The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions at one of the pages linked to above.
quadratic equation
who has given the method to solve the quadratic equation?
bumped from October 2 - I'm still open to requests
This seemed like the ideal place to ask because I know the right people are going to read this.
I've been very bored lately and I feel like contributing with some nice images to Wikipedia. Animations and 3D images are more fun to do, so those are preferred. Anything is cool as long as they're maths\physics related. Oh, and here are some sample images I created to replace the ugly old ones or to add new content:
Also, check
my gallery to see what sort of thing I can do.
So, do you guys have any suggestions for images, graphics or diagrams that need to be improved or replaced? :) ☢
Ҡi∊ff⌇↯00:19, 2 October 2006 (UTC)reply
Thanks, I joined the project. I see they want an animation of a torus morphing into a coffee mug. I'll give it a shot. :) ☢
Ҡi∊ff⌇↯02:59, 2 October 2006 (UTC)reply
Hey Kieff! Well, since you asked, I've had
Flexagons on my wishlist for almost a year. It would be really neat if you could show an animation of how to fold one, say, a hexaflexagon. --
HappyCamper03:46, 2 October 2006 (UTC)reply
Interesting. I'll have to make one of those to see how the folding goes, but I'll give it a shot. Oh, and to the right you'll see the coffee mug morphing into a torus that I just finished doing. :) ☢
Ҡi∊ff⌇↯06:09, 2 October 2006 (UTC)reply
Hey thats good stuff. I cant even draw a straight line. What I been looking fo for months is just a simple (scalable) circle that can be put onto pages. There is one page in particular that desperately needs some good diagrams and Im not capacble of doing them. THat page is
Relativistic electromagnetism!--
Light current21:10, 2 October 2006 (UTC)reply
When using a formula, we often know the value of one variable to a greater degree of accuracy than we know the others. In your opinion, what affect, if any, does it make on our use of a formula if we know the value of one variable to a greater degree of accuracy than another?
-13:37, 7 October 2006 (UTC)~***
I suggest trying out all the minimum and maximum values, given the error range on each, to see how they effect the result.
StuRat20:37, 7 October 2006 (UTC)reply
In
modular arithmetic, isn't two integers and always equivalent (by definition)? I believe so, and therefore think that your statement follows directly from that definition, and does probably not have a name. But, I could be wrong, of course. :-) —
Bromskloss17:11, 7 October 2006 (UTC)reply
Mm, doesn't help. :( Busy computing both sides of the equation using Haskell and getting wrong results. :( Never mind, thanks for the help. x42bn6Talk11:04, 8 October 2006 (UTC)reply
You could call it the binomial theorem. Let where . Then, but all terms of this sum except for the one with is divisable by , and this term is so indeed .
On the other hand, that's not IMO the "real" reason why this is true. I think of this as just a direct consequence of that if and then . Applying this repeatedly with , , and , we get by induction that . –
b_jonas11:59, 8 October 2006 (UTC)reply
Permutations with and without repetitions
Shuffle a standard deck of cards. What is the probability that two cards of the same value (e.g. two aces) are next to each other? More generally, suppose you have X equivalent copies each of Y different kinds of objects. How many of the permutations have repetitions? Is there a closed–form solution? —
Keenan Pepper17:26, 7 October 2006 (UTC)reply
The problem is, whether one card has an adjacent match isn't an independent event with whether another card has an adjacent match. So, it would necessary to look at every possible deck arrangement which has a match, and divide by the total number of possible deck arrangements (52!).
StuRat20:29, 7 October 2006 (UTC)reply
Yup, that's what makes it hard. Anyone have an answer, a reference, or just a glimmer of insight? BTW, we're ignoring suits, so the total number of permutations is not 52! but .
For the case Y = 2 this is the number of ways X couples can sit in a row without any spouses next to each other (sequence A007060 in the
OEIS). There are no OEIS entries for Y = 3 or Y = 4. –
LambiamTalk22:10, 7 October 2006 (UTC)reply
I think I may be switching X and Y here compared to the intention of the question. I'm also not sure I interpreted "ignoring suits" correctly. –
LambiamTalk22:30, 7 October 2006 (UTC)reply
Yes, I said X was the number of copies of each kind, and Y the number of kinds, so (sequence A007060 in the
OEIS) is for X = 2. For Y = 2 there are always exactly 2 permutations without repetitions. By "ignoring suits" I meant that permutations of the cards with the same sequence of values (ace, two, three...) are considered equivalent, even if the suits of the cards (clubs, diamonds...) are different. It doesn't matter for the probability, because both the numerator and denominator would change by the same factor of (4!)^13. —
Keenan Pepper22:42, 7 October 2006 (UTC)reply
For a card deck, X = 4, Y = 13, I find that 63394531038905867912088 out of 92024242230271040357108320801872044844750000000000 permutations are repetition–free, giving a probability of 0.68888946545492·10–27. –
LambiamTalk02:34, 8 October 2006 (UTC)reply
First, I managed to get confused again about the roles of X and Y; the number 63394531038905867912088 above is for X = 13, Y = 4. The correct number for X = 4, Y = 13 is 4184920420968817245135211427730337964623328025600, which gives a much higher probability for a stutter–free random shuffle: about 4.5%.
