Here's my problem:

We have two points A and B, and based on their positions (We could supply any two pairs of coordinates), we need to draw a line that is centered on, and perpendicular to a line connecting A and B. This line needs to be 10 pixels long, and if possible, can be moved up and down the line connecting A and B by adjusting some kind of percent as a variable.

http://img151.imageshack.us/img151/2341/linesb.png

Hopefully the picture should make my written explanation more clear. I am looking for a formula for the coordinates of the endpoints to the above mentioned line (C and D).

Input of Function

F(x1,y1,x2,y2,percent)

I hope i have made the problem clear enough. Thanks in advance for the help.

74.117.245.62 (talk) 03:43, 31 December 2010 (UTC)[reply]

If you have three points p = (p₁,p₂), q = (q₁,q₂) and r = (r₁,r₂) (p ≠ q) then the line at right angles to the chord joining p and q that passes through r is given by the points (x,y) which satisfy the equation

${\displaystyle (r_{1}-x)(p_{1}-q_{1})+(r_{2}-y)(p_{2}-q_{2})=0\,.}$

You can assume that r lies on the midpoint of the chord joining p and q so that

${\displaystyle \left(r_{1},r_{2}\right)={\frac {1}{2}}\left(p_{1}-q_{1},p_{2}-q_{2}\right)\,.}$

In fact, you can construct a one-parameter family of lines given by

${\displaystyle \left(r_{1},r_{2}\right)=\lambda \left(p_{1},p_{2}\right)+(1-\lambda )(q_{1},q_{2})\,,}$

where 0 ≤ λ ≤ 1. When λ = 0 you have the line passing through q that is perpendicular to the chord joining p and q. When λ = 1 you have the line passing through p that is perpendicular to the chord joining p and q. So you just need to plot all of the lines

${\displaystyle (r_{1}-x)(p_{1}-q_{1})+(r_{2}-y)(p_{2}-q_{2})=0\,,}$

${\displaystyle {\text{where }}\left(r_{1},r_{2}\right)=\lambda \left(p_{1},p_{2}\right)+(1-\lambda )(q_{1},q_{2})\,,}$

and ½ – ε ≤ λ ≤ ½ + ε for a number 0 ≤ ε ≤ 1 of your choosing. This will give you a strip of carpet at right angles to the chord joining p and q that is centred at the midpoint of that chord. I hope that helps. — Fly by Night (talk) 04:20, 31 December 2010 (UTC)[reply]

Okay.... I understand MOST of this, I basically select my lambda which locates your point "R" by mathematically setting the "contribution" of my named points "A" and "B". Then i can use the other equation you gave to get a general equation going through point R with the appropriate vector (perpendicular to the line AB).... but then whats the rest of it? Is it some sort of restriction on the domain of this general equation? or.... ?

74.117.245.62 (talk) 04:28, 31 December 2010 (UTC)[reply]

You talk about a "strip of carpet".... seeming to indicate a rectangular area or a parallelogram.... what i need is a line segment... so i need to know how to restrict x somehow in the equation right?

74.117.245.62 (talk) 04:30, 31 December 2010 (UTC)[reply]

Pardon my multiple posts, im just trying to work this out mathematically. Since we are working in pixels, this creates a special situation i think (All x,y should be integers). We have the point R = (r1,r2), and when x = r1 in the equation of the perpendicular line, then y = r2. Therefore to get the points "C" and "D" in my picture i need to plug in x = r1 +- 5 and round off/up/down to get the respective y coordinates? This seems to work well except for in the special case where the line is vertical. Is this correct?

74.117.245.62 (talk) 04:41, 31 December 2010 (UTC)[reply]

I see. I misunderstood the problem slightly. I thought you wanted the width of the blue line in your link to be fixed. Changing the range of values of λ changes the thickness of that line. Changing r changes where the red line crosses the blue line. What you need to do is fix a λ to give a single line and then restrict the domain of x and y. As for the pixelation, well, I wanted to give a mathematical solution and was hoping that you would be able to do the computer side. I'm useless with programming; so I can't help with that. Sorry. — Fly by Night (talk) 18:36, 31 December 2010 (UTC)[reply]

If I understand you, then no, that's not quite right. It sounds like that would only work for a 45 degree line, and then the length would be 10*SQRT(2), not 10. StuRat (talk) 07:19, 31 December 2010 (UTC)[reply]

If processing time is not an issue, you can do this a rather simplistic way. The algorithm is: Calculate distance from A to B. Imagine a line of that length from the origin (0,0) as A and a new B at (Length, 0) - so it is perfectly horizontal. The X coordinate of C and D will be percent*length. The Y coordinate for C will be 5. For D it will be -5. You now have all the coordinates for a horizontal bar with A at the origin. You need to rotate it to the proper angle, leaving A at the origin. The angle of rotation is easily calculated with trignometry. The cosine of the angle is the x offset between A and B (x for B minus x for A) divided by the distance between them (which we already calculated). Sine would be the y offset between A and B divided by the distance. With cosine and sine, given any coordinate (x,y), the rotated coordinate is (x*cos - y*sin, y*cos + x*sin) -- if I am remembering it correctly. Now, you've rotated it to the correct angle, but A is at the origin. Add A's x to very x and A's y to every y and you'll offset it to the proper location. -- kainaw™ 13:52, 31 December 2010 (UTC)[reply]

Fly by Night, It's okay, what you gave me was useful still. :) Kainaw, your method seems like it would work... but i have difficulty understanding clearly how to proceed with the rotation part. I will do my best and bust out the distance formula etc on this and see where i get....

74.117.245.62 (talk) 19:22, 31 December 2010 (UTC)[reply]

Okay, i THINK i get it...

${\displaystyle T=Theta=arccos((x_{b}-x_{a})/Distance)}$

${\displaystyle Rotation(x_{1},y_{1})(x_{2},y_{2}):(x_{2},y_{2})=(x_{1}*cosT-y_{1}*sinT,y_{1}*cosT+x_{1}*sinT)}$

Is this correct?

74.117.245.62 (talk) 19:53, 31 December 2010 (UTC)[reply]

You don't need to use arccos because you don't need the angle. You only need the cosine and sine (you are calculating the angle from cosine and then calculating the cosine from the angle). Distance will be sqrt((bx-ax)²+(by-ay)²). Before rotation, C will be at coordinate (pct*distance, 5). D will be at (pct*distance,-5). I'm going to call pct*distance P. So, C is at (P,5) and D is at (P,-5). Now, rotate. C will be at (P*cos-5*sin, 5*cos+P*sin). Note, cos is (bx-ax)/distance and sin is (by-ay)/distance. You don't actually use a cos or sin function. D will be at (P*cos+5*sin, -5*cos+P*sin). Then move C and D by adding A to them. C will be at (P*cos+5*sin+Ax, 5*cos+P*sin+Ay). -- kainaw™ 20:18, 31 December 2010 (UTC)[reply]

Right, i just calculated θ for simplified looking equations, that's all. Your solution makes sense, thanks! 74.117.245.62 (talk) 20:23, 31 December 2010 (UTC)[reply]

I'm not sure that we need to use trigonometry; even though it is very natural. In your picture, you have points A and B. Now imagine you set off in a straight line from A to B. We can express your position as ( 1 – t ) A + t B, where for t = 0 you're at A and at time t = 1 you're at B. The line joining C and D can be thought of as you carrying a pole at right angles to your direction of travel. The vector B – A has its tail at A and its head at B. Let us write A = (a₁, a₂) and B = (b₁, b₂). A vector perpendicular to B – A is given by

${\displaystyle (b_{1}-a_{1},b_{2}-a_{2})^{\perp }=(a_{2}-b_{2},b_{1}-a_{1}).}$

This vector is parallel to the line joining C and D in your picture. We need to normalise it, meaning that we'll make it have unit length. That will make the calculations easier. Let

${\displaystyle {\mathbf {N} }={\frac {\left(a_{2}-b_{2},b_{1}-a_{1}\right)}{\sqrt {(a_{1}-b_{1})^{2}+(a_{2}-b_{2})^{2}}}}\ .}$

Let's make ( 1 – t ) A + t B. Let p(t) denote the point between A and B after times t. When t = 0 you're at A and at time t = 1 you're at B.

