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March 10

Error in the fundamental theorem of calculus?

I integrate $f(x,y)={\sqrt {y}}$ over

$\Omega =\{(x,y)\in R^{2}:0\leq y\leq 1,-{\sqrt {y}}\leq x\leq 0\}=\{-1\leq x\leq 0,x^{2}\leq y\leq 1\}$

On one hand,

$\int _{0}^{1}\int _{-{\sqrt {y}}}^{0}{\sqrt {y}}\,dx\,dy=\int _{0}^{1}ydy=\left.{\frac {y^{2}}{2}}\right|_{0}^{1}={\frac {1}{2}}$

On the other hand, the same integral is equal to

$\int _{-1}^{0}\int _{x^{2}}^{1}{\sqrt {y}}\,dy\,dx=\int _{-1}^{0}\left[\left.{\frac {y^{3/2}}{3/2}}\right|_{x^{2}}^{1}\right]dx={\frac {2}{3}}\int _{-1}^{0}(1-x^{3})dx=\left.{\frac {2}{3}}(x-x^{4}/4)\right|_{-1}^{0}={\frac {5}{6}}$

Have fun.(Igny (talk) 00:05, 10 March 2008 (UTC))

Since $x<0$, ${\sqrt {x^{2}}}=-x$ and thus $(x^{2})^{3/2}=-x^{3}\;\!$ rather than $x^{3}$. ${\tfrac {1}{2}}$ is correct. -- Meni Rosenfeld (talk) 00:17, 10 March 2008 (UTC)

Oh well, that was fast. I didn't hide it well enough. (Igny (talk) 00:22, 10 March 2008 (UTC))

In general, the value of a double integral may depend on the order you do the integration (though not in this case) 163.1.148.158 (talk) 10:27, 10 March 2008 (UTC)

Can you give an example? Our Multiple integral doesn't say a lot about this, but I recall that the equality holds under fairly mild conditions. -- Meni Rosenfeld (talk) 10:38, 10 March 2008 (UTC)

From Hilary Priestley's book: the function ${\frac {x-y}{(x+y)^{3}}}$ on the unit square will do it. It's pretty clear what's going to happen I think. Or on the same region, ${\frac {x^{2}-y^{2}}{(x^{2}+y^{2})^{2}}}$. Even $e^{-xy}$ on $[0,1]\times [1,\infty )$ may do it. These are examples when Fubini's theorem tells you the function is not in $L(\mathbb {R} ^{2})$. One thing they tell you is that a function can look fairly innocuous, and still fail to be integrable over a compact region. Edit: a sufficient condition, from Fubini/Tonelli, for the double integrals of $f$ to be equal is that one of the repeated integrals of $|f|$ exists.163.1.148.158 (talk) 11:01, 10 March 2008 (UTC)

Another example, from Reed and Simon, is $f(x,y)$ defined on $\mathbb {R} ^{2}$ as,

$f(x,y)={\begin{cases}1,&x>0,\quad y>0,\quad 0\leq x-y\leq 1\\-1,&x>0,\quad y>0,\quad 0<y-x\leq 1\\0,&{\mbox{otherwise}}\end{cases}}$.

This function is pretty easy to visualize. It is only non-zero in the first quadrant, between the lines $y=x-1$ and $y=x+1$. It is 1 between $y=x+1$ and $y=x$, and -1 between $y=x-1$ and $y=x$. It's not hard to check that $|f(x,y)|$ is not integrable, and that the double integrals are different. 134.173.93.127 (talk) 06:07, 11 March 2008 (UTC)

Question

What is the correct pronounciation of "kilometre": "ki-loh-mee-tre" or "ki-lo-ma-ta"? 58.168.209.250 (talk) 01:20, 10 March 2008 (UTC)

You are likely to receive more helpful responses at Wikipedia:Reference_desk/Language. Michael Slone (talk) 01:29, 10 March 2008 (UTC)

I'd go for kill-om-e-tur. -mattbuck (Talk) 09:32, 10 March 2008 (UTC)

Some think it should be keel-o-meet-ur or keel-om-eat-ur.87.102.94.48 (talk) 15:56, 10 March 2008 (UTC)

And some pronounce it "stupid" :-) --Carnildo (talk) 19:49, 10 March 2008 (UTC)

Kill-'em-eat-her? —Keenan Pepper 20:00, 10 March 2008 (UTC)

Kill-'im, eat her - the cannibals wedding...87.102.94.48 (talk) 22:31, 10 March 2008 (UTC)

As he said, you're more likely to receive a helpful response at the Language desk. :) Black Carrot (talk) 00:20, 11 March 2008 (UTC)

Try this and this link (available at Merriam-Webster online dictionary through two red loudspeaker icons at this page). --CiaPan (talk) 15:12, 11 March 2008 (UTC)

Rings

I was wondering why mathematical rings are called "rings"? I can't think of any way in which rings are more "ringlike" than other algebraic systems. What's the history behind the name? Thanks. --Bmk (talk) 04:55, 10 March 2008 (UTC)

According to Ring theory#History, "The term ring (Zahlring) was coined by David Hilbert in the article Die Theorie der algebraischen Zahlkörper, Jahresbericht der Deutschen Mathematiker Vereinigung, Vol. 4, 1897." —Bkell (talk) 06:17, 10 March 2008 (UTC)

Ah, thanks - I was looking in the Ring (mathematics) article, and I didn't notice the article on Ring theory. Anyone know why Hilbert called them rings? I don't think I have access to that article. --Bmk (talk) 06:54, 10 March 2008 (UTC)

A review of the English edition of Hilbert's article[1] contains the phrase: "even though Hilbert uses the word "(Zahl)ring" for orders on algebraic number fields, this must not be taken as evidence that Hilbert employs here parts of our current algebraic terminology the way we would do it; rather than referring to a general algebraic structure, the word "ring" is used for sets of algebraic integers which form a ring in our modern sense of the word." I'm not quite sure what to make of this; it sounds a bit like the statement that the works of Shakespeare were actually not written by Shakespeare but by another person of the same name. It also does not clarify why Hilbert chose to use the word "Zahlring" for these sets of algebraic integers, but it may be a piece in the puzzle. My first speculation on reading the question was that it might have something to do with the cyclic structure of the rings Z/nZ for n > 1, but that is less likely in view of the quotation.  --Lambiam 08:50, 10 March 2008 (UTC)

Dictionary.com doesn't have an etymology on it, but I'm impressed they even have the definition. Black Carrot (talk) 09:25, 10 March 2008 (UTC)

Searching for -ring group etymology- on Google, however, does turn up this, where he says, "Short for Zahlring (German for number ring). Think of Z[2^(1/3)]. Here the generating element loops around like a ring." Black Carrot (talk) 09:26, 10 March 2008 (UTC)

Sounds like that's probably the explanation for the name. Thanks folks! --Bmk (talk) 17:35, 10 March 2008 (UTC)

f(x)=1^x&g(x)=(-1)^x

I have to questions here,is the function,f(x)=1^x,afixed point function?what is the value of g(x)if ,x=an irrational number like,2^1/2?thank you.Husseinshimaljasimdini (talk) 13:36, 10 March 2008 (UTC)

Exponentiation over the complex numbers is inherently a multivalued function. In some cases we can choose a nice branch and it will be single-valued; in other cases we cannot. For $1^{x}$ the obvious choice of branch is $f(x)=1$ which is a constant function. For $(-1)^{x}$ there is no such obvious choice. Its values are $(-1)^{x}=(e^{(2k+1)\pi i})^{x}=e^{(2k+1)x\pi i}=\cos((2k+1)x)+i\sin((2k+1)x)\;\!$ For $k\in \mathbb {Z}$. -- Meni Rosenfeld (talk) 14:00, 10 March 2008 (UTC)

Learning Calculus fast

Anyone have a strategy of getting a decent understanding of Calculus within about a months time? 131.91.80.75 (talk) 15:46, 10 March 2008 (UTC)

Get a decent tutor.(Igny (talk) 16:20, 10 March 2008 (UTC))

I grabbed my Additional Mathematics textbook and did every single problem in it but I don't think it's for everyone. x42bn6 Talk Mess 16:27, 10 March 2008 (UTC)

If you’re just looking for the concept as to what it is, and not the computational ability for doing stuff with it, I’d recommend the book “Calculus for cats.” It’s short, and uses a bit of humor, and I found it to be exceptional at getting across the concepts of calculus. Again, though, it is not a textbook: it’s more for somebody that’s curious about calculus, but that doesn’t actually plan to use it, or a supplement to a textbook to explain concepts that textbooks don’t do well. GromXXVII (talk) 23:18, 10 March 2008 (UTC)

Special functions

Consider:

$\,{d \over dx}f(x)=f(x)$

$f(x)=e^{x}\,$

The function is its own derivative.

