# Wikipedia:Reference desk/Archives/Mathematics/2006 December 24

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# December 24

## Double Helix

Yesterday, I up loaded this image for the article Double helix, however, I'm having second thoughts; Have I misunderstood the article? is my image not depicting a double helix? I want to correct this if it isn't.--HoneymaneHeghlu meH QaQ jajvam 03:49, 25 December 2006 (UTC)

Nope, the image is fine. My only request would be to show a couple more turns. — Kieff 04:20, 25 December 2006 (UTC)
Thank you very much, I'll upload a more twisty one today.--HoneymaneHeghlu meH QaQ jajvam 04:29, 25 December 2006 (UTC)
There you go, I hope this version is better.--HoneymaneHeghlu meH QaQ jajvam 04:43, 25 December 2006 (UTC)
There are three color problems:
1. The recommended background color is white or transparent, not black.
2. The blue on black is low contrast, so hard to decipher.
3. Thin blue lines challenge our eyes.
You might also want to experiment to see if a short fat helix shows the shape better than a long skinny one.
For future questions of this nature, we have two projects designed to help:
The first is for general images, while the second concentrates on mathematical images. --KSmrqT 07:59, 25 December 2006 (UTC)
I took liberty to edit the original: high contrast (to give a better sense of depth), transparent background, PNG. Results at right. — Kieff 23:03, 26 December 2006 (UTC)
Alright, to answer your color problems; I felt that black made the objects stand out more, because the material 'glows'. I felt that the blue and red complemented one another, if you can suggest another color I'd be happy to change it for you. Also, it's not that thin of a line :). I can change it so it renders a white back ground, but not a transparent one, sorry.--HoneymaneHeghlu meH QaQ jajvam 00:05, 28 December 2006 (UTC)

I will place my questions there then.--HoneymaneHeghlu meH QaQ jajvam 06:11, 26 December 2006 (UTC)

Here's another twist: I think you have a perspective view there, perhaps an isometric view would be easier to interpret ? StuRat 12:41, 25 December 2006 (UTC)

On second thought, I think it may be an isometric view, after all. Never mind. StuRat 12:47, 25 December 2006 (UTC)
I think it is a perspective view: you see the imaginary cylinder in which the helices fit snugly visually taper off in the distance. And I think a slight perspective helps us to interpret a 2D image as representing a 3D image. A lower pitch (more turns per distance travelled) might further make the image clearer. It would be cool if the image showed the mathematical abstraction of the actual shape of the DNA double helix. Does anyone have the values for the parameters ω and Δα in the parametric representation modelling the DNA double helix? It is:
xi = cos (ωt + αi)
yi = sin (ωt + αi)
zi = t
for i = 1, 2.  --LambiamTalk 17:22, 25 December 2006 (UTC)
The world may have enough DNA images already. (Try this interactive 3D tutorial.) As for the parameters, they're not quite fixed because this is a biological molecule that changes structure depending on its environment. (Note we have A, B, and Z forms, with Z-DNA winding left.) But it seems there will be roughly 10 nucleotide units per turn, and our DNA article says each unit is 0.33 nanometers long, and the width of the helix is about 2.3 nanometers. The two components are not separated by a 180° "phase shift", so there is a 2.2 nanometer major groove and a 1.2 nanometer minor groove. --KSmrqT 03:07, 26 December 2006 (UTC)

## Progressions

Please give me a hint to solve the following problem-

A Geometric progression consists of 2n terms.If the sum of the terms occupying the odd places is S and the sum of the terms occupying the even places is T then what is the common ratio of the progression(in terms of S and T)?

Thank you.

-Sruthi —The preceding unsigned comment was added by 59.93.65.189 (talk) 05:12, 25 December 2006 (UTC).

