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February 19

Orthogonal

Hi, is an orthogonal a T shape or an L shape? ~ R.T.G 06:43, 19 February 2009 (UTC)[reply]

In mathematics, orthogonal is an adjective meaning "perpendicular to" (in other fields, it is still an adjective, but may have a slightly different meaning). So you can have a pair of orthogonal lines or orthogonal planes, or you can say that one line is orthogonal to another, but you can't have "an orthogonal". In both a T shape and in an L shape, the two lines are orthogonal to one another in each case. Gandalf61 (talk) 09:09, 19 February 2009 (UTC)[reply]

You can have an orthogonal relationship but clears it up now thanks ~ R.T.G 14:40, 19 February 2009 (UTC)[reply]

Surd Questions

Maybe its just the lack of sleep or something but i havent done these in a while and i have an assessment on these soon..im probably missing something pretty simple but any help would be appreciated =) !

[Ill show the answers i received]

Q.1

${\frac {2{\sqrt {8}}a^{3}}{a}}+4{\sqrt {2a}}-{\frac {5{\sqrt {3}}2a^{5}}{a^{2}}}$

$={\frac {4{\sqrt {2}}a^{3}}{a}}+4{\sqrt {2a}}-{\frac {20{\sqrt {2}}a^{5}}{a^{2}}}(Factoring)$

$=4{\sqrt {2a^{2}}}+4{\sqrt {2a}}-20{\sqrt {2}}a^{3}(Division/Subtracttopfrombottom)$

(i also thought about factoring the pronumerals[and move them infront of the radical then subtract/divide]but that didnt work out either..

$=4a{\sqrt {2a}}+4{\sqrt {2a}}-20{\sqrt {2}}a$

Answer is: $-12{\sqrt {2a}}$

Q.2

$1{\frac {4}{5}}{\sqrt {7}}-{\frac {2{\sqrt {7}}}{3}}$

i dont have much of an idea what to do here..anything to help me learn would be appreciated

Answer is ${\frac {19}{12}}{\sqrt {7}}$

Q.3

Just a few small things here...

I expand this by multiplying the equation in the first set of brackets by whats out side then.. multiplying it by the other one? Or by expanding the brackets first and..doing ..something i dont know

$2{\sqrt {3}}(1+4{\sqrt {3}})(1-{\sqrt {3}})$

$(2{\sqrt {3}}+7{\sqrt {3}})(1-{\sqrt {3}})$

$2{\sqrt {3}}-3{\sqrt {3}}+7{\sqrt {3}}-7{\sqrt {3}}$

Correct answer is $18-22{\sqrt {3}}$

(also how do i expand if theres 3 brackets :S)

Any help would be appreciated.

You're doing some rather strange things. You seem to be equating

2{\sqrt {3}}(1+4{\sqrt {3}})\,

with

2{\sqrt {3}}+7{\sqrt {3}}.\,

That is wrong. You need the distributive law:

2{\sqrt {3}}(1+4{\sqrt {3}})=(2{\sqrt {3}}\cdot 1)+(2{\sqrt {3}}\cdot 4{\sqrt {3}})\,

The first term in the last expression simplifies to

2{\sqrt {3}}.\,

The second term is

2{\sqrt {3}}\cdot 4{\sqrt {3}}=2\cdot 4\cdot {\sqrt {3}}\cdot {\sqrt {3}}=2\cdot 4\cdot 3=24.\,

And you've done something similar repeatedly throughout what you've done above. I haven't figured out how you got 7, but it shouldn't be there.

If you don't know that

{\sqrt {3}}\cdot {\sqrt {3}}=3,\,

then you have no idea what square roots are. That is what they are. Michael Hardy (talk) 11:37, 19 February 2009 (UTC)[reply]

Also, you wrote:

{\frac {2{\sqrt {8}}a^{3}}{a}}+4{\sqrt {2a}}-{\frac {5{\sqrt {3}}2a^{5}}{a^{2}}}

Are you ABSOLUTELY sure that it didn't say

{\sqrt {8a^{3}}}\,

rather than

{\sqrt {8}}a^{3}\,

and

{\sqrt {32a^{5}}}\,

rather than

{\sqrt {3}}2a^{5}\,

or

{\sqrt {32}}a^{5},\,

the former being what you wrote?

