I need some raw data to try some different things and comparisons...

How many murders have been committed in Minneapolis, Minnesota in year 2010?

If available the same data for Hennepin County and the state as a whole

How many murders have been committed in the Greater London Area (boroughs plus The Square Mile) in the year 2010?

How many in England & Wales?

I am really curious as to WHERE to find this data so I can directly do my own research in the future. My intent is to play around with the population and educational level variables and just experiment with this rather depressing data. Thanks I have a reference question (talk) 05:25, 16 August 2010 (UTC)[reply]

The miscellaneous reference desk would be better for a question like that. Also have you tried searching using Google or a suchlike search engine? Dmcq (talk) 08:37, 16 August 2010 (UTC)[reply]

Crime statistics for Greater London are available from the Metropolitan Police here - for example, there were 125 homicides (murder, manslaughter, corporate manslaughter and infanticide) in Greater London in the 12 months to July 2010. Crime statistics for England and Wales are available from the Home Office - their report on Crime in England and Wales 2009/2010 is available here. Gandalf61 (talk) 12:39, 16 August 2010 (UTC)[reply]

Hello, is there a laurent series expanded around z=0 for ln(z)? Thanks.Rich (talk) 11:54, 16 August 2010 (UTC)[reply]

ln(z) around 0 is multi valued or discontinuous, while laurent series are single valued and continuous. So no. Bo Jacoby (talk) 12:58, 16 August 2010 (UTC).[reply]

Hmm thanks that's interesting. Does that mean ln(z) has an essential singularity at 0?-Rich Peterson24.7.28.186 (talk) 13:36, 17 August 2010 (UTC)[reply]

No, because that would require the function to be holomorphic in a neighbourhood of 0 minus 0 itself, and as Bo explained, this condition fails for whatever branch of ln(z) you choose. If the function had an essential singularity in 0, it would also have a Laurent series around 0.—Emil J. 13:43, 17 August 2010 (UTC)[reply]

This type of singularity is called a "branch point singularity". Note that functions with such a singularity at some point can have well defined values at that point. The word "singularity" refers to non-analytic behavior. So, e.g. the function z^p for positive non-integer p has a branch point singularity at z = 0, even though the function is well defined at z = 0. Count Iblis (talk) 14:43, 17 August 2010 (UTC)[reply]

We have an article titled branch point (I haven't looked at it recently, so I can't promise anything about what it says. Michael Hardy (talk) 15:02, 17 August 2010 (UTC)[reply]

Thanks everyone.Rich (talk) 02:46, 19 August 2010 (UTC)[reply]

I think I understand this discussion of why SO(3) is not simply connected. I gather that Spin(3) is simply connected, but I'm curious what this exactly means. As I understand it, it should be possible, with any simply connected manifold, to continuously deform any path between two points on the manifold to any other path with the same end points. So, suppose I represent a rotation changing over time using Slerp. Suppose further that I use Slerp, or an extension like Squad to mediate a rotation of ${\displaystyle 2\pi }$ about some axis. Is it possible to continuously "deform" this rotation until it represents no rotation over time whatsoever? Thanks.--Leon (talk) 13:51, 16 August 2010 (UTC)[reply]

Slerp does not explain why the circle S¹ is not simply connected but the spheres Sⁿ for n>1 are simply connected. Note that Spin(3) is the 3-sphere. The main point you should try to understand is the difference between the circle and the 2-sphere with regard to simple connectivity. Tkuvho (talk) 14:23, 16 August 2010 (UTC)[reply]

I'm not too clear on what you mean. And I'm not sure why the circle S¹ is relevant.--Leon (talk) 14:31, 16 August 2010 (UTC)[reply]

Thinking in terms of Slerp, what would be your guess concerning the circle: does Slerp suggest it is simply connected, or not? Tkuvho (talk) 14:39, 16 August 2010 (UTC)[reply]

Well I believe the answer is "no", but I'm still not entirely sure what you mean.

What I'm after is an intuitive geometrical understanding of the difference between the two groups when used to represent rotations in ${\displaystyle \mathbb {R} ^{3}}$, and for a theoretical engineering application. Also, does the statement

"Surprisingly, if you run through the path twice, i.e., from north pole down to south pole and back to the north pole so that φ runs from 0 to 4π, you get a closed loop which can be shrunk to a single point"

mean that if I were to construct a rotation of ${\displaystyle 4\pi }$ about some axis, I could continuously deform such a rotation to no rotation at all?--Leon (talk) 14:48, 16 August 2010 (UTC)[reply]

Exactly. More precisely, you can deform such a family of rotations, to the trivial (i.e. constant) family. You are unlikely to be able to do that with something as canonical as the Slerp deformations, but perhaps you could. Tkuvho (talk) 15:01, 16 August 2010 (UTC)[reply]

To summarize what's involved: rotations of 3-space can be represented by unit quaternions q. Two "opposite" quaternions, q and -q, represent the same rotation of 3-space. Hence the space of rotations is the quotient of the unit quaternions by the antipodal map. Thus, while unit quaternions form a 3-sphere (simply-connected), the rotations form a manifold which is its antipodal quotient, and therefore non-simply connected. Tkuvho (talk) 15:05, 16 August 2010 (UTC)[reply]

Cool, thanks. But, with regards to the earlier bit, is there any "intuitive, geometrical" difference between the two groups when used to represent rotations? To me it seems that the significance is one can continuously deform a path described by elements of Spin(3) to any other such path, keeping the same end points the same, and this isn't true of SO(3). As a ${\displaystyle 2\pi }$ rotation in Spin(3) is not an example of an element going to itself (quaternions: 1->i->-1 is a ${\displaystyle 2\pi }$ rotation but it isn't an element travelling some path to end where it started) it couldn't be contracted to a point in any case. Or am I missing something...--Leon (talk) 15:13, 16 August 2010 (UTC)[reply]

The path you have described illustrates the point very nicely. As you mentioned, the path is not closed. However, under the antipodal quotient, it does descend to a closed path in the quotient space, namely SO(3)=RP³. This is because 1 and -1 define the same rotation, namely the trivial one. By general covering space theory, whenever a non-closed path descends to a loop, the resulting loop cannot be contracted to a point. Tkuvho (talk) 15:18, 16 August 2010 (UTC)[reply]

Really, I struck it out because I felt like an idiot for writing it! In any case, thanks for your help.--Leon (talk) 15:21, 16 August 2010 (UTC)[reply]

Hi, Does anybody know hwat is the meaning of the ${\displaystyle \pi -weight}$ and ${\displaystyle \pi -base}$ ? —Preceding unsigned comment added by 212.199.96.133 (talk) 13:56, 16 August 2010 (UTC) Topologia clalit (talk) 13:58, 16 August 2010 (UTC)[reply]

A π-base is a collection B of nonempty open sets such that every nonempty open set contains some member of B. I guess that the π-weight of a space is the cardinality of its smallest π-base.—Emil J. 14:01, 16 August 2010 (UTC)[reply]

Since it may not be immediately obvious what is the difference between a base and a π-base, here's some basic info: any base is also a π-base, but not vice versa. Consider the Sorgenfrey line S. Since every nonempty basic set in S contains a nonempty open set in the standard topology of the reals, the set {(p,q): p,q in Q, p < q} is a π-base of S, and in particular, S has countable π-weight. However, this set is not a base. In fact, if B is any base of S, then for each real a there is a set U_a in B such that ${\displaystyle a\in U_{a}\subseteq [a,a+1)}$. Since this makes a = min(U_a), all these sets have to be distinct, hence S has weight ${\displaystyle 2^{\aleph _{0}}}$.—Emil J. 15:10, 16 August 2010 (UTC)[reply]

Thanks! Topologia clalit (talk) 07:19, 18 August 2010 (UTC)[reply]

Hi there - I hope someone can help me out - it's been a long time since college stats classes...