But what about:

${\frac {|f''(x)|}{(1+(f'(x))^{2})^{3/2}}}=f(x).$

The function is its own curvature????? --Goon Noot (talk) 21:43, 10 March 2008 (UTC)

Well yes. But you don't really have that function as yet.. only a differential equation - I wonder what solution(s) would look like......87.102.94.48 (talk) 22:29, 10 March 2008 (UTC)

Yes, I think that's the question that's being asked. We all know a function which is equal to its derivative, but is there a function which is equal to its curvature? The answer: I have no idea! --Tango (talk) 22:40, 10 March 2008 (UTC)

Of course there is a function (infinitely many, actually). There's the trivial zero function; another starts with $1+{\frac {t^{2}}{2}}+{\frac {t^{4}}{6}}+{\frac {79t^{6}}{720}}+\cdots$. Whether it has a closed form is another matter entirely. Does anyone know of a version of Plouffe's inverter which gives a function based on its Taylor coefficients? -- Meni Rosenfeld (talk) 23:05, 10 March 2008 (UTC)

The zero function would be a dot at (0,0)? There's also a straight line at y=infinity. Is that right?87.102.14.194 (talk) 10:47, 11 March 2008 (UTC)

Well, yes, I was ignoring the 0 function. Any idea what the radius of convergence is for that power series? A function that's defined over the whole real line/complex plane would be nice. --Tango (talk) 00:34, 11 March 2008 (UTC)

I have no idea about this question other than if it has a full Taylor series (i.e., a positive radius of convergence), I would try to express the problem of finding it as one of recurrence relations. But otherwise no idea. Neat question though. Baccyak4H (Yak!) 01:56, 11 March 2008 (UTC)

I tried entering a power series expansion, and amazingly enough it turned ugly within the first couple of terms. If a_0 and a_1 are the constant and linear coefficients respectively (and are used to take care of the two degrees of freedom the DE allows) the expansion starts with (assuming I got the algebra right):

$a_{0}+a_{1}x+{\tfrac {1}{2}}a_{0}\left(1+a_{1}{}^{2}\right)^{\tfrac {3}{2}}x^{2}+{\tfrac {1}{3}}a_{1}\left(1+a_{1}{}^{2}\right)^{\tfrac {3}{2}}\left(1+3a_{0}\left(1+a_{1}{}^{2}\right)^{\tfrac {1}{2}}\right)x^{3}+{\mathcal {O}}\left(x^{4}\right)$

Don't ask me what the radius of convergence is like, though. I'm going to go out on a limb and suggest that the function doesn't have a nice closed form. However, like Baccyak4H says, you could always get the recurrence relation going so you at least know something about the coefficients. Confusing Manifestation(Say hi!) 03:53, 11 March 2008 (UTC)

It occurred to me that it might be more useful to work with the parametric form $y^{2}(x'y''-y'x'')^{2}=(x'^{2}+y'^{2})^{3}\,\!$ together with some extra constraint to fix the parameterization. With x'=1 you get the f(x) form above. With y'=1 you get $y^{2}x''(y)^{2}=(x'(y)^{2}+1)^{3}\,\!$, which has a closed-form solution for x' (not x): $x'(y)=\pm {\frac {\ln(y/y_{0})}{\sqrt {1-[\ln(y/y_{0})]^{2}}}}=\pm \tan \sin ^{-1}\ln(y/y_{0})$. Numerically integrated and plotted sideways (x(y),y) it looks like this:

for the case y₀ = 1, x(1) = 0. Other values of y₀ and x(y₀) just scale and translate it. I assume the general solution can be assembled from pieces of this. -- BenRG (talk) 16:10, 11 March 2008 (UTC)

Oops, I think I got this backwards—f was supposed to be the reciprocal of its radius of curvature. In that case the parametric form is $(x'y''-y'x'')^{2}=y^{2}(x'^{2}+y'^{2})^{3}\,\!$ and I get $x'(y)=\pm {\frac {y^{2}+C}{\sqrt {4-(y^{2}+C)^{2}}}}=\pm \tan \sin ^{-1}{\tfrac {1}{2}}(y^{2}+C)$. This does seem to have an antiderivative in terms of elliptic integrals, but it's fairly nasty and I'm not sure if it works for all C. I think Meni Rosenfeld's series is the case C = −3. For C > 0 you get a sinusoidal curve that looks like it could repeat across all of $\mathbb {R}$. -- BenRG (talk) 18:09, 11 March 2008 (UTC)

I'm guessing that the sinusoidal curve is very roughly like the cycloid curve, except more curvy at y=zero .. anyway how about using f(x)=a0+a1 sin x + a2 sin 2x +etc and attempting to solve in a similar fashion to the methods above that gave the first few terms easily.. 87.102.74.53 (talk) 19:07, 11 March 2008 (UTC)

Might just be me but I'm getting x'³(x'y''-y'x'')=y²(x'²+y'²)³ (where x' = dx/dt) - can someone point out where and if I'm,going wrong.. kindly please.87.102.74.53 (talk) 18:03, 11 March 2008 (UTC)

I don't think this can be right because it's not symmetric in x' and y'. -- BenRG (talk) 18:09, 11 March 2008 (UTC)

I just used the equation for curvature and put in the parametric derivatives.. maybe I made an obvious mistake - could you give a first step of what you did so I know I'm not barking up the wrong tree?87.102.74.53 (talk) 18:13, 11 March 2008 (UTC)

OOPS sorry I get x'³(x'y''-y'x'')²=y²(x'²+y'²)³.. hang on a minute —Preceding unsigned comment added by 87.102.74.53 (talk) 18:15, 11 March 2008 (UTC)

Ignore (boloks) I get the same, msut be drunk or getting old ignore previous87.102.74.53 (talk) 18:18, 11 March 2008 (UTC)

Mathematica churns for a few minutes, spits out several warnings exhorting us (humans) to check the answer by hand and various other terrible diagnostics, then says:

       {{y[x] ->    InverseFunction[(\[ImaginaryI] Sqrt[
        2] ((1 + 
             C[1]) EllipticE[\[ImaginaryI] ArcSinh[
              Sqrt[1/(2 - 2 C[1])] #1], (-1 + C[1])/(1 + C[1])] - 
          EllipticF[\[ImaginaryI] ArcSinh[
             Sqrt[1/(2 - 2 C[1])] #1], (-1 + C[1])/(
           1 + C[1])]) Sqrt[(2 + 2 C[1] - #1^2)/(1 + C[1])] Sqrt[(
        2 - 2 C[1] + #1^2)/(1 - C[1])])/(Sqrt[1/(1 - C[1])] Sqrt[
        2 - 2 C[1] + #1^2]
         Sqrt[-2 (1 + C[1]) + #1^2]) &][-\[ImaginaryI] x + 
    C[2]]}, {y[x] -> 
  InverseFunction[(\[ImaginaryI] Sqrt[
        2] ((1 + 
             C[1]) EllipticE[\[ImaginaryI] ArcSinh[
              Sqrt[1/(2 - 2 C[1])] #1], (-1 + C[1])/(1 + C[1])] - 
          EllipticF[\[ImaginaryI] ArcSinh[
             Sqrt[1/(2 - 2 C[1])] #1], (-1 + C[1])/(
           1 + C[1])]) Sqrt[(2 + 2 C[1] - #1^2)/(1 + C[1])] Sqrt[(
        2 - 2 C[1] + #1^2)/(1 - C[1])])/(Sqrt[1/(1 - C[1])] Sqrt[
        2 - 2 C[1] + #1^2]
         Sqrt[-2 (1 + C[1]) + #1^2]) &][\[ImaginaryI] x + C[2]]}}

Enjoy, Robinh (talk) 08:48, 11 March 2008 (UTC)

Could someone convert that back into maths? Or maybe not - it really looks like mathematica(TM) has 'gone insane' over this question - specifically 'square root insanity' would be my diagnosis... Poor old computer.87.102.14.194 (talk) 08:57, 11 March 2008 (UTC)

What does the graph look like????--Goon Noot (talk) 10:06, 11 March 2008 (UTC)

If you mean meni rosenfeld's answer .. it looks like a 'steep' parabola, or a cosh function.. that sort of shape (assuming the curvature is always considered positive ie magnitude.87.102.14.194 (talk) 10:35, 11 March 2008 (UTC)

Someone asked about a "A function that's defined over the whole real line/complex plane" technically menirosenfeld's function (as I'm now calling it) or in general functions that satisfy the equations given by Confusing Manifestation will work for imaginary numbers. But what if the complex function used to describe a scalar value ie f(x)=a+ib g(x)=sqrt(aa+bb) and the curvature of that scalar at x is equal to the the scalar at x.. That's impossible to analyse right? or am I missing some more maths to learn?87.102.14.194 (talk) 11:02, 11 March 2008 (UTC)

What does complex curvature mean??--Goon Noot (talk) 15:48, 11 March 2008 (UTC)

difficult to give a 'real world' equvivalent - as curvature is as you know the radius of a circle corresponding to the rate of change of slope at a point, then a complex curvature means the circle has a complex radius.. If the slope is changing complexely (ie the complex part is changing) then the radius of curvature will have a complex coefficient. Obviously this won't happen if x and y are always real.. Did that explain at all, oe help?87.102.74.53 (talk) 18:09, 11 March 2008 (UTC)

Is it just me or Jakob Bernoulli's "Spira Mirabilis" is its own evolute? So the equation $r=e^{a\theta }$ is this curve that he is asking about, right?A Real Kaiser (talk) 05:02, 12 March 2008 (UTC)

Mmh I assumed they meant radius of curvature = y (rectangular), but if radius or curvature = r (polar) you are probably right.(or very close) luckily Spira_mirabilis#Properties saves me the bother of having to work it out, you are right it is indeed its own evolute.87.102.17.32 (talk) 13:40, 12 March 2008 (UTC)

As far as I can tell im certain that the radius of curvature of (sin(x)e^x,cos(x)e^x) is not e^x and can't find an exact solution87.102.17.32 (talk) 17:21, 12 March 2008 (UTC)

The functions e^z, e^kz have (eg the function used as radius when z is angle in polar coords) have radius of curvature sqrt(2)e^z, sqrt(1+k²)e^kz ie the radius of curvature is always proportional to the function, but not the same.. 87.102.32.239 (talk) 23:15, 12 March 2008 (UTC)

To GOON NOOT - I realised the answers I've been giving were based on radius of curvature not curvature - ie the reciprocal, apologies for any confusion.87.102.17.32 (talk) 16:15, 12 March 2008 (UTC)

One way of looking at the problem is to consider a parametric representation of the curve, in which the point (x(s), y(s)) is a function of the arclength s. If we also introduce φ(s), giving the direction of the curve expressed as the angle of the tangent with the x-axis, then, writing x =x(s) etc. as usual,

${\frac {dx}{ds}}=\cos \varphi ,\,\,\,{\frac {dy}{ds}}=\sin \varphi ,\,\,\,{\frac {d\varphi }{ds}}=\kappa ,\,$

where κ is the curvature, which may itself be a function of s, x, y, and φ. Here it is given that |κ| = y, which is a bit indeterminate: whenever y reaches 0, in general two different continuations are possible.