${\displaystyle {\mathbf {p} }(t)=(1-t)(a_{1},a_{2})+t(b_{1},b_{2})=((1-t)a_{1}+b_{1},(1-t)a_{2}+b_{2}).}$

Then your lines is parametrised by the following:

${\displaystyle {\mathbf {p} }(t)+s{\mathbf {N} },}$

where 0 ≤ t ≤ 1 choses where the line joining C to D hits the line joining A to B. In the picture, 10% means t = 0.1, 50% means that t = 0.9 and 90% means that t = 0.9. The variable s tells you how far, on either side of the line from A to B, you want the line from C to D to extent. If you chose s = 0 then you'll just have a point on the line from A to B. If you chose –2.5 ≤ s ≤ 2.5 then the line will be of length 5 with half on either side. I hope this makes sense. — Fly by Night (talk) 02:35, 1 January 2011 (UTC)[reply]

so here is an array:
123
456
789

you can rotate it 90 degrees clockwise
741
852
963

can you give me pseudocode to rotate a two-dimensional array clockwise or counterclockwise by 90 degrees?

This question is being posted in three steps, to slowly cook the frog and get what I'm really interested in. (I'm the fractal guy from above). Please bear with me! —Preceding unsigned comment added by 87.91.6.33 (talk) 10:06, 31 December 2010 (UTC)[reply]

Pseudocode is more the computing reference desk, what's wrong with Rotation matrix or Group (mathematics)for your purpose? Dmcq (talk) 10:35, 31 December 2010 (UTC)[reply]

the second answer is simple enough: I can't wrap my head around the idea of a group :( :(. I'm just too dumb. Maybe I shouldn't be doing what I'm doing then, but whatever. Look, I know my question above seems simplistic, but I am just laying the groundwork for my 2nd and 3rd followup which introduce something that I simply can't wrap my head around. remember, I don't need real code, just pseudocode, something like (for 90 degree clockwise):

for each row and each column in the new matrix:
New Matrix Row i, Column j = Old matrix Row ???,column ????...

even though it is simplistic, could someone tell me that? 87.91.6.33 (talk) 11:25, 31 December 2010 (UTC)[reply]

Sorry I see, that was an actual array and you want a bit of programming. Okay

for each row and each column in the new matrix:
New Matrix Row i, Column j = Old matrix Row (upper bound - j + lower bound) ,column i

cheers Dmcq (talk) 12:30, 31 December 2010 (UTC)[reply]

okay, so let's step through it with a 2x2 matrix:1
12
34

meaning: 
old row 1 column 1 = 1
old row 1 column 2 = 2
old row 1 column 1 = 3
old row 1 column 2 = 4

now let me use your formula
new row 1 column 1 = old row (2 - 1 + 1=) 2 column 1
new row 1 column 2 = old row (2 - 2 + 1=) 1 column 1
new row 2 column 1 = old row (2 - 1 + 1=) 2 column 2
new row 2 column 2 = old row (2 - 2 + 1=) 1 column 2
yielding
31
42

great it looks good!  So, onward and upward!!

now let's have a 3-dimensional "cube" array just as the above is a 2-dimensional "square" array. For simplicity I'll just use 2x2x2 cube. front slice (facing you):

12
34

the slice behind it:

56
78

Now could you please give me the pseudocode for rotating THAT by 90 degrees clockwise. There are a few possibilities: clockwise just like the above (ie the side facing you) clockwise from side to side as it remains flat on the table (if it is resting on the table, not lifting it from the table, just spinning it clockwise 90 degrees) clockwise from front to back, so that the right and left side just spin around without changing from staying on your right and on your left respectively.

so, what are the respective three pseudocodes for those three manipulations? Thanks! 87.91.6.33 (talk) 13:46, 31 December 2010 (UTC)[reply]

For clarity, are you writing a Rubik's cube program? If that is the case, we can provide more specific help.

As for rotation - the code for each rotation is unique. You can't have a single "90 degree rotation" function that handles any rotation possible. You need one function to rotate the face. You need one to rotate each side. You need one to rotate the top and one to rotate the bottom. It is very easy to write these functions. It is a very trivial matter of looking at the array and writing new[x][y] will have value of old[...][...]. -- kainaw™ 15:06, 31 December 2010 (UTC)[reply]

Nooononono, this isn't where it's going!!! ONE function to rotate the cube by 90 degrees (obviously the function can take the axis of rotation, x, y, or z) is EXACTLY what I'm looking for! Or, if not one function that takes the axis of rotation, then three exmaples of pseudocode, one for each function. 87.91.6.33 (talk) 15:28, 31 December 2010 (UTC)[reply]

also, just to be explicit: this is a solid cube (like playing dice), not a Rubik's cube or something else. 87.91.6.33 (talk) 15:33, 31 December 2010 (UTC)[reply]

So, you want to rotate the whole cube. You have a 3-dimensional array. I will use computer-based arrays since you are referring to pseudocode. The old array looks like {{{1,2},{3,4}},{{5,6},{7,8}}}. Index 0,0,0 is 1. Index 0,0,1 is 2. index 0,1,0 is 3. Index 0,1,1 is 4... and so on. You want to rotate it so you get {{{5,1},{7,3}},{{6,2},{8,4}}} (as one example of a rotation). As such, the new index 0,0,0 has a 5. That used to be in the old index 1,0,0. So, new(0,0,0) = old(1,0,0). The new index 0,0,1 has a 1. That used to be the old index 0,0,0. So, new(0,0,1) = old(0,0,0). It is very trivial to do this for every index. You do not need a "formula". However, a computer language that allows for advanced list manipulation (like Lisp) will make the process much easier. -- kainaw™ 15:48, 31 December 2010 (UTC)[reply]

Um, I do need the formula. The formula is exactly what interests me about this whole thing. Even though you say I need to do it by hand for 27 numbers, obviously there is a formula for what I am calculating in my head... could you write it down for me explicitly, i.e. as a pseudocode loop... —Preceding unsigned comment added by 87.91.6.33 (talk) 16:53, 31 December 2010 (UTC)[reply]

Okay I'll bite

Choose a,b,c to be a permutation of 1,2,3
ub =upper bound and ul = lower bound

for every index a and index b and index c
 New Matrix i of index a, j of index b, k of index c =
   Old matrix ub+lb-j of index a, i of index b, k of index c

cheers Dmcq (talk) 22:14, 31 December 2010 (UTC)[reply]

First of all, thank you for taking my request seriously. I migh be able to answer this myself by stepping through your pseudocode, but it's probably faster to just ask you: could you explain in words what happens when you choose a permutation of 1,2,3? Thank you for your response, your time and effort, and taking my request seriously. Oh, and: Happy New Year!!! 87.91.6.33 (talk) 01:57, 1 January 2011 (UTC)[reply]