I have five business groups, for each one I have a performance metric (a percentage score). I am looking for factors which might explain that metric (I understand that correlation <> causation!)

The groups have offices (between 20 and 90, each group has a different number), they have revenue (varies with each group), and the offices are in different locations. Some of the offices are in locations that are on a list of 20 places that are problematic - we'll call this the bad location list. All the metrics are for the group as a whole, not the individual offices.

OK, so, number of offices correlates well with performance - r2 - .87. Percentage of offices that are on the list correlates well (but negatively) about -.8. So far so good - the more locations, the better, and the fewer of them that are on the bad list the better. What I'm worried about is that the larger the number of offices, the lower the percentage on the bad list CAN be - that is to say there are only 20 'bad' locations, so if you have more than 20 office locations you are automatically going to have a lower percentage on the bad list - how can I deal with this? I want to disagregate the effect of size per-se from the effect of bad locations.

Thanks so much! —Preceding unsigned comment added by 97.115.64.81 (talk) 14:54, 16 August 2010 (UTC)[reply]

Where did this list of 20 bad locations come from? You might be better off going back to the starting point and using whatever metric was used to compile the list rather than using the list itself. --Tango (talk) 08:18, 18 August 2010 (UTC)[reply]

5 data points is very few, so whatever results you'll find will probably be highly insignificant. When there's more data, one thing you can do is to look at all points where the total number of offices is some fixed value, and see how the performance varies with number of bad offices.

Perhaps you can use some form of ANOVA where you do multiple regression for various choices of variables and see which contribute the most to explaining the variance. If you use a baseline model using only the total number of offices, and a second model including both that and the number of bad offices, then a large difference in the RSS between them will tell you that the number of bad offices indeed correlates with performance.

This is of course assuming that a linear model is at all appropriate. -- Meni Rosenfeld (talk) 07:40, 19 August 2010 (UTC)[reply]

It's easy to identify multiples of 2 (last digit even), 3 (digits sum to 3 or a multiple of 3), 5 (last digit 0 or 5) and 11 (alternate digits sum to 11). I seem to remember that there's also a trick to spotting multiples of 7, but I can't remember what it is. Can anyone help? Thanks Rojomoke (talk) 16:26, 16 August 2010 (UTC)[reply]

Divisibility rule#Divisibility by 7. Algebraist 16:38, 16 August 2010 (UTC)[reply]

The powers of 10 are 1, 10, 100, 1000 and so on. Modulo 7 these powers are 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, and so on. So the rule is: if the ones plus thrice the tenths plus twice the hundreds minus the thousands minus thrice the tenthousands minus twice the hundredthousands and so on, is a multiple of seven, then so is the original number. Bo Jacoby (talk) 16:52, 16 August 2010 (UTC).[reply]

If you do the same thing as for 11 except with thousands you can make the checks for 7, 11 and 13 easier. For instance with 134,768,345,558 add 134+345 and 768+558 to give 479 and 1326. Subtract the smaller from the larger to give 847 and then just check this number 847 instead of the original. It is divisible by 7 but not by 11 or 13 so the original number is divisible by 7 but not 11 or 13. This is because 1001 = 7.11.13 Dmcq (talk) 19:12, 16 August 2010 (UTC)[reply]

I see there is an article about all this Divisibility rule Dmcq (talk) 19:18, 16 August 2010 (UTC)[reply]

Are there any algorithms for finding appropriate starting values for algorithms such as Brent's method? I think what would be needed would be 1. some approximation of the highest and lowest possible roots, and 2. what resolution to search at to make sure you aren't skipping any roots. —Preceding unsigned comment added by 70.162.15.58 (talk) 00:54, 17 August 2010 (UTC)[reply]

In general, there is no easy way to do this, but in case of polynomial roots, you can always use Sturm's theorem. (Igny (talk) 03:14, 17 August 2010 (UTC))[reply]

See Durand-Kerner method. Bo Jacoby (talk) 14:59, 17 August 2010 (UTC).[reply]

I am exposing my ignorance of trig here... When I do u substitution with u = sinx I get (1/2)(sinx)^2 but my ti89 gives me -(1/2)(cosx)^2. What am I missing here? are these two results equivalent? -- 99.20.118.197 (talk) 04:02, 17 August 2010 (UTC)[reply]

What happens when you do the substitution u = cosx? What is sin²x + cos²x (the most important trig identity)? How does C (the constant of integration) fit into this? -- 111.84.234.215 (talk) 05:05, 17 August 2010 (UTC)[reply]

The easiest check is to differentiate both of them and with both of them you end up with ${\displaystyle \sin {x}\cos {x}}$. We can therefore safely say they are both integrals of ${\displaystyle \sin {x}\cos {x}}$. Are they equivalent? Let's plug in some numbers. ${\displaystyle {\frac {1}{2}}\sin ^{2}{0}=0}$.   ${\displaystyle -{\frac {1}{2}}\cos ^{2}{0}=-{\frac {1}{2}}}$. So clearly they are not equivalent. Look at the coefficient of integration next and see if you spot anything. Readro (talk) 08:21, 17 August 2010 (UTC)[reply]

The integral can be looked at in a number of different ways.

• Substitute u = sin x, with (du/dx) dx = cos x dx:

${\displaystyle \int \sin x\cos xdx=\int udu={\frac {1}{2}}u^{2}+C={\frac {1}{2}}\sin ^{2}x+C.}$

• Substitute u = cos x, with (du/dx) dx = -sin x dx:

${\displaystyle \int \sin x\cos xdx=-\int udu=-{\frac {1}{2}}u^{2}+C=-{\frac {1}{2}}\cos ^{2}x+C=-{\frac {1}{2}}(1-\sin ^{2}x)+C={\frac {1}{2}}\sin ^{2}x-{\frac {1}{2}}+C.}$

• Multiply by 2, take 0.5 outside the integral, and convert to sin 2x:

${\displaystyle \int \sin x\cos xdx={\frac {1}{2}}\int 2\sin x\cos xdx={\frac {1}{2}}\int \sin 2xdx={\frac {1}{2}}(-{\frac {1}{2}}\cos 2x)+C={\frac {1}{4}}(2\sin ^{2}x-1)+C={\frac {1}{2}}\sin ^{2}x-{\frac {1}{4}}+C.}$

However you look at it, though, the antiderivative is the same up to a constant – and this difference is accounted for by the constant of integration C. —Anonymous Dissident^Talk 10:39, 17 August 2010 (UTC)[reply]

Can't somebody mention the trigonometric identity that says

${\displaystyle \sin x\cos x={\frac {\sin(2x)}{2}}}$?