Assuming κ = y, we have

${\frac {dy}{d\varphi }}={\frac {\sin \varphi }{y}},$

which is separable and can be solved to give

${\tfrac {1}{2}}y^{2}+\cos \varphi =C.\quad (\ast )$

This is the generic form of the integral curves of the vector field assigning to the point (φ,y) the directional vector (y,sin φ). Each value of C > −1 in (*) gives one curve. Assume we start in an area where −1 < C < 1. By inspection of the vector field diagram, we see anti-clockwise cycles around (π,0), which correspond to a sinus-like curve in the x-y-plane wiggling around the x-axis in the negative direction. As we move away from φ = π and pass φ = π/2 (or φ = 3π/2), the wiggling gets more pronounced and becomes like meandering.

If we start in a position with C > 1, y cannot vanish, so the sign of y is invariant. According to the sign, φ will monotonically increase or decrease, and in the x-y-plane we get to see cycles. I haven't analyzed the critical case C = 1.

Another approach I have not further looked into is to put v = dy/ds, so that on the one hand

${\frac {d^{2}y}{ds^{2}}}={\frac {dv}{ds}}={\frac {dy}{ds}}\,{\frac {dv}{dy}}=v{\frac {dv}{dy}},$

while also

${\frac {d^{2}y}{ds^{2}}}={\frac {d\,\sin \varphi }{ds}}=\cos \varphi \,{\frac {d\varphi }{ds}}=\pm y{\sqrt {1-v^{2}}},$.

which combine to give another separable equation:

$v{\frac {dv}{dy}}=\pm y{\sqrt {1-v^{2}}}.$

I hope I did not make many mistakes, since this was scribbled on a (too small) napkin. --Lambiam 18:12, 13 March 2008 (UTC)

division by zero

We all learn in elementary school that any number divided by itself is 1. Later, we learn that division by zero is undefined. I wondered why it is undefined, because if any number divided by itself is one, shouldn't $0/0$ be 1? I thought about this...

Consider $f(x)=r/x$ where r is any real number.
As lim_x→0⁺, x_→+∞
Also, as lim_x→0^–, x_→-∞

So does $0/0$=+∞ and -∞?
J.delanoy^gabs_adds 23:28, 10 March 2008 (UTC)

We have an article on Division_by_zero. Black Carrot (talk) 23:29, 10 March 2008 (UTC)

Sorry, I didn't realize that... J.delanoy^gabs_adds 23:31, 10 March 2008 (UTC)

No problem at all. Feel free to expand on your question if the article isn't detailed enough. Black Carrot (talk) 00:06, 11 March 2008 (UTC)

March 11

Topology

I was curious about the definition of a topology. Going by the article, it says after the definition that "All sets in T are called open; note that not all subsets of X are in T".

Considering this:

$X=\lbrace 1,2,3\rbrace \,$

$T=\lbrace \emptyset ,\lbrace 1\rbrace ,\lbrace 2\rbrace ,\lbrace 3\rbrace ,\lbrace 1,2\rbrace ,\lbrace 1,3\rbrace ,\lbrace 2,3\rbrace ,\lbrace 1,2,3\rbrace \rbrace$

1) Isn't T equivalent to the powerset of X? Or is this just the case for simple examples like this?

2) Isn't "not all subsets of X are in T" incorrect in the above example? Damien Karras (talk) 09:24, 11 March 2008 (UTC)

It means 'not all subsets of X need be in T'. If they all are, we call T the discrete topology on X. Algebraist 10:31, 11 March 2008 (UTC)

Ok, I think I get you. Apologies for the informality, I'm trying to establish a "visual" ideal of what a topology is. Starting with a set of points (ordered pairs (m,n)) on a plane. If we have a set of points, T within X that are bounded within say, a circle, we can have a smaller subset of points not in T - i.e. a 2 dimensional projection of a torus (or a topologically equivalent shape)? Damien Karras (talk) 12:28, 11 March 2008 (UTC)

$T=\lbrace \emptyset ,\mathbb {R} ^{2},\lbrace (x,y)|x^{2}+y^{2}<1\rbrace \rbrace$ is a valid topology on the plane. Does that answer your question? Any collection of subsets satisfying the axioms of a topology is a topology. --Tango (talk) 12:52, 11 March 2008 (UTC)

Also note that that is only one way of defining it; some texts take T as a family of neighborhoods and derive the open sets from there. GromXXVII (talk) 11:35, 11 March 2008 (UTC)

Well, a topology just describes what open sets look like in your space (call it X). In general, a topology will not include all possible subsets of your space X. The power set of X is one of the possible topologies on X but it is not the only one. So in the power set, all subsets are open but in a general topology, this may not be true. For example, here is another topology on X, {the empty set, X} with only two elements. You can verify the axioms of a topology and prove that this is a perfectly valid topology (called the trivial topology). In fact this is the smallest topology and the power set topology is the largest topology in the sense that any other topology will be contained in the power set topology and contains the trivial topology. Another not obvious example of a topology would be that a set in R is open if it contains the number zero. Obviously, not all sets are open under this topology. Hope this helps!A Real Kaiser (talk) 04:55, 12 March 2008 (UTC)

It says in the article "all sets in T are open", yet what about $T=\lbrace \emptyset ,\mathbb {R} ^{2},\lbrace (x,y)|x^{2}+y^{2}<=1\rbrace \rbrace$? Is that not a valid topology? I cannot include a point that exceeds a distance of one, even by the tiniest amount, so it's closed? Damien Karras (talk) 08:28, 12 March 2008 (UTC)

I believe you’re assuming a metric on the real numbers, which induces a different topology from the one you want to look at. GromXXVII (talk) 11:07, 12 March 2008 (UTC)

Don't confuse open sets in topology with open sets in real analysis. In real analysis you start with a distance function (a metric) and define an open set to be one in which every point in the set has a neighborhood in the set. Then you can prove various things about open sets: the empty set is open, the whole space is open, a union of open sets is open, a finite intersection of open sets is open, a function is continuous iff the preimage of any open set under the function is open. In topology you turn that on its head: you define the open sets as any set of subsets that contains the empty set and the whole space and is closed under arbitrary union and finite intersection, you define continuity of a function by the open-preimage condition, and so on. The reason is that in topology you're studying properties that are invariant under homeomorphisms, and (as a nontrivial theorem in analysis and a trivial theorem in topology) the open sets are exactly the structure on the space that's preserved by homeomorphisms. Of course, there turn out to be topologies that don't arise from metric spaces. -- BenRG (talk) 11:19, 12 March 2008 (UTC)

Put simply, all sets in T are open because that's what "open" means. What you generally think of as open is "open with respect to the topology induced by the Euclidean metric", but in topology "open" is just the name we give to members of the set T. --Tango (talk) 14:13, 12 March 2008 (UTC)

This may be nitpicking, but in the "open if it contains the number zero" example, doesn't the empty set have to be open (as well as closed) in any topology ? Should that be "open if it contains the number zero or is the empty set" ? Gandalf61 (talk) 07:08, 12 March 2008 (UTC)

Yes, the empty set is an open set, so that would need to be included as well. In the example, to be even more nitpicky, he didn’t say if and only if, and I bet you could derive that the whole space was closed, and so the empty set would be open anyway. I don’t think there are any other open sets though. GromXXVII (talk) 11:07, 12 March 2008 (UTC)

I did wonder whether specifically saying the empty set was open would be redundant in the "open if it contains the number zero" example, but I don't see how that can be. If you are not told that the empty set is open or (equivalently) that R is closed then I don't see how you can derive either of these propositions because:

1. You can't construct R from the intersection of closed sets because one of these sets must be R itself, which you don't know is closed.
2. You can't construct R from the union of closed sets because one of these sets must contain 0, and so will be open.
3. You can't construct the empty set from the intersection of open sets because all the open sets that are given contain 0.
4. You can't construct the empty set from the union of open sets because the empty set is only the union of other copies of itself, and you don't know that it is open.

