Wikipedia:Reference desk/Archives/Mathematics/2008 March 7

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March 7

Math Vocab

I am having some trouble finding these words:

1. Terms that have the same variable part are called______?
2. For a function f(x), a(n)______is an x-value for which f(x)=0 —Preceding unsigned comment added by 64.119.61.7 (talk) 01:10, 7 March 2008 (UTC)

I'm not sure about the first, but the second is probably either "zero" or, at least for polynomials, "root". Black Carrot (talk) 01:37, 7 March 2008 (UTC)

Also "x-intercept", when graphing the function. Confusing Manifestation(Say hi!) 02:40, 7 March 2008 (UTC)

I hope it is not your homework. The first one should be "like terms" and ConMan is right, the second is "x-intercept". Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) —Preceding comment was added at 11:01, 7 March 2008 (UTC)

I'd say "zero" is better than "x-intercept". The latter, as ConMan says, refers to the graph of the function, the former refers to the function itself. --Tango (talk) 11:27, 7 March 2008 (UTC)

Sine

Let's say, just for argument's sake, that explicit use of any trigonometric function besides the sine was outlawed. (Maybe the cosine is allowed too.) Exactly who would this inconvenience, and how? Black Carrot (talk) 01:40, 7 March 2008 (UTC)

Assuming the general meaning of the word inconvenient I would say that it would make trigonometric calculations very messy for everyone. Probably high school students would be the most confused. Other then that I can't think of any problems.--Shahab (talk) 05:57, 7 March 2008 (UTC)

$\cos \theta =\sin({\frac {\pi }{2}}-\theta )$

$\tan \theta ={\frac {\sin \theta }{\cos \theta }}={\frac {\sin \theta }{\sin({\frac {\pi }{2}}-\theta )}}$

So everything can be done using just sine, but it's certainly not convenient. --Tango (talk) 11:32, 7 March 2008 (UTC)

It was actually high school students I was thinking it might simplify things for. How does it help to have a specialized notation for the reciprocal of each function, for instance? This more than doubles the number of formulas involved, without saving any trouble anywhere else. Similarly, everything I've seen the tangent used in, with the exception of formulas translating from one trigonometric function to another, would be as easily expressed as the ratio of sine and cosine. Without those four, the entire class could be a study of the symmetry properties of a single curve (and perhaps an important translation of that curve), which seems simpler. In fact, in the class itself we were officially advised to replace the other functions with sines and cosines before trying to prove an identity, since it makes it so much more convenient. What do you think? Black Carrot (talk) 17:14, 7 March 2008 (UTC)

Take a look at Image:Circle-trig6.svg for an idea of why the reciprocals are useful in their own rights - they have their own geometric interpretations. Tan comes up very frequently, for example in reference to the gradient of a curve, and while you could just say "sin/cos" it's easier to say "tan". --Tango (talk) 17:28, 7 March 2008 (UTC)

Why have we stopped teaching the exsecant and versine, then, which also show up on that diagram? If it's a matter of covering all bases, they seem equally important. Black Carrot (talk) 18:38, 7 March 2008 (UTC)

Well, they're not particularly useful. They've just sec-1 and 1-cos respectively - they don't seem to come up often enough to warrant learning the names. If you look carefully, all the other functions appear as the sides of 3 similar triangles (sin, cos and 1; tan, sec and 1; and cot, cosec and 1) - those triangles are what make the functions useful. --Tango (talk) 21:27, 7 March 2008 (UTC)

Interesting. I think I see what you're saying. You said that tan comes up when you're talking about the gradient of a curve. Is there any similarly common use for, for instance, csc? Black Carrot (talk) 21:56, 7 March 2008 (UTC)

Nothing that springs to mind, but it's the length of the hypotenuse of a right angled triangle with the length of the side opposite the angle in question being 1. That situation is sure to appear from time to time. --Tango (talk) 22:33, 7 March 2008 (UTC)

Don't stop there! The whole field of trigonometry, and the whole notion of numerical angle, can be rather neatly excised from the study of geometry. Euclid didn't need it; he only cared about congruence of angles and their sums and differences, which can just as easily be expressed as properties of the sides of triangles containing the angles as of the numerical angles themselves. Classical geometry is algebraic, and the trigonometric functions, being transcendental, don't fit nicely into it. If you never use an angle in your problem statement or require an angle in your answer then you can eliminate them from your intermediate work too, just as you can ignore imperial units if they're never supplied or demanded. Trig and inverse trig functions are simply conversions between angle units and length units, so they disappear too. A guy named Norman Wildberger has written a book about this; see rational trigonometry. Unfortunately he writes like a crackpot (his book is even self-published!) but there's nothing wrong with the mathematics. A more widely used example of this idea is the quaternions, which are a fundamentally non-trigonometric representation of relative spatial orientation. There are some things you really need transcendentals for, but exp and log are good enough for that. -- BenRG (talk) 14:08, 9 March 2008 (UTC)

