question and answer thread.

But if hot air rises, why is it so much cooler the higher up you go?
Yes, hot air does rise (an explanation could be given). Its not exactly true that the higher up you go the cooler it gets, though. If you go up really high you will see that the temperature gets cooler then warmer then cooler again and there are actually different levels to the atmosphere (an introduction to the atmosphere's stratification could be given here). But those other levels are much higher than even the highest mountains, and here at the lowest level, things do get cooler as you go up. Why? Its because of the way that things behave as they expand.

The air above us is held on to the Earth by a force called gravity which causes things to have weight depending only on how much of the thing (mass, not number, which could be introduced briefly) there is. If you look at any height, the air at that height is pressed down towards the ground by the gravity of the earth (which has alot of mass) as well as by the weight of all of the air above it. Since there is less air above as you go higher (and since gravity gets weaker as you move away from the earth), air at higher places is forced down on less and less as you go up higher. This causes the air to be "squished" more really close to the eath and less squished as you move up (an exploration of pressure and density would go here).

Now the important thing here is that the lower pressure of the air as you go higher causes any denser air from below that might be rising to expand (adiabatic expansion), and also any air from above that may be falling gets squished as it goes down. And the interesting thing is that when a gas expands it cools down and when a gas get squished, its temperature rises (toss in an explanation of this basic gas law here, keeping it simple but the explanation correct in terms of kinetic energy and such). The heat is still there, it is just not the same temperature. This cooling occurs faster than any equilibrium of temperatures might happen, and so as you go up, the temperature of the air decreases.

At very high elevations, there are places where the atmosphere actually absorbs alot of the sun's radiation due to the presence of things like ozone and ions (which are also created by such absorptions), and it is the fact that these species are stratified (because of which radiation frequency actually passes to which level), that causes the different temperature profiles in the various atmospheric strata. It also explains why only certain intervals of light are able to pass through the atmosphere (mainly the visible spectrum and some extremes like certain radio) and many others are absorbed at various levels. Of course, that explanation is a bit technical and wouldrequire some elaboration for use in telling it to kids. One could also explain to kids the "coincidence" of why the light that is transmitted is the light that animals, plants, and photoreceptive bacteria "see" is precisely because that is the light that is transmitted and they could then evolve to receive it, if one wanted to go there.

But why isn't the ground of mountains warmed like the rest of the ground?
Well, it is. But the light is hitting at an angle for most of the mountain, and this spreads the light out more (insert explanation of how this is seen for different latitudes as well). But also, the air here is also cooler, so equilibrium is reached at a lower temperature. And also, it tends to rain or snow more at these higher elevatin pieces of land (which continues the rain shield discussion earlier), and snow tends to reflect more light than rocks and sand, so less of the light is absorbed as heat as well. All of these reasons add up to why the mountains are cooler than the land lower down.

So....
So, this is the kind of explanation for this question you can give to a kid. I didn't expect this kind of detail as a response to my question, I just wanted to show how one would wrap it all up to turn that kind of question from a kid into an opportunity to give the kid a deeper insight and physical inuition about the world and how things work. Notice that the responses that were given to the question implied alot of background and would not have clarified the situation at all to a child that didn't already posess such a background. If you add alot of visuals to the presentation, maybe break up the discussion into a few days worth of explanations that went into the bits and pieces in a systematic way, and introduced the question at the very beginning as the goal, you then have a very nice lesson plan that could be given to a kid. There is no need to introduce the equations, or even give the kid the expectation that they must memorize or learn every piece of information, because long term potentiation requires repeated exposure to material before it is learned (to paraphrase Natural Born Killers, "repitition works!"). Every year (or whatever interval might be found optimal to learning) similar materials can be introduced, slowly adding on the technical details as the child's ability to understand and use mathematical representations for physical events matures. It's kind of like souldog's questions leading up to the Banach-Tarski paradox (which is what inspired me to ask this question). Of course, I am sure that everyone who read the question probably already knew most if not all of this already, which is why it may have seemed like I was trying to ask a trickier question, or why it wasn't worth answering. But I sometimes think that some of the teachers out there who do know this stuff still would not have taken such a question as the educational opportunity I believe it is. I think they probably would have two different reasons for that as well; first, it might seem too difficult to explain in words and concepts that the child understands (which really underestimates children, or even animals, for that matter, who have highly developed physical intuitions), or because they do not realize that succinct answers are often more confusing to the child than detailed ones and may not realize all of the details being assumed by their answer, because most of the pieces are left unexplained. Of course, in galathaea's case, the teacher just didn't know the answer....