Considering the array as a 3-D cube which is what I believe you mean with the first two indexes mapping to horizontal axis and the third to up and down then the array is rotated a right angle clockwise round the vertical axis through the centre. Dmcq (talk) 09:32, 1 January 2011 (UTC)[reply]

Okay, so you just decided on the axis of rotation for me, and wrote the code with that in mind? How did you write it? (If I wanted to generalize to rotations about the other two axis...) And also, below, the much-anticipated followup!! —Preceding unsigned comment added by 87.91.6.33 (talk) 09:38, 1 January 2011 (UTC)[reply]

The code doesn't change index c so it is the axis rotated about. The various permutations of 1,2,3 will give six different rotations, either a clockwise or anticlockwise 90° rotation around each axis. There's other possible rotations for instance along an axis between opposite corners but this doesn't give them, you only asked for 90° rotations. Dmcq (talk) 09:51, 1 January 2011 (UTC)[reply]

And now comes the hard part. Please give me code to rotate the following FOUR dimensional cube:

12
34

and behind it

56
78

and atwixt the cube above is the cube below:

9A
BC

and behind it (atwixt 56/78):

DE
FG

NOTE: I only changed to letters near the end for the sake of "ascii art" - A, B, C, etc is just 10, 11, 12, etc.

The word "atwixt" is the word I'm using for the fourth dimension, like behind or next to or above for the first three dimensions.

So, could someone give me formulas for rotating the 4-dimensional cube as written, about the front-back axis, the up-down axis, the side-to-side axis, and the atwixt-to-atwixt axis.

Forget four: please give me the pseudocode for rotating an n-dimensional "cube" array as described above. I would specify which axis, i.e. dimension number 1, 2, 3, 4, ... and the pseudocode should tell me how to apply a 90 degree clockwise rotation on that dimension. This is what I was really getting at throughout my question, but I think my gradual, boil-the-frog approach has the greatest chances of getting me a correct answer! Thank you!! 87.91.6.33 (talk) 09:46, 1 January 2011 (UTC)[reply]

For rotations of 90° where two axis are given and you want to leave the other indexes along the pseudocode above just need extra letters added and the various end indexes unchanged. However note that for 4-D such a rotation is a rotation with a plane as an axis - you have two indexes which don't change. A second independent rotation within the first rotations axis plane could be done at the same time. Dmcq (talk) 09:59, 1 January 2011 (UTC)[reply]

wait, what. I don't understand any of that. where do planes come in, or TWO axis? Doesn't one axis and the words "90 degrees clockwise" define a unique rotation? If it doesn't, could you give a counterexample? Then I'll ask about the rest of your paragraph! (which I very much appreciate). 87.91.6.33 (talk) 10:02, 1 January 2011 (UTC)[reply]

When you rotated your 2-d array everything moved. When you rotated your 3d array one of the indexes in the new matrix was the sane as in the old one, it defined a line with increasing values of the 3rd index going up the line. In 4D two of the indexes would be the same if you used the same pseudocode. The two unchanged indexes define a 2d plane. So the rotation is about a plane, not a line. You can rotate the 2D plane the first rotation is about - so one can do two independent rotations at the same time each about an axis formed by the plane of the other rotation. Dmcq (talk) 10:53, 1 January 2011 (UTC)[reply]

In fact on the tesseract page at the top it shows a simple rotation and halfway down it shows a double rotation. Not all that easy to see but nice to look at. Dmcq (talk)

To illustrate what Dmcq is saying, here is your tesseract, dissected to show its 2x2 layers in a 2x2 grid:

1 2 3 4	5 6 7 8
9 A B C	D E F G

Here is how it appears after we rotate the first and second co-ordinates through 90 degrees, keeping the third and fourth co-ordinates fixed:

3 1 4 2	7 5 8 6
B 9 C A	F D G E

Here is how it appears after we instead rotate the third and fourth co-ordinates through 90 degrees, keeping the first and second co-ordinates fixed:

9 A B C	1 2 3 4
D E F G	5 6 7 8

And here is how it appears after we do both of these rotations at the same time:

B 9 C A	3 1 4 2
F D G E	7 5 8 6

Gandalf61 (talk) 14:28, 1 January 2011 (UTC)[reply]

Thanks guys!!! I'm getting my head around the answers. Meanwhile: Gandalf, is there a transformation ("rotation") you could apply to your Grid that would cause "1234" to be rotated so that 1 is in upper-left grid square (which starts with 1234), 2 goes to upper right (which had held "5678"), 3 to lower-left, 4 to lower-right? I don't honestly know what part of these grid locations the numbers should go into, or what should happen to the other numbers. It just seems to make sense for me that there should be a way to "rotate" the dimensions so that 1234, move to the outer dimension... Is there such a "rotation"? What is the complete result? Thanks! 87.91.6.33 (talk) 22:28, 2 January 2011 (UTC)[reply]

You can rotate it to this:

1 5 9 D	2 6 A E
3 7 B F	4 8 C G

but you can't get 1,2,3,4 to the far corners, they'd be too far from each other. If you express the numbers as binary starting from 0 it becomes more obvious, so 7 for instance is 0111. Plus there is an orientation constraint. Dmcq (talk) 22:58, 2 January 2011 (UTC)[reply]

Indeed. Note that this transformation swaps the first and third co-ordinates and also swaps the second and fourth co-ordinates. So it is the result of reflections in two hyper-planes, and is therefore a rotation. And we can see that it is a rotation of order 2 - if you do the rotation twice, you get back to the starting position. The two hyper-planes intersect in the fixed plane of the rotation, which passes through the cells labelled 1, 6, B, G. Gandalf61 (talk) 09:25, 3 January 2011 (UTC)[reply]

You will enjoy learning the J (programming language). Consider this:

   i. 2 2 2 2
 0  1
 2  3

 4  5
 6  7 


 8  9
10 11

12 13
14 15

   |."1 i. 2 2 2 2
 1  0
 3  2

 5  4
 7  6


 9  8
11 10

13 12
15 14

   |."2 i. 2 2 2 2
 2  3
 0  1

 6  7
 4  5


10 11
 8  9

14 15
12 13

   |."3 i. 2 2 2 2
 4  5
 6  7

 0  1
 2  3


12 13
14 15

 8  9
10 11

   |."4 i. 2 2 2 2
 8  9
10 11

12 13
14 15


 0  1
 2  3

 4  5
 6  7

Bo Jacoby (talk) 17:02, 1 January 2011 (UTC).[reply]

How many unique letters games are possible? --Omniintelligentia (talk) 10:40, 31 December 2010 (UTC)[reply]

It would be better if you just asked the question directly, rather than send us to a web site where we try to determine what you mean. As best as I can figure, you are asking this:

How many combinations of 9 letters are possible, where:

1) Repeats are allowed.

2) Each combo must have at least 4 consonants and 3 vowels.

I assume they mean that AEIOU are the 5 vowels and the remaining 21 letters are consonants. So, this gives us these 3 possibilities:

A) 3 vowels, 6 consonants: Chances are 5³ × 21⁶

B) 4 vowels, 5 consonants: Chances are 5⁴ × 21⁵

C) 5 vowels, 4 consonants: Chances are 5⁵ × 21⁴

Just add those together to get the final answer. StuRat (talk) 15:55, 31 December 2010 (UTC)[reply]

That's wrong, StuRat. You have to consider the order of the letters. Eg. in the first case the 3 vowels and 6 consonants can be in any order, which would mean 5³·12⁶·${\displaystyle \textstyle {\binom {9}{3}}}$ possibilities. – b_jonas 18:10, 1 January 2011 (UTC)[reply]

That depends on what you mean by "unique". Dbfirs 21:40, 2 January 2011 (UTC)[reply]

1. How can one prove that the subgroup of SL(2,F5) of order 24 is normal in SL(2,F5)?

2. Is f: SL(2,F5)/{+/- I} --> A5 defined by mapping g/{+/-I} (where g is an element in SL(2,F5)) to gK/{+/-I} a valid isomorphism?