You have two series each with n terms and common ratio R = r^2. You know that the sum of first series s1 is S, and the sum of second series s2 is T. This gives you two equations. The two unknowns are R and A0, the first term in the series with sum S. The first term in the series with sum T is simply r*A0. Just find R and A0 using the equations sum(s1) = S and sum(s2) = T. Then just take the square root of R. Hope that helps. deeptrivia (talk) 07:05, 25 December 2006 (UTC)
There is no need to over complicate things. A simple hint is to write out terms explicitly.
${\displaystyle a,ar,ar^{2},ar^{3},\dots ,ar^{2n-1}\,\!}$
Next write the two sums.
${\displaystyle a+ar^{2}+\cdots +ar^{2n-2}\,\!}$
${\displaystyle ar+ar^{3}+\cdots +ar^{2n-1}\,\!}$
Both sums have n terms. Compare corresponding terms; notice anything? Use the distributive law and the conclusion is manifest. --KSmrqT 07:44, 25 December 2006 (UTC)
Yeah, that's much clearer. My LaTeXphobia sucks. Hats off! deeptrivia (talk) 09:18, 25 December 2006 (UTC)
This assumes that the parity (evenness/oddness) of the "places" is understood, but there are at least two reasonable ways of indexing the terms:
${\displaystyle x_{i}=a^{i},~~~~\!i=0,\ldots ,n-1\,\!}$
${\displaystyle x_{i}=a^{i-1},~i=1,\ldots ,n\,\!}$
If I encountered this on a test, I would guess that the second is meant, with the parity of index i being the evenness/oddness of the problem formulation. But I think the first way of indexing is usually more convenient, mathematically.  --LambiamTalk 09:25, 25 December 2006 (UTC)
Ah, but I cleverly (?) did not say which sum was even terms and which odd. After all, the request was for a hint. :-D --KSmrqT 11:07, 25 December 2006 (UTC)
The word "This" in my comment referred directly to the original posting (and therefore only obliquely to yours); otherwise I would have used a deeper indentation. But I'm indeed impressed by how cunningly you managed to work your way around the issue. :-)  --LambiamTalk 16:19, 25 December 2006 (UTC)

## Four color theorem

I wrote a proof of this theorem. It´s on the talk page of its article. People have discussed another proof of the theorem on the same talk page, but they are not discussing mine. As I read it this morning, it looked a little pedant and someone reading it might think I´m just kidding. My question is: what is wrong with the proof? Can someone actually read it, understand it and then show me where did I get it wrong?FoCoTh 15:46, 25 December 2006 (UTC)

(I just started to make pictures to illustrate my proof. You can read the proof and see the pictures at my user page. I´ve just started, I want to make as many pictures as necessary for the proof to be very understandableFoCoTh 05:21, 27 December 2006 (UTC))

Can I read it: Yes. Can I understand it: No. In any case, talk pages of articles are for discussion on how to improve the article. A talk page is considered research for the article, and the policy of "no original research" also applies to talk pages.  --LambiamTalk 16:02, 25 December 2006 (UTC)
Without getting into this specific attempt at a proof, I do want to clarify that WP:NOR doesn't pertain to talk pages in the same way it does the actual articles. For example, let's say I find an unusual bit of information that might pertain to an article, but I don't have a reference. It would be perfectly appropriate for me to mention what I found on that article's talk page and ask if any other editors have an outside published reference we could use to include that information in the actual article.
Another example would be if something in the article appears to be incorrect, but the evidence for that error is original research. In that case, it's appropriate to mention your finding on the talk page, in hopes that your finding has already been published and another editor can find an appropriate citation.
Thus it's acceptable to include original research on a talk page, assuming you are doing so in a good faith attempt to improve the article itself. Dugwiki 17:07, 26 December 2006 (UTC)
I read it and understood it. The place where you went wrong was your lack of mathematical rigour. In mathematics, you can't prove something by examples, and you can't prove it by common sense (a.k.a. "it's obvious that..."). One of my former lecturers in maths told me, "a mathematical proof is that which is required to convince an unreasonably doubtful person." Whilst your argument (with a bit of cleaning up to make it easier to read) might be great for convincing everyday people, it doesn't qualify as a mathematical proof, firstly because it's informal (things aren't concrete - you rely on intuitive notions like "sectors" without formally defining them), and secondly because it lacks rigour.
You make a lot of common-sense claims in your argument, like "if you swallow a group of sectors with a new sector, and then add other sectors onto that swallowing sector by the two-points rule, you can reuse the colours from the swallowed sectors on sectors that attach to the swallowing sector" (heavily paraphrased). A claim like that is perfectly sensible and reasonable, and any moderately intelligent person would see its truth right away. But mathematical proofs can't rely on that kind of "it's quite obvious, if you think about it, that this is the case" argument. They must be formal and rigourous.
Nonetheless, I think it would do quite well as a fairly convincing argument for informal purposes, although, as Lambiam noted above, Wikipedia is not the right place for it either way. I recommend learning some Graph theory, if you haven't yet, since the Four Colour Theorem can be translated into an isomorphic problem of graph theory, and we have already developed lots of mathematical techniques and suitable terminology for graph theory (no point reinventing the wheel and all that). Best of luck! Maelin (Talk | Contribs) 16:27, 25 December 2006 (UTC)
Thank you! If rigour is all that is lacking, then I should be happy about it. I guess it wouldn´t be too hard to write it all again in a very rigorous way. FoCoTh 17:25, 25 December 2006 (UTC)