So really, the main answer was: its warmer on the ground because that is where the light gets absorbed (and hotter in deserts because of specific heat) and cools as one goes higher in the atmosphere because of adiabatic cooling. But that sounds like an answer that a kid memorizes for a test and then promptly forgets...

So I guess the points would go to both solarflare and SeventhStar! The thread is now open again for questions!

I hope you didn't expect us to give THAT answer!!!

Anyway I have a question:

Suppose you have an unlimited number of cells that are linked together like this
() cell, - link
...-()-()-()-()-...
Each cell has two conditions on -(1)- and off -(0)-
Each cell checks it's neighbours at every priod of time T and:
if BOTH neighbours are on or both are off it turns (stays) off
if only ONE of the nighbours is on it turns (stays) on

you have only One cell turned on at the beggining:
...all off...-(0)-(0)-(1)-(0)-(0)-...all off...
how many periods of time T would pass in order to have X - cells ON

Well, each round will always have only a power of two cells on, and each power of two will occur first exactly on the same number round (1 on round 1, 2 on round 2, 4 on round 4, etc...). So if the question is asking for the round where exactly X are ON, then it is equal to X iff X is a power of two, otherwise it is never. If it is not asking for exact number on, just enough, then it will occur on the round of the first power of two greater than or equal to X. It looks like an inductive proof suffices here.

The pattern is that the cells tend to switch between two states:
10000...00001
and
101010...010101

From the first position, say there are 2^(n)-1 zeros between the ones. They will each form a separate pattern of 2^(n-1) ones, stretching 2^(n-1)-1 cells in each direction from the initial one, and taking 2^(n-1)-1 turns to do so. This means that the first pattern will change into two small versions of the second pattern - but guess what, they're side by side, so it's really now one version of length 2^(n+1)-1, containing 2^(n) ones. I think it is clear enough that the second pattern will then change into the first pattern (twice as long) after just one time interval.

Notice that to get 2^(n) ones from 2^(n-1) ones, it takes 2^(n-1)-1+1 turns. Simple manipulation shows that (since the first time interval doesn't count) to get X ones from ...000010000... takes X-1 time intervals.

This thread seems to have lost its head, so I decided to figure out some other basic stuff about the past problem to fill time...

First start with a notation. Write 1-1 for the sequence 1, 1-3 for the sequence 101, and generically 1-(2n-1) for the alternating sequence 1010...01 with n 1's. Write 0-1 for the sequence 0, and generally 0-n for the sequence of n 0's. So the first few iterations are therefore:
0-Inf 1-1 0-Inf (Inf is short here for infinite)
0-Inf 1-3 0-Inf
0-Inf 1-1 0-3 1-1 0-Inf
...

Basically there are three basic relations that come in handy. There are two basic transformation types, where we focus on a 1-x group and see how it alters its surrounding zero group and how it changes form "internally". Obviously the neighboring 0-x's can potentially be altered by two 1-x groups. Using S to indicate the successor function, we get the two transformations
S(0-m | 1-n | 0-m) -> 0-(m-1) | 1-1 0-n 1-1 | 0-(m-1) (for m>2, n>1)
S(0-m | 1-1 | 0-m) -> 0-(m-1) | 1-3 | 0-(m-1) (for m>2)
and the one direct equivalence relation (I call it a collapse :))
0-x | 1-n 0-1 1-n | 0-x = 0-x | 1-(2n+1) | 0-x
that settles all the case of m, n.