If you're not paying close attention to that sort of thing, then it's useless to get into the other parts of the problem.

The following are two DIFFERENT things:

{\sqrt {32}}a^{5}\,

{\sqrt {32a^{5}}}\,

Michael Hardy (talk) 11:43, 19 February 2009 (UTC)[reply]

Akin to absolute value

Is there a name, or standard notation, for the function $f:(0,\infty )\rightarrow (0,\infty )$ defined by

f(x)={\begin{cases}x,&{\mbox{if }}x\geq 1,\\1/x,&{\mbox{otherwise}}\end{cases}}

?

I'm thinking of it as being to multiplication what absolute value is to addition, also noting that the two are connected through exponentials and logarithms:

f(x)=\mathrm {e} ^{|\ln x|}

—Bromskloss (talk) 16:18, 19 February 2009 (UTC)[reply]

Help with LaTeX

Hey everyone. I'm having trouble getting something to compile correctly and my google-fu is failing me. Perhaps you can help.

What I want to have is a sentence appear on the next line, centered, and I want to be able to "label" the line all the way at the right with an asterisk (where an equation number might normally show up, say). Can anyone tell me how to do this?

Thanks in advance. –King Bee ^{(τ • γ)} 19:27, 19 February 2009 (UTC)[reply]

\hfill is basically what you want.

Wrapped up nicely in a command:

 \newcommand{\kingbee}[1]{~\newline\phantom{($\ast$)}\hfill$\displaystyle{#1}$\hfill($\ast$)\newline}

And a demo tex document:

\documentclass{article}
\newcommand{\kingbee}[1]{~\newline\phantom{($\ast$)}\hfill$\displaystyle{#1}$\hfill($\ast$)\newline}
\begin{document}
Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum.
\kingbee{x^2+1=0}
Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum.

\kingbee{x^2+1=0}

Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum.
\kingbee{x^2+1=0}

Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum.

\kingbee{x^2+1=0}
Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum.
\end{document}

Probably other ways to do it too, but the latex code eqnarray and equation was complicated. JackSchmidt (talk) 19:56, 19 February 2009 (UTC)[reply]

I recall using something from CTAN some time ago which allowed for arbitrary equation labels, say an asterisk or daggar, but it escapes me now. You could check there. Baccyak4H (Yak!) 19:59, 19 February 2009 (UTC)[reply]

If you want to do the stuff with math material, then you don't need any fancy macros, just the basic TeX primitive

$$ equation text \eqno (*)$$

will work, as well as LaTeX equivalents like

\[\tag{$*$} equation text\]

(unless you use options like leqno or fleqn). If you want to do it with text, you can either wrap it in a box:

$$\hbox{your sentence}\eqno (*)$$

\[\tag{$*$}\hbox{your sentence}\]

or you can play with extensible glue, like

\par\hbox to\hsize{\hss your sentence\hss\llap{$(*)$}}

The \par is there only to get you to vertical mode, simply leaving a blank line before the \hbox works just as well. — Emil J. 11:45, 20 February 2009 (UTC)[reply]

I just noticed that \tag appears to be an AMSLaTeX feature, so you need to \usepackage{amsmath} for that (or simply use the primitive $$ version). — Emil J. 11:48, 20 February 2009 (UTC)[reply]

Thank you all for your aid in this matter! –King Bee ^{(τ • γ)} 19:49, 23 February 2009 (UTC)[reply]

Finite multiplicative subgroup of direct product of fields

A finite subgroup of the group of units of a commutative field is cyclic. Is a finite subgroup of the group of units of a direct product of n commutative fields generated by at most n elements? It is fine to assume all the fields are equal, so just a direct power.