3. Would I be right in thinking that there are no subgroups of SL(2,F5) isomorphic to A5?

Thanks! —Preceding unsigned comment added by 131.111.222.12 (talk) 13:23, 31 December 2010 (UTC)[reply]

Why, when you take the imaginary unit i and take it to the power of itself, you get approximately 0.2077, or something to that effect? 75.73.225.224 (talk) 14:54, 31 December 2010 (UTC)[reply]

iⁱ is an example at Exponentiation#Computing complex powers. The value you have in mind corresponds to k = 0 there: e^−π/2 = 0.20787957635... PrimeHunter (talk) 15:14, 31 December 2010 (UTC)[reply]

${\displaystyle i=e^{i{\frac {\pi }{2}}}}$ so ${\displaystyle i^{i}=(e^{i{\frac {\pi }{2}}})^{i}=e^{i^{2}{\frac {\pi }{2}}}=e^{-{\frac {\pi }{2}}}}$. Proof that ${\displaystyle i=e^{i{\frac {\pi }{2}}}}$ follows directly from the identity ${\displaystyle e^{ix}=\cos {x}+i\sin {x}}$ for x=π/2 --Omniintelligentia (talk) 16:31, 31 December 2010 (UTC)[reply]

That's one solution. The real answer is multi-valued because

${\displaystyle i=e^{i(\pi /2+2k\pi )}\,,}$

where k is a whole number. So, in fact,

${\displaystyle i^{i}=e^{-\pi /2+2n\pi }\,,}$

where n is a whole number. Take a look at Branch point. — Fly by Night (talk) 20:19, 31 December 2010 (UTC)[reply]

Fascinating. I just asked because that's what my calculator said, but then again, the answer to 78-94 is not "Low battery". 75.73.225.224 (talk) 21:21, 31 December 2010 (UTC)[reply]

You can't trust your calculator. Ask it what sin^–1(0) is and it'll return zero. This is because it gives you the principal value. But sin^–1(x) is another multi-valued function because sin(πn) = 0 for any whole number n. This means that sin^–1(x) is multi-valued:

${\displaystyle \sin(x)=0\ {\text{ for all }}\ x\in \{\ldots ,-2\pi ,-\pi ,0,\pi ,2\pi ,\ldots \}.}$

All the best! — Fly by Night (talk) 23:54, 31 December 2010 (UTC)[reply]

I think that sin${\displaystyle ^{-1}}$ in the calculator refers to arcsin which is defined as the unique solution to sin(x)=0 on interval of ${\displaystyle [-\pi /2,\pi /2]}$ So with such restriction, the calculator is correct, however since we are talking about complex numbers here, you are right about the multivalued inverse sin. (Igny (talk) 01:42, 1 January 2011 (UTC))[reply]

To be useful, a calculator is supposed to give you answers, even if they're not entirely correct, instead of riddles and excuses. That's why a simple four-banger calculator (that cannot handle values greater than 10⁸), when you enter an operation that would overflow, signals an error but also gives you the correct value divided by 10⁸. – b_jonas 18:05, 1 January 2011 (UTC)[reply]

Suppose a is a complex number with |a| > 1. Show that any analytic function f on ${\displaystyle \mathbb {C} ^{*}=\mathbb {C} -\{0\}}$ with ${\displaystyle f(az)=f(z)\forall z\in \mathbb {C} ^{*}}$ must be constant, but that there is a non-constant meromorphic function f on ${\displaystyle \mathbb {C} ^{*}}$ with f(az) = f(z) ${\displaystyle \forall z\in \mathbb {C} ^{*}}$.

The first approach I can see is to use something like the identity principle or principle of isolated zeros - everything bunches up at zero as f(z)=f(z/a), but sadly the function is not defined at the origin so it isn't that simple.

My second thought was to use something like the logarithm to change the periodicity f(z)=f(az) to something of the form g(z)=g(z+b), where I have some more useful theorems for this situation. Could anyone suggest anything? And hopefully once I've come up with a solution to the first part, an example for the second may present itself. Thankyou! Estrenostre (talk) 21:57, 31 December 2010 (UTC)[reply]

Since you only ask for suggestions, I'm only going to do some prodding: for the first part, you've thought about theorems which require entire functions; what must f do in order to not be entire? In particular, if f is not entire, what must be true of values near the origin, and what implications does this have for its holomorphicity a bounded distance from the origin? While this won't solve the second part, it should give you an intuition for why allowing meromorphicity instead is helpful. Invrnc (talk) 06:59, 1 January 2011 (UTC)[reply]

For the second part, you may find a review of elliptic functions inspiring. –Henning Makholm (talk) 15:12, 1 January 2011 (UTC)[reply]

But can we get use out of the concept of entirety when we've punctured the plane at the origin? The two ways to get around the issue I can see are either to ignore the missing point at the origin (but I'm not sure this is valid; we could take something like 1/z and cause some problems I expect) or to 'define the function at the origin' and then work in the whole complex plane. I've never really seen 'entire' except in all of C so I'm not sure what goes and what doesn't sorry! Estrenostre (talk) 03:18, 2 January 2011 (UTC)[reply]

No, as you note, the function isn't entire; that is, it's not continuous at the origin. But when it comes to analytic functions, discontinuities only come in a few flavors. You might think about the implications of each of these cases separately.--71.175.63.136 (talk) 07:49, 2 January 2011 (UTC)[reply]

You did suggest yourself to "use something like the logarithm to change the periodicity f(z)=f(az) to something of the form g(z)=g(z+b)". So consider ${\displaystyle g(z)=f(exp(z))}$. That's entire and encodes everything there is to know about ${\displaystyle f}$. Furthermore, it has period ${\displaystyle 2\pi i}$ as well as ${\displaystyle \log(a)}$. Now a few compactness arguments will let you apply the maximum modulus principle to g. Once you do that, step a few paces back from your proof and squint a little ... then you can see the outline of one that works for f directly. 217.74.211.98 (talk) 22:14, 3 January 2011 (UTC)[reply]

I was solving problems from a Graph theory text book and came across this problem: (1) Prove that removing opposite corner squares from an 8 by 8 checkerboard leaves a subboard that cannot be partitioned into 1 by 2 and 2 by 1 rectangles. (I assume the problem asks for removing only 1 pair of corner squares). (2) Using the same argument, make a general statement about all bipartite graphs. Can someone help me with this please. Thanks -Shahab (talk) 14:29, 1 January 2011 (UTC)[reply]

Explicitly making this a question about a checkerboard (rather than an arbitrary 2n by 2n grid) actually makes it easier, since the checkerboard comes pre-equipped with a helpful colouring. Algebraist 14:33, 1 January 2011 (UTC)[reply]

I keep thinking of Ike somehow... --Stephan Schulz (talk) 14:42, 1 January 2011 (UTC)[reply]

You think he saw everything as black and white? ;-) Dmcq (talk) 15:22, 1 January 2011 (UTC)[reply]

To expand on Algebraist's comments:

1) Opposite corners come in the same color on an 8×8 checkerboard, and initially there are the same number of black squares as white (32 each).