As far as I can see, this is the most common kind of mistaken attempt at a 4-color-theorem proof: confusing the four-color theorem (difficult to prove) with the statement that there cannot be more than four regions all touching each other (much easier to prove, and what you seem to be proving). There are (nonplanar) maps in which the number of colors is much larger than the number of regions that mutually touch each other; see e.g. Grötzsch graph for an example. Your argument doesn't rule out the existence of a map requiring more than four colors but having more than five regions. —David Eppstein 17:05, 25 December 2006 (UTC)

I´m sorry to say I couldn´t understand what you meant, but I´m sure if you explain it a little better, I´ll be able to. Doesn´t the fact that there cannot be more than four regions all touching each other prove that all planar maps can be colored using only four colors and even so no region of one color touches a region of the same color? How so? FoCoTh 17:22, 25 December 2006 (UTC)
There exist maps (on nonplanar surfaces) for which no more than four regions can all touch each other, but which nevertheless require more than four colors. So it isn't good enough to prove that no more than four regions of a planar map can touch. It's a true statement, but not the same as the four color theorem. —David Eppstein 17:35, 25 December 2006 (UTC)
I can see there are such maps. But nonplanar surfaces have three dimensions and so a lot of other things are allowed on them which are not allowed on planes. On planes, no more than four regions can touch each other plus all the lines that go on one direction will never reach their origin, like for instance if they had been on a sphere... FoCoTh 18:13, 25 December 2006 (UTC)
David Eppstein, I think I know what´s the difference between a planar surface and those other surfaces you are talking about. In planar surfaces, no more than four regions of the map can mutually touch each other. The same rule applies to some nonplanar surfaces which are not four-colorable. However, in planar surfaces no region can touch four regions that mutually touch each other, while on those nonplanar surfaces you talk about this rule does not hold. Indeed, on those nonplanar surfaces, more than four regions cannot mutually touch each other, but one region can touch all of those four regions that mutually touch each other. This last fact, when it happens, makes necessary the use of a fifth color.FoCoTh 05:17, 27 December 2006 (UTC)
The more you react like this the more you convince me that you do not understand the problem well enough to construct a proper proof. Of course nonplanar surfaces can contain more than four mutually adjacent regions; that is not the point. The point is the following: let A="map is planar", B="map doesn't have more than four mutually adjacent regions", and C="map is four-colorable". Your proof, if it can be made rigorous, appears to be proving that A implies B, a known and easy to prove fact. You appear to be assuming that B implies C, and therefore concluding that A implies C. But B does not imply C. In fact there are maps for with B is true and C is false. So your logic is broken. Telling me that there are maps for which A and B are both false (as you have just done) doesn't address the gap in reasoning between B and C. —David Eppstein 05:35, 27 December 2006 (UTC)
I am not saying B implies C. I didn´t realise the nonplanar map I was talking about had five mutually touching regions. I am happy anyway because I found out on my own that there could be a map with five mutually touching regions. I didn´t understand why doesn´t B imply C and I didn´t understand what implies C then. Now, "easy" is a very subjective concept. What you may find "easy to prove" today may not have been sometime in the past. A "known" fact would not be so until someone had discovered it. I´m happy to have done it. I´m happy to have found out this all by my own (that on the plane there are no more than four mutually touching regions), and I don´t know if you would be able to find this out when you were younger and hadn´t formally studied graph theory, using only your imagination and a sheet of paper. Obviously I don´t understand the problem well enough to construct a proper proof. Do you? Does someone?
My proof doesn´t show in a clear way why does the fact happen, but it shows it does. I was so amazed when I first learned about the four color theorem that I imagined myself drawing a map and I wanted to find out when the four-colorable property would emerge in the drawing process, when would I be able to see the map actually four-coloring itself (or making itself a four-colorable one). I thought of all the ways I could move my pen in constructing a map, and it is intuitive that the three actions are enough to do it, even though I haven´t formally and rigorously proven it. It is also intuitive that none of the three actions will ever cause a fifth color to appear, and the process to show this is intuitive as well. Isn´t that useful? I know it is not the so called B property that implies C, but I can see intuitively C being implied somehow while drawing the map in the way I suggest. And it is not B that implies C, I never said that, you only assumed I did. I can´t understand yet how can there be a map in which B holds and C does not, however. I clicked your link and didn´t get it. Neither do I know if I must understand it to understand the real reason why C is true. You haven´t shown me what is wrong with my proof because you didn´t even understand what my proof is about. My proof shows all configurations a map can possibly have on a planar surface. In other surfaces my proof doesn´t work because the actions to create maps on those surfaces must be so different._FoCoTh 07:24, 27 December 2006 (UTC)
I'm intrigued, David. If the statement "B implies C" is false, can you give a counterexample of a map with no more than four mutually adjacent regions that is not four-colourable? Would, instead, the statement "A & B imply C" be true (whether it is proven in Focoth's proof or not)? Maelin (Talk | Contribs) 14:22, 27 December 2006 (UTC)