Using these relations, its fairly easy to prove some simple statements:

• 1 sequences are the same length throughout the string on a given round
• # of 1 sequences is always a power of 2
• collapse occurs on steps after the existence of 0-3 in a string, therefore every four rounds (though complete collapse occurs only on the powers of 2 that are used for the proof of the original problem)

Okay, that's enough for that completely useless exploration... I was curious about the bipartite graph problem series. As I've looked over my notes and tried to use it to study along with some other graph problems (the book by Lovasz, if it edifies), I came across some other proofs of the perfect matching and they all seem to focus in on the proof of what is called the Konig-Hall theorem. I was just curious, souldog (if you are still peeking at this thread), if my interpretation is correct that the first problem was indeed a version of this theorem (well, I mean it is fairly simple to set the problem so that the Konig-Hall theorem gives the perfect matching existence immediately, so maybe I really am asking if it was intentional or known this connection)?

Ahh... maybe I'll go bug some other thread now...

Yes, I am aware of it. I usually relate it to the Hall-Rado theorem and am not aware off the top of my head what the distinction between the two is. I didn't want to bring in any theorems at that point in the discussion.

Ok, I am going to use my power as granted me by the United Association of Solipsists to declare the previous question dead, since as far as I can tell it has been answered correctly. I use my power to grant the point to solarflare who gave the inductive proof (though he may have forgotten to clarify that X was only powers of two since it was already clarified previously). Solarflare also correctly answered the question in terms of intervals instead of the rounds I used.

But I'll ask the next question. Its about one of my favorite areas of recreational math: series manipulations. Now I would like to use the S notation, but I don't want to go to all the trouble of putting the indices so I will leave them blank and use these two rules:

• All series are infinite from 0 on up.
• The series variable will be "i".

Definitions
We will denote by (not the standard notation, vBulletin ain't TEX)
(a; n)
with a e C and n e Z >= 0, to be the shifted or rising factorial which is defined as
a (a + 1) (a + 2) ... (a + n - 1) or P (a + i)
where the product is over i from i = 0 to i = n - 1. By defintion we take (a; 0) = 1 for all a. Notice that unlike the normal factorial, a may be any complex number.

Now you can build infinite series with these things. One defines a hypergeometric series pFq(a1, a2, ..., ap; b1, b2, ..., bq; x) as
S ( [(a1; i) (a2; i) ... (ap; i)] / [(b1; i) (b2; i) ... (bq; 1)] x^i / i! ).
In other words, it is a sum of terms each with p rising factorials in the numerator and q rising factorials in the denominator all multiplied by x to the ith power divided by the normal factorial of i (which can also be represented as (1; i)). All the a1, ..., bq are parameters to the function just like x. There are many common functions that have representations as hypergeometric functions. For example, the simplest hypergeometric function, 0F0(x), is just the common exponential function (to Euler's base).

On a different line of definitions, say you have a given series
Q = S f(i, ...)
over a function f that takes the series variable i as one of its parameters (it may have other parameters). Now I will make an exception to my previous series rules and give the range of the series variable i explicitly after the sum.
We define the (m, n) - multisection of the series Q over the basis f as the sum
S f(i, ...) (where i = m (mod n) and i >= 0).
In other words, the sum is over all nonnegative i that have remainder m when divided by n. For example, look again at the exponential series with f(i, x) = x^i / i!. The (0,2) - multisection of the exponential function over f is then
1 + x^2/2! + x^4/4! + x^6/6! + ...
which is just the hyperbolic cosine. Similarly, the (1, 2) - multisection is just the hyperbolic sine.

Problem
Express the general (m, n) - multisection of the exponential function over the basis f given above as a hypergeometric function. The hypergeometric function's parameters may vary with m, n and even the numbers p and q of rising factorial terms can vary with m, n.