Now about the method. Calling the card "values" A, B, C, ..., M, we need to count the number of strings containing exactly 4 of each of those that are stutter-free. One such string is CIFBAIBEALKCHGDGMLELIFCFMJDGDEMJFEKJBMLJDHKGIHBHACKA. Let fv(a,b,c,...,m) stand for the number of stutter-free sequences containing a As, b Bs, c Cs, ..., and m Ms, and ending in the value v. So the length of the sequence is a+b+c+...+m. The string above contributes for one to fA(4, 4, 4, ..., 4). By summing fv(4, 4, 4, ..., 4) for all 13 possible values of v we get the desired result.
How much is fA(a,b,c,...,m)? If a is 0 (or less), the count is 0: there are no sequences that end in A but contain no As. If a = 1 and all of b ... m are 0, we need to count the stutter-free sequences of length 1 that end in A. That is easy: there is precisely one such sequence. Otherwise. if a is at least 1 and the sequences have length > 1, each sequence counted can be turned into a shorter non-empty stutter-free sequences by removing the last element. Each such shortened sequence can end on any element except A. Conversely, each of the sequences counted by fB(a–1,b,c,...,m) – or any other element than B, except of course A itself – can be lengthened by appending A to it. So then fA(a,b,c,...,m) = fB(a–1,b,c,...,m) + ... + fM(a–1,b,c,...,m).
But wait, there are many symmetries. For one, if we permute the numbers b,c,...,m in fA(a,b,c,...,m) it will not affect the outcome. So, using ((...)) for a multiset notation, in which ((1,3,3)) = ((3,1,3)), which is different from ((1,3)), we can, instead of fA(a,b,c,...,m), consider fA(a, ((b,c,...,m))). Similarly fB(b, ((a,c,...,m))), and so on. But surely fA(p, ((q,c,...,m))) is the same as fB(p, ((q,c,...,m))). We don't need the subscript to f. Just find f(4, ((4,4,...,4))) – with 12 elements in the multiset – and multiply the result by 13.
Now, in general, how to find f(a, M), in which M is a multiset? We apply the earlier considerations to the new representation making use of the symmetries. If a = 1 and M contains no positive elements, the result is 1. Otherwise, f(a, M) is the sum over the elements v in M of f(v, M – ((v)) + ((a–1))), in which + and – are (also) used for multiset subtraction and addition in the obvious way. So f(a,((b,c,...,m))) = f(b,((a–1,c,...,m))) + ... + f(m,((b,c,...,a–1))).
We now have a recurrence relation that allows us to compute the function values. Doing that naively takes time proportional to the result value. The crucial step is to use the technique of
dynamic programming, or tabulation, or
memoization, or function
caching, or whatever you wish to call it, to avoid repeatedly computing the function value for the same arguments. This can be finessed by discarding any number 0 from the multiset, and by recognizing that in the summing process equal values taken from the multiset will give equal results. But also without that the process is now fast enough to produce the sought result quickly enough. --
LambiamTalk04:41, 9 October 2006 (UTC)reply
I think you had a slight mixup in the second paragraph - it should read: "...we need to count the number of strings containing exactly 4 of each of those...", and the following string should have 4 of each of A through M. --
Meni Rosenfeld (
talk)
07:33, 9 October 2006 (UTC)reply
You are absolutely right. I've applied the correction in situ. There must be some Freudian reason why I don't mind having the Jack of Hearts and the Jack of Spades next to each other, but don't want the Queen of Hearts next to the King of Hearts. I guess I'm confused by the misleading analogy to the spouse-seating problem. --
LambiamTalk11:10, 9 October 2006 (UTC)reply
So, the questions remain: Is there a closed-form solution? Can it be shown that there is not a closed-form solution? Is it an open problem? —
Keenan Pepper22:57, 9 October 2006 (UTC)reply
I think it is an open problem. If there is a closed-form solution, it's not going to look pretty, offering little incentive to look really hard for one. --
LambiamTalk20:16, 10 October 2006 (UTC)reply
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions at one of the pages linked to above.
quadratic equation
who has given the method to solve the quadratic equation?
bumped from October 2 - I'm still open to requests
This seemed like the ideal place to ask because I know the right people are going to read this.