The trig functions are just shorthand for the ratio of the lengths of certain sides of a triangle, so yeah, you can just work with the ratios directly if you want. It's often easier to use the trig functions, though. --Tango (talk) 14:18, 9 March 2008 (UTC)

Well, I don't know about that. We never used trigonometry in geometry class anyway, or at least not much. We didn't hit trig until precal, and then I think it was more so we'd have some background for the functions' differential properties. Black Carrot (talk) 07:41, 10 March 2008 (UTC)

Log equation

Is there any way to solve for x in an equation in the form a*ln(x)=b*x? Imaninjapirate^{talk to me} 03:41, 7 March 2008 (UTC)

Looks similar to Example 1 in the Lambert W function article. --tcsetattr (talk / contribs) 04:27, 7 March 2008 (UTC)

Yeah that helps a lot! Thanks! Imaninjapirate^{talk to me} 03:45, 8 March 2008 (UTC)

Cauchy-Riemann equations

The article on the above contains the line: We can write $x=(z+{\bar {z}})/2$ and $y=(z-{\bar {z}})/(2i)$. Differentiating ${\mathit {x}}$ and ${\mathit {y}}$ gives:

${\partial x \over \partial z}={1 \over 2}\ \mathrm {and} \ {\partial y \over \partial z}={1 \over 2i}$

What I want is to prove that ${\partial {\bar {z}} \over \partial z}=0$. Can someone also please explain the definition of ${\partial {\bar {z}} \over \partial z}$. --Shahab (talk) 04:35, 7 March 2008 (UTC)

The problem is that ${\bar {z}}$ is not a Holomorphic function so it is not complex-differentiable, you can look at the limit ${\partial {\bar {z}} \over \partial z}=\lim _{h\rightarrow 0}{{\overline {z+h}}-{\bar {z}} \over h}=\lim _{h\rightarrow 0}{{\bar {h}} \over h}$ and see you get different values as h → 0 along the real or imaginary axes. So ${\partial {\bar {z}} \over \partial z}$ is not well defined.

In the section Cauchy-Riemann equations#Alternative formulation in complex coordinates note the very last line: Observe that in doing so, $z$ and ${\bar {z}}$ are regarded as independent variables for all the formal manipulations. Treated as independent variables you will trivially have ${\partial {\bar {z}} \over \partial z}=0$. --Salix alba (talk) 08:55, 7 March 2008 (UTC)

Notationally, ${\frac {\partial }{\partial {\overline {z}}}}$ is often simply defined to be ${\frac {1}{2}}\left({\frac {\partial }{\partial x}}+i{\frac {\partial }{\partial y}}\right)$ (and yes, it's a plus sign and not a minus sign). From here, ${\frac {\partial z}{\partial {\overline {z}}}}$ is defined and equals zero.

Note that in this setup, ${\frac {\partial f}{\partial {\overline {z}}}}=0$ if and only if f is holomorphic (indeed, this equation is seen to be equivalent to the Cauchy-Riemann equations if you simply expand it out). kfgauss (talk) 20:09, 12 March 2008 (UTC)

I seem to have told you how to compute ${\frac {\partial z}{\partial {\overline {z}}}}$ instead of ${\frac {\partial {\overline {z}}}{\partial z}}$, which is what you asked for. For the latter, use ${\frac {\partial }{\partial z}}={\frac {1}{2}}\left({\frac {\partial }{\partial x}}-i{\frac {\partial }{\partial y}}\right)$. kfgauss (talk) 20:12, 12 March 2008 (UTC)

Lines

Not sure whether to post this in Language, or Math. As I have been pestering the Language folk much lately, I have opted to pester the Math folk tonight. Why is it so common that people use the term "a straight line" ... as any "line" is "straight" by definition, no? If one were to say, "take out a piece of paper and draw a straight line" ... that means exactly the same thing as ... "take out a piece of paper and draw a line" ... correct? (Hereby acknowledging that line "segment" is truly correct in the previous sentence --- but you wouldn't hear "line segment" used, generally speaking, in everyday life.) (Nor do you actually "draw" a line, in that you merely draw a representation of a line.) I'm just curious about this. Mathematics aside, most laypeople understand that "line" means "straight" --- right? Any thoughts? No complex non-Euclidean mumbo jumbo, please. I'm just referring to regular folk. Thanks. (Joseph A. Spadaro (talk) 05:44, 7 March 2008 (UTC))