2) Therefore, removing one opposite set leaves you with a color imbalance (30 white squares and 32 black or vice-versa).

3) Since removing a 1×2 rectangle will always remove both a black and a white square, that will maintain the color balance. So, after one rectangle is removed, a full checkerboard would have 31 black squares and 31 whites, for example.

4) If you do the math on removing 1×2 squares from a checkerboard with one pair of opposite corners removed, you find you eventualy end up with only two black squares or two whites, which can't posibly form a rectangle, due to the way the board is laid out.

OK, so how do we generalize this ?

A) The general rule appears to be that you can't remove balanced color rectangles (or squares) from a larger imbalanced-color square (or rectangle), and end up with nothing.

B) Note that this doesn't necessarily mean that you always can remove balanced color rectangles (or squares) from a larger balanced-color square (or rectangle), such as removing 6×1 rectangles from a chessboard. At a minimum, the larger product must be a multiple of the smaller product. So, in our example, the 62 squares we had left did divide evenly by the 2 squares we removed at a time, while it would not divide evenly by the 6 squares in a 6×1 removal scheme. Of course, being a multiple, alone, isn't sufficient to ensure that the configuration allows them to be removed exactly. There's the color imbalance issue, but also the problem of physically fitting the removal patches on the board so they don't overlap or fall off an edge.

C) Think of which rectangular or square boards have opposite corners the same color (it helps to have a checkerboard in front of you when you do this, so here's one: chessboard). Now, can you come up with a formula to describe the relationship between the X and Y directions that gives you opposite corners the same color ? Hint: It's more complex than just X=Y.

D) Similarly, think of squares or rectangles that can be removed which have "balanced color". Is there any square or rectangle that has more blacks than white or vice-versa ? (And, if so, is it always imbalanced in the same direction, or not ?).

StuRat (talk) 16:11, 1 January 2011 (UTC)[reply]

Could anyone direct me to a proof, using Zermelo-Fraenkel axioms, that a countable union of countable sets cannot have cardinality ${\displaystyle \aleph _{2}}$? I've searched online but can't manage to track one down! Thankyou :) Simba31415 (talk) 03:24, 2 January 2011 (UTC)[reply]

Suppose ω₂ (the smallest ordinal of size ${\displaystyle \aleph _{2}}$) is a countable union of countable sets. If we had choice, we could simultaneously count all the sets to get an injection from ω₂ to ω×ω, which of course is impossible, since ω×ω is countable. Without choice, we can't do that, but what we do have is explicit well-orders for all the sets (since they're subsets of the well-ordered set ω₂). The Mostowski collapse lemma gives explicit bijections from these well-ordered sets to the corresponding ordinals. Since the sets are countable, these are countable ordinals, and so we have a injection from ω₂ to ω×ω₁, which is impossible since the latter set has cardinality ${\displaystyle \aleph _{1}}$. Algebraist 04:31, 2 January 2011 (UTC)[reply]

I really like this proof! 86.205.29.53 (talk) 23:30, 3 January 2011 (UTC)[reply]

I am trying, without success, to understand the method of characteristics for solving PDEs. I am working with the following example.

"Consider ${\displaystyle e^{x}u_{x}+u_{y}}$ with ${\displaystyle u(x,0)=cosh(x)}$. The initial curve B is the x axis (x=t, y=0) and the initial data along b is h(t)=cosh(t). The characteristics satisfy

${\displaystyle {\frac {dx}{ds}}=e^{x}}$ and ${\displaystyle {\frac {dy}{ds}}=1}$, ${\displaystyle x(0)=t}$ and ${\displaystyle y(0)=0}$." (There is more to the problem but this should be sufficient for my question.)

What I cannot understand is where the initial conditions for x(0) and y(0) come from. I have looked through my notes and other examples countless times but just cannot see it. I have every confidence I am being incredibly dim but could someone please enlighten me? Thank you. asyndeton talk 20:16, 2 January 2011 (UTC)[reply]

The initial curve B is the x axis (x=t, y=0)... :) Hence, your initial conditions. 86.164.58.246 (talk) 12:07, 3 January 2011 (UTC)[reply]

I'm such a fool. Thank you. asyndeton talk 17:01, 4 January 2011 (UTC)[reply]

Let's say you hold an election and the sample size is s₁. You achieve an approval rating of A%. Then, you analyse the results from randomly selected subsets of the electorate. Assume that you randomly selected s₂ people and they tell you how they actually voted.

• What is the probability of getting an approval rating of A% in this smaller sub-sample?
• What is the probability of the approval rating being within ±d% of the original approval rating?

— Fly by Night (talk) 00:19, 3 January 2011 (UTC)[reply]

If you want an exact result, you need to sum the values of the Hypergeometric distribution. Note that if ${\displaystyle s_{1}}$ and ${\displaystyle s_{2}}$ are coprime and ${\displaystyle 0<A<1}$, then the probability of having exactly A is 0.

For an approximation, you can use the normal distribution with ${\displaystyle \mu =A}$ and ${\displaystyle \sigma ={\frac {A(1-A)}{s_{2}}}}$. -- Meni Rosenfeld (talk) 09:25, 3 January 2011 (UTC)[reply]

Thanks for that Meni. — Fly by Night (talk) 15:13, 3 January 2011 (UTC)[reply]

I'm having difficulty with my thesis. can "knowledge or awareness of people to statistics" be subjected to study? what statistical tool can be used? — Preceding unsigned comment added by Rionsgeo (talk • contribs) 02:06, 4 January 2011 (UTC)[reply]

If I understand you, you want to measure people's knowledge and awareness of statistics, using statistical methods. I suppose you could do a poll/quiz where you ask people how often they use statistics, then ask them to solve some statistics problems, and use that data to calculate standard deviations, confidence intervals, etc.

One suggestion for a refinement of your thesis: "Resolved, that people who are ignorant or distrustful of statistics tend to engage in statistically unhealthy habits and thus shorten their lives." You could design a way to test this assertion and either prove or disprove it. StuRat (talk) 05:43, 4 January 2011 (UTC)[reply]

How can I compute the number of ways to choose n elements in sets of size k (with replacement), so that no element occurs in each set more than x times? 70.162.9.144 (talk) 07:30, 4 January 2011 (UTC)[reply]

I don't know if this helps, but it should be equal to the coefficient of ${\displaystyle y^{k}}$ in ${\displaystyle (y^{x+1}-1)^{n}/(y-1)^{n}}$, if I followed your notation correctly. I'd guess there isn't any nice closed-form solution. Are you looking for a way to efficiently compute it? Eric. 82.139.80.114 (talk) 01:39, 5 January 2011 (UTC)[reply]

Sorry, could you explain what you mean by coefficient and how it is derived from ${\displaystyle (y^{x+1}-1)^{n}/(y-1)^{n}}$? 70.162.9.144 (talk) 04:37, 5 January 2011 (UTC)[reply]

${\displaystyle {\frac {(y^{x+1}-1)^{n}}{(y-1)^{n}}}=\left(\sum _{k=0}^{x}y^{k}\right)^{n}}$ is expanded by the Multinomial theorem. Bo Jacoby (talk) 12:56, 5 January 2011 (UTC).[reply]

Let A be free abelian on 3 generators a,b,c and K the subgroup (n+m)a+(n-m)b+(m-n)c for all integers n,m. Is A/K just Z x Z/2Z? This seems like a very trivial task a computer should be able to do, is there any software to identify stuff like this? Money is tight (talk) 08:26, 4 January 2011 (UTC)[reply]