Am I missing something here ? Gandalf61 (talk) 11:46, 12 March 2008 (UTC)

The best thing to do would be to specify that it is open. I still think the construction is possible, although not as easy as I thought it was. The empty set and the whole set are always open and closed for a reason – it’s not arbitrary. So something should break down by not having the empty set open: and the construction could come from that. GromXXVII (talk) 12:45, 12 March 2008 (UTC)

They are always open because the first axiom of a topology says so. If they could be proven to be open, you wouldn't need to take that as an axiom. What would break down is that you wouldn't have a topology so none of the results we have about topological spaces would apply. There is one way of "proving" the first axiom: The empty set is the empty union of open sets and the whole space is the empty intersection of open sets. That's just a matter of defining the empty union and intersection appropriately rather than defining a topology appropriately - it's still a matter of definition. --Tango (talk) 14:09, 12 March 2008 (UTC)

Well, you guys are right that I didn't specify the empty set being open. But I didn't need to say if and only if because I am defining my open sets to be all subsets of R which contain the number zero. In definitions, we don't "have to" say if and only if, because a definition already includes that. A set is open therefore it contains zero, by definition. A set contains zero therefore it is open by definition. So, in a definition, I don't need to specify if and only if. So here is my formal definition of a topology on R. A set in R is said to be open if it is either empty or if it contains the number zero.A Real Kaiser (talk) 20:02, 12 March 2008 (UTC)

math

how does a polynomial divided by a binomial be used in real life situations? —Preceding unsigned comment added by Lighteyes22003 (talk • contribs) 16:42, 11 March 2008 (UTC)

Rational functions#Applications has a few uses. I can't immediately think of any direct real world applications, but rational functions (of which a polynomial divided by a binomial is a special case) are very useful in various areas of maths which do have real world applications. --Tango (talk) 16:50, 11 March 2008 (UTC)

You can use this to help you find the roots of a polynomial. If you have, say, a fifth-degree polynomial, there is no general algebraic way to find its roots. But if you can discover one root somehow (maybe by guessing), say $x=3$, you can divide the fifth-degree polynomial by $(x-3)$ to get a fourth-degree polynomial, which you can solve algebraically (though it's messy; see quartic equation). —Bkell (talk) 18:31, 11 March 2008 (UTC)

Yes, that's the most obvious use of polynomial division. I was thinking of the case where the binomial doesn't divide the polynomial, otherwise you actually just have a polynomial written oddly and not really a polynomial divided by a binomial. (The question say "divided by" not "dividing by", but I may be reading to much into the wording.) --Tango (talk) 18:35, 11 March 2008 (UTC)

Mmmh bit of a long shot - but for inverse square relationships somtimes there are equations such as

e^-kx/x² or e^-k(x-a)/(x-a)²

which could be of theoretical interest to physicists.. does that count?87.102.74.53 (talk) 19:04, 11 March 2008 (UTC)

It could be of use in astrophysics where atronomers need to acurately calculate parabolic paths of bodies. I suppose this could also be applied to ballistics of any sort PiTalk - Contribs 20:01, 11 March 2008 (UTC)

Wouldn't a parabolic path be the solution to a quadratic?87.102.74.53 (talk) 21:11, 11 March 2008 (UTC)

That would count if an exponential were a polynomial... --Tango (talk) 21:53, 11 March 2008 (UTC)

It is, sort of - an infinite one..87.102.17.32 (talk) 13:43, 12 March 2008 (UTC)

It can be expressed as a power series, sure. A polynomial has a finite number of terms, otherwise it isn't a polynomial - polynomials and power series have some very different properties. --Tango (talk) 18:54, 12 March 2008 (UTC)

You've never heard the term 'infinte polynomial' then or Euler's claim that "what holds for a finite polynomial holds for an infinite polynomial".

Also try searching for "finite polynomial" - it's a common phrase So just stop posting wrong stuff please.87.102.17.32 (talk) 19:41, 12 March 2008 (UTC)

"A polynomial has a finite number of terms" is not wrong. All due respect to Euler, but I reckon the vocabulary was quite limited in his time and he thus resorted to using this odd language (and of course, his statement is dead wrong). "Finite polynomial" is definitely not common, based both on my experience and a quick google test:

• "Polynomial" - 5000000 ghits.
• "Finite polynomial" - 5000 ghits, most seem to be taken out of context.
• "Infinite polynomial" - 2000 ghits.

When people speak of polynomials they mean something with certain properties, most of which unsatisfied by a general power series. -- Meni Rosenfeld (talk) 20:17, 12 March 2008 (UTC)

I don't see much difference in properties between an infinite and finite polynomial, (or to change the semantics on its head - "infinte and finite power series"). Are there really any major differences I should be aware of, excluding the number of 'nomials' in each? 87.102.17.32 (talk) 20:31, 12 March 2008 (UTC)

For a start, with power series you have to worry about convergence, which you obviously don't for polynomials. --Tango (talk) 21:00, 12 March 2008 (UTC)

Right, but as far as you're concerned a the exponential function is not a polynomial.87.102.17.32 (talk) 21:02, 12 March 2008 (UTC)

By the standard definition of polynomial (which is pretty much universal), an exponential is not a polynomial. --Tango (talk) 21:04, 12 March 2008 (UTC)

"Polynomial of infinite degree not a polynomial."?? where is this standard definition that excludes the infinite case87.102.17.32 (talk) 21:26, 12 March 2008 (UTC)

[outdent]See polynomial. One trademark characteristic of polynomials is that if you differentiate one enough times, you end up with zero. Another is that a polynomial can only grow so fast asymptotically (at a rate which is called, unsurprisingly, "polynomial"). Another is that it (for a positive degree) always has a complex root, in fact, as many as the degree of the polynomial if you count multiplicities. Polynomials exist over any ring and do not depend on a topology (being composed of just additions and multiplications, not limits). The list goes on, while your "infinite polynomial", a term which is virtually never used, is basically just any analytic function. -- Meni Rosenfeld (talk) 22:16, 12 March 2008 (UTC)

I get that these are properties of finite polynomials - again if you include polynomials of infinite order you lose that 'trademark characteristic' . or at least you would have to differentiate infinite times (I know that is meaningless)

Don't quite know what you meant by 'for positive degree always has a complex root' - I assume that was a reference to Fundamental theorem of algebra, though it sounded like you were saying polynomials always have complex numbered solutions for polynomial=0?

Look at Polynomial#Elementary_properties_of_polynomials this extends to polynomials of infinite order. I'm just trying to suggest not making a semantic distinction between polynomials and 'power series' - as far as I see it infinite series are a subset of polynomials and the exponential function, sin etc are in that subset, and as such inherit the properties of finite polynomials - being careful to note of course that when an operation on the function is dependent on the degree, that operation will never reach a final state in the case of degree=infinity.

I'm convinced that it's productive to treat both finite and infinite power series as examples of the same set. And yes all members of that set are analytical functions.

87.102.32.239 (talk) 23:43, 12 March 2008 (UTC)

Yes, no doubt finite power series are a special case of power series. In fact, this case is so special that it was even given a special name - "polynomial". But that's really backwards. Polynomials come first as a composition of the basic operations that exist in any ring - addition and multiplication. You can then discuss what happens when you take limits - a nontrivial feature of reals and complexes, not shared with most rings - of polynomials, and end up with power series. But almost everything which makes polynomials what they are is lost in the limiting process. -- Meni Rosenfeld (talk) 13:22, 13 March 2008 (UTC)

It could be used to approximate a more complicated function with a simple pole (i.e. something that behaves like it's been divided by (x+a)), though for what real life purpose I'm not sure.

It's not really an answer or anything, but this came up on my watchlist and made me laugh:

Wikipedia:Reference desk/Mathematics‎; 22:17 . . (+192) . . ConMan (Talk | contribs) (→math: possible answer)

No clue about the function. -mattbuck (Talk) 00:10, 12 March 2008 (UTC)

March 12

Least Squares Question

Let $v_{1}={\begin{bmatrix}1\\1\\0\\1\end{bmatrix}}$ and $v_{2}={\begin{bmatrix}0\\0\\1\\0\end{bmatrix}}$ and $b={\begin{bmatrix}0\\1\\0\\-1\end{bmatrix}}$. Let $V=span\{v_{1},v_{2}\}$. I am trying to find the vector in V that is closest to the vector b. Now, my questions is that I am trying to solve the system Ax=b and I know that b is my b above and $x={\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}}$ is a four dimensional vector but what would my matrix A be? Would it be just $A={\begin{bmatrix}1&1&0&1\\0&0&1&0\end{bmatrix}}$? But then I can't multiply both sides by transpose of A and them multiply by the inverse of (Transpose(A)*A)? Basically I want to have that $x=(A^{*}A)^{-1}*A^{*}$ but I can't figure out what my A will be. Any help would be appreciated! Thanks!A Real Kaiser (talk) 05:27, 12 March 2008 (UTC)

Because you're searching for the best approximation in V, you want the range of A to be V. The range of a matrix is the span of its columns, and so A is the 4 x 2 matrix with $v_{1}{\mbox{ and }}v_{2}$ as columns. Your x is actually a length 2 vector, representing the coefficients of $v_{1}{\mbox{ and }}v_{2}$. In this case, $A^{*}A$ will be square, and invertible. Check out Moore-Penrose inverse. 134.173.93.127 (talk) 07:46, 12 March 2008 (UTC)

Ironically, I can see by inspection that b is orthogonal to the vs. So the closest vector is just the zero vector. Baccyak4H (Yak!) 18:41, 12 March 2008 (UTC)

Actually, that was my intuition also that zero would be the answer (because that is the one and only point both spaces share so obviously it is the closest one). My I just couldn't understand how to setup my matrix A. And when I tried to do it (the way you basically said), I was wondering as to why my matrix x was only two dimensional. Now it makes sense the the entries of matrix x represent the coefficients for linear combination of v1 and v2. So this means that the x1v1+x2v2 would be the vector that I am looking for, right? In span{v1,v2}?A Real Kaiser (talk) 04:50, 13 March 2008 (UTC)

That's right. 134.173.93.127 (talk) 07:41, 13 March 2008 (UTC)

Which two spaces? There's the space V, and the vector b which is not in it. You may have meant span{b}, which indeed intersects V only at 0. But this is not enough for the closest vector to be zero - for this you need orthogonality. Consider v1 = {1, 0, 0}, v2 = {0, 1, 0} and b = {1, 1, 1}. Only {0, 0, 0} is common to span {v1, v2} and span {b}, but the closest vector in V to b is {1, 1, 0}. -- Meni Rosenfeld (talk) 14:36, 13 March 2008 (UTC)

Probabilities and this year's Champions League quarter finals

OK, I wasn't too shabby at Maths when I last studied it, but I always struggled with Probabilities. I thought of a nice bunch of Probabilities questions based on the forthcoming Champs Lg draw. Can you help me understand how to solve them, and thereby make up for either a duff teacher all those years ago, or, more likely, my inattention in class at crucial moments:

• There are 8 remaining teams in the competition
• 4 are English
• The draw will put them into 4 pairings that will decide the semi-final line up
• Assume all teams are of equal ability and AGF that UEFA don't rig their draws!