Michael Hardy (talk) 14:57, 17 August 2010 (UTC)[reply]

OK, I see someone did mention that. But why wasn't it the first thing mentioned here? To anyone who knows trigonometry, that would be the first thing that comes to mind. Michael Hardy (talk) 14:59, 17 August 2010 (UTC)[reply]

Probably because this isn't what the OP asked. He asked how to reconcile his (correct) computation for the integral with the output of the calculator. -- Meni Rosenfeld (talk) 17:50, 17 August 2010 (UTC)[reply]

Thanks alot everyone. I worked out your examples on paper and read the article on C. Before I posted my question I did get the feeling that I could do some pythagorean substitution to change sine to cosine ... but the piece of the puzzle I was missing was an understanding of +C. Thanks again -- 99.20.118.197 (talk) 15:31, 17 August 2010 (UTC)[reply]

Hi. The rate of my dorm for 38 weeks is 1900, so it gives 50 every week (it's even written on the page that it is 50). I have to pay each month 237,5. On the average, a month has 30,42 days, which gives a month 4,35 weeks. 4,35 weeks times 50 is 217,5. Where are those 20?! (I didn't give the currency, but I think it's ok). 83.31.113.33 (talk) 14:05, 17 August 2010 (UTC)[reply]

It looks like you are right. Either you're paying for some extra service, there's an additional tax, there's a mistake in their billing or published numbers, or they're just ripping you off. -- Meni Rosenfeld (talk) 14:31, 17 August 2010 (UTC)[reply]

237.5 is exactly one eighth of 1900. 38 weeks is just under 9 months. So maybe it is simpler for your bank and whoever is letting the dorm to process 8 equal monthly payments than it would be to process 8 equal payments and then an odd amount for the "short" month at the end. Sensible thing is to ask whoever is letting the dorm to clarify how many monthly payments you will be making. Gandalf61 (talk) 14:40, 17 August 2010 (UTC)[reply]

Hm, I think you are right. If anyone wants to check their information, it's here: [1] (Lyon St, twin). 83.31.113.33 (talk) 14:42, 17 August 2010 (UTC)[reply]

If you read the rates details [2] it says you get the 38 weeks for 8 monthly payemnts of £237.50. It says that the weekly amount is for info only. It all matches up because you aren't paying for 8.7 months. -- SGBailey (talk) 16:29, 18 August 2010 (UTC)[reply]

Hobbyist filmer here. I'd like to make mind-blowing ultra wide-angle images like Terry Gilliam, but in the more economical format of Super 8 mm film. What I do know is that I'm not even remotely looking like Gilliam with any focal length above 18mm...but that's my desired upper length in 35mm only. The focal length to achieve a particular angle of view (which is the thing that makes for the mind-blowing images with wide-angle images) is different with any format and sensor size you use, hence there's articles such as 35 mm equivalent focal length and crop factor. In other words, if you change the format (i. e. size of your film or sensor) but wanna have the same angle of view, you need a different focal length.

Now, I have a chance of acquiring a lens (for a Super8 camera made by the Austrian Eumig brand) which is labeled as ultra wide-angle, according to trade press this lens is guaranteed to be entirely rectilinear (no barrel aka fish-eye distortion, as I don't want this), and its focal length in Super8 is 4mm.

So what I'd like to know is, what's the 35mm equivalent of these 4mm in Super8, according to crop factor? Or in other words: If my desired upper limit is an 18mm focal length in 35mm, what equivalent focal length would that be in Super8?

I guess what might help are the dimensions of the Super8 frame area: 5.97mm horizontal x 4.01mm vertical, compared to 22mm horizontal x 16mm vertical in 35mm.

My second choice would be a 3CCD miniDV with a 1" chip size. What's the equivalent to 18mm there? --79.193.41.61 (talk) 06:50, 18 August 2010 (UTC)[reply]

I don't think this is the right forum for the question, unless there is a mathie who happens to be photography hobbyist as well. You might try the science desk.--RDBury (talk) 21:11, 18 August 2010 (UTC)[reply]

I've responded at science. Thegreenj 22:42, 18 August 2010 (UTC)[reply]

I have always been absolutely terrible at this type of problem. Any hints for making it easier to solve would be appreciated. The basic problem is, given two functions, which is asymptotically greater than the other? In other words, given function f and g, to be asymptotically greater than g, it must be proven that f is O(g). Here is an example and how I solve it (which is surely the most difficult way to solve it):

Given ${\displaystyle f=2^{\sqrt {\lg {n}}}}$ and ${\displaystyle g=n(\lg {n})^{3}}$.

I solve this by estimating a value for ${\displaystyle {2^{\sqrt {\lg {n}}}} \over {n(\lg {n})^{3}}}$. I will show my steps because I am sure I do not remember the log rules correctly:

${\displaystyle {{2^{\sqrt {\lg {n}}}} \over {n(\lg {n})^{3}}}={{\lg {2^{\sqrt {\lg {n}}}}} \over {\lg {(n(\lg {n})^{3})}}}={{\sqrt {\lg {n}}} \over {\lg {n}+\lg {((\lg {n})^{3})}}}}$

Now, if I look at this as ${\displaystyle {\sqrt {x}} \over {x+lg(x^{3})}}$, it is obvious that result will be less than 1. Therefore, I claim that g = O(f) and g is asymptotically greater than f. Correct? -- kainaw™ 20:48, 18 August 2010 (UTC)[reply]

Looks OK to me.--RDBury (talk) 21:05, 18 August 2010 (UTC)[reply]

Be careful,

${\displaystyle {\frac {x}{y}}}$ and ${\displaystyle {\frac {\lg {x}}{\lg {y}}}}$ are not always equal. What you have proven is that ${\displaystyle \,\!\lg {f}=O(\lg {g})}$. This in general does not imply that ${\displaystyle \,\!f=O(g)}$ (consider for example, ${\displaystyle \,\!f=x}$ and ${\displaystyle \,\!g=x^{1/2}}$).

In any case, your problem is a bit clearer if you substitute ${\displaystyle \,\!x=\lg {n}}$. Invrnc (talk) 03:45, 19 August 2010 (UTC)[reply]

If by "asymptotically greater" you mean "greater for every value starting at some point", then this is not what big O notation is. ${\displaystyle f=O(g)}$ means that f is less than (not greater than, as you wrote) g up to a constant. In your example, the (absolute value of the) ratio between f and g is bounded (it doesn't have to be less than 1), so ${\displaystyle f=O(g)}$. This is equivalent to showing that its logarithm is bounded from above. The correct log rule is ${\displaystyle \lg \left({\frac {x}{y}}\right)=\lg x-\lg y}$, so you just need to show that ${\displaystyle \lg {{2^{\sqrt {\lg {n}}}} \over {n(\lg {n})^{3}}}={{\lg {2^{\sqrt {\lg {n}}}}}-{\lg {(n(\lg {n})^{3})}}}<M}$ eventually (this follows from the fact that it goes to ${\displaystyle -\infty }$, which is easy to see). -- Meni Rosenfeld (talk) 07:27, 19 August 2010 (UTC)[reply]

You are correct. I stated it backwards. I do have a related question... Given that ${\displaystyle {\frac {x}{y}}}$ and ${\displaystyle {\frac {\lg {x}}{\lg {y}}}}$ are not always equal, is it valid to use that reduction here? If x is asymptotically greater than y, won't lg x be asymptotically greater than lg y? -- kainaw™ 15:43, 19 August 2010 (UTC)[reply]

Yes. Assuming everything is positive, ${\displaystyle {\frac {x}{y}}<1\iff x<y\iff \lg x<\lg y\iff {\frac {\lg x}{\lg y}}<1}$. However, as Invrnc explains, this will not work if you are talking about big O. -- Meni Rosenfeld (talk) 18:08, 19 August 2010 (UTC)[reply]

I believe that I found an example where it doesn't work. Proving that lg(f/g) is bounded from above did work. Thanks. -- kainaw™ 19:06, 19 August 2010 (UTC)[reply]

Consider a smooth space curve, parametrised by arc length. The Frenet frame {T,N,B} defines a rigid body at each point of the curve. This rigid body is the cube with sides T, N and B. The infinitesimal axis of symmetry of this rigid body is generated by τT + κB where κ is the curvature, and τ is the torsion, of the space curve. This means that infinitesimally the cube is rotating about the line spanned by τT + κB. Can someone show me how to calculate the infinitesimal angular frequency of the cube about this infinitesimal axis of symmetry? — Fly by Night (talk) 22:55, 18 August 2010 (UTC)[reply]

I'm not sure if the question is well posed as infinitesimally the box will be moving along the line as well as spinning. You also need to specify how fast the box is moving along the curve, you could say its a unit speed curve.