K is generated by the elements a + b − c and 2a, it thus consists of elements of the form na + mb + kc where k = −m and ${\displaystyle n\equiv m{\pmod {2}}}$. From this it follows easily that K = Ker(f), where f: A → Z × (Z/2Z) is defined by f(na + mb + kc) = (m + k, n + m mod 2). Since f is clearly onto, A/K is indeed isomorphic to Z × (Z/2Z).—Emil J. 12:41, 4 January 2011 (UTC)[reply]

Thanks, I'm used to doing "show/prove" questions and this question wasn't one of those, just needed some confirmation to check my understanding is correct. Money is tight (talk) 00:56, 5 January 2011 (UTC)[reply]

I think the convention is to use the symbol ⊕ to denote the direct sum of abelian groups. (Instead of using the direct product, which is often used with non-abelian groups.) So it would be normal to write Z ⊕ Z/2Z. If A₁, …, A_n are abelian groups then the direct sum A₁ ⊕ … ⊕ A_n is the set of n-tuplues (a₁, …, a_n), where a₁ ∈ A₁, …, a_n ∈ A_n, under the binary operation

${\displaystyle (a_{1},\ldots ,a_{n})+(a_{1}',\ldots ,a_{n}')=(a_{1}+a_{1}',\ldots ,a_{n}+a_{n}').}$

This turns A₁ ⊕ … ⊕ A_n into an abelian group. — Fly by Night (talk) 13:43, 5 January 2011 (UTC)[reply]

I'm reading a book that has a chapter on magic squares, and it gives the following special magic square:

 5 22 18
28 15  2
12  8 25

If you write the names for the numbers out in English and then count the number of letters in each name, you get another magic square:

 4  9  8
11  7  3
 6  5 10

It then says that, for totals of less than 200, English has seven of these squares, while French only has one. What are they? --75.60.13.19 (talk) 03:08, 5 January 2011 (UTC)[reply]

Hello,

I am struggling with what should be an easy exercise...

let G be a group acting primitively on a set ${\displaystyle \Omega }$. Let ${\displaystyle \chi }$ be the permutation character( let's say over ${\displaystyle \mathbb {C} }$). Hence ${\displaystyle \chi }$ maps every${\displaystyle g\in G}$ to the number of elements in ${\displaystyle \Omega }$ it fixes.

Let${\displaystyle \chi _{1}}$ be any irreducible constituent of ${\displaystyle \chi }$, different from the trivial character. Prove that ${\displaystyle \chi _{1}}$ is faithful (i.e. the corresponding representation of ${\displaystyle G}$ maps only the trivial element of ${\displaystyle G}$ to the trivial matrix).

I think it should be easy, but it appears I am missing a crucial observation. I already know why primitivity is important, because otherwise the non-faithful permutation character on the blocks of imprimitivity would be contained in the character.

Do all the 15 groups of order 24 have a subgroup of order 12? How can I prove or disprove this.-Shahab (talk) 09:41, 5 January 2011 (UTC)[reply]

Take a look at the Sylow theorems article. As the article says, they are "a collection of theorems... that give detailed information about the number of subgroups of fixed order that a given finite group contains.". Since 24 = 2³⋅3, Burnside's theorem tells us that G is solvable. There's lots more information to be had in the Solvable group article. — Fly by Night (talk) 15:48, 5 January 2011 (UTC)[reply]

I think that the semidirect product ${\displaystyle G=C_{3}\ltimes Q_{8}}$ has no subgroup of order 12, where Q₈ is the quaternion group, and C₃ acts on Q₈ by cyclic permutation of i, j, k. Any subgroup of G of index 2 would be normal, hence it would induce a nontrivial homomorphism f: G → C₂. However, f vanishes on C₃ (as it has order 3) and on at least one of i, j, k (as ij = k), and therefore on all of i, j, k (as they are conjugated in G, and C₂ is abelian). Thus f is trivial, a contradiction.—Emil J. 16:00, 5 January 2011 (UTC)[reply]

This is a central extension of A₄ by C₂. A₄ has no subgroups of order 6 and this implies that the extension has no subgroups of order 12. This is also the only group of order 24 that has no subgroup of order 12. I got this by checking generators and relations for all 15 groups, these are well known, but I suppose it wouldn't be too hard to prove it from scratch using Sylow etc.--RDBury (talk) 16:54, 5 January 2011 (UTC)[reply]

What were the definitions and axioms used in Principia Mathematica that required such a long proof of 1+1=2 and other basic arithmetic facts? A link to a page or website describing the foundations is sufficient; I just don't know where to find such a reference. --24.27.16.22 (talk) 15:54, 5 January 2011 (UTC)[reply]

At the very end of that page about Principia Mathematica you can find proposition 54.43 which is that 1+1=2 validated using a modern theorem checker Metamath. Dmcq (talk) 16:06, 5 January 2011 (UTC)[reply]

Qualitatively speaking, the issue is that the axioms are very low-level. For instance, 1+1=2 does not require proof if you are using the Peano axioms. But if you don't want to do that, you could start with Zermelo_Fraenkel_set_theory and derive the Peano system from there. This would be a much longer proof than that quoted on the PM page. Both Peano and ZFC schemes pre-date PM, and Whitehead even used much of Peano's notation. I think that PM axioms are lower-level / weaker than ZFC, but hopefully someone else can elaborate on that. I can't easily find an short itemized list of axioms in the PM, but you can download the whole book through google books here [1]. SemanticMantis (talk) 17:03, 5 January 2011 (UTC)[reply]

If we start with ZFC, and identify natural numbers with finite cardinals and define addition as a disjoint union of sets (using some silly trick like A+B = (Ax∅)∪(Bx{∅})), would a complete formal proof of 1+1=2 require a similar length? --24.27.16.22 (talk) 17:49, 5 January 2011 (UTC)[reply]

No, but that's an awful lot of "if"s. Russell and Whitehead were working at a time when there was no settled notions of things such as a logical theory, and so they had to build up everything from the very ground. Many tricks and tools that would be used in a modern work to simplify and streamline the presentation simply hadn't been invented yet (or had been used once or twice but their general applicability not appreciated). Furthermore, the theory they designed is/was markedly more cumbersome to work with and reason about than ZFC, and as a result didn't really catch on.

Also, nobody's saying that all of the 360 pages that preceded *54.43 were necessary prerequisites to that particular proposition. As a comparison with a newer text, Mendelson Introduction to Mathematical Logic (4th ed., 1997) reaches page 258 before it defines "${\displaystyle X{+_{c}}Y\mathrm {~for~} (X\times \{\emptyset \})\cup (Y\times \{1\})}$". However, that includes 70 pages that develop formal number theory from the Peano axioms and are not used for the axiomatic set theory, so "number of pages before addition is defined" is not really a meaningful metric. –Henning Makholm (talk) 22:08, 5 January 2011 (UTC)[reply]

Does anyone know any source where real everyday mathematics is developed formally in ZFC? By real math I mean calculus, algebra, analysis etc. Money is tight (talk) 23:22, 5 January 2011 (UTC)[reply]

http://us.metamath.org/mpegif/mmset.html 67.122.209.190 (talk) 08:50, 6 January 2011 (UTC)[reply]

One of my profs made an interesting remark today. He was doing an example on the board, and the answer came out to ${\displaystyle x={\sqrt {2}}}$. But he stopped, and said that he didn't feel this was a complete solution, and that the fullest answer possible would be to say that ${\displaystyle x=1.4142...}$.

