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March 13

Generating random numbers

How would I generate random numbers with a distribution matching Zipf's law? --Carnildo (talk) 04:18, 13 March 2008 (UTC)

Could you start with a probability distribution that is linear ie each value occurs equally often and scale that to the zipf distribution? Would you need help scaling the fnuction?

Can it be assumed that you already have a method of generating random numbers in general?87.102.8.240 (talk) 09:10, 13 March 2008 (UTC)

According to the classic version of Zipf's law, the number of occurrences of the word that has rank k in a large corpus of words is proportional to 1/k, which means it is of the form c/k for some large constant c. But actually the value of c depends on the size of the corpus, where c is the number of different words in the corpus, and the size n of the corpus is roughly n = c·log c. Therefore it makes more sense to generate a sequence of numbers for which each initial segment obeys Zipf's law.

In pseudocode:

Let z₁, z₂, ... be the sequence to be generated.

Set c to 1

For n from 1 to whatever:

Set ν to 1/((log c)+1)

Select one alternative from:

(A) with probability ν:

Set z_n to c

Set c to c+1

(B) with probability 1 − ν:

Pick a random number uniformly from the range {1 ... (n−1)}

Set z_n to z_r

Obviously the beginning will have few different numbers, so you may want to discard a large initial part. It is then not actually necessary to store the segment to be discarded itself, as long as the number of occurrences of each value is kept track of. --Lambiam 09:58, 13 March 2008 (UTC)

The above is surely right - but looked a little mysterious to me - here's another version of 'psuedo-code' - in this case I don't explicitly normalise the probabilities but instead generate a sum "Total" of 1/k and find the position of a random number between 0 and "Total" ie the first 'word' is between 0 and 1/1, the second between 1/1 and 1/1+1/2 etc

For different functions simply change all instances of 1/value to the new thing eg 1/value^s I've marked these positions with a ^*

n=number of words

Total = Sum to n of 1/k ^* !ie 1/1+1/2+1/3 etc

LOOP 1

R=Total x RND() ! ie Generate a random number between 0 and Sum total called R {RND() is a random function between 0 and 1} !

Length=0

m=1 !a loop counter

LOOP 2

Length=Length+1/m ^*

if R<=Length then print/output "m"  ! m is the m^th word ! : GOTO LOOP 1 !start again!

increase m by 1 : goto LOOP 2

Just added in case others found the first example a little confusing.. apologies to any offended by the GOTOS.. the !!'s are comments. not factorials!87.102.8.240 (talk) 10:57, 13 March 2008 (UTC)

A counting variable can be added so that the program would stop after a certain number of random values had been produced. Would suggest adding this inside LOOP 1 eg Count=Count+1:If Count=10000 then END/STOP>87.102.8.240 (talk) 11:03, 13 March 2008 (UTC)

Note - what I've done here shares some similarities with Arithmetic coding or Entropy encoding - at least in small part.87.102.8.240 (talk) 11:15, 13 March 2008 (UTC)

(it could be extended to non-zipf distributions that have finite states from a single event, also the above could be speeded up slightly computer wise if that was neccessary - ask if you want further details on these.)83.100.138.116 (talk) 16:28, 13 March 2008 (UTC)

Financial calculator question

I have a financial calculator question. I have the APY for a Certificate of Deposit; I am looking for a calculation that will give me its dividend rate if the dividends are paid Quarterly or if they are paid monthly.

For example, for a 12 month Certificate the APY is 3.25% what is the calculation to find out what the dividends are? —Preceding unsigned comment added by 207.109.247.177 (talk) 17:58, 13 March 2008 (UTC)

Comment: The poster is referring to the annual percentage yield. Pallida  Mors 19:16, 13 March 2008 (UTC)

Different institutions may use different definitions of the Annual Percentage Yield, which may be different from the effective rate, the Annual percentage rate, because certain costs have not been accounted for yet. Assuming the formula in our article holds for the definition actually used, and assuming dividend rate may be equated with the nominal interest rate, then:

• for monthly dividend rate (12 periods) you get over a year:

  $12\times \left(\left(1+{\frac {APY}{100\%}}\right)^{\frac {1}{12}}-1\right)\times 100\%=12\times \left(1.0325^{\frac {1}{12}}-1\right)\times 100\%=12\times 0.00267\times 100\%=3.20\%;$

• for quarterly dividend rate (4 periods) you get over a year:

  $4\times \left(\left(1+{\frac {APY}{100\%}}\right)^{\frac {1}{4}}-1\right)\times 100\%=4\times \left(1.0325^{\frac {1}{4}}-1\right)\times 100\%=4\times 0.00803\times 100\%=3.21\%.$

Note the caveats. Your mileage may vary.  --Lambiam 19:18, 13 March 2008 (UTC)

PDE -> ODE

Are there a set of "standard" methods one could try to use to transform a PDE in several variables to a set of coupled ODEs? —Preceding unsigned comment added by 12.196.44.226 (talk) 20:22, 13 March 2008 (UTC)

maybe separation of variables#Partial differential equations? --Spoon! (talk) 03:49, 14 March 2008 (UTC)

I need help with pi & circumference

If you calculated the circumference of a circle the size of the known universe, requiring that the answer be accurate to within the radius of one proton, how many decimal places of pi would you need to use?