Those caverts taken into account you can consider rotation given by

${\displaystyle {\frac {d\mathbf {v} }{dt}}={\begin{pmatrix}0&\kappa &0\\-\kappa &0&\tau \\0&-\tau &0\end{pmatrix}}\mathbf {v} }$

The axis of this rotation is ${\displaystyle (\tau ,0,-\kappa )\,}$. To find the angular frequency take a vector perpendicular to the axis, say ${\displaystyle (0,1,0)}$ its tangental velocity is just ${\displaystyle (\kappa ,0,-\tau )\,}$, so the angular frequency is just ${\displaystyle \omega =|(\kappa ,0,-\tau )|={\sqrt {\kappa ^{2}+\tau ^{2}}}}$. --Salix (talk): 09:30, 19 August 2010 (UTC)[reply]

Thanks Salix. I think I mentioned that the curve was parametrised by arc length. Wouldn't that make it unit speed? You said that the axis of rotation is (τ,0,–κ) = τT – κB. Isn't there a sign error? I read that the axis of rotation is τT + κB. If you take a vector v = aT + bN + cB and solve dv/ds = 0 then you get, assuming v ≠ 0, that a = τ, b = 0 and c = κ; meaning that v = τT + κB. I've also seen it calculated by working out a vector perpendicular to dT/ds, dN/ds and dB/ds, i.e. perpendicular to κN, τB – κT and –τN. If κ and τ aren't both zero then you get τT + κB. Also, what did you mean by the "tangential velocity" of (0,1,0)? Do you mean the tangential velocity of N? And if so, what is it's tangential velocity? (It seems to be –dN/ds from your answer). I would be very pleased if you could fill in the details and answer some of the whys as well as the hows. Thanks in advance. — Fly by Night (talk) 12:35, 19 August 2010 (UTC)[reply]

Looks like brain not fully working this morning. Yes being parametrised by arc length does make it unit speed. Yes it is a sign error axis should be (τ,0,κ) = τT + κB. Tangential velocity comes from the angular frequency article, I'm meaning work out component of the velocity of the point (0,1,0) = N which is tangental a circle lying in a plane normal to the axis. The velocity is dN/ds = (–κ,0,τ) (another sign error on my part). Note that this has no component in the N direction and no component along the axis, so it is already the tangental velocity. The radius is 1 so the angular frequency is simply |dN/ds|. --Salix (talk): 18:57, 19 August 2010 (UTC)[reply]

Thanks again Salix! I appreciate your help. Let me check that I understand what I need to do. If the axis of symmetry is spanned by the vector v then I need to choose a unit vector, say w, at a right angle to v and then work out || dw / ds || to give the angular frequency? (Like you said, this all depends on the parametrisation of the curve, but once it's unit speed we can relax.) — Fly by Night (talk) 19:44, 19 August 2010 (UTC)[reply]

Yep. You may find the Angular velocity article more helpful as it has a few more details than angular frequency.--Salix (talk): 20:32, 19 August 2010 (UTC)[reply]

Does anyone know who won the fields medal? I can't seem to find this information anywhere! ... If it hasn't been released, when will it be? I heard that it would be released on 19 August,2010 but I don't want to be waiting till midnight for the declaration of the world's best mathematicians! Please guys, put me out of my misery and tell me how many more hourse I need to wait? I've waited for 4 years already ... Thanks! Signed:THE DUDE. (BTW, I fixed my typsetting bug!) —Preceding unsigned comment added by 110.20.26.228 (talk) 08:37, 19 August 2010 (UTC)[reply]

I think I know now ... Damn, why do applied mathematicians always get the awards? Where are the algebraists, topologists and geometers (and number theorists for that matter)? And why on Earth hasn't anyone received it for logic and set theory yet? (FOrget the the first question in above paragraph, I think I know who won it.) Thanks guys ... —Preceding unsigned comment added by 110.20.26.228 (talk) 08:37, 19 August 2010 (UTC)[reply]

There is already a Wikipedia User:THE DUDE, which apparently is not you. Both impersonating other users and pretending to be a user which doesn't really exist is forbidden. Your continued failure to sign your posts, followed by vandalizing your automatically-supplied signatures, is disruptive. Please stop. You can create an account if you wish. -- Meni Rosenfeld (talk) 08:16, 19 August 2010 (UTC)[reply]

Meni Rosenfeld, I just don't want to be known to the public. I want to be anonymous. You have a name, Meni Rosenfeld. If THE DUDE is taken, can I have another name. I never vandalized signatures. I just didn't want to be made public. Isn't that a simple rquest? Or are the people here at Wikipedia so serious. What's a change in signature between friends? —Preceding unsigned comment added by 110.20.26.228 (talk) 08:37, 19 August 2010 (UTC)[reply]

Dmcq has advised you to create an account if you want to remain anonymous. You ignored his advice, lied about your motivation, and now you are playing dumb. We have no reason to believe your intentions are in any way friendly. -- Meni Rosenfeld (talk) 09:44, 19 August 2010 (UTC)[reply]

OK, I fixed up all signatures? Now can you please answer my question Meni Rosenfeld? —Preceding unsigned comment added by 114.72.209.200 (talk) 11:34, 19 August 2010 (UTC)[reply]

I don't know the answer, but I'm sure that now those who do will be much happier to assist you. (Your next step is to sign your posts by typing 4 tildes, ~~~~ ). -- Meni Rosenfeld (talk) 11:45, 19 August 2010 (UTC)[reply]

114.72.244.249 (talk) 08:35, 20 August 2010 (UTC) Is this right? 114.72.244.249 (talk) 08:35, 20 August 2010 (UTC) How did you type 4 tildas without a signature coming up?[reply]

Creating an account is the only way to hide you IP address. No matter what edit you do to this page its and easy matter just to look at the history to find your IP. You can create an account at Special:UserLogin/signup --Salix (talk): 08:46, 19 August 2010 (UTC)[reply]

Using the nowiki tag. -- Meni Rosenfeld (talk) 13:24, 20 August 2010 (UTC)[reply]

Why isn't anyone answering my questions? Please answer guys ... Thanks guys ... 114.72.244.249 (talk) 08:37, 20 August 2010 (UTC)[reply]

Can't speak for anyone else, but I haven't attaempted to answer your original questions because (a) they call for opinion, which is not the best use of Wikipedia's reference desks and (b) you come across as bad-tempered and argumentative, so even if I did venture an opinion, I think there is a good chance you would just use it as a springboard for an argument, so it would be a waste of my time. Gandalf61 (talk) 08:53, 20 August 2010 (UTC)[reply]

Please answer my question? I won't attack you. What makes you think I'm bad tempered? I just like to express my opinion 'is all. I'm as polite as I can be and yet I'm misunderstood.I won't argue. Come on? —Preceding unsigned comment added by 110.20.55.15 (talk) 11:38, 20 August 2010 (UTC)[reply]