I'd say in laymen's terms, a line is any thin drawing like that, whether straight or not. Hence the fairly common term "curved line." While I think most people understand that a mathematical line IS straight, they also consider curves to be "lines". For example, Slur (music) markings would be a called a line over notes: indeed, the article uses the phrase "curved line" --Evan ¤ Seeds 06:17, 7 March 2008 (UTC)

In language, a "line" is considered to be "straight". However, in Maths., a "straight line" should be stated to be "straight" as you can't prove a line to be "straight" or "curved" if it is not given. "Straight line" is an important concept of Maths. since some theorems (eg: vertically opposite angles, adjacent angle on a "straight line", etc.) works only if a given line is straight. Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 11:16, 7 March 2008 (UTC)

I'd say it's the other way round. In layman's terms, a line can be straight or curved, in maths, a line is straight and you would say "curve" if you want to allow it to be curved (it should be noted that a straight line is a type of curve). --Tango (talk) 11:36, 7 March 2008 (UTC)

I think what Ftbhrygvn meant was a geodesic, such as the "straight lines" in elliptic geometry, which can also be interpreted as curved. Black Carrot (talk) 17:07, 7 March 2008 (UTC)

Geodesics in non-flat space are not straight, they are, in a sense, the closest you can get to straight lines, but they are certainly not straight. When people describe geodesics as "straight lines" it is almost always in quotes, because it's not strictly true. --Tango (talk) 17:25, 7 March 2008 (UTC)

How are they not straight? Take a geodesic on a hyperbolic plane of constant curvature, for instance, just one axiom different from the Euclidean plane. A line, along with being the shortest path between two points, also forms two right angles (180 deg) on each side of itself. That is, it at no time deviates from "straight ahead". Judged entirely on its own terms, straight line doesn't seem like a bad word for it. Black Carrot (talk) 18:36, 7 March 2008 (UTC)

Mathematical definitions are purely a matter of convention, and the convention I've been taught is that geodesics are not (generally) considered straight. That's why we have the word "geodesic". If geodesics were considered straight, we would just call them straight lines. --Tango (talk) 20:22, 7 March 2008 (UTC)

An editor above (EvanSeeds) made reference to the musical notation (Slur (music)) that is sometimes called a "curved line". The correct mathematical (geometry) term for that musical slur is what, exactly ... an arc? (Joseph A. Spadaro (talk) 20:46, 7 March 2008 (UTC))

An arc is a curve of constant curvature (ie. part of a circle). The general term is just "curve" (or "smooth curve", or "regular curve" to remove the chances of sharp points or self intersections, respectively). You could have a parabolic curve or a hyperbolic curve or various other specific curves - I don't know what a slur would be, probably nothing specific. --Tango (talk) 21:23, 7 March 2008 (UTC)

My response would be that language is often redundant, and there's nothing wrong with that--it's a feature, not a bug, of language. Chuck (talk) 21:00, 7 March 2008 (UTC)

Please see follow up question below (Follow up on question of "straight lines") ... on March 8 posting. Thanks. (Joseph A. Spadaro (talk) 20:49, 8 March 2008 (UTC))

RIEMANN ZETA FUNCTION

i have viewd riemann zeta function and i thought i understand it but i am really confused about two things.

• FIRST, [ζ(0)=-1\2]. when i tried to compute it, it shoud be equals, 1+1+1+... i mean is not any number to power zero should equal one or what??? i really dont get it.
• SECOND, riemann zeta function has zeros at the negative even integers. is it the same function,[ζ(s)=∑1\n^s]??? because obviously, for example, ζ(-2)=1+4+9+.... or what?

Husseinshimaljasimdini (talk) 11:30, 7 March 2008 (UTC)

The function defined on the half plane Re(s)>1 by the series ∑1/n^s has a unique analytic continuation to the whole complex plane except the point 1. However, the series is divergent when Re(s) is less or equal to 1, so other representations are necessary if you want to compute values there. See Riemann zeta function#Representations. Bikasuishin (talk) 12:04, 7 March 2008 (UTC).Oh!i got it now.thank you sir.Husseinshimaljasimdini (talk) 10:32, 8 March 2008 (UTC)

simple σ-algebra product identity

The answer to this must be trivial but I can't quite see it.

Let X,Y be sets and ${\mathcal {A}},{\mathcal {B}}$ algebras of subsets in X,Y respectively. By a rectangle with respect to ${\mathcal {A}},{\mathcal {B}}$ we mean a product $A\times B$ with $A\in {\mathcal {A}}$ and $B\in {\mathcal {B}}$. We let ${\mathcal {A}}\times {\mathcal {B}}$ denote the collection of all finite disjoint unions of rectangles with respect to ${\mathcal {A}},{\mathcal {B}}$. Then ${\mathcal {A}}\times {\mathcal {B}}$ is an algebra in $X\times Y$.