I have an abelian groups proof, but I was interested in something with polynomials or varieties or tori or something. JackSchmidt (talk) 19:36, 19 February 2009 (UTC)[reply]

Your question is equivalent to: for all positive integers

m_{1},\ldots ,m_{n}

, any subgroup of

\mathbb {Z} /m_{1}\mathbb {Z} \times \cdots \times \mathbb {Z} /m_{n}\mathbb {Z}

can be generated by at most n elements. But I do not know if that is true or not. Eric. 131.215.158.184 (talk) 07:35, 20 February 2009 (UTC)[reply]

~~I am not sure those are equivalent since you are now making claims about infinite subgroups too.~~ It is ~~however~~, as you say, equivalent to every subgroup of a direct product of n finite cyclic groups being generated by at most n elements. One just takes torsion subgroups of the unit groups of the fields to get a direct product of locally cyclic groups. A finite subgroup of exponent E is then contained in the direct product of n cyclic subgroups of exponents dividing E. JackSchmidt (talk) 12:13, 20 February 2009 (UTC)[reply]

Can't you get the same equivalence by just considering the projections of your finite subgroup G onto the n fields (which are cyclic by the original result), and noting that G is a subgroup of the product of these projections? Then show that the finite group statement is true by considering G as a quotient of Zⁿ and recalling that every submodule of Zⁿ is generated by at most n elements? Algebraist 12:27, 20 February 2009 (UTC)[reply]

I think we said exactly the same thing, but yes I agree, there is a reasonably straight-forward abelian groups proof. My question is mostly about using the structure of the fields in the proof. Like every finite subset, they are a (possibly reducible) algebraic variety, and I somewhat hope that there are ways of proving the result on the generators that might give more insight into the structure of the algebraic group. JackSchmidt (talk) 13:12, 20 February 2009 (UTC)[reply]

Algebraist: First question: yes, that was my reasoning. Second question: Can you elaborate on that reasoning? I assume by module you mean Z-module. Isn't G not a submodule of Zⁿ? Or does your argument proceed in a different direction? Eric. 131.215.158.184 (talk) 21:29, 20 February 2009 (UTC)[reply]

G is a quotient module of a submodule H of Zⁿ. Since H (is free over a commutative ring and so) has at most n generators, G also has at most n generators. JackSchmidt (talk) 21:51, 20 February 2009 (UTC)[reply]

JackSchmidt: Isn't that what I said? I'm certainly not talking about infinite groups in any case. Eric. 131.215.158.184 (talk) 21:29, 20 February 2009 (UTC)[reply]

Oh, right. I meant positive integers. I've gone back and fixed that now. Eric. 131.215.158.184 (talk) 21:36, 20 February 2009 (UTC)[reply]

Oh, no, this is my mistake (or at best a math rendering bug). The version of your comment that I read/imagined had some Z/mZ × Z × Z/mZ, that is, some bare Z not in a quotient that might be gotten from the structure theorem of finitely generated abelian groups, rather than of finite abelian groups. No wonder my comment was confusing. Sorry, JackSchmidt (talk) 21:49, 20 February 2009 (UTC)[reply]

Showing my working

Find the coordinates of the stationary point on the curve $y=xe^{-3x}$

I began by finding dy/dx: $dy/dx=(1-3x)e^{-3x}$

Then to find the x coordinate of the stationary point $dy/dx=(1-3x)e^{-3x}=0$

So for the expression on the left to equal 0, either (1-3x) or e^{-3x} must be equal to 0. One obvious solution is (1-3x)=0 when x=1/3 but what about the e^{-3x}? I know it doesn't equal 0 because no value of x could make it equal to 0.

So my question really is whether it's ok to show my working as follows: ...

$(1-3x)e^{-3x}=0$

$(1-3x)=0$

$3x=1$

$x=1/3$

...