Note that a surface is, by definition, two-dimensional. Do not confuse the properties of the surface itself with a higher-dimensional space in which it might be embedded. I think David Eppstein's point was that if there is an example on any surface of a map in which no more than four regions all touch each other, but which nevertheless requires more than four colours to colour it then this shows that the property "no more than four regions all touch each other" does not imply the property "can be coloured with only four colours". This would show that your approach to a proof of the four colour theorem cannot work because it depends on a logical implication that does not hold. Gandalf61 18:29, 25 December 2006 (UTC)
(after edit conflict:) But this is enough to invalidate the idea that one proves the other in some obvious way. For G a graph, let Planar(G) mean: G is planar, let Cliq(G) mean: the size of largest clique in G, and Chrom(G): the number of colours required to colour G. Then perhaps you originally had a proof of the following form in mind:
Planar(G) implies Cliq(G) ≤ 4.
Cliq(G) = n implies Chrom(G) = n for all n.
Therefore Planar(G) implies Chrom(G) ≤ 4.
However, the middle premise does not hold in general; David Eppstein already mentioned a counterexample. Now you argue: "Yes, but you did not take into account that we know that G is planar". OK, true by itself, but that then needs to figure into your argument in some other way than it already did, and you haven't told us how.  --LambiamTalk 18:32, 25 December 2006 (UTC)
It looks to me that it already figures in my argument (the fact that the map is being drawn in a plane that does not show those properties needed in order to the property of four regions not being enough to prove the theorem). Well, I admit it lacks mathematical rigour and it may not be clear how it is in my argument. The fact is that, when adding a new region to the map in the plane, it´s impossible for the region to start to be drawn in the center of the sheet, then being drawn until the edge of the sheet of paper is reached and then suddenly appearing in the other edge of the sheet again (as if the sheet of paper were representing the surface of a sphere or cylinder, for instance). I don´t know how to insert it in a very clear manner in my argument yet, but it does not look too difficult. (By the way, I know that a map on the surface of a sphere or cylinder is in fact four colorable, it was just an example) FoCoTh 01:29, 26 December 2006 (UTC)
We appreciate your enthusiasm and your interest in mathematics, but we need you to understand that Wikipedia is not the right place to discuss a proposed new proof of the four color theorem.
But wait, it gets worse. The first proof accepted by mathematicians was slow in coming, and slow to be accepted. It used a combination of subtle ideas, and ultimately required a massive computerized enumeration of cases even after reduction. No correct "simple" proof has ever been found (unless you consider this simple), but many incorrect proofs have been proposed. Especially since you are unfamiliar with the necessary mathematics, the chances that you have found a correct simple proof are vanishingly small. (Even with the background it would be unlikely.)
This is an excellent opportunity to learn more mathematics. First, read our article thoroughly. Second, study the references linked at its end. Third, read the Google archives of the newsgroup news:sci.math, here. Fourth, raise any further questions in the newsgroup. --KSmrqT 02:41, 26 December 2006 (UTC)
I would appreciate solving the theorem as much as I would appreciate learning why can´t I solve it. I think a good start to learn this is actually trying to solve the theorem, and trying to see why can´t it be done. Instead of spending a lot of time reading a lot of articles and learning about concepts that I wouldn´t know how to use immediately in order to better understand the theorem would be rather boring. I want to learn those concepts, if I see they can really help me. And I will do so. Now, the fact is I am convinced that all maps in a plane are four-colorable, and I am not convinced because someone said it, but because I´ve imagined all kinds of "actions" you can do to a map in order to add new regions to it, and I saw that none of these actions have the capacity to generate a situation in which a fifth color is needed, and I kind of understood why. I would imagine that, as my "proof" is very simple, someone would just point out what´s wrong, and it would be simple to understand. Doesn´t look the case. By the way, I´m not using Wikipedia to try finding a proof of the theorem, I´ve instead asked a legitimate mathematics question, i.e., "what´s wrong with my proof?"FoCoTh 14:35, 26 December 2006 (UTC)
I'm sorry, but several of us did point out what is wrong with your proof. We are all convinced that the theorem is true; it is just that what you appear to think is a valid proof method is not considered valid by mathematicians. Your reaction to my explanation was not logically appropriate and makes me think that you do not understand what it takes for something to be an acceptable proof.  --LambiamTalk 15:33, 26 December 2006 (UTC)
Lambian, thank you for your help, but I didn´t know what a clique is. I´m gonna try to understand it. It would be good for me however if you could rewrite your explanation, without the term clique, if it is possible at all.
I understand your explanation has two premises and a necessary conclusion, and you said one of the premises is not proved. I really am going to try to learn what a clique is. I think what you meant is the same thing Meni Rosenfeld is saying below, that I didn´t prove the three "actions" include all possible actions.
Maybe the premise which "doesn´t hold in general" is the one that says all maps in which four regions cannot mutually touch each other are four-colorable. This really does not hold in general, as there are maps in nonplanar surfaces which are not four-colorable. My answer (not my proof) to this is that the property of nonplanar surfaces which allows this to happen is not present on planar surfaces. If this is proven, do you think the theorem will also be proved?FoCoTh 01:19, 27 December 2006 (UTC)
As Maelin said, "a mathematical proof is that which is required to convince an unreasonably doubtful person." I know as a fact I´m not convincing people, and what I thought to be doing is, like Meni rosenfeld said, "to give a correct proof which omits some technical details (ones that anyone can fill in)." FoCoTh 00:56, 27 December 2006 (UTC)