 A. Two million
 B. 39
 C. 48,000
 D. Six billion

Thanks in advance.207.69.140.24 (talk) 21:48, 13 March 2008 (UTC)

I seem to recall the answer is 39, though I don't remember why. Strad (talk) 21:56, 13 March 2008 (UTC)

Assuming you’re calculating it from the radius of the known universe I think you could find an upper bound by dividing the smallest distance in question (radius of a proton) by twice the largest distance in question (radius of the universe), and using pi with as many decimal places as this number: because then would be accurate to the radius of a proton when multiplied by twice the radius of the known universe. GromXXVII (talk) 22:11, 13 March 2008 (UTC)

[ec] According to Observable universe, its diameter is $9\cdot 10^{10}$ ly, so its circumference is $3\cdot 10^{11}$ ly or $3\cdot 10^{24}$ m. The radius of a proton is roughly $10^{-15}$ m. The ratio is roughly $10^{39}$, thus roughly 39 digits of π are required. But don't let anyone fool you into thinking that this means that more digits are not required for science and applications. -- Meni Rosenfeld (talk) 22:18, 13 March 2008 (UTC)

A question similar to this has been asked before. It was to find the circumference of the known universe to the accuracy of a planck length.

And the answer is that 62 digits of Pi was required, preferably 64 digits. I'll see if I can dig up a link.

http://en.wikipedia.org/wiki/Wikipedia:Reference_desk_archive/Science/2006_September_27#How_many_digits_of_pi_for_the_known_universe.3F

202.168.50.40 (talk) 00:08, 14 March 2008 (UTC)

We have no data to base the calculation on, so knowing the decimal expansion of π to many places is not going to help. Even if we knew the radius of the observable universe to within the radius of a proton, and we could halt the universe so that it stopped expanding while we are doing the calculation, we don't even know if the universe is flat.  --Lambiam 08:40, 14 March 2008 (UTC)

Well, the question didn't ask about the circumference of the observable universe, but rather of a circle the size of it. I guess what this means exactly is open to interpretation, but I think "fix a number of arbitrarily high precision and magnitude on the same order as the observable universe, and use it as a radius" is a valid one. -- Meni Rosenfeld (talk) 10:38, 14 March 2008 (UTC)

See Cosmic View: The Universe in 40 Jumps by Kees Boeke - a classic. Gandalf61 (talk) 14:21, 14 March 2008 (UTC)

March 14

Happy Pi Day

Just wishing all of you a happy Pi Day! Please celebrate responsibly. --Kinu ^t/_c 04:12, 14 March 2008 (UTC)

What should I do at 1:59:26? I've only got an hour! HYENASTE 04:22, 14 March 2008 (UTC)

Um... run around in a circle? —Keenan Pepper 05:54, 14 March 2008 (UTC)

Just returned from my circle! Happy Pi Day!!! HYENASTE 05:59, 14 March 2008 (UTC)

getting dizzy! this is so exciting87.102.83.204 (talk) 10:05, 14 March 2008 (UTC)

Damn, I've just fallen over.--88.109.224.4 (talk) 13:41, 14 March 2008 (UTC)

Happy 3.14... Day! --Mayfare (talk) 15:19, 14 March 2008 (UTC)

I've just drank 3.1 pina coladas oh goddd.... Damien Karras (talk) 15:23, 14 March 2008 (UTC)

I've never liked the idea of having a celebration based on an arbitrary and unnatural number system. What say we move pi day to the third of July (or the seventh of March for transatlantic types)? Algebraist 15:48, 14 March 2008 (UTC)

Let's just make every day pi-day see Big yellow disc in sky87.102.83.204 (talk) 16:49, 14 March 2008 (UTC)

You may be an algebraist, but is that an excuse for taking a rational approach to pi day?  --Lambiam 16:56, 14 March 2008 (UTC)

How are continued fractions less transcendental than decimals? And that username's out of date: I'm currently applying for a DPhil in logic. Algebraist 20:54, 15 March 2008 (UTC)

That explains a lot.  --Lambiam 07:27, 16 March 2008 (UTC)

"Give three point one four one six cheers for the Science Officer with pointed ears." —Tamfang (talk) 00:24, 19 March 2008 (UTC)

Percentages

Here I am at a very advanced age totally ashamed to say that I cannot work out percentages! Please advise me how to work out the percentage of one sum against the other; for example what percentage has my pension increased from last year's income to this year's. How to I do that ? Thanks in anticipation.--88.109.224.4 (talk) 08:47, 14 March 2008 (UTC)

Simple a percentage is one over another times 100.

