\sqrt[{5}]{12}}-a)-({\sqrt[{12}]{5}}+b)=0.5$; where both a and b are positive.

${\sqrt[{5}]{12}}-a-{\sqrt[{12}]{5}}-b=0.5$

${\sqrt[{5}]{12}}-{\sqrt[{12}]{5}}=0.5+a+b$; where both a and b are positive.

${\sqrt[{5}]{12}}-{\sqrt[{12}]{5}}>0.5$.

--NorwegianBlue ^talk 21:38, 8 June 2016 (UTC)

I found two fractions with smaller numerators and denominators for which the same logic holds: 166/101 and 143/87. --NorwegianBlue ^talk 19:41, 9 June 2016 (UTC)

Maybe you meant to subtract 1/2 from the smaller of your two fractions? (Your first example has the smallest max denominator for any pair of rational numbers differing by at least 1/2 in the relevant interval.) --JBL (talk) 20:23, 9 June 2016 (UTC)

June 9

Differentiable multivariate function with discontinuous partial derivatives

As far as I know, having continuous partial derivatives is strictly a stronger condition than differentiability. Does anyone have a simple example of a function of more than one real variable that is differentiable, but whose partial derivatives are discontinuous somewhere?--Jasper Deng (talk) 10:07, 9 June 2016 (UTC)

Not at all sure of myself on this, but I start from the second (univariate) example in Smoothness#Examples:

The function

$f(x)={\begin{cases}x^{2}\sin {({\tfrac {1}{x}})}&{\mbox{if }}x\neq 0,\\0&{\mbox{if }}x=0\end{cases}}$

is differentiable, with derivative

$f'(x)={\begin{cases}-{\mathord {\cos({\tfrac {1}{x}})}}+2x\sin({\tfrac {1}{x}})&{\mbox{if }}x\neq 0,\\0&{\mbox{if }}x=0.\end{cases}}$

Because cos(1/x) oscillates as x → 0, f ’(x) is not continuous at zero. Therefore, this function is differentiable but not of class C¹.

If you multiply the function f by y to get g(x, y) and treat y times the f ′(x) expression as the partial of g with respect to x, does that give you what you're looking for? Here the partial of g wrt y equals the above expression for f (x).Loraof (talk) 14:27, 9 June 2016 (UTC)

The question is if that resulting function is differentiable, which is stronger than the existence of the partial derivatives.--Jasper Deng (talk) 19:48, 9 June 2016 (UTC)

No, the resulting function, the partial derivative, is not. As the quote from the article says, f ′(x) (hence partial g partial x) is not continuous at zero, so it is not differentiable there. Loraof (talk) 20:38, 9 June 2016 (UTC)

But like I said, that is not strictly necessary for differentiability of the multivariable function (in general; to be clear I'm talking about g(x, y) = yf(x))... so the example as-is still may or may not be differentiable.--Jasper Deng (talk) 20:42, 9 June 2016 (UTC)

f′, and thus g_x, is not even defined at x=0, since as the quote says cos(1/x) oscillates as x→0. If it's not defined there, it can't be differentiable there. Loraof (talk) 21:52, 9 June 2016 (UTC)

The quote says that f' is 0 at x = 0. It has no limit as it approaches 0, but is certainly defined there.--Jasper Deng (talk) 22:06, 9 June 2016 (UTC)

Is there susch a number as "reachable-attainable"?

(I think there is at least two) (I will not surely be back) 49.135.2.215 (talk) 00:59, 10 June 2016 (UTC)Like sushi