More specifically, how do you know that the "actions" you have considered really include any possible action? My guess is that you do not currently know that, and that if you attempted to find a simple proof for that, you would probably not succeed. It's one thing to give a correct proof which omits some technical details (ones that anyone can fill in), and quite another to assume that something is obvious when in fact it is not. -- Meni Rosenfeld (talk) 17:54, 26 December 2006 (UTC)
Do you mean that if I prove that those actions include all possible ones, then the proof would be correct?FoCoTh 00:56, 27 December 2006 (UTC)
I've tried to reread your attempted proof, and noticed that you used this argument:
just try to do it and you will see that, in maps that use "no points rule" and "two points rule" only, there are never four colors on the border!
Using the same kind of argument, you could have given a much simpler and more direct proof of the Four-colour Theorem:
just try to do it and you will see that, in planar maps, you never need more than four colors!
In a mathematical proof, however, this kind of argument has no place. You must prove that if your construction method is followed, at no point of the construction process are there four colours on the border. Incidentally, I think that statement is false, so you should not be able to prove it. But since you describe your method "by example" instead of by unambiguous rules – another no-no in mathematical proofs – I cannot give a concrete counterexample. (Nor do I need to; it is upon you to give a real proof.) If you want to continue your investigation, you should at least study how mathematicians turn the problem into a graph-theoretical problem. Above you wrote that you did not know the concept of clique. In what I wrote, I hyperlinked all technical terms to Wikipedia articles, so it should be easy enough to follow these and read up. I suggest that you take a break in this discussion until you've studied and tried to digest some graph theory – a fascinating subject with many interesting problems, the study of which should be rewarding enough in and by itself.  --LambiamTalk 07:58, 27 December 2006 (UTC)
I saw it know that it may look like I was reaching my conclusion from a non-verifiable fact, that all maps drawn with those rules can be colored with only three colors on the border. That´s not true. I spent the whole proof showing how this becomes true, and that´s why I said "maps using those rules" and not "all maps", which I could, because I expected the reader to have at this point understood this fact already. The reason why they can be colored with only three colors on the regions at the border is that, whenever the need of a fourth color appears, one region of the map must be closed inside other three regions at least and will not touch the border. Since hte map didn´t show the need for a fourth color until that point, and one of the regions will not be on the border anymore, there will be only three colors on the border. Saying that is different from telling the person to draw all possible maps, which is impossible, but proves exactly the same thing as if you had drawn all possible maps (it proves the fact is true, but doesn´t prove "why" it is)FoCoTh 08:59, 27 December 2006 (UTC)
Please read carefully what I wrote. By the words "if your construction method is followed", I mean of course the same as your "maps using those rules". It is not intuitively clear to me, and I believe it is a false statement. You assume that, when "the need of a fourth color appears", the new region will shield at least one colour on the old border so that it is no longer on the new border. But you do not take account of the possibility that all three colours occur on the old border both in the part that will become adjacent to the new region and thus get shielded, and in the part that will remain exposed on the new border. If that happens, then you have got four colours on the border. Now you may think that this is maybe the first time I've looked at a purported amateur proof of the Theorem, but is more like the twelfth time – I've lost track – and I know that a little problem like this can be repaired, except that then there is another little problem, which luckily can get fixed, except for a tiny detail, which however can... and so on. So if you are serious (but consider that many great minds have tried to prove this theorem for over a century and failed, so somehow it is extremely unlikely that there is a simple proof), first fix the proof so that there is not even the tiniest hole left you could put a pin in, and only then present it. And do read up on graph theory.  --LambiamTalk 09:46, 27 December 2006 (UTC)
If there were a map with three colors on the border regions, being that one color appears both on the part that would be shielded and on other part which would not be shielded, and then such an action was undertaken that would create the need for a fouth color (therefore shielding one of those two regions which have the third color, but not shielding the other) the non-shielded region would simply change it´s color to the same color as the new region (color 4). This region which had the same color of the shielded region but was on another part of the map didn´t touch a region with the same color as its own (!), and could so have its color changed to color 4. This happens in any case when the need of a fourth color appears, the border of the map can be colored with only 3 colors because the region which had color 3 and will be shielded will be shielded and not on the border, and the other regions which were colored 3 would be adjacent only to regions colored 1 and 2 (because there were only three colors on the map so far), and could have its color changed to 4. (Note that I have stated a rigid construction method, which must be used to understand the proof, but I didn´t state a coloring method. You can change the colors as the map grows)FoCoTh 16:43, 27 December 2006 (UTC)
This is a perfect illustration of the problem I mentioned above with amateur proofs. Forgive me for not further responding.  --LambiamTalk 17:44, 27 December 2006 (UTC)