As an example, he said, suppose we had to solve the equation ${\displaystyle x^{3}=4}$. Saying that ${\displaystyle x={\sqrt[{3}]{4}}}$ is just tautological. The only meaningful solution would be to write out cube root of 4 as a decimal expansion.

At first I agreed, but thinking about it now, I'm not so sure. A decimal expansion of a number is, by definition, the expression of the number as an finite or infinite series of fractions with a denominator of a power of ten. How is this any more valid of an answer? —Preceding unsigned comment added by 74.15.138.87 (talk) 16:56, 5 January 2011 (UTC)[reply]

What constitutes a "complete solution" in mathematics is somewhat arbitrary, and most professors would be perfectly happy to accept ${\displaystyle x={\sqrt {2}}}$ in my experience, since that is the most precise answer one can provide. It's really not clear exactly what your professor is looking for in a complete solution, except possibly that he expects final answers to be written out in decimal notation? --COVIZAPIBETEFOKY (talk) 17:14, 5 January 2011 (UTC)[reply]

√2 is exact, whereas the decimal form is approximate. Of course all solutions of equations are in a sense tautological. Suppose the equation had been

${\displaystyle 3x-5=0.\,}$

Would he object that x = 5/3 was "just tautological? Michael Hardy (talk) 18:30, 5 January 2011 (UTC)[reply]

Right on Michael. Much of math can be seen as a search for tautologies. In my experience, "just a tautology" is used as a dismissal much more commonly in fields outside of math. SemanticMantis (talk) 19:02, 5 January 2011 (UTC)[reply]

Incidentally: I took two classes taught by Alphonse Vasquez, and remember he used to say that everything that's been proven in math is tautological, and if one doesn't see something as such then he hasn't sufficiently wrapped his head around it.—msh210℠ 19:16, 5 January 2011 (UTC)[reply]

When I teach math, I ask students for exact answers. Partially this is to make it easier on the grader, and partially because I think it is important to grasp that root two cannot be expressed exactly in decimal notation. But this is just personal preference. Also, asking for decimal approximations implicitly encourages students to use calculators when they are not required, and I believe this is counterproductive to really learning math. --But on to your prof's comments. I disagree completely that the decimal approximation is the "fullest possible answer", but perhaps this is not a direct quote from the instructor. Rather than talk of 'meaningful' or 'valid', we may consider whether an answer is *informative*. Consider an application where two quantities are to be compared. If solved exactly, it is not easy to see how ${\displaystyle {\sqrt[{5}]{2}}}$ compares to ${\displaystyle {\sqrt[{10}]{5}}}$. However, it's quite easy to see that 1.1487... < 1.1746... So really, the best form for an answer depends on why you're quantifying something in the first place. SemanticMantis (talk) 18:55, 5 January 2011 (UTC)[reply]

There's an important point that my prof made that I forgot to mention. In the final solution, he doesn't want decimals; he wants radicals, because decimals aren't exact. His point was that the only reason ${\displaystyle x={\sqrt {2}}}$ is an acceptable answer is because someone could look up the decimal expansion to the desired precision. Presumably, he also means to say that if mathematicians were dumber, and had made the notation ${\displaystyle x={\sqrt {2}}}$ but couldn't figure out how to expand it, then ${\displaystyle x={\sqrt {2}}}$ would be meaningless. But I don't see why decimal representation should have validity than other representations. At the same time, if we accept, as Michael Hardy said (and which I agree with) that all solutions are tautological, then that would mean that when we say ${\displaystyle x={\sqrt {2}}}$, we are really saying "I have found that the solution to such-and-such equation also happens to be the positive solution to the equation ${\displaystyle x^{2}-2=0}$". I guess this makes sense, but it makes the whole business of solving equations kinda...arbitrary, no?74.15.138.87 (talk) 20:53, 5 January 2011 (UTC)[reply]

It depends why you're solving something. Do you need to know how many people it will take to change your lightbulb? Then you need a decimal representation (actually, the ceiling of one, usually). Do you need a number you can substitute into another expression? Then a form like${\displaystyle {\sqrt {2}}}$ is usually best. Do you need it as an element of ℂ? Then you want (perhaps) ${\displaystyle {\sqrt {2}}+0i}$. Etc.—msh210℠ 21:00, 5 January 2011 (UTC)

That's very dismissive of tautologies. I try and make everything I say logical and as nearly tautological as possible. Like for instance 'If I don't get some sleep I'll never wake up in the morning' ;-) Dmcq (talk) 21:30, 5 January 2011 (UTC)[reply]

I grappled with the same problem when initially introduced to square roots and logarithms at school (and arcsin...in fact, ANY inverse function). It seemed retarded and meaningless to say that the solution of ${\displaystyle 10^{x}=2}$ is ${\displaystyle x=\log(2)}$ (decimal log obviously). What's the point of just inventing notation and calling these answers "solutions"? They are no more useful than the original equations. In a sense, ALL inverse functions are just tautologies, useless for ACTUALLY solving anything. :That was until I reached university and learned about Taylor series, which allows you to actually compute and evaluate those expressions in a meaningful way. That was when it all made sense. Zunaid 12:20, 6 January 2011 (UTC)[reply]

How many cases should you consider until you come up to the conclusion that an ethnic group has this or that feature? For example, if you take nations with 300 millions or 100 millions, is my personal experience of 200 interactions each year enough? Quest09 (talk) 17:03, 5 January 2011 (UTC)[reply]

There's a question above about elections that is related. What if the 200 people you meet all had a property that no-one else in the population had? I guess you want to answer the following question: If p% of a sample has a certain property, then what's the probability that (p±d)% of the whole population has that property. You need to decide what percentage of the sample/population needs to posses a property before you call it a characteristic. Take a look at my question, and Meni's answer, here. — Fly by Night (talk) 17:19, 5 January 2011 (UTC)[reply]

This is a question of statistics. A simple random sample of 1000 or so people is generally enough to establish with reasonable certainty the rough proportion of people in a given population who possess a feature/hold a particular opinion/etc, independently of the size of the population (assuming only that it is significantly larger than 1000). Your personal experiences most likely do not constitute a simple random sample, and therefore cannot be used for this purpose. --COVIZAPIBETEFOKY (talk) 17:24, 5 January 2011 (UTC)[reply]

What COVIZAPIBETEFOKY said. We all have social interactions strongly biased by our income, locality, profession, etc. Our own experience is never a good source from which to extrapolate to the general population. See Sampling bias#Historical examples for some famous cases where samples of millions that were not truly random and unbiased, were insufficient. Ray^Talk 18:35, 5 January 2011 (UTC)[reply]

Let me put it this way. Let's say your IQ is 300, and you've just made a handful of major breakthroughs in your chosen field, however you are just an undergraduate at a huge state school. Let's say that you are able to prove yourself to be an extremely valuable researcher to a professor at your University, if he will speak to you for 10 minutes. Then he would be convinced by your ideas, be instantly swayed, and want to publish with you. As a result of this, you would be able to get admitted to the graduate program of your choice, even if you did nothing more than flesh out the ideas you just published as an undergrad, you would be set to get tenure based on that, if not at Harvard, then at least in some respectable state school such as the one you're attending. There's just one little problem: your state school is in Misouri, has low standards of admission, you are of a minority race, and the professor has had a LOT of experience with semi-literate members of that minority!! He might refuse the 10-minute interview on that grounds alone, just from remembering your face among the 300 faces he teaches at any one time!! So, let me ask you the question this way: how many members of your race that he had experiences with, who were semi-literate and had an IQ of more like 60 (one fifth of yours), would make you say: You know what, he shouldn't waste 10 minutes on an interview with me, it's just not a reasonable request. 100 such people? 1000? A million? How about if you're Indian, and there are one BILLION people who are all different from you, and you're the only Indian who can do Italian opera in all the world? Would you agree that the director of La Scala should refuse to even listen to you on that basis? 87.91.6.33 (talk) 19:05, 5 January 2011 (UTC)[reply]

I would argue that such an interview should never be refused. However, there are cases where the "judge" is in such demand that he can't spend the time, so then underlings should be enlisted to do a "pretest", to determine if you are anything worth bothering the "judge" about. StuRat (talk) 21:56, 5 January 2011 (UTC)[reply]

Exacrtly. the OP should come to the same conclusion, and, therefore, realize that he should not become racist even after a billion examples confirming his suspicions about a group. 87.91.6.33 (talk) 22:02, 5 January 2011 (UTC)[reply]

(ec) Please, there is no need for emotional language; just state your assertion clearly and with justification (and without fallacies like appeal to consequences).