I will let you play around with it for a little while. Its really not a difficult problem once you realize that you need a couple of relations for (a; n). I'll give a hint in a day unless it is solved or somebody says they don't want the hint because they are making progress and don't want it spoiled.

Bravo Galathaea for taking the initiative. SeventhStar has been very lax in his responsibilities.

Nice Question. How do you create the mathematical notation with vB? Could you direct me to a link with the needed keywords?

I actually just use [font=symbol]S[/font] for the sigma and with capital P for pi. I also increase the size. To check out how people do things that I haven't seen before, I usually will hit the quote button to see the whole vBulletin commands used. There's a few that aren't documented in the help, which is annoying. Sam Hobbs has a thread in the testing forum where I recently saw another way to get extra characters.

ooops I'm really sory guy. I forgot abou this thread :rolleyes:

(NOW WHEN THE SUBJECTS OF THE CODEGURU MAILS ARE OFF I AM I LITTLE CONFUSED :mad: )

But galathaea's got it (first). So he got the point and the right to ask (which he did).
I've been trying to perform a general counting of the points :) why dindt you do it before me?...

So if you'd let me ask I have a question that would be next on the thread:

Who has how many points in this thread ( a tough one :D ) ?

oh and think that we should stop with this strictly mathematical questions... They are not nice (:D)
They scare people!
And most importantly I, personally dont have time to think over them. Try simpler (as a condition) questions. If it hard but it's said in a simple way I would remeber it and think over it. I may be too lazy to write down these LOOOONG questions :rolleyes:
If anyone dissagrees let me here his point.

You will need to figure out formulas for:
(a; x + y) = ? (something in terms of rising factorials without sums)
(a; xy) = ? (something in terms of rising factorials without products)

If you look at the (m, n) - multisection sum given in a slightly different way, you get to a point where you have to apply these two transformations and right away a little touch up gives you the hypergeometric series. Its really quite a short manipulation once you figure out how to apply the the two formulas above.

The harder forrmula to figure out is the product, but I'll give a graphical hint for multiplying by 2 to get you started on that one.

Code:

---

............
||
\/
. . . . . .      . . . . . .
(and shrink the spaces by extracting a coefficient)

---

I'll just wrap this one up so we can move on to a different question. First the addition formula follows from breaking up the multiplication chain. Its fairly easy to see visually:
...........
or written out:
(a; i + m) = (a; m) (a + m; i)

The multiplication formula takes a little more consideration, but it follows when you realize a series of facts on the multiplication chain. First, since the chain is of length a multiple of the multiplier, you can break the chain apart into separate multiplier number of chains. The particular choice one wants to make is to keep the distance between elements the same for each chain, and picking the chains in an interlocked fashion like the visual clue given accomplishes this constant size difference (for the clue, it is 2). Notice how this decomposition of the chain is similar to how one may take a buffer and view it as an matrix (vector of vectors) in a column decomposition. Now notice that if you were to take a normal rising factorial chain (which has common difference 1) and multiply each of the elements by some number n, you get a multiplication chain with common difference n. So you can do this reversed and extract a coefficient from each term of a chain with some common difference not 1 and you get a standard rising factorial multiplied by a coefficient. Doing this for the separate chains when you break apart the multiplication factorial gives you this formula:
(a; n i) = n^(n i) P ((a + j) / n; i)
(for j from 0 to n - 1)

With thos two formulas, its basically two steps to the solution, with a little cleaning up each time. I'll break it up somewhat to make it clear what is happening each step.
S x^i / (1; i) (with i = m (mod n), i >= 0)
||
|| (addition formula and switch summation range)
\/
S x^(i + m) / (1; i + m)
= x^m / (1; m) S x^i / (1 + m; i)
(with i = 0 (mod n), i >= 0)
||
|| (multiplication formula and final range switch)
\/
x^m / (1; m) S x^(n i) / (1 + m; n i)
= x^m / (1; m) S (x^n)^i / [n^(n i) P ((1 + m + j) / n; i)]
= x^m / (1; m) S ((1; i) / P ((1 + m + j) / n; i)) ((x / n)^n)^i / (1; i)
= x^m / (1; m) 1Fn(1; {(1 + m + j) / n}j; (x / n)^n)
(with i >= 0; 0 <= j < n and {...}j meaning the set over j )
Notice how the two manipulations were just what was needed to get the right hand side to simplify to i when we switched the bounds over to that of standard hypergeometric form.