From Wiki: Each region must be contiguous: that is, it may not consist of separate sections like such real countries as Angola, Azerbaijan, and the United States. Does it mean that we may use two colors for United States? Thank you in advance for your clarification. Twma 22:32, 27 December 2006 (UTC)

Yes, it does. You're allowed to use different colours for each of: (1) the contiguous United States, (2) Alaska, and (3) Hawaii. For the purpose of the theorem they count as separate regions. If you use your microscope you may see that actually the state Hawaii consists of many islands, each of which you can colour with its own colour – but as they all border only the ocean, you can use one colour for all.  --LambiamTalk 23:19, 27 December 2006 (UTC)

Is it acceptable to claim that a map M is a two dimensional pseudomanifold (collection of triangles) on the plane with integer label on each of the triangle (simplex)? The lebelling is a function from M into integers. A country (region) is the collection of triangles with the same label. Is it acceptable to consider the region outside the pseudomanifold as a special country S with fixed color 4? Pick one outmost triangle N and apply induction to the the deleted pseudomanifold (old map), i.e. colored already. This new triangle N has one edge adjacent to S and the other two edges may be shared with at most two triangles TA,TB from the old map. If N is part of the existing country, its color is fixed. Does it produce any contradiction? I expect that the contiguous condition to play a part. If N is not part of any existing countries, is it possible to claim that TA,TB are ADJACENT to S and hence cannot have color 4? As a result, TA,TB can have either color 1,2 or 3. In this case, we can alway pick a color for N different from those of TA,TB. This is NOT a proof but is to formalize the problem in terms of pseudomanifolds and induction. Many cases (less than 600??) are overlooked but they seem to be within the capacity of human being without the usage of computers. Perhaps people in collaborative way might be interested to dig in by more detailed explanation of terminologies such as pseudomanifolds and by developing the idea into an acceptable proof by induction and by more careful case-studies. Are there any volunteers who would draw some triangulations in order to help the development of idea along this line? Are there any examples to show that the contiguous condition is essential? I shall be on holiday. HOW CAN I FOLLOW UP THIS TRACK AFTER 10 days when it is removed from here? My dear FoCoTh, are you going to copy part of this track to your TALK so that people like me can read after the new year holiday? Thank you in advance. Twma 01:28, 28 December 2006 (UTC)

The answers to your first few questions are yes. If you consider S a region in its own right but you want to maintain the (in my opinion unnecessarily complicated) view of a region being a collection of triangles, then you will also have to decompose S into triangles. If N is part of an existing region R that already has been assigned some colour C in the inductive process, yes, then it is possible that when you now try to add N to the "inductively" coloured map, it borders another region than R that also has already been assigned the same colour C.
Unlike many mathematicians, I consider computer proofs acceptable if sufficiently high standards are met. On the contrary side, I would find it hard to trust a human-produced proof considering several hundred cases. I'd like to see a computer-verifiable proof then that no case has been overlooked.
When this section disappears from this page, you'll still be able to access it at Wikipedia:Reference desk/Archives/Mathematics/2006 December 24#Four color theorem. However, Wikipedia is not meant to be a tool for collaborative original research.  --LambiamTalk 07:54, 28 December 2006 (UTC)
Thank you. I should shut up although based on what I said and based on what you ACCEPTED, I might have found a ridiculously simple proof (probably among many false ones) for public scrutiny. Need blessing. Twma 08:19, 28 December 2006 (UTC)