Q1 What's the probability of exactly 1 all-English quarter-final?
Q2 What's the probability of exactly 2 all-English quarter-finals?
Q3 What's the probability of at least 1 all-English quarter-final?
Q4 What's the probability of exactly 1 all-English semi-final?
Q5 What's the probability of exactly 2 all-English semi-finals?
Q6 What's the probability of at least 1 all-English semi-final?
Q7 What's the probability of an all-English final?

Oh, and leaving my probability of inattention in class aside (which, frankly is nearly 100%), if my maths teacher had used problems like this, instead of boys choosing socks in the dark (when they could just switch a ruddy light on) I might be able to do this without your help!

Cheers, --Dweller (talk) 12:23, 12 March 2008 (UTC)

Well, I will try to provide some of the answers, but I don't guarrantee that they are right :)

First, let's count how many combinations of quarter-final ties there are (only accounting that Team1 will be drawn against Team2, without accounting who is drawn to host the first leg). It should be $C_{2}^{8}={\frac {8!}{2!(8-2)!}}=28$.

A1. Out of these there is a $C_{2}^{4}={\frac {4!}{2!(4-2)!}}=6$ possibilities that exactly 2 English clubs are drawn together, thus making a probability of ${\frac {6}{28}}={\frac {3}{14}}\approx 0.21$, i.e. 21%.

A2. I suppose, therefore, that the probability of 2 all-English quarter-finals should be $({\frac {3}{14}})^{2}={\frac {9}{196}}\approx 0.05$, i.e. 5%.

A3. And, therefore, the probability of at least 1 all-English quarter-final should be the sum of above two probabilities, i.e. ${\frac {3}{14}}+{\frac {9}{196}}={\frac {51}{196}}\approx 0.26$, i.e. 26%.

It's hard to go on from here because we have to account for different cases.

A5. Of course, there can't be exactly 2 all-English semi-finals if there's at least 1 all-English quarterfinal. The probability that no English teams will be drawn together in the quarter-finals should be, according to the above information, $1-{\frac {51}{196}}={\frac {145}{196}}\approx 0.74$, i.e. 74%. Assuming that all clubs are of equal ability to qualify for the semi-finals, the probability that all English teams will win their ties in this case is $({\frac {1}{2}})^{4}={\frac {1}{16}}=0.0625$, i.e. 6.25%. Multiplying these two would give us the answer: ${\frac {145}{196}}\times {\frac {1}{16}}={\frac {145}{3136}}\approx 0.05$, i.e. 5%.

Well, the other questions are more complicated than these, maybe I will come back to them later... maybe not :). Again, I'm not sure if what I did above is right, but it sure makes sense. To me at least! Hope that helps!  ARTYOM  13:18, 12 March 2008 (UTC)

Urk. Sorry to be picky after you've done all that work, but you'll need to slow down. I think I remember what the ! is (but can't remember what it's called - is it "factorial" or something?) but haven't a clue what the big C thingy is, nor how/why you populate the first equation that leads to 28. Remember, I want to understand how to do this, not just learn the answers! --Dweller (talk) 13:47, 12 March 2008 (UTC)

For the benefit of people similarly mathematically illiterate to me who come here (I advertised this thread at WT:FOOTY) the ! is indeed Factorial --Dweller (talk) 13:52, 12 March 2008 (UTC)

Yes, it is, and the C stands for choose. $C_{2}^{8}$ is the number of ways of choosing 2 members from a set of 8 elements. --Tango (talk) 14:43, 12 March 2008 (UTC)

Thanks for that. So remembering about ! and learning that C notation means I've already learned something. --Dweller (talk) 15:10, 12 March 2008 (UTC)

I don't think that's right - only a 26% chance of an all-England matchup doesn't sound right at all. Your ${\frac {6}{28}}$ is the probability that any given tie, taken independently of the others, will be an all-England clash. But you can't just square that to get the probability of two, because the ties are not independent.

By my reasoning the probability that there are no all-English quarter-finals is given by placing the English teams into the draw first; so the first team can be placed in any of the 8 spots, the second in 6 of the remaining 7 spots (i.e. not against the first), the third in 4 of 6, and the last in 2 of 5, giving ${\frac {8}{8}}\times {\frac {6}{7}}\times {\frac {4}{6}}\times {\frac {2}{5}}={\frac {8}{35}}\approx 22.86\%$. So the probability of at least one all-English tie is the remainder, $\approx 77.14\%$. I can't think how to do the rest right now but I'll get back to you! — sjorford++ 14:53, 12 March 2008 (UTC)

In these things, it is often helpful to do the extreme cases first:

• Question 3 is easiest to answer, since we just want to find the probability of NOT having an english QF. As above, this is ${\dfrac {8}{8}}\times {\dfrac {6}{7}}\times {\dfrac {4}{6}}\times {\dfrac {2}{5}}={\dfrac {8}{35}}$, and thus P(at least 1 english QF) = ${\dfrac {27}{35}}$.
• Question 2. EXACTLY TWO english QF. Let us pick the english teams first. We can choose any position for the first team, so that is irrelevant. Let the 2nd team be a match (1/7). Then 6 choices for the 3rd, and 1 for the 4th. Thus we have ${\dfrac {8}{8}}\times {\dfrac {1}{7}}\times {\dfrac {6}{6}}\times {\dfrac {1}{5}}={\dfrac {1}{35}}$ chance of this. Now, suppose the 2nd team does not get a match (6/7), then the 3rd team has 2 options, and the 4th team just 1. Thus chances of 2 matches is ${\dfrac {1}{35}}+{\dfrac {8}{8}}\times {\dfrac {6}{7}}\times {\dfrac {2}{6}}\times {\dfrac {1}{5}}={\dfrac {1}{35}}+{\dfrac {2}{35}}={\dfrac {3}{35}}$
• Question 1. Now, there were 3 possible cases with 4 teams - 1 match, 2 matches, or no matches. So, now $P(1)=1-{\dfrac {3+8}{35}}={\dfrac {24}{35}}.$
• Question 5. For 2 english semi-finals, we require that the english teams all have different matches (8/35), and that they all win (1/2). Thus, P(2 E SF) = ${\dfrac {8}{35}}\left({\dfrac {1}{2}}\right)^{4}={\dfrac {8}{35}}\times {\dfrac {1}{16}}={\dfrac {1}{70}}$.
• Question 4. Here it gets difficult, as we need to consider many cases.
  • The easiest is if there are 2 english QFs (3/35) - then we can guarantee exactly two english teams in the SFs. The chance of them facing each other is ${\dfrac {1}{3}}$, which when combined with the probability of 2 english QFs is ${\dfrac {1}{35}}$.
  • Now suppose that there are no english QFs (8/35). Then to get 1 english QF we require at least 2 teams to win.
    • Suppose exactly 2 do win - there are $^{4}C_{2}=6$ ways to do this, each of which has a probaility of $\left({\dfrac {1}{2}}\right)^{4}={\dfrac {1}{16}}$ (2 teams must win, 2 lose). Now, we require that the 2nd team face the 1st team. This has a chance of 1/3 as above. Thus here we have ${\dfrac {8}{35}}\times {\dfrac {6}{16}}\times {\dfrac {1}{3}}={\dfrac {1}{35}}$
    • Suppose 3 English teams win - $^{4}C_{3}=4$ ways, each with P = 1/16. Any draw we are bound to have 2 english teams facing off, thus we have ${\dfrac {8}{35}}\times {\dfrac {4}{16}}\times {\dfrac {1}{1}}={\dfrac {2}{35}}$
  • Now, suppose there is one english QF (24/35). We are bound to have 1 english team in the SF.
    • Suppose both other english teams win (P = (1/2)^2 = 1/4). Then we have 3 teams and 4 spots - there's bound to be an english SF, with $P={\dfrac {24}{35}}\times {\dfrac {1}{4}}={\dfrac {6}{35}}$
    • Now, suppose that only one of the other enlish teams win. There are 2 ways to do this, each with probability 1/4. Now the chances of them facing each other in the SF is 1/3. Thus $P={\dfrac {24}{35}}\times {\dfrac {2}{4}}\times {\dfrac {1}{3}}={\dfrac {4}{35}}$
  • Thus, we now sum these probabilities: $P={\dfrac {1+1+2+6+4}{35}}={\dfrac {14}{35}}={\dfrac {2}{5}}$
• Question 6 - this is now easy, we just sum the probabilities of 1 and 2 english SFs. $P={\dfrac {2}{5}}+{\dfrac {1}{70}}={\dfrac {29}{70}}$
• Question 7 - Almost there, but again several cases to consider.
  • The most obvious place to start is with 2 all-english SFs. We know this had probability 1/70. We must have an all-english final here.
  • Next consider there are exactly 3 english teams in the SFs. This happens if either there is one english QF and both other teams win (P = 6/35), or no english QFs and 3 teams win (P=2/35). Thus we have a total P=8/35. We're bound to have 1 english team in the final, and the other has a 0.5 chance. Thus we have a 4/35 chance from this case.
  • We could take forever finding the probability of 2 english teams in the semis. However, note that there are an equal number of english and foreign teams, and that they all have equal chances of winning. Thus, the chances of two no-english SFs is 1/70, and the chance of only one english team in the SFs is 8/35. Thus, we have used up 17/35 options, and the remaining 18 must be exactly 2 teams facing each other, of which 1/3 are going to have the english teams facing each other and thus can be discarded. So we have a 12/35 chance of having exactly 2 english teams in different matches. Now, we require that both win. This has P = 1/4. Thus the chance of an english final here is ${\dfrac {12}{35}}\times {\dfrac {1}{4}}={\dfrac {3}{35}}$.
  • Now, sum these: $P={\dfrac {1}{70}}+{\dfrac {4}{35}}+{\dfrac {3}{35}}={\dfrac {3}{14}}$