Or is there a way to show why the e^{-3x} cannot equal 0? --RMFan1 (talk) 20:45, 19 February 2009 (UTC)[reply]

It is, of course, possible to prove that e^x never takes the value zero, but whether you're expected to do so will depend on the situation. In my education, for example, at school level (age 16-18 say) I was expected to take the important properties of e^x for granted, while as a first year undergraduate (age 19 or so) I was expected to be able to prove them all from the definitions, and ever since then I've been expected to take them for granted again. At a guess, if you're allowed to assume the chain rule and the fact that the derivative of e^x is e^x, you're probably allowed to assume it's never zero. Algebraist 21:03, 19 February 2009 (UTC)[reply]

In "showing your work", I would at least state that the step where e^x disappears is justified by the fact that e^x ≠ 0. I wouldn't do that in a research paper since "everybody knows" it, but in a homework assignment you're supposed to demonstrate that you know what you're doing. In an exposition within a Wikipedia article I might explicitly state it—that depends on the context. Michael Hardy (talk) 22:57, 19 February 2009 (UTC)[reply]

While we're on the subject of showing work, I'll repeat something that one of my supervisors drummed into me: in a fully written-out argument, in all but the most obvious contexts, every line should begin with actual English text, even if it's just 'thus' or 'so'. If you habitually just write down a sequence of equations or similar, it can be very unclear whether the equations are being asserted, assumed, inferred, claimed to be equivalent, taken as a definition, or whatever. Algebraist 00:06, 20 February 2009 (UTC)[reply]

If you don't like words, though, you can always use

\implies

and

\therefore

and similar. --Tango (talk) 00:38, 20 February 2009 (UTC)[reply]

Theory about graphs

Is there any theory about, or perhaps is it known, whether any graph can be described mathematically? For example, if I was to randomly scribble something onto some graph paper, could any such a graph definitely be described mathematically? --RMFan1 (talk) 23:56, 19 February 2009 (UTC)[reply]

When you say "graph", which of these two kinds are you referring to?

—Bromskloss (talk) 00:15, 20 February 2009 (UTC)[reply]

And what do you mean by "described mathematically"? --Tango (talk) 00:17, 20 February 2009 (UTC)[reply]

Im referring to graphs of functions. And I mean is it known whether anything someone could possibly scribble on paper could be defined by an equation. Or if there's a theory, general belief among mathematicians etc. on this --RMFan1 (talk) 00:26, 20 February 2009 (UTC)[reply]

I'm thinking kind of. Assuming you draw a continuous function, there is a finite polynomial that approximates the function to the point that the polynomial will never leave your graphite line. However, in the more mathematical sense, no. The number of possible mathematical equations is denumerable, and the number of possible functions that can be visualized is nondenumerable. In other words, there are more graphs you can make than there are equations to describe them. Indeed123 (talk) 03:54, 20 February 2009 (UTC)[reply]

It is too bad we don't have an article on mechanical curves. These were studied in the 16th century or so, and were found not to be described well by the standard mathematics of the time which was based either on classical geometry or polynomial/rational functions. I think an example curve is a catenary, but I don't have a reference handy. At any rate, these curves helped usher in the use of calculus into geometry and made analytic geometry more analytic and less algebraic. JackSchmidt (talk) 04:20, 20 February 2009 (UTC)[reply]

Possibly also the cycloid. JackSchmidt (talk) 04:25, 20 February 2009 (UTC)[reply]

The "number of possible mathematical equations" is *not* countable. 163.1.148.158 (talk) 20:40, 20 February 2009 (UTC)[reply]

I'm thinking it's more of a physics question: Are all possible movements of the pen such that they lie within what can be described by a mathematical expression? I have no answer to that and it seems to me that you would need an infinitely accurate description of physics to answer it. Do we even necessarily have well-defined paths of the pen in quantum mechanics? (I know to little about it.) —Bromskloss (talk) 07:52, 20 February 2009 (UTC)[reply]

See the article on Weierstrass Approximation Theorem. Bo Jacoby (talk) 09:39, 20 February 2009 (UTC).[reply]

A little while ago I saw a formula which, when graphed under the proper conditions, produced itself. I wonder if that would be relevant here. After all, every formula you scrawl on a piece of paper constitutes a piecewise continuous curve in the plane, which may or may not be described by a nice formula. What if it was, though, is there an infinite sequence of formulas where each graphs the previous one? Black Carrot (talk) 03:55, 21 February 2009 (UTC)[reply]

..."a formula which, when graphed under the proper conditions, produced itself" ... possibly Tupper's self-referential formula ? Gandalf61 (talk) 18:25, 21 February 2009 (UTC)[reply]