Fields Medal winners are listed at Fields Medal. Dragons flight (talk) 08:54, 20 August 2010 (UTC)[reply]

The argument of a given binary function is an (ordered) pair, so the argument of a given unary function is a singleton, isn't it? Eliko (talk) 08:45, 19 August 2010 (UTC)[reply]

Yes--Shahab (talk) 10:38, 19 August 2010 (UTC)[reply]

Unfortunately, it's mentioned nowhere in Wikipedia, as far as I know. Eliko (talk) 10:47, 19 August 2010 (UTC)[reply]

Well, technically, it's an individual rather than a singleton. But you could just as easily take it to be a singleton, and nothing important would change. --Trovatore (talk) 10:47, 19 August 2010 (UTC)[reply]

I'm looking for the formal definition. "technically" (as you've put it), a binary function doesn't have two arguments each of which is an individual, but rather has an argument being an (ordered) pair of individuals (because the order of both individuals must be preserved), so I would expect that, "technically", a unary function won't have one argument which is an individual, but rather will have an argument being a singleton of an individual. Eliko (talk) 12:33, 19 August 2010 (UTC)[reply]

There isn't necessarily a "the" formal definition. You're talking about details at the "implementation" level, which may vary from formalization to formalization, but don't change the conceptual content for most purposes.

I would say that a unary function takes an individual, and a binary function, at the most fundamental level, takes two individuals (distinguishing between them, as you note). For many purposes it's convenient to package those two individuals into a single individual, that we then call an ordered pair. But at the very fine level you seem to be examining this, that technically makes it a unary function that takes as its sole argument an ordered pair, rather than a binary function with two arguments. --Trovatore (talk) 22:11, 19 August 2010 (UTC)[reply]

You seem to distinguish between "the most fundamental level" and the "very fine level". So would you agree with the following? "At the most fundamental level", a binary real function takes two arguments each of which is a real number, while a unary real function takes one argument which is a real number. On the other hand, i.e. "at the very fine level", every real function is unary, the following way: a function that would be called "binary" at "the most fundamental level", is in fact (i.e. "at the very fine level") a unary function that "takes as its sole argument an ordered pair" of real numbers, while a function that would be called "unary" at "the most fundamental level", is in fact (i.e. "at the very fine level") a unary function that "takes as its sole argument a singleton" of a real number. Eliko (talk) 07:15, 20 August 2010 (UTC)[reply]

Dude, fair warning: If you persist in trying to parse my rhetorical flourishes for some precise technical meaning, I'm going to get upset with you and leave the conversation. I doubt I'm the only one who'll react that way. Phrases like "at a very fine level" have a function, which you as a presumably competent social actor should be able to discern, but the indication of some exact mathematical or philosophical category is not it.

That said, here's how I see it: A binary function, conceptually, is a way of (but not necessarily an expressible rule for) taking two things and getting a third thing, in such a way that the third thing depends only on the two inputs (with the slots for the two inputs being distinguishable).

How you code that concept into some specific framework, such as set theory, is another matter. You have lots of choices, and most of the time it doesn't matter even slightly which choice you pick. Occasionally it will matter (say, when you don't have a pairing function available), and for those instances it's good to practice mental flexibility and not get too hung up on the formalism. --Trovatore (talk) 08:54, 20 August 2010 (UTC)[reply]

In 2001: A Space Odyssey, the Monoliths are 1:4:9. That would be 1 by 4 by 9. But 1 what by 4 what by 9 what? Centimetres? Metres? Yards? Feet? Inches? Can somebody help? --Editor510 ^{drop us a line, mate} 08:58, 19 August 2010 (UTC)[reply]

PS: I know this is about an entertainment film but it is a mathematical question, so please don't ask me to move the question.

The author carefully avoided that question by using the smaller side of the monolith as unit of measurement, and then noticing that the larger side was 9 units and the intermediate side was 4 units. Bo Jacoby (talk) 10:56, 19 August 2010 (UTC).[reply]

(EC) Monolith (Space Odyssey)#Appearance and capabilities discusses this, the monoliths are not all the same size.81.98.38.48 (talk) 10:58, 19 August 2010 (UTC)[reply]

Ratios don't require units. HiLo48 (talk) 11:09, 19 August 2010 (UTC)[reply]

Hello! Let d(n) be a number of positive divisors of n, for example, d(6)=4. In 1907 S. Wigert proved that for any ${\displaystyle \epsilon >0}$ holds two statements:
1) for infinitely many n ${\displaystyle d(n)>2^{(1-\epsilon ){\frac {\ln n}{\ln \ln n}}}}$;
2) for all sufficiently large n ${\displaystyle d(n)<2^{(1+\epsilon ){\frac {\ln n}{\ln \ln n}}}}$;

Does anybody know more precise estimations than these?
Actually I am able to prove better estimation than 1), but I don't have any information about better results.
Thank you!
RaitisMath (talk) 12:52, 19 August 2010 (UTC)[reply]

If you don't get an answer here, you might try MathSciNet.—msh210℠ 17:32, 20 August 2010 (UTC)[reply]

Our divisor function function article discusses estimates a little bit and has some references. I'm not sure if there's enough to help you. 67.122.209.167 (talk) 09:28, 21 August 2010 (UTC)[reply]

Hi, I have recently encountered this proposition that seems somewhat vague to me: Proposition: Let X be a topological space without isolated points having countable ${\displaystyle \pi }$-weight and such that every nowhere dense subset in it is closed. Then it is a Pytkeev space.

The thing which is not clear to me is this, if every nowhere dense subset is closed, doesn't that means that it has to be discrete? and doesn't discrete means that every point is isolated? So, is the condition given in this proposition is that, there arn't any nowhere dense subsets in X? Which means that every subset of X is dense somewhere? which doesn't make sense.. I mean, what about subsets of X that contain one point for instance? Thanks! Topologia clalit (talk) 21:55, 19 August 2010 (UTC)[reply]

Consider an infinite set with the cofinite topology. This space is not discrete, but every nowhere dense set is closed: if a set is infinite, its closure is the entire space, so the only nowhere dense sets are finite. (This space also has no isolated points, and, if the underlying set is countable, has countable ${\displaystyle \pi }$-weight.) —Preceding unsigned comment added by 130.195.5.7 (talk) 00:09, 20 August 2010 (UTC)[reply]

Alternatively, you can start with any topology and make a finer one by defining all the nowhere dense sets to be closed (and then taking appropriate unions). This new space will have isolated points precisely if the old space did, will have no greater ${\displaystyle \pi }$-weight (maybe the same?), and all nowhere dense sets will be closed. —Preceding unsigned comment added by 130.195.5.7 (talk) 00:35, 20 August 2010 (UTC)[reply]

I see.. Thanks for your example. But I am still confused. I mean, here is the proof of this proposition. If we take your example of the cofinite topology, under consideration, how can I explain the emphesized remark in brackets? I mean, the nowhere dense subsets in your example are closed and not discrete.. Proof: Let ${\displaystyle x\in Cl(A)\setminus A}$. Then ${\displaystyle x\in Cl(Int(Cl(A)))}$, because every nowhere dense set is closed (and hence discrete). Let ${\displaystyle P_{n}:n\in \omega }$ be a list of elements of a countable ${\displaystyle \pi }$-base in the space, which are contained in ${\displaystyle Int(Cl(A))}$. Let ${\displaystyle M_{n}\cap P_{n}}$. Then ${\displaystyle \{M_{n}:n\in \omega \}}$ is a countable ${\displaystyle \pi }$-net, at x, and each ${\displaystyle M_{n}}$ is infinite.