Now this is a simple little series of transformations that can be applied to any hypergeometric series to produce its (m, n) - multisection. And since there is an alternate form for expressing multisections in terms of the function's value on the n nth-roots of unity when the basis is equivalent to the exponential basis x^i (as the hypergeometric functions are), you have an entire class of hypergeometric identities which form a particular basis of identities on which one can classify a lot of information on the standard identities. Additionally, the (m, n) - multisection of the exponential series over the given basis obeys the function d^n / (dx)^n = id, so we have a particular basis over differential equations of order n much like one uses normal fourier analysis over the trignometric decomposition to get the two element symmetries and related information. Also, one has almost identical ability to do these transformations on basic (or q-) hypergeometric series, and filling in the details gives all sorts of relations in group theory over this really nice basis of relations. Finally, because the rising factorial and hypergeometric functions in general have formulations in terms of integrals, you get integral relationships over this entire basis.

So you can see that this little problem can be used as the starting point for an exposition into many pieces of analysis, combinatorics, and group theory. It also has the appearance of an identity probably found by thousands of undergrads or grads each year when they first start messing around with hypergeometric series. However, although I haven't made a detailed search yet, I have yet to see it actually mentioned in any of the standard texts like the various Dover textbooks, Andrews/Askey/Roy, the various works of Gauss, Jacobi, etc. An ex-professor of mine had mentioned that she thought she had seen an exposition of these bases under multisections and the theory, but she was quite busy and I have yet to find any references, though many special cases of the results are quite classical and it seems likely Gauss would have known of these results since the multiplication theorem is due to him (though I haven't seen explicitly the addition formula, it is obviously used in many proofs, often too obvious to be spelled out as a theorem). I spent a summer working out the various directions this could go a few years ago because I felt I could really use this basis well as a stepping stone into learning about all the areas mentioned. There are some very strange looking inversion requirements as well, and this whole area still interests me intensely. I now use alot of the experience gained from studying this one problem and its ramifications to work out many of the entires in Ramanujan's Notebooks as published by Berndt as exercises and try to figure out related problems for myself. I thought it would be nice to introduce here and think that anyone following this thread that wants to get some really nice exercise in the fields mentioned should follow through with the calcuations. I really think it can be quite an educational experience, and it is quite interesting once you get started.

Well, SeventhStar mentioned that he would enjoy it better if the questions weren't quite so in-depth so everyone can work on them more easily without needing to get all the definitions straight. I've been working more towards the in-depth questions with the possibility of extending the knowledge learned over various fields of knowledge, as I though that would be a better way to spend my time. Maybe there is a point somewhere in the middle that we can all work on and enjoy, or maybe some other format entirely would be better? I'll just leave this thread open, and the next question can be proposed in the format that the posting person desires, and we'll see where we can meet :)?

I will give this thread a brief resuscitation attempt. I'll just throw out a problem and let it sit. No time limits. No hints unless someone is actually interested and asks. Just a little problem. Its a mathematical problem, but it probably has no need for definitions to anyone who has followed this thread as it is covered usually at the high-school level, so maybe it falls within SeventhStar's criteria. I just don't like to see this thread die without an open problem...

Lets say you are given two vectors (3 dimensional) A and B and a constant b. Now, you are given the information that

A x X = B (x is cross-product)
A . X = b (. is vector dot product)

Solve for X in terms of A, B, b, and the magnitude of A.

Try it out. There are nice little pieces to this puzzle, and it doesn't involve any higher mathematics. If you understand the problem, you'd understand the solution.