I've been very bored lately and I feel like contributing with some nice images to Wikipedia. Animations and 3D images are more fun to do, so those are preferred. Anything is cool as long as they're maths\physics related. Oh, and here are some sample images I created to replace the ugly old ones or to add new content:
Also, check
my gallery to see what sort of thing I can do.
So, do you guys have any suggestions for images, graphics or diagrams that need to be improved or replaced? :) ☢
Ҡi∊ff⌇↯00:19, 2 October 2006 (UTC)reply
Thanks, I joined the project. I see they want an animation of a torus morphing into a coffee mug. I'll give it a shot. :) ☢
Ҡi∊ff⌇↯02:59, 2 October 2006 (UTC)reply
Hey Kieff! Well, since you asked, I've had
Flexagons on my wishlist for almost a year. It would be really neat if you could show an animation of how to fold one, say, a hexaflexagon. --
HappyCamper03:46, 2 October 2006 (UTC)reply
Interesting. I'll have to make one of those to see how the folding goes, but I'll give it a shot. Oh, and to the right you'll see the coffee mug morphing into a torus that I just finished doing. :) ☢
Ҡi∊ff⌇↯06:09, 2 October 2006 (UTC)reply
Hey thats good stuff. I cant even draw a straight line. What I been looking fo for months is just a simple (scalable) circle that can be put onto pages. There is one page in particular that desperately needs some good diagrams and Im not capacble of doing them. THat page is
Relativistic electromagnetism!--
Light current21:10, 2 October 2006 (UTC)reply
When using a formula, we often know the value of one variable to a greater degree of accuracy than we know the others. In your opinion, what affect, if any, does it make on our use of a formula if we know the value of one variable to a greater degree of accuracy than another?
-13:37, 7 October 2006 (UTC)~***
I suggest trying out all the minimum and maximum values, given the error range on each, to see how they effect the result.
StuRat20:37, 7 October 2006 (UTC)reply
In
modular arithmetic, isn't two integers and always equivalent (by definition)? I believe so, and therefore think that your statement follows directly from that definition, and does probably not have a name. But, I could be wrong, of course. :-) —
Bromskloss17:11, 7 October 2006 (UTC)reply
Mm, doesn't help. :( Busy computing both sides of the equation using Haskell and getting wrong results. :( Never mind, thanks for the help. x42bn6Talk11:04, 8 October 2006 (UTC)reply
You could call it the binomial theorem. Let where . Then, but all terms of this sum except for the one with is divisable by , and this term is so indeed .
On the other hand, that's not IMO the "real" reason why this is true. I think of this as just a direct consequence of that if and then . Applying this repeatedly with , , and , we get by induction that . –
b_jonas11:59, 8 October 2006 (UTC)reply
Permutations with and without repetitions
Shuffle a standard deck of cards. What is the probability that two cards of the same value (e.g. two aces) are next to each other? More generally, suppose you have X equivalent copies each of Y different kinds of objects. How many of the permutations have repetitions? Is there a closed–form solution? —
Keenan Pepper17:26, 7 October 2006 (UTC)reply
The problem is, whether one card has an adjacent match isn't an independent event with whether another card has an adjacent match. So, it would necessary to look at every possible deck arrangement which has a match, and divide by the total number of possible deck arrangements (52!).
StuRat20:29, 7 October 2006 (UTC)reply
Yup, that's what makes it hard. Anyone have an answer, a reference, or just a glimmer of insight? BTW, we're ignoring suits, so the total number of permutations is not 52! but .
For the case Y = 2 this is the number of ways X couples can sit in a row without any spouses next to each other (sequence A007060 in the
OEIS). There are no OEIS entries for Y = 3 or Y = 4. –
LambiamTalk22:10, 7 October 2006 (UTC)reply
I think I may be switching X and Y here compared to the intention of the question. I'm also not sure I interpreted "ignoring suits" correctly. –
LambiamTalk22:30, 7 October 2006 (UTC)reply
Yes, I said X was the number of copies of each kind, and Y the number of kinds, so (sequence A007060 in the
OEIS) is for X = 2. For Y = 2 there are always exactly 2 permutations without repetitions. By "ignoring suits" I meant that permutations of the cards with the same sequence of values (ace, two, three...) are considered equivalent, even if the suits of the cards (clubs, diamonds...) are different. It doesn't matter for the probability, because both the numerator and denominator would change by the same factor of (4!)^13. —
Keenan Pepper22:42, 7 October 2006 (UTC)reply
For a card deck, X = 4, Y = 13, I find that 63394531038905867912088 out of 92024242230271040357108320801872044844750000000000 permutations are repetition–free, giving a probability of 0.68888946545492·10–27. –
LambiamTalk02:34, 8 October 2006 (UTC)reply
First, I managed to get confused again about the roles of X and Y; the number 63394531038905867912088 above is for X = 13, Y = 4. The correct number for X = 4, Y = 13 is 4184920420968817245135211427730337964623328025600, which gives a much higher probability for a stutter–free random shuffle: about 4.5%.