The comment is saying that the nowhere dense subsets are discrete. I gave you an example where the total space isn't discrete. It's not hard to see that every finite set in my example is discrete, however.

To see that nowhere dense sets are discrete under the hypothesis, suppose one wasn't. Then it would have an accumulation point. Remove this accumulation point, and the resulting set would still be nowhere dense, but would not be closed, contrary to assumption.

Also, something in your second remark bothers me. Suppose for example that I take ${\displaystyle \mathbb {R} ^{2}}$ with the usual topology and try to define all the nowhere dense sets to be closed. Then, for example, the nowhere dense set ${\displaystyle \{{1 \over n}\}_{n\in \mathbb {N} }}$ will turn closed? But it cant be since it doesn't contain it's accumulation point 0.. what am I missing here? Thanks! Topologia clalit (talk) 08:10, 21 August 2010 (UTC)[reply]

Could you please supply an explicit reference to the text? — Fly by Night (talk) 16:11, 21 August 2010 (UTC)[reply]

Ya sure, it's "Weakly Frechest-Urysohn and Pytkeev spaces" by V.I. Malykhin and G. Tironi. From Topology and its Applications 104 (2000) 181-190 Let me know if you can't find it. Thanks! Topologia clalit (talk) 16:20, 21 August 2010 (UTC)[reply]

No, I can't find it. If it's from a journal then it's highly unlikely that I'll be able to find it. I was thinking more along the lines of a book. — Fly by Night (talk) 16:35, 21 August 2010 (UTC)[reply]

When you define the nowhere dense sets to be closed, you're changing the topology. Not everything which was an accumulation point still is.

I see.. OK thanks. I'll think about it. Prove to myself that this is a topology... Here is a direct link to the article: http://www.f2h.co.il/307676563971 Proposition 2.1 there.. What do you think? Topologia clalit (talk) 17:07, 21 August 2010 (UTC)[reply]

I don't speak Hebrew, and I don't want to download the application because I don't understand the site. Sorry. — Fly by Night (talk) 17:14, 21 August 2010 (UTC)[reply]

Nothing to worry about, just your average infested-with-extremely-intrusive-ads-and-possibly-malware file-sharing site...

Transferbigfiles below is more friendly. Anyway what the OP linked to is a pdf, not an application. -- Meni Rosenfeld (talk) 18:34, 21 August 2010 (UTC)[reply]

Sorry, I haven't even noticed that this link is in Hebrew.. Here is another link in: https://www.transferbigfiles.com/53c3b8c5-f0f5-4bbc-93d2-a0030eaae50a?rid=gcIEkCL2J4VQPfWN7RipdA%3d%3d It's in English.. —Preceding unsigned comment added by Topologia clalit (talk • contribs) 17:51, 21 August 2010 (UTC) Topologia clalit (talk) 18:07, 21 August 2010 (UTC)[reply]

${\displaystyle {\begin{bmatrix}1&2&0&1\\0&1&1&2\\0&0&0&1\end{bmatrix}}.}$
115.178.29.142 (talk) 03:25, 20 August 2010 (UTC)[reply]

Subtract twice the second row from the first row

${\displaystyle {\begin{bmatrix}1&0&-2&-3\\0&1&1&2\\0&0&0&1\end{bmatrix}}.}$

Subtract multiples of the third row from the other rows

${\displaystyle {\begin{bmatrix}1&0&-2&0\\0&1&1&0\\0&0&0&1\end{bmatrix}}.}$

This matrix is in reduced row echelon form. Bo Jacoby (talk) 06:42, 20 August 2010 (UTC).[reply]

Do one more step: add twice the second row to the first row:

${\displaystyle {\begin{bmatrix}1&0&0&0\\0&1&1&0\\0&0&0&1\end{bmatrix}}.}$ Staecker (talk) 11:47, 20 August 2010 (UTC)[reply]

That doesn't work since the (2,2)th entry x 2 + (1,2)th entry isn't 0; it's 2. [ personal attack removed ] —Preceding unsigned comment added by 110.20.55.15 (talk) 12:00, 20 August 2010 (UTC)[reply]

Your correction of Staecker's error (who does know the basic math but made a momentary mistake) is welcome. Your personal attacks are not welcome. Please stop this. -- Meni Rosenfeld (talk) 13:08, 20 August 2010 (UTC)[reply]

Heh- yeah my mistake. That's what I get for adding answers first thing when I wake up. Thanks Meni- Staecker (talk) 13:10, 20 August 2010 (UTC)[reply]

I've always wanted to find "the" Linear algebra book. A book that would serve me for my entire math life. But I never could find it. Hence I ask the gurus at this reference desk.

My ideal linear algebra book would: (a) Cover all the fundamentals of linear algebra such as what's covered in Herstein's Topics in Algebra in a nice and elegant way (canonical forms, bilinear forms, hermitian matrices, you know advanced topics + the basics) (b) Be significantly more advanced than this and cover topics in linear algebra that pop up deeply in other branches of math (say tensor algebra) (c) Be a pure math textbook.

So is there such a book? I like the style of Herstein's topics in algebra so a book with this style would be desirable. My problem with Herstein's topics in algebra is: it's great and solid but I want a more advanced book. I like my book to give some calculations but also lots of nice theory in a elegant way. + the book should cover as many advanced topics as possible. This is important. It should be comprehensive. If a series of texts is required, that's fine. I don't mind if my book requires math background in general algebra, in fact I want a book like that. Thanks guys ... Oh and I forget to add: this book should be primarily on finite dimensional vector spaces. It's alright if other vector spaces are covered, but I want the emphasis on finite. Thanks guys again. .. —Preceding unsigned comment added by 114.72.248.27 (talk) 07:29, 20 August 2010 (UTC)[reply]

Please answer my question? I request you as politely as I can. Why do people avoid me? What have I done? Please answer my question. I've made my question sound as polite as possible Thanks guys ... 114.72.244.249 (talk) 08:37, 20 August 2010 (UTC)[reply]

Presumably no one can think of any such book ATM. This is not a mathematical question, hence it's not possible to ponder it and figure out a solution; if no one here happens to have encountered a book meeting your rather high demands, you may well get no answer.—Emil J. 13:12, 20 August 2010 (UTC)[reply]

I can't think of any myself but you might find something at List of important publications in mathematics.--Salix (talk): 00:50, 21 August 2010 (UTC)[reply]

Find the values of ${\displaystyle k}$ for which the simultaneous equations:
${\displaystyle (k+1)x-3y=0}$
${\displaystyle 2x-ky=0}$
have solutions other than x=0, y=0
--220.253.222.146 (talk) 09:05, 20 August 2010 (UTC)[reply]

Geometrically, what is going on here is that each equation represents a line through the origin (0,0). A solution to the simultaneous equations is the co-ordinates of a point at which the lines cross. The lines will not cross at any other point apart from (0,0) unless the two equations actually represent the same line, in which case any pair of co-ordinates that satisfies one equation will also satisfy the other. From the condition that the two equations represent the same line, you can get a quadratic equation in k which has two (real) roots. The same quadratic equation falls out if you take an algebraic approach and think in terms of matrices and determinants instead. Gandalf61 (talk) 09:41, 20 August 2010 (UTC)[reply]

Or simply solve for one variable in the one and then substitute into the other. ${\displaystyle k=-3,2}$. —Preceding unsigned comment added by 203.97.79.114 (talk) 10:32, 20 August 2010 (UTC)[reply]

