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December 31

Drawing Lines.....

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

(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

\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

\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

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

{\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...

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

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

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

(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

{\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.

{\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:

{\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]

pseudocode to rotate an array by 90 degrees as a square

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!!

Question Part 2

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]

Question Part 3

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.

Hard Part 2

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:

   |."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]

Number of possible letters games

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·

\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]

Nature of SL(2,F5)

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]

iⁱ

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]

i=e^{i{\frac {\pi }{2}}}

so

i^{i}=(e^{i{\frac {\pi }{2}}})^{i}=e^{i^{2}{\frac {\pi }{2}}}=e^{-{\frac {\pi }{2}}}

. Proof that

i=e^{i{\frac {\pi }{2}}}

follows directly from the identity

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

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

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

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:

\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

^{-1}

in the calculator refers to arcsin which is defined as the unique solution to sin(x)=0 on interval of

[-\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]

Behaviour of a function similar to a periodic function

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

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

g(z)=f(exp(z))

. That's entire and encodes everything there is to know about

f

. Furthermore, it has period

2\pi i

as well as

\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]