On the offchance that you haven't figured it out yet, and since nobody's taken the trouble to do so, I'd like to provide a counterexample that doesn't ignore the point of the proof. I have this because I started off exactly where you are a little over a week ago, and found the error almost immediately. Imagine three mutually adjascent sectors labelled 1, 2, and 3. From the edge of 2 to the edge of 3 (the short way around), draw a line, and color the new sector 1. Finally, from the edge of 2 to the edge of 3 (enclosing the last sector), draw another line, and color the new sector 4. I believe this may have been mentioned above, but it got buried. Following the exact rules you specified, it is possible to reach infinitely many similar maps that must have on the boundary 4 colors. If you allow for recoloring, that can be fixed, which is what my (independently reached) proof below takes into account, but you haven't allowed for that. If you had colored the fourth sector 4 in the first place, and the last one 1, you would wind up with an acceptable map, but your system doesn't allow you to at any stage have a map with four colors on the boundary. Black Carrot 22:46, 30 December 2006 (UTC)

Correction: You did respond to one like this, but missed the point. The way you've formulated it, each sector is colored when it's made. You made no allowance for recoloring, or for later coloring. If you do, you have to explain how it would be done, and it's not the case that recoloring individual sectors along the boundary would be enough, because with larger maps many of those sectors will border sectors with the same color they're changing to, so those would have to be changed, which means others would have to be changed, and how do you know that doesn't work its way all the way thorough and out the other side to the boundary again and screw things up for you? In fact, I'm pretty sure it usually does. If you want to explain what you'll do about that, great, but without a thorough recoloring system the proof isn't finished and can easily be wrong. If you wanted to wait to color things until you knew what they would be touching, again fine, but how? It's harder than it sounds, since on most maps you wind up with a lot of options and no way to find the right ones. Black Carrot 22:57, 30 December 2006 (UTC)

## How to make Financial Models?

### Proluge

• It would be useful if you told us what precisely you wanted to model, how detailed the model has to be and whether you want to make a continuous or discrete model, but I can recommend the article on the Black-Scholes Model to get you started. JChap2006 03:53, 21 December 2006 (UTC)

### Part 1 - (Precisely what I wanted in the model)

Say we know some items of the Model. In this example:

DECISION ANALYSIS FACTORS: years 0 through 11

• Real Cash Flow to Owner
• Present Value of Real Cash Flow
• Net Present Value of Real Cash Flow:
• After Tax Real Internal Rate of Return:
• PROFORMA INCOME STATEMENT: years 0 through 11
• Annual Gross Rental Income
• Vacancy and Collection Losses
• Effective Rental
• Operating Expenses
• Net Operating Income
• Interest Expense
• Depreciation (cost recovery)
• Taxable Income
• Income Tax Liability
• Net Income After Tax
• PROFORMA CASH FLOW STATEMENT: years 0 through 11
• Annual Gross Rental Income
• Vacancy and Collection Loses
• Effective Rental
• Operating Expenses
• Net Operating Income
• Debt Service
• Income Tax Liability
• Equity Dividend (cash to owner)
• Down Payment/Reversion
• Total Cash Flow to Owner
• Real Cash Flow to Owner

MORTGAGE LOAN AMORTIZATION SCHEDULE: years 0 through 11

• Balance Owed, beginning of year
• Annual Mortgage Payment
• Interest Portion of Payment
• Amortization of principal
• Balance Owed, end of year

ANALYSIS OF REVERSION ON SALE:

• Net Operating Income Projected, Year After Sale (Year 11)
• Cap Rate At Sale Date
• Capitalized Value (Sale Price)
• Transaction Cost
• Net Sales Price
• Book Value At Sales Date (cost-dep)
• Capital Gain ( Net Price - BV)
• Capital Gains Tax
• Mortgage Balance Owed
• Reversion in nominal dollars to owner at sales date
• What is the next step? --Foundby 05:59, 22 December 2006 (UTC)
• The first step is to determine what it is you want to model: The expected revenue of a ski lift? When to declare bankruptcy? Is this a "what if" exercise, and what conditions are under your control to vary? Then you determine what quantities figure into the equation (inflation rate, number of baby boomers retiring, whatever). At that point you realize you're never going to be able to find sufficiently reliable forecasts and give up. If, however, you insist on continuing, then the next thing to do is to examine the relationships between all relevant quantities and express them as mathematical equations (revenue = income − expenses; income = quantity × price; expenses = fixed + quantity × variable; that kind of stuff). Possibly you have a time involvement (revenue2006, revenue2008, balance2008 = balance2006 + revenue2008). Then you do the calculations, or solve the equations, or determine the optimal value of something, depending on the nature of the model. Perhaps you could first try something a bit simpler and less ambitious than you seem to have in mind.  --LambiamTalk 11:36, 22 December 2006 (UTC)

### Part 2 - (Net Operating Income Year 1 & Property Valuation)

• In the first equation you need a minus sign instead of a plus sign. It is basically the equation Net profit = Gross revenue − Expenses. The revenue goes into your wallet, the expenses are paid out of your wallet, and what remains is the net profit. For the second equation, per definition Cap rate = Income / Value. So then by elementary algebra Value = Income / Cap rate. The two factors 100 cancel.  --LambiamTalk 05:11, 23 December 2006 (UTC)