This looks like a homework question. Wikipedia Reference Desks do not do people's homework for them. We are happy to point people in the direction of suitable information on Wikipedia so they can read and learn, but there is no benefit to you if we do your homework. Dolphin (t) 11:33, 20 August 2010 (UTC)[reply]

[ vulgarisms removed ]

For the record, the answer is to use matrixes. It's as imple as 1,2,3 ... [ personal attack removed ] Your system of equations is

k+1 -3

2 -k

and your goal is to determine when this matrix above kills something to 0. (like maps (x,y) to 0) But it does so when it's determinent is 0 and this is like when -k (k+1) = -6 or -k^2 - k + 6 = 0. But the quadratic has a solution = 1 +,- sqrt(1 + 24)/-2. So k = 3 or k = -2. I repeat

<<<<<k = -3 or k = 2>>>>>

Write that on your homework assignment with the advanced matrix stuff and 'all. Your teacher will be very impressed and you'll get an A+. If she's a hot lady, maybe she'll go a bit further than that for your own pleasure ...—Preceding unsigned comment added by 110.20.55.15 (talk) 12:01, 20 August 2010

Please stop disrupting this reference desk, or you will be reported and subsequently blocked. -- Meni Rosenfeld (talk) 13:13, 20 August 2010 (UTC)[reply]

I'm just speaking my voice. Maybe I was vulgar in the 80's but nowadays the words I used (I won't state them cause clearly people have problems with it) are common. If you get out more Meni Rosenfeld, you'll see that. And yes, the guy who asked the question is a ******* if he wants his homework done for him. But ******** are ******** and so we might as well give the ******* the answers. I CAN'T see what I've done wrong. I CAN see why other people *THINK* I've done wrong. For some odd reason society frowns down upon telling the truth ... But why shouldn't I voice my opinions? BTW, now that you've told me I'll stop, but all I'm saying is that you guys are all against me. First some dude called Tango removes my question for no particular reason. Then some annonymous guy tries to block me, and then you tell me to "sign my post". Well, I complied. But now people don't answer my questions anymore. I mean what do I need to do to get some service around here? Don't you guys get paid for answering my questions? I'll report you guys to Wikipedia for not doing your job right ... I mean I don't want to, but I just lost my temper. I'm ignored. And some guy called Gandalf insults me above. So everyone hates me. I think I should just leave and save you all the miseries. Goodbye. Block me, I don't really care. But I'm not posting here anymore until you guys start treating me with some respect. —Preceding unsigned comment added by 110.20.55.15 (talk) 13:28, 20 August 2010 (UTC)[reply]

You seem to misunderstand a lot of aspects of how the reference desk operates. One of these is that no one here gets paid for anything. Another is that we don't do homeworks, which you'd know if you read the prominent box on top of this page.—Emil J. 13:50, 20 August 2010 (UTC)[reply]

right: homework exists to help keep class distinctions in a country (United States) that has all too many social upstarts and "mobility". The only thing keeping any uppity declasse' from rising through the ranks of education is the assignment of homework that other kids with better-educated parents will have help in completing but which that kid will not. I think it is the last barrier to total communist anarchy dictatorship of the proletariat, and any cheating, or attempts by students to coopt other classes' parents (such as the parents who would contribute to the refernece desk) must be smitten strongly, quickly, and decisively. If someone can't ask their own parent this question, then maybe that person should accept his lot in life instead of trying to pull down the average in a higher class to which he, of rights, would never accede. 92.230.70.110 (talk) 21:23, 20 August 2010 (UTC)[reply]

Stop talking bullcrap 92.230.70.110. If you want to see bullshit in it's finest form look above LOL. —Preceding unsigned comment added by 110.20.11.94 (talk) 00:57, 21 August 2010 (UTC)[reply]

In a Yoplait Yogurt commercial, the development of a woman's bone strength is represented as a curve that starts at (0,0) and rises quickly before dropping at a slightly slower rate which decreases, and then approaching the X-axis. Whether this is accurate or not I don't know, particularly since the curve breaks like a bone when it reaches the point where a woman's bones would break easily. But I recall seeing a function like this, which for values of x less than 0, F(-x) = -F(x). What might this function be?Vchimpanzee · talk · contributions · 21:36, 20 August 2010 (UTC)[reply]

${\displaystyle xe^{-x^{2}}}$ ? --Qwfp (talk) 21:56, 20 August 2010 (UTC)[reply]

Thanks. That might be it. Where could I find a graph of that, because Google has failed me.Vchimpanzee · talk · contributions · 17:59, 21 August 2010 (UTC)[reply]

http://www.wolframalpha.com/input/?i=x*e^%28-x^2%29. Note that this is just one example, there are infinitely many functions matching your description. -- Meni Rosenfeld (talk) 18:18, 21 August 2010 (UTC)[reply]

Hi all!

I'm trying to solve the following:

Let K be a field and c ∈ K. If m, n ∈ Z>0 are coprime, show that X^mn−c is irreducible if and only if both X^m − c and Xⁿ − c are irreducible. (Hint: Use the Tower Law.)

Problem is, I don't really understand how you can use the tower law when you don't know something is irreducible: surely you require irreducibility of f(x) for K[x]/(f(x)) to actually be a field? In which case, you can't really say anything about the degree of the relevant field extensions before you know they're irreducible? And obviously you can't say they're irreducible until you've proved the result, at which point it's moot.

For example, if K_mn, K_m and K_n represent K[x]/(x^mn-c), and so on, respectively, and we assume K_m, K_n are irreducible, then [K_n:K][K_m:K]=[K_mn:K_m][K_m:K]=mn I believe, but we can't then say '[K_mn:K]=mn' because we don't know that's a field necessarily, do we? I'd appreciate a nudge in the right direction, as I've been stuck on this for a while and I think it's probably a very short argument, just one I'm not getting - I believe I've proved X^mn−c is irreducible implies both X^m − c and Xⁿ − c are irreducible via the contrapositive, but I didn't use the tower law, or the fact that m, n are coprime: simply write Y=X^m (assuming Xⁿ-c reducible), then X^mn - c = Yⁿ - c is reducible. Am I doing something wrong here? Or just missing the obvious?

Thankyou very much for any responses, much appreciated indeed - 84.45.219.231 (talk) 22:52, 20 August 2010 (UTC)[reply]

Just let α be any root of X^mn-c and consider K[α]. This is definitely a field; it's isomorphic to K[x]/(f), where f is the minimal polynomial of α (which could in general be any irreducible factor of X^mn-c). Algebraist 23:02, 20 August 2010 (UTC)[reply]

Right, well I got the half of the proof which I was missing, so thankyou! (Something along the lines of 'm divides this, n divides this, they're coprime so mn divides this and it's less than or equal to mn, so equals mn'...) - Out of interest, does this argument also enable us to prove the direction I succeeded in proving without the tower law etc? I can't see how you could use it to prove '<=>', rather than just '<=', but I could be overlooking something. Either way, thanks for the help, I suppose I should probably press on with this fiasco which is teaching myself Galois theory, though I'm quite convinced I'm totally useless at it! Cheers :) 84.45.219.231 (talk) 00:09, 21 August 2010 (UTC)[reply]

If you can't do Galois theory what hope do you have to suceed in math? So to help you, take the following advice: Compute as many Galois groups as possible. That should aid your disorder. No need to thank me. —Preceding unsigned comment added by 110.20.11.94 (talk) 01:01, 21 August 2010 (UTC)[reply]

I'm not sure to what extent that's sarcastic, but if I intended to remain useless at Galois theory, I probably wouldn't be here trying to improve :) 84.45.219.231 (talk) 01:40, 21 August 2010 (UTC)[reply]