Now about the method. Calling the card "values" A, B, C, ..., M, we need to count the number of strings containing exactly 4 of each of those that are stutter-free. One such string is CIFBAIBEALKCHGDGMLELIFCFMJDGDEMJFEKJBMLJDHKGIHBHACKA. Let fv(a,b,c,...,m) stand for the number of stutter-free sequences containing a As, b Bs, c Cs, ..., and m Ms, and ending in the value v. So the length of the sequence is a+b+c+...+m. The string above contributes for one to fA(4, 4, 4, ..., 4). By summing fv(4, 4, 4, ..., 4) for all 13 possible values of v we get the desired result.
How much is fA(a,b,c,...,m)? If a is 0 (or less), the count is 0: there are no sequences that end in A but contain no As. If a = 1 and all of b ... m are 0, we need to count the stutter-free sequences of length 1 that end in A. That is easy: there is precisely one such sequence. Otherwise. if a is at least 1 and the sequences have length > 1, each sequence counted can be turned into a shorter non-empty stutter-free sequences by removing the last element. Each such shortened sequence can end on any element except A. Conversely, each of the sequences counted by fB(a–1,b,c,...,m) – or any other element than B, except of course A itself – can be lengthened by appending A to it. So then fA(a,b,c,...,m) = fB(a–1,b,c,...,m) + ... + fM(a–1,b,c,...,m).
But wait, there are many symmetries. For one, if we permute the numbers b,c,...,m in fA(a,b,c,...,m) it will not affect the outcome. So, using ((...)) for a multiset notation, in which ((1,3,3)) = ((3,1,3)), which is different from ((1,3)), we can, instead of fA(a,b,c,...,m), consider fA(a, ((b,c,...,m))). Similarly fB(b, ((a,c,...,m))), and so on. But surely fA(p, ((q,c,...,m))) is the same as fB(p, ((q,c,...,m))). We don't need the subscript to f. Just find f(4, ((4,4,...,4))) – with 12 elements in the multiset – and multiply the result by 13.
Now, in general, how to find f(a, M), in which M is a multiset? We apply the earlier considerations to the new representation making use of the symmetries. If a = 1 and M contains no positive elements, the result is 1. Otherwise, f(a, M) is the sum over the elements v in M of f(v, M – ((v)) + ((a–1))), in which + and – are (also) used for multiset subtraction and addition in the obvious way. So f(a,((b,c,...,m))) = f(b,((a–1,c,...,m))) + ... + f(m,((b,c,...,a–1))).
We now have a recurrence relation that allows us to compute the function values. Doing that naively takes time proportional to the result value. The crucial step is to use the technique of
dynamic programming, or tabulation, or
memoization, or function
caching, or whatever you wish to call it, to avoid repeatedly computing the function value for the same arguments. This can be finessed by discarding any number 0 from the multiset, and by recognizing that in the summing process equal values taken from the multiset will give equal results. But also without that the process is now fast enough to produce the sought result quickly enough. --
LambiamTalk04:41, 9 October 2006 (UTC)reply
I think you had a slight mixup in the second paragraph - it should read: "...we need to count the number of strings containing exactly 4 of each of those...", and the following string should have 4 of each of A through M. --
Meni Rosenfeld (
talk)
07:33, 9 October 2006 (UTC)reply
You are absolutely right. I've applied the correction in situ. There must be some Freudian reason why I don't mind having the Jack of Hearts and the Jack of Spades next to each other, but don't want the Queen of Hearts next to the King of Hearts. I guess I'm confused by the misleading analogy to the spouse-seating problem. --
LambiamTalk11:10, 9 October 2006 (UTC)reply
So, the questions remain: Is there a closed-form solution? Can it be shown that there is not a closed-form solution? Is it an open problem? —
Keenan Pepper22:57, 9 October 2006 (UTC)reply
I think it is an open problem. If there is a closed-form solution, it's not going to look pretty, offering little incentive to look really hard for one. --
LambiamTalk20:16, 10 October 2006 (UTC)reply