We denote by ${\mathcal {A}}\otimes {\mathcal {B}}$ the σ-algebra generated by ${\mathcal {A}}\times {\mathcal {B}}$. Also we denote by ${\mathcal {A}}^{\sigma }$ the σ-algebra generated by ${\mathcal {A}}$ in X.

Claim: ${\mathcal {A}}^{\sigma }\otimes {\mathcal {B}}^{\sigma }=({\mathcal {A}}\times {\mathcal {B}})^{\sigma }.$

Since $({\mathcal {A}}\times {\mathcal {B}})\subset ({\mathcal {A}}^{\sigma }\times {\mathcal {B}}^{\sigma })\subset ({\mathcal {A}}^{\sigma }\otimes {\mathcal {B}}^{\sigma })$ it follows that $({\mathcal {A}}\times {\mathcal {B}})^{\sigma }\subset {\mathcal {A}}^{\sigma }\otimes {\mathcal {B}}^{\sigma }.$

But why does the other inclusion follow?  — merge 14:18, 7 March 2008 (UTC)

To show ${\mathcal {A}}^{\sigma }\otimes {\mathcal {B}}^{\sigma }\subset ({\mathcal {A}}\times {\mathcal {B}})^{\sigma }$, it's enough to show ${\mathcal {A}}^{\sigma }\times {\mathcal {B}}^{\sigma }\subset ({\mathcal {A}}\times {\mathcal {B}})^{\sigma }$. So it's enough to show that if $X\in {\mathcal {A}}^{\sigma }$ and $Y\in {\mathcal {B}}^{\sigma }$ then $X\times Y\in ({\mathcal {A}}\times {\mathcal {B}})^{\sigma }$. But then it's enough to show that $X\times B$ and $A\times Y$ are in $({\mathcal {A}}\times {\mathcal {B}})^{\sigma }$, and this is obvious. Algebraist 15:45, 7 March 2008 (UTC)

Thank you. The first two sentences are clear, but neither part of the third one is—this is where I'm befuddled. I guess I just need to stare at it some more.  — merge 16:03, 7 March 2008 (UTC)

Well, $({\mathcal {A}}\times {\mathcal {B}})^{\sigma }$ is closed under (countable) intersections, so if it contains $X\times B$ and $A\times Y$ then it certainly contains $(X\times B)\cap (A\times Y)=X\times Y$. To show that $({\mathcal {A}}\times {\mathcal {B}})^{\sigma }$ contains $X\times B$, consider the collection of subsets X' of A such that $(X'\times B)\in ({\mathcal {A}}\times {\mathcal {B}})^{\sigma }$. It's easy to see that this collection is a σ-algebra, and it contains everything in ${\mathcal {A}}$, so it contains X. Algebraist 01:22, 8 March 2008 (UTC)

Wow, this completely lost me until I realised we had our notational wires crossed! Your A and B were what I had called X and Y, and to me A and B were arbitrary elements of ${\mathcal {A}}$ and ${\mathcal {B}}$. But the dust is starting to clear now. Thank you very much.  — merge 12:40, 8 March 2008 (UTC)

Sorry about that, I didn't read your question closely enough. For me any structure called ${\mathcal {M}}$ or whatever always has carrier set M. Algebraist 13:08, 8 March 2008 (UTC)

line to point reflection formula

The equation for generating a reflection curve for a point to point reflection is of course the equation for the ellipse. What is the equation for generating a line to point reflection curve or is this not possible, and if not, what would be the equation for generating a line to line (all points of the lines on the same extended line) reflection curve?

Elliptical point to point Reflection Curve

Are you asking what shape would focus parallel incoming lines to a single point? If so, you're looking for a parabolic reflector, as shown below:

jeffjon (talk) 15:30, 7 March 2008 (UTC)

Not parabolic. See 2nd drawing.

question dans l'infini debutons avec quel est la difference entre un nombre et un numero . see us. —Preceding unsigned comment added by Lhaadi (talk • contribs) 18:06, 7 March 2008 (UTC)

Traduction n'est pas intelligible. S’il vous plaît dire en anglais.

Line to line reflection curve

I don't think a prefect solution is possible. Consider the set of lines generated by reflection in some curve, the Envelope of the lines can be found. For the parabola/ellipse case this envelope will have a cusp at the focus, which is what makes the point particularly bright. I seem to recall that you can only get a sharp focus at points corresponding to vertex of the curve. Such vertices are isolated implying that the foci must be isolated. --Salix alba (talk) 20:13, 7 March 2008 (UTC)

Seems that would rule out a point to line or line to point reflection curve as well.