Damn that was a lot of work. We'd better fkn win. -mattbuck (Talk) 16:16, 12 March 2008 (UTC)

Simpler approach for Q7: Forget the route to the final - there are (8x7)/2=28 possible pairs of teams in the final, of which (4x3)/2=6 consist of a pair of English teams. So probability of an all-English final is 6/28, which is 3/14. Gandalf61 (talk) 16:32, 12 March 2008 (UTC)

Damn my looking for the complicated route. I thought there should be an easy way to do it. -mattbuck (Talk) 16:46, 12 March 2008 (UTC)

Math: real-life examples of polynomial division?

what are some examples from real life in which you i might use polynomial division? —Preceding unsigned comment added by Lighteyes22003 (talk • contribs) 13:38, 12 March 2008 (UTC)

Finding roots of polynomials is the obvious one. It comes up all the time in various disciplines, since lots of things are described (at least approximately) by polynomials and it's often useful to know when they are zero. --Tango (talk) 13:45, 12 March 2008 (UTC)

See Linear response function for one example. It is used in radio and television design. Bo Jacoby (talk) 13:56, 12 March 2008 (UTC).

I saw your previous question and felt I should answer it, but it's hard to come up with an answer at what I think is your level of understanding. As others have mentioned, it's used to find roots of polynomials (e.g. to find more roots after the first is known). I doubt you are familiar with eigenvalues or root locus plots, but those require polynomial roots. Let's say I'm designing a control system for an aircraft. I've taught this using the roll control system of an F-18 as one example. First, you write out the equations of motion of the aircraft. The result is a system of differential equations. You then apply a Laplace transform to that system, so you get a system of polynomial equations in 's' (the Laplace variable). If you can find one root, by numerical or other means, you can use polynomial division to simplify the equation, getting the entire set of real and complex roots. Depending on those roots (are they real or complex? are their real parts negative or positive? how close to the origin are they?) you can tell a lot about the behaviour of the aircraft. For instance, if the real parts are negative, the aircraft is stable, otherwise it is unstable. Adding a control system to the loop modifies the set of differential equations and allows us to set the roots where we want them, ensuring the behaviour of the aircraft is how we want it, and that the aircraft is stable, maneuverable, and other desirable properties. moink (talk) 14:37, 12 March 2008 (UTC)

And the reference desk serves its function of helping us improve articles: Root locus needs some serious work. moink (talk) 14:44, 12 March 2008 (UTC)

One thing is for sure - polynomial division is a much less important mathematical skill for an average individual than others which are unfortunately not studied enough (or at all) in school, such as logic and probabilistic thinking. -- Meni Rosenfeld (talk) 15:08, 12 March 2008 (UTC)

That's for sure -- I agree 100%. Logic especially needs to be taught more. It is completely foreign to most students, unfortunately. (Joseph A. Spadaro (talk) 08:18, 13 March 2008 (UTC))

Chi-square test

When calculating a chi-square test what are the steps between calculating the differences between the expected and observed values and obtaining the chi-square value? Then, knowing the degrees of freedom, how is the p-value of the chi-square obtained (other than by looking it up in a table)? Thanks. The chi-square test article does not explain these points. Itsmejudith (talk) 15:47, 12 March 2008 (UTC)

If you are talking about a contingency table, then the expected value for each entry is

(column total / table total) × (row total / table total) × table total.

I know this expression can be simplified, but in this form it is easier to see conceptually what is being estimated. Subtract this from the observed value, square the difference, and divide by this expected value again. The sum over all the table is the value of the χ² statistic. Call that value X. Then the p-value is just the probability that a χ² distribution with the appropriate degrees of freedom is at least X (that is, one minus the CDF of the χ² distribution, at X). Baccyak4H (Yak!) 18:33, 12 March 2008 (UTC)

Can this detail be added to the worked example in the article? Thanks. Itsmejudith (talk) 08:50, 13 March 2008 (UTC)

OK, done. Although I can see the point someone might make that it is too much "How to...". Baccyak4H (Yak!)

March 13

Generating random numbers

How would I generate random numbers with a distribution matching Zipf's law? --Carnildo (talk) 04:18, 13 March 2008 (UTC)

Could you start with a probability distribution that is linear ie each value occurs equally often and scale that to the zipf distribution? Would you need help scaling the fnuction?

Can it be assumed that you already have a method of generating random numbers in general?87.102.8.240 (talk) 09:10, 13 March 2008 (UTC)

According to the classic version of Zipf's law, the number of occurrences of the word that has rank k in a large corpus of words is proportional to 1/k, which means it is of the form c/k for some large constant c. But actually the value of c depends on the size of the corpus, where c is the number of different words in the corpus, and the size n of the corpus is roughly . Therefore it makes more sense to generate a sequence of numbers for which each initial segment obeys Zipf's law.

In pseudocode:

Let z₁, z₂, ... be the sequence to be generated.

Set c to 1

For n from 1 to whatever:

Set ν to 1/((log c)+1)

Select one alternative from:

(A) with probability ν:

Set z_n to c

Set c to c+1

(B) with probability 1 − ν:

Pick a random number uniformly from the range {1 ... (n−1)}

Set z_n to z_r

Obviously the beginning will have few different numbers, so you may want to discard a large initial part. It is then not actually necessary to store the segment to be discarded itself, as long as the number of occurrences of each value is kept track of. --Lambiam 09:58, 13 March 2008 (UTC)

The above is surely right - but looked a little mysterious to me - here's another version of 'psuedo-code' - in this case I don't explicitly normalise the probabilities but instead generate a sum "Total" of 1/k and find the position of a random number between 0 and "Total" ie the first 'word' is between 0 and 1/1, the second between 1/1 and 1/1+1/2 etc

For different functions simply change all instances of 1/value to the new thing eg 1/value^s I've marked these positions with a ^*

n=number of words

Total = Sum to n of 1/k ^* !ie 1/1+1/2+1/3 etc

LOOP 1

R=Total x RND() ! ie Generate a random number between 0 and Sum total called R {RND() is a random function between 0 and 1} !

Length=0

m=1 !a loop counter

LOOP 2

Length=Length+1/m ^*

if R<=Length then print/output "m"  ! m is the m^th word ! : GOTO LOOP 1 !start again!

increase m by 1 : goto LOOP 2

Just added in case others found the first example a little confusing.. apologies to any offended by the GOTOS.. the !!'s are comments. not factorials!87.102.8.240 (talk) 10:57, 13 March 2008 (UTC)

A counting variable can be added so that the program would stop after a certain number of random values had been produced. Would suggest adding this inside LOOP 1 eg Count=Count+1:If Count=10000 then END/STOP>87.102.8.240 (talk) 11:03, 13 March 2008 (UTC)

Note - what I've done here shares some similarities with Arithmetic coding or Entropy encoding - at least in small part.87.102.8.240 (talk) 11:15, 13 March 2008 (UTC)

(it could be extended to non-zipf distributions that have finite states from a single event, also the above could be speeded up slightly computer wise if that was neccessary - ask if you want further details on these.)83.100.138.116 (talk) 16:28, 13 March 2008 (UTC)

Financial calculator question

I have a financial calculator question. I have the APY for a Certificate of Deposit; I am looking for a calculation that will give me its dividend rate if the dividends are paid Quarterly or if they are paid monthly.

For example, for a 12 month Certificate the APY is 3.25% what is the calculation to find out what the dividends are? —Preceding unsigned comment added by 207.109.247.177 (talk) 17:58, 13 March 2008 (UTC)

Comment: The poster is referring to the annual percentage yield. Pallida  Mors 19:16, 13 March 2008 (UTC)

Different institutions may use different definitions of the Annual Percentage Yield, which may be different from the effective rate, the Annual percentage rate, because certain costs have not been accounted for yet. Assuming the formula in our article holds for the definition actually used, and assuming dividend rate may be equated with the nominal interest rate, then:

• for monthly dividend rate (12 periods) you get over a year:

  $12\times \left(\left(1+{\frac {APY}{100\%}}\right)^{\frac {1}{12}}-1\right)\times 100\%=12\times \left(1.0325^{\frac {1}{12}}-1\right)\times 100\%=12\times 0.00267\times 100\%=3.20\%;$

• for quarterly dividend rate (4 periods) you get over a year:

  $4\times \left(\left(1+{\frac {APY}{100\%}}\right)^{\frac {1}{4}}-1\right)\times 100\%=4\times \left(1.0325^{\frac {1}{4}}-1\right)\times 100\%=4\times 0.00803\times 100\%=3.21\%.$

Note the caveats. Your mileage may vary.  --Lambiam 19:18, 13 March 2008 (UTC)

PDE -> ODE

Are there a set of "standard" methods one could try to use to transform a PDE in several variables to a set of coupled ODEs? —Preceding unsigned comment added by 12.196.44.226 (talk) 20:22, 13 March 2008 (UTC)

maybe separation of variables#Partial differential equations? --Spoon! (talk) 03:49, 14 March 2008 (UTC)

I need help with pi & circumference

If you calculated the circumference of a circle the size of the known universe, requiring that the answer be accurate to within the radius of one proton, how many decimal places of pi would you need to use?