The question Quest09 asked ("is my personal experience of 200 interactions each year enough [... to] come up [with] the conclusion that an ethnic group has this or that feature") is a simple question of statistics for which COVIZAPIBETEFOKY gave a direct answer ("Your personal experiences most likely do not constitute a simple random sample, and therefore cannot be used for this purpose."). Eric. 82.139.80.114 (talk) 22:05, 5 January 2011 (UTC)[reply]

Yes, this was a question about mathematics in the math RD. I was not asking about moral implications and actually not even thinking about discriminating people. Quest09 (talk) 23:04, 5 January 2011 (UTC)[reply]

Of course there also is an implicit assumption about why the interview was rejected. Maybe the professor can only grant so many interviews and cuts it off at an arbitrary number. Or maybe he grants no non-class-related interviews to undergrads at all. Or maybe he has already seen common misconceptions during the initial request for the interview. All of these are arguably beyond the realm of mathematics, of course.Unless you consider "There are only 16 working hours per day, if I grant one interview to every of my 600 students I'll never finish that research" as maths ;-) --Stephan Schulz (talk) 10:41, 6 January 2011 (UTC)[reply]

600 student won't ask you for an interview. And even if they do, if you spend 15 minutes with each, that would only make 30'/day each year. Quest09 (talk) 12:10, 6 January 2011 (UTC)[reply]

With 600 what I'd do is I wouldn't even bother reading the first line of the application for most. I'd do a random draw of a selection to check through and then winnow down to an even smaller number to interview. Actually I think interviews are very overrated so its mainly to eliminate the unsuitable rather than to pick the best. Dmcq (talk) 13:15, 6 January 2011 (UTC)[reply]

"the conclusion that an ethnic group has this or that feature" is the definition of racism. Sorry. That thought is the definition of racism. In other words, this is a question about what level of statistical confidence justifies racism. The answer is: none. Even if you have a hundred billion examples of a member of an ethnic group with a certain feature, you still can't come to "the conlcusion that an ethnic group has this or that feature". Is that clear enough for you? How about this way: I grew up in a very poor part of Boston. I met literally thousands of black kids who were way below the required level in their grade. How many should I have met before I concluded that a black kid has features that make them, say, not qualified for a Presidential-track education? The answer is, there is no such number. (Math: Not a Number, NaN). Because even if you have 300,000,000 black kids who can't be president, because they're too stupid and all the ritalin in the world would not make them smart enough: it only takes one. You can never induce the rule. The rule is the definition of racism. Clear enough for you? 87.91.6.33 (talk) 20:38, 6 January 2011 (UTC)[reply]

in yet different terms, an ethnic group can't have any features (besides the tautological ones*). Except in the mind of a racist. A racist can list dozens of features for any given ethnic group. Go try a racist sometime, I am not making this up. 87.91.6.33 (talk) 20:43, 6 January 2011 (UTC)[reply]

* tautological = obviously if you group people in races based on criteria, the "race" will have member meeting that criteria... but this says nothing about them, and only about you and your grouping choices -- it's "begging the question".

If I take a regular tetrahedron and draw segments from each of the vertices to the center, what's the angle between two of those segments? --75.60.13.19 (talk) 00:54, 6 January 2011 (UTC)[reply]

If you look at the Tetrahedron article, i.e. click here, then you'll find out everything you need to know, any many things you don't. — Fly by Night (talk) 01:33, 6 January 2011 (UTC)[reply]

What books would you recommend for learning the general techniques of mathematical proofs? 74.14.111.188 (talk) 07:17, 6 January 2011 (UTC)[reply]

I suppose it really depends on the level of mathematics you're concerned with, but one course I found very helpful in the early years of my undergraduate studies used the textbook "A Transition to Higher Mathematics," by Smith, Eggen, and St. Andre. I found it be very illuminating at the time, particularly in regards to learning methods of proof. Nm420 (talk) —Preceding undated comment added 15:41, 6 January 2011 (UTC).[reply]

I need to find out when the sites listed below first established a website/page on the internet. Thanks

Taylor & Francis Group: an informa business - http://www.taylorandfrancisgroup.com/

- www.tandf.co.uk

Association for Childhood Education International - http://acei.org/

- www.acei.org/cehp.htm

National Council of Teachers of Mathematics - http://www.nctm.org/

- http://my.nctm.org

National Association for the Education of Young Children - http://www.naeyc.org/yc/

- www.journal.naeyc.org —Preceding unsigned comment added by 24.210.25.124 (talk) 13:30, 6 January 2011 (UTC)[reply]

Question reformatted for legibility. The Wayback machine should be able to help - e.g. it suggests that www.tandf.co.uk first appeared in early 1997. AndrewWTaylor (talk) 20:50, 6 January 2011 (UTC)[reply]

If the Taylor series becomes arbitrarily close to the original function for all analytic functions (in other words, all function that we are normally interested in), what is the purpose of the remainder function? 24.92.70.160 (talk) 21:34, 6 January 2011 (UTC)[reply]

Analytic functions define a very small subset in the space of all function. Asking a function to have continuous derivatives of all orders and then asking for a series to converge is a very, very big ask. High school functions like 1/x aren't analytic; it fails to be continuous at x = 0. Take a look at the article on flat functions. The exponential ${\displaystyle \scriptstyle f(x)\,=\,e^{-1/x^{2}}}$ is an example of a flat function. It is well defined at x = 0 because both the positive and negative limits x → 0 give ƒ(x) → 0. In fact, it is a smooth function because each of its derivatives exist and are continuous for all x (particularly at x = 0). But you will find that each and every derivative vanishes at x = 0. So the function is always contained in the remainder function; no matter how far along the Taylor series you go. As for non-smooth functions, only the first few derivatives may be continuous, so we can only define the Taylor series up to a certain, finite order. Then the remainder term takes up the slack. — Fly by Night (talk) 22:03, 6 January 2011 (UTC)[reply]

Can someone suggest a good LaTeX writing program for Windows. I mean a program where I type the code and it compiles it, turns it into .dvi, .ps or .pdf. I use Kile on Linux but it needs some fiddling with to run on Windows and I don't know how. A nice user friendly interface would be perfect, with some symbol buttons that substitute the LaTeX code when you click them, etc. — Fly by Night (talk) 22:25, 6 January 2011 (UTC)[reply]

there aren't any. just install miktex like everyone else. 87.91.6.33 (talk) 22:42, 6 January 2011 (UTC)[reply]