### Part 3 - (Loan Amount & Equity Required)

• Loan Amount = Loan to Value ratio X Property Valuation is just a rearrangement of Loan to Value ratio = Loan Amount / Property Valuation, which is a definition of Loan to Value ratio.
• Equity Required = Property Valuation - Loan Amount is a definition of Equity Required - it says that the purchaser's equity in a property is what is left after subtracting the amount of any loans or mortgages from the value of the property.
If you are asking questions like these, you need to study some financial mathematics, especially around commercial mortgages - you are a long way from constructing the type of complex financial model that you seem to be aiming for. Oh - and moving your question around on the Reference Desk won't get it more attention ! Gandalf61 13:06, 23 December 2006 (UTC)

### Part 4 - (Frivilous Comments part)

• Time for a christmas present to yourself. Get hold of a book on small business accounting. This would go into much more detail of how to drawup ballance sheets and cash flows than we will be able to do justice with here. The cost of purchace will really pay in the long term. Do not rely on wikipedia for financial advice! --Salix alba (talk) 12:14, 23 December 2006 (UTC)
• Whatever happened to You can give the gift of knowledge by donating to Wikimedia?--Foundby 12:58, 23 December 2006 (UTC)
• I'm sorry to hear that. Maybe a library in your vicinity has helpful books. I'm also sorry to see that you considered Salix' or my responses frivolous. I can assure you they are meant to be entirely serious.  --LambiamTalk 12:34, 24 December 2006 (UTC)

### Part 5 ( 1 + i )

Anyone who has studied business has at least a passing familiarity with “time value of money” formulas. Yet while many people can comprehend explanations of loan repayment schedules or retirement savings accounts, or even use financial calculators, their level of time value understanding may not include any sense of where the time value formulas come from, or how the various time value applications are related. This brief article presents a derivation of all of the standard (and some less well-known) time value of money formulas from the future value of one dollar factor.

Basic future value applications are easily understood at an intuitive level. Suppose you have \$1,000 in an account today. If this amount earns a rate of return of 10% for a year, the balance grows to \$1,100 (the original \$1,000 plus \$100 in earnings). Symbolically, this time value adjustment can be represented as

(1) PV0 (1 + i) = FV1 ,

an equation in which PV0 is the \$1,000 in the present, i = .1 or 10%, and FV1 is the \$1,100 future amount one year hence.

This process can be repeated, in the sense that we can now view the \$1,100 as the initial balance in a second future value analysis. If this amount grows at 10% for one year, the account balance grows to \$1,210 by the end of that second period. Symbolically,

(2) FV1 (1 + i) = FV2 .

Substituting equation (1)’s representation of FV1 into equation (2) reveals how the initial present value can be transformed into a future value two years later:

PV0 (1 + i)(1 + i) = FV2 or

(3) PV0 (1 + i)2 = FV2 .

Generalizing for any number of years n, we have

(4) PV0 (1 + i)n = FVn .

Thus, we have discovered the first of the standard time value of money factors, the future value of one dollar factor: FVFn = (1 + i)n .

• In formula 1. PV0 (1 + i) = FV1, how did they come up with 1 + i ? --Foundby 16:41, 24 December 2006 (UTC)
i is the interest rate - in the example it is 10% or 0.1. So the interest earned on an initial balance of PV0 in one year is PV0 times i. But you also have your capital of PV0 at the end of the year. So the future value of your investment at the end of year 1, FV1, is given by
FV1 = Capital + Interest = PV0 + PV0.i = PV0(1+i)
Take a look at our article on time value of money. Gandalf61 18:37, 24 December 2006 (UTC)

### Part 6 - (Mortgage Loan Constant)

• Mortgage Loan Constant is the annualised ratio of the loan repayments to the initial amount of the loan - it looks like an interest rate, but it is higher than the interest rate of the loan because repayments also include capital repayments as well as interest. This link might help, although it is quite mathematical.Gandalf61 13:06, 23 December 2006 (UTC)

• If you just take the ratio of a single loan repayment to the initial value of the loan, you get very different figures if the repayments are, say, monthly than if they are, say, quarterly. To establish a common yardstick you need to annualise the ratio by calculating an annual equivalent rate. For example, if your monthly repayment is 1% of the original loan value, the annual equivalent of 1% monthly is (1.01)^12 - 1 = 12.68% (approximately), so the Mortgage Loan Constant would be 12.68%. (Feel free ask any follow up questions on my talk page). Gandalf61 10:25, 26 December 2006 (UTC)