 A. Two million
 B. 39
 C. 48,000
 D. Six billion

Thanks in advance.207.69.140.24 (talk) 21:48, 13 March 2008 (UTC)

I seem to recall the answer is 39, though I don't remember why. Strad (talk) 21:56, 13 March 2008 (UTC)

Assuming you’re calculating it from the radius of the known universe I think you could find an upper bound by dividing the smallest distance in question (radius of a proton) by twice the largest distance in question (radius of the universe), and using pi with as many decimal places as this number: because then would be accurate to the radius of a proton when multiplied by twice the radius of the known universe. GromXXVII (talk) 22:11, 13 March 2008 (UTC)

[ec] According to Observable universe, its diameter is $9\cdot 10^{10}$ ly, so its circumference is $3\cdot 10^{11}$ ly or $3\cdot 10^{24}$ m. The radius of a proton is roughly $10^{-15}$ m. The ratio is roughly $10^{39}$, thus roughly 39 digits of π are required. But don't let anyone fool you into thinking that this means that more digits are not required for science and applications. -- Meni Rosenfeld (talk) 22:18, 13 March 2008 (UTC)

A question similar to this has been asked before. It was to find the circumference of the known universe to the accuracy of a planck length.

And the answer is that 62 digits of Pi was required, preferably 64 digits. I'll see if I can dig up a link.

http://en.wikipedia.org/wiki/Wikipedia:Reference_desk_archive/Science/2006_September_27#How_many_digits_of_pi_for_the_known_universe.3F

202.168.50.40 (talk) 00:08, 14 March 2008 (UTC)

We have no data to base the calculation on, so knowing the decimal expansion of π to many places is not going to help. Even if we knew the radius of the observable universe to within the radius of a proton, and we could halt the universe so that it stopped expanding while we are doing the calculation, we don't even know if the universe is flat.  --Lambiam 08:40, 14 March 2008 (UTC)

Well, the question didn't ask about the circumference of the observable universe, but rather of a circle the size of it. I guess what this means exactly is open to interpretation, but I think "fix a number of arbitrarily high precision and magnitude on the same order as the observable universe, and use it as a radius" is a valid one. -- Meni Rosenfeld (talk) 10:38, 14 March 2008 (UTC)

See Cosmic View: The Universe in 40 Jumps by Kees Boeke - a classic. Gandalf61 (talk) 14:21, 14 March 2008 (UTC)

March 14

Happy Pi Day

Just wishing all of you a happy Pi Day! Please celebrate responsibly. --Kinu ^t/_c 04:12, 14 March 2008 (UTC)

What should I do at 1:59:26? I've only got an hour! HYENASTE 04:22, 14 March 2008 (UTC)

Um... run around in a circle? —Keenan Pepper 05:54, 14 March 2008 (UTC)

Just returned from my circle! Happy Pi Day!!! HYENASTE 05:59, 14 March 2008 (UTC)

getting dizzy! this is so exciting87.102.83.204 (talk) 10:05, 14 March 2008 (UTC)

Damn, I've just fallen over.--88.109.224.4 (talk) 13:41, 14 March 2008 (UTC)

Happy 3.14... Day! --Mayfare (talk) 15:19, 14 March 2008 (UTC)

I've just drank 3.1 pina coladas oh goddd.... Damien Karras (talk) 15:23, 14 March 2008 (UTC)

I've never liked the idea of having a celebration based on an arbitrary and unnatural number system. What say we move pi day to the third of July (or the seventh of March for transatlantic types)? Algebraist 15:48, 14 March 2008 (UTC)

Let's just make every day pi-day see Big yellow disc in sky87.102.83.204 (talk) 16:49, 14 March 2008 (UTC)

You may be an algebraist, but is that an excuse for taking a rational approach to pi day?  --Lambiam 16:56, 14 March 2008 (UTC)

How are continued fractions less transcendental than decimals? And that username's out of date: I'm currently applying for a DPhil in logic. Algebraist 20:54, 15 March 2008 (UTC)

That explains a lot.  --Lambiam 07:27, 16 March 2008 (UTC)

Percentages

Here I am at a very advanced age totally ashamed to say that I cannot work out percentages! Please advise me how to work out the percentage of one sum against the other; for example what percentage has my pension increased from last year's income to this year's. How to I do that ? Thanks in anticipation.--88.109.224.4 (talk) 08:47, 14 March 2008 (UTC)

Simple a percentage is one over another times 100.

So if income was 110 this year, and 95 last year this gives (110/95)*100 = 115.8%

So the increase is 15.8% (since 100% equals 1:1 ratio or no change I subtract 100%)

Or if my wage (77) increases by 2% my new wage is 77 + ( 2/100 x 77 ) = 77+1.54 = 78.54 —Preceding unsigned comment added by 87.102.83.204 (talk) 10:05, 14 March 2008 (UTC)

Did you check out the article Percentage?  --Lambiam 16:11, 14 March 2008 (UTC)

Thank you both, esp. Lambiam. The article is excellent and brings me up to speed. Thanks again.--88.109.224.4 (talk) 16:57, 14 March 2008 (UTC)

Also if you're struggling to do it 'mentally' (i.e. without a calculator/spreadsheet) I find it easiest to break things int0 10% and 1% then just do simple addition...E.g. If I want to know what 22% of 721 is I just go.... ok 10% is 72.1 and 1% is 7.21...I have 2 x 72.1 = 144.2 and then 2 x 7.21 so that's 14.42 add the two together and I get 158.62 which a quick google search (http://www.google.co.uk/search?hl=en&q=22%25+of+721&btnG=Search&meta= suggest is correct. This is particularly useful as a technique when you just want to get a ball-park amount really quite rapidly. ny156uk (talk) 17:50, 14 March 2008 (UTC)

Another thing. A lot of people get confused or thrown off by the percent symbol (%). Whenever you see that symbol, you can erase it altogether ... and simply replace it with the fraction 1/100 or, if you prefer, the decimal 0.01. That is, the % symbol simply means to take whatever number you are dealing with and to multiply it by the fraction 1/100 or (equivalently) the decimal 0.01 -- whichever you prefer or whichever is easier in a given situation. Thus:

• 7% means ( 7 * 1/100 ) = 7/100 --- if you need / want the fraction version of the answer
• 7% means ( 7 * 0.01 ) = 0.07 --- if you need / want the decimal version of the answer

Thanks. (Joseph A. Spadaro (talk) 22:25, 14 March 2008 (UTC))

Group Theory

The theory of groups interest me. I have a knowledge of mathematics up to:

a) Complex number arithmetic

b) Basic differentiation / integration

c) Basic knowledge of matrices (inverse, determinants)

d) Basic Logic and set theory

I understand the brief introductions and historical accounts of its development; and I'm tempted to pick up a book- although I don't know if I'm qualified to handle even the simplest introductory text.

It's basically a crude engineering mathematics background. Can you suggest a path of study to lead me up to investigations into group theory? Can I step into it with the basic knowledge I have or would you suggest deep study into which particular areas of mathematics? —Preceding unsigned comment added by 81.187.252.174 (talk) 14:12, 14 March 2008 (UTC)

Basic set theory is all you need, really. Find a book called something like "A first course in groups" and see how you find it. At a basic level, groups are pretty easy to understand. They are just sets with a way of combining 2 elements to get another element in a well-behaved way (for example, adding 2 integers to get a 3rd integer, or reflecting a polygon and then rotating it and realising that it's the same as having just reflected it in a different axis). --Tango (talk) 14:26, 14 March 2008 (UTC)

A textbook that is not expensive (used) is An introduction to the theory of groups by George W. Polites.[2] I don't know the book, so I can't vouch for it, but it is thin (80 pages), and if it is a mismatch to your needs, at least you did not waste lots of money.  --Lambiam 16:05, 14 March 2008 (UTC)

To hell with "not expensive"; here are three free online books for you to choose from: [3] [4] [5] —Keenan Pepper 21:59, 14 March 2008 (UTC)

Origin of asymmetry in conformation of symmetrical shape in a higher dimension (+ origami!)

Ok, this one has me totally stumped, and bear with me, as I'm not sure I'll explain it wonderfully well. There is a certain "mathematical" origami model, called a hyperbolic parabola (or paraboloid). Which is folded using a crease-pattern that is four-way symmetric under both reflectional and rotational transformations. (See: http://www.math.lsu.edu/~verrill/origami/parabola/ )

Now, here's the weirdness: when the crease-pattern is concertina-ed (fanned-up), the model assumes the three-dimensional shape for which it is named -- which is not completely symmetrical! The shape once conformed possesses chirality, or handedness, which is not possessed by the tetrahedron into which it will fit perfectly. Where does this rotational asymmetry come from? It is not there in the two-dimensional crease-pattern, but somehow emerges from the conformation into three dimensions, despite there being no way to distinguish between what is done to any four-way division of the model.