Please ignore the comments of 110.20.11.94, who unfortunately is now trolling this reference desk. I am sorry that you had to see this. -- Meni Rosenfeld (talk) 17:26, 21 August 2010 (UTC)[reply]

Much appreciated Meni, I shall do! Always glad there are dedicated 'refdesk-ees' such as yourself and Algebraist about to help. 84.45.219.231 (talk) 20:18, 21 August 2010 (UTC)[reply]

What functions f(x) are there such that f[f(x)]=x? Obviously excluding f(x)=constant of course. 76.230.213.76 (talk) 00:13, 21 August 2010 (UTC)[reply]

Assuming you can graph these, have a look at functions which are symmetric about the line y=x: for example, f(x)=1/x, f(x)=1-x etc. Your example is incorrect - for example, f(x)=0 gives f(f(1))=f(0)=0 which does not equal 1. Maybe you were thinking of f(x)=x, which is trivially symmetric about y=x (since it 'is' y=x). 84.45.219.231 (talk) 00:31, 21 August 2010 (UTC)[reply]

Such a function is called an involution. Algebraist 00:36, 21 August 2010 (UTC)[reply]

(also called a self-inverse function) Dbfirs 11:51, 21 August 2010 (UTC)[reply]

Another interesting concept to think about is the functional square root. That is, given a function f, can we find or create a function g such that g(g(x))=f(x)? After that you might consider the idea fractional derivatives. Not all these avenues of thought have proved fruitful but they are interesting to think about nonetheless. Zunaid 07:01, 21 August 2010 (UTC)[reply]

This is similar to Fixed point (mathematics).Smallman12q (talk) 02:13, 22 August 2010 (UTC)[reply]

What would be the best way to graphically display a flight course given data (per second) of elements such as airspeed, altitude, vertical acceleration, heading, pitch, etc?

Thank You —Preceding unsigned comment added by 74.113.218.132 (talk) 00:16, 21 August 2010 (UTC)[reply]

You mean on a graph or chart?Smallman12q (talk) 02:15, 22 August 2010 (UTC)[reply]

If your data consists only of speeds and accelerations then you need to integrate to be able to plot the course. Someone might point you towards some special software, but you could easily set up the approximate calculations on a spreadsheet, and then graph the resulting displacements to display a flight course. Errors would accumulate over a long course. Dbfirs 06:41, 22 August 2010 (UTC)[reply]

Does anyone know a proof that if A is a square matrix then there's a matrix B s.t. AB = BA = det(A) I. I know B is called the adjugate of A, but how do you find such a B or proof such a B exists/ Thanks. —Preceding unsigned comment added by 110.20.59.41 (talk) 12:29, 21 August 2010 (UTC)[reply]

You might want to take a look at the adjugate matrix article. -- The Anome (talk) 12:49, 21 August 2010 (UTC)[reply]

(edit conflict) Try the adjugate matrix article. It's mentioned in this subsection. It's a consequence of the Laplace expansion. — Fly by Night (talk) 12:51, 21 August 2010 (UTC)[reply]

I know but how do you PROVE that the adjugate exists? Sorry I wasn't clear and thanks for youre answer. —Preceding unsigned comment added by 110.20.9.174 (talk) 12:51, 21 August 2010 (UTC)[reply]

It has an explicit definition, and so must exist. The definition involves taking determinants of sub-matrices. That's just elementary arithmetic and algebra. Try this subsection of the adjugate matrix article. The only thing that needs to be proved is that a square matrix M, multiplied by its adjugate matrix adj(M), is the determinant of M multiplied by the identity matrix: i.e. M⋅adj(M) = det(M)⋅I. — Fly by Night (talk) 12:52, 21 August 2010 (UTC)[reply]

To be precise, note that the existence of such a ${\displaystyle B}$ is fairly obvious, and the relation does not characterize the adjugate. Precisely, if ${\displaystyle A}$ is not invertible, ${\displaystyle B:=0}$ obviously verifies the relation, although in general it's not the adjugate of ${\displaystyle A.}$ On the other hand, if ${\displaystyle A}$ is invertible, ${\displaystyle B:=\det(A)A^{-1}}$ obviously verifies the relation too (and in this case, it is the adjugate). Just to say that the existence of such a ${\displaystyle B}$ is fairly obvious, even though it does not characterize the adjugate. --pma 21:53, 21 August 2010 (UTC)[reply]

Thanks pm. Bt can someone give the explciit calculation? I can't find this in linear algebra books. In one book, it says the existence of teh adjugate is another form of the Cayley-Hamilton theorem. Is this true? If so how? Thanks for you answer. —Preceding unsigned comment added by 114.72.245.87 (talk) 00:11, 22 August 2010 (UTC)[reply]

I want to show that any arithmetic progression if continued long enough wouldn't contain primes. How should I proceed. Thanks-Shahab (talk) 07:46, 22 August 2010 (UTC)[reply]

[personal attack removed] If it contained primes over and over you'd clearly get a contradiction. Think about it and maybe you'll see it —Preceding unsigned comment added by 114.72.236.126 (talk) 08:10, 22 August 2010 (UTC)[reply]

Do you mean that any arithmetic progression eventually contains a number which is not prime? This follows from the prime number theorem. -- Meni Rosenfeld (talk) 08:53, 22 August 2010 (UTC)[reply]

Ya don't need any [profanity removed] prime number theory to see it. Besides, it doesn't follow from th eprime number theorem since ya could get primes appearing in the sequence over and over but less frequently as n goes to INFINITY WITHOUT contradicting any bullcrap prime number theory. So no, think harder Meni Rosenfeld and Shahb. —Preceding unsigned comment added by 114.72.192.74 (talk) 09:02, 22 August 2010 (UTC)[reply]

Yes I did mean so Meni, like the AP 5,11,17,23,29 has primes only but then the next term is non-prime. Uptil now the biggest such prime AP that has been found has around 20 terms. Not sure how the prime number theorem applies. -Shahab (talk) 09:05, 22 August 2010 (UTC)[reply]

Prove by contradiction. If there was a sequence with only primes, what would a lower bound for ${\displaystyle \lim _{n\to \infty }{\frac {\pi (n)}{n}}}$ be? What is the limit in light of the PNT? -- Meni Rosenfeld (talk) 09:19, 22 August 2010 (UTC)[reply]

[irrelevant vulgarities and personal attack removed] —Preceding unsigned comment added by 110.20.58.220 (talk) 10:25, 22 August 2010 (UTC)[reply]

[personal attacks removed] —Preceding unsigned comment added by 114.72.192.74 (talk) 09:09, 22 August 2010 (UTC)[reply]

Because if an arithmetic progressions with difference d contained only primes then the asymptotic density of primes would be at least 1/d, which contradicts the prime number theorem. Gandalf61 (talk) 09:19, 22 August 2010 (UTC)[reply]

For a constructive proof, recall that an arithmetic progression looks like ${\displaystyle a+bn}$ for some choice of ${\displaystyle a}$ and ${\displaystyle b}$. You want a choice of ${\displaystyle n}$ that will make this be obviously non-prime. Since all you have to work with are ${\displaystyle a}$ and ${\displaystyle b}$, try choosing one of those for ${\displaystyle n}$ and seeing what you can get. —Preceding unsigned comment added by 203.97.79.114 (talk) 09:20, 22 August 2010 (UTC)[reply]

... but make sure your proof covers the case a = 1. Gandalf61 (talk) 09:30, 22 August 2010 (UTC)[reply]