Can someone explain this puzzling phenomenon? (Preferably without using arcane technical terms). Many thanks, 85.194.245.82 (talk) 20:47, 14 March 2008 (UTC)

(original poster): Still be not-overly-confident of how well I explained my confusion, here's an addendum I just thought of: from the perspective of the paper as it is twisting into three-dimensions, both of its (indistinguishable) diagonals curve into circular arc-segments, one "up", the other "down" -- the puzzle is what logic does the paper use in conspiring with three-dimensional space to decide which of these diagonals goes up and which goes down?! 85.194.245.82 (talk) 20:53, 14 March 2008 (UTC)

I must admit I had no idea what you were talking about until I actually made one of these. Do you have one on hand? If so, try this: grab two opposite sides of the square piece of paper (which are two opposite edges of the tetrahedron) and twist in such a way as to flatten out the square. Then keep twisting in the same direction far past that point. Try to do it in one smooth motion. If you do it right, the thing will flip inside out and become an identical shape, but this time the diagonals go the opposite directions. So really it's not that a symmetrical pattern leads to an asymmetrical shape, it's that a symmetrical pattern leads to two possible stable shapes which are opposites, and they're symmetrical if you consider them as a pair. In between them is an unstable equilibrium which is a corrugated flat square (it's unstable because you effectively give the paper a negative Gaussian curvature, so it doesn't like to be flat anymore). Neat. —Keenan Pepper 21:41, 14 March 2008 (UTC)

Technical nitpick with one of your statements: The shape it makes is not chiral. It has point group $D_{2d}$, which is a proper subgroup of the point group of the flat square you start out with ($D_{4h}$), so it does have "less symmetry" in a specific sense, but it still has mirror symmetry, so it's not chiral. —Keenan Pepper 21:51, 14 March 2008 (UTC)

Hah, just realized I used many "arcane technical terms". Sorry about that. I'll be happy to explain them to you. —Keenan Pepper 21:53, 14 March 2008 (UTC)

Aye, empirical experimentation shortly after framing the question lead me to the "the paper does it" conclusion you point to, which at first made me think, "awww, reality's boring", as I was hoping for some kind of crazy metaphysical implication about the non-independence of spacial dimensions relative to one another (I have an overactive imagination that way). But then, I thought about it some more and had a little discussion on IRC, and decided it was still a very interesting phenomenon despite the "choice" of axis-curving resulting from environmental factors. Interesting because a purely geometrical consideration lead to an amplification of information. That is to say, although really only one bit of information is added to the system in choosing the subset of the point group, the effect is to cause a large difference in eventual topology. Someone said this was an example, or at least analogous, to the theoretical-physical concept of spontaneous symmetry breaking (I personally think the term "spontaneous" is a terrible misnomer, as it implies that the information arrives randomly, ex nihilo, from nowhere, which is not how mathematics or the universe works, no matter how hard we try to pretend otherwise). 85.194.245.82 (talk) 22:09, 14 March 2008 (UTC)

(thanks for the terminology info, btw. I had a feeling chirality wasn't right, but couldn't think of any other asymmetry. Is there a specific term for it? Will look at the article you linked to.) 85.194.245.82 (talk) 22:09, 14 March 2008 (UTC)

The phenomenon is an instance of spontaneous symmetry breaking. An even much simpler example is when you put a small ball (as from a ball bearing) exactly in the middle on top of a larger ball. Perfect symmetry. Turn around and the little ball will have dropped off, in an asymmetric way (since all ways of dropping off are asymmetric). If it does not drop off instantaneously, you may have to wait for a butterfly or Heisenberg to assist the process.  --Lambiam 23:43, 14 March 2008 (UTC)

March 15

A measurable set?

Let $(X,\mu )$ and $(Y,\nu )$ be measure spaces and $\{A_{k}\},\{B_{k}\}$ sequences of sets of finite measure in X and Y respectively. Let the "rectangles" $R_{k}=A_{k}\times B_{k}$, and assume that

$\sum _{k=1}^{\infty }(\mu \otimes \nu )(R_{k})<\infty .$

Let $R_{k,x}=\{y\ |\ (x,y)\in R_{k}\}$ and

$T_{n}=\{x\ |\ 1/n\leq \sum _{k}\nu (R_{k,x})\}.$

Why is it obvious that T_n is measurable?  — merge 17:01, 15 March 2008 (UTC)

Well, $R_{k,x}$ is just B_k if x is in A_k, and empty otherwise. So $\sum _{k}\nu (R_{k,x})$ is the some of the measures of the B_k such that x is in A_k. Thus whether x is in T_n is determined by which of the A_ks x is in, and T_n is a union of intersections of the A_ks. Algebraist 17:45, 15 March 2008 (UTC)

Oh, I think I see how it works out. If $\{f_{k}\}$ is a sequence of nonnegative measurable real-valued functions and α is a real number, the sets

$V_{k}=\{x\ |\ f_{1}(x)+\cdots +f_{k}(x)>\alpha \}$

are measurable, and so are

$W_{\alpha }=\bigcup _{k}V_{k}=\{x\ |\ \sum _{k=1}^{\infty }f_{k}(x)>\alpha \}$

and

$\bigcap _{j}W_{\alpha -1/j}=\{x\ |\ \sum _{k=1}^{\infty }f_{k}(x)\geq \alpha \}$.

In this case $f_{k}(x)=\nu (R_{k,x})=\chi _{A_{k}}\nu (B_{k})$ and $\alpha =1/n$.  — merge 21:05, 16 March 2008 (UTC)

March 16

grasping math to an intermediate level

hi, I like to get to an intermediate level of understanding of different subjects and I want to get a decent grasp of mathematics, but I'm having a hard time with it. I have a BS in math and an MS in statistics. I feel as if I'm wandering blindly in a mathematical universe, feeling the obvious, stumbling over the less-obvious, but without confidence in my travels. When I started to learn typography, say, I had that blind feeling at first but with study soon found the fundamentals and understood the basic ideas and questions in that field.

I want to get to that level of confidence with math, but I'm not sure how to get there. Obviously it's a big world out there, but is there a field that I should concentrate study on (analysis maybe?) to get an intermediate grasp (i.e., not expert and not blind beginner) ? thanks 72.150.136.11 (talk) 14:24, 16 March 2008 (UTC)

In all honesty, it's pretty much impossible to get an intermediate grasp of everything, since there is so much. I mean, I am doing an MMath, and I'd say that by doing so I was intermediate in the subjects I do, but there are others I simply have no clue about - topology, fluid mechanics, relativity, anything related to statistics... maths is so vast, and a lot of it is so remote from anything you do at school as to be an entirely different subject. Group theory is probably a useful thing to try and understand, as is basic vector calculus and mathematical analysis. I personally like complex analysis and combinatorics. Of course, combinatorial game theory is also fun to try. -mattbuck (Talk) 15:00, 16 March 2008 (UTC)

Linear algebra, while rather boring, is also a useful subject to have a working knowledge of - it seems to pop up all over the place. However, if you've got a BS in Maths I would expect you've already covered the basics of all these areas. If you want to do more Maths, you need to specialise, there's no way to have a better than undergraduate understanding of more than one or two areas. --Tango (talk) 15:19, 16 March 2008 (UTC)

I disagree with the last statement, there are many researchers who are active in several quite distinct areas. It's not easy, of course.

Mathematical logic is very important to know the basics of (say, up to the level of understanding Godel's theorems). Likewise for topology, which at the very least one needs to know to the level of feeling the difference between metric features and topological ones. I recall that when I have been applying for a math MSc, virtually all schools frowned upon me not having taken a topology course in my BA. Fortunately I had learnt the material independently - my peers who hadn't had some trouble with many of the courses.

Of course, there's the obvious stuff, like analysis (real, complex, functional, differential geometry...), algebra (groups, rings, fields, Galois theory...), discrete (combinatorics, graphs, ...), and of course computer science (computational complexity, information theory, machine learning, ...) and applied math (numerical methods, mathematical physics, game theory...). Disclaimer: This categorization is based on my own personal view and the subjects are not equally divided. -- Meni Rosenfeld (talk) 17:37, 16 March 2008 (UTC)

Well, I guess you run into trouble defining an "area" of maths. There are plenty of people that do research in what would usually be considered distinct areas, but they're usually working in some kind of overlap between them - pretty much all branches of maths overlap somewhere. Are there many people that have a solid understanding of the whole of several areas? I would say most understand just the parts of those areas that are relevant to their research. It does depend very much on how large you consider an area to be (is algebra a single area or is ring theory a distinct area to Galois theory? - obviously, you can understand both ring theory and Galois theory to a high level, but understanding the whole of algebra to a high level is pretty challenging without even trying to start on any non-algebraic areas). Put simply: I think we're both right for appropriate definitions of "area". --Tango (talk) 18:12, 16 March 2008 (UTC)

Thanks for these answers. I thought it might be impossible for a not-particularly-gifted person to walk with confidence in the math spaces. I just hate that feeling of flailing around blindly, holding only to those proofs I really really understand. Sometimes I feel like I'm missing a sense that some others have; which I suppose is true for everyone on some level. I'll explore mathematical logic and topology and see how that goes. thanks everyone 72.150.136.11 (talk) 19:00, 16 March 2008 (UTC)

While we all strive for perfection, nobody gets there. However far you go, there's always going to be more you don't understand, so you'll probably always be walking blindly, you'll just be walking in more advanced parts of the world of maths. I think the important thing is to enjoy it - study the areas of maths that interest you and that you find fun. There's little point studying anything else. --Tango (talk) 19:31, 16 March 2008 (UTC)

Local Sidereal Time Conversion

How do I convert local sidereal time to something that is more commonly used? Is this actually a measure of time? I read the article about sidereal time and it appeared to be an angle measurement, a measure of place. So when someone tells me "do this action at 13:20" local sidereal time, is this something that makes sense, or is it just astrological mumbo jumbo? And if it makes sense, how do I convert it to Eastern Standard Time or Hawaii Standard Time or GMT or anything else that is more commonly used? —Preceding unsigned comment added by RastaNancy (talk • contribs) 20:45, 16 March 2008 (UTC)