So if income was 110 this year, and 95 last year this gives (110/95)*100 = 115.8%

So the increase is 15.8% (since 100% equals 1:1 ratio or no change I subtract 100%)

Or if my wage (77) increases by 2% my new wage is 77 + ( 2/100 x 77 ) = 77+1.54 = 78.54 —Preceding unsigned comment added by 87.102.83.204 (talk) 10:05, 14 March 2008 (UTC)

Did you check out the article Percentage?  --Lambiam 16:11, 14 March 2008 (UTC)

Thank you both, esp. Lambiam. The article is excellent and brings me up to speed. Thanks again.--88.109.224.4 (talk) 16:57, 14 March 2008 (UTC)

Also if you're struggling to do it 'mentally' (i.e. without a calculator/spreadsheet) I find it easiest to break things int0 10% and 1% then just do simple addition...E.g. If I want to know what 22% of 721 is I just go.... ok 10% is 72.1 and 1% is 7.21...I have 2 x 72.1 = 144.2 and then 2 x 7.21 so that's 14.42 add the two together and I get 158.62 which a quick google search (http://www.google.co.uk/search?hl=en&q=22%25+of+721&btnG=Search&meta= suggest is correct. This is particularly useful as a technique when you just want to get a ball-park amount really quite rapidly. ny156uk (talk) 17:50, 14 March 2008 (UTC)

Another thing. A lot of people get confused or thrown off by the percent symbol (%). Whenever you see that symbol, you can erase it altogether ... and simply replace it with the fraction 1/100 or, if you prefer, the decimal 0.01. That is, the % symbol simply means to take whatever number you are dealing with and to multiply it by the fraction 1/100 or (equivalently) the decimal 0.01 -- whichever you prefer or whichever is easier in a given situation. Thus:

• 7% means ( 7 * 1/100 ) = 7/100 --- if you need / want the fraction version of the answer
• 7% means ( 7 * 0.01 ) = 0.07 --- if you need / want the decimal version of the answer

Thanks. (Joseph A. Spadaro (talk) 22:25, 14 March 2008 (UTC))

Group Theory

The theory of groups interest me. I have a knowledge of mathematics up to:

a) Complex number arithmetic

b) Basic differentiation / integration

c) Basic knowledge of matrices (inverse, determinants)

d) Basic Logic and set theory

I understand the brief introductions and historical accounts of its development; and I'm tempted to pick up a book- although I don't know if I'm qualified to handle even the simplest introductory text.

It's basically a crude engineering mathematics background. Can you suggest a path of study to lead me up to investigations into group theory? Can I step into it with the basic knowledge I have or would you suggest deep study into which particular areas of mathematics? —Preceding unsigned comment added by 81.187.252.174 (talk) 14:12, 14 March 2008 (UTC)

Basic set theory is all you need, really. Find a book called something like "A first course in groups" and see how you find it. At a basic level, groups are pretty easy to understand. They are just sets with a way of combining 2 elements to get another element in a well-behaved way (for example, adding 2 integers to get a 3rd integer, or reflecting a polygon and then rotating it and realising that it's the same as having just reflected it in a different axis). --Tango (talk) 14:26, 14 March 2008 (UTC)

A textbook that is not expensive (used) is An introduction to the theory of groups by George W. Polites.[1] I don't know the book, so I can't vouch for it, but it is thin (80 pages), and if it is a mismatch to your needs, at least you did not waste lots of money.  --Lambiam 16:05, 14 March 2008 (UTC)

To hell with "not expensive"; here are three free online books for you to choose from: [2] [3] [4] —Keenan Pepper 21:59, 14 March 2008 (UTC)

Origin of asymmetry in conformation of symmetrical shape in a higher dimension (+ origami!)

Ok, this one has me totally stumped, and bear with me, as I'm not sure I'll explain it wonderfully well. There is a certain "mathematical" origami model, called a hyperbolic parabola (or paraboloid). Which is folded using a crease-pattern that is four-way symmetric under both reflectional and rotational transformations. (See: http://www.math.lsu.edu/~verrill/origami/parabola/ )

Now, here's the weirdness: when the crease-pattern is concertina-ed (fanned-up), the model assumes the three-dimensional shape for which it is named -- which is not completely symmetrical! The shape once conformed possesses chirality, or handedness, which is not possessed by the tetrahedron into which it will fit perfectly. Where does this rotational asymmetry come from? It is not there in the two-dimensional crease-pattern, but somehow emerges from the conformation into three dimensions, despite there being no way to distinguish between what is done to any four-way division of the model.

Can someone explain this puzzling phenomenon? (Preferably without using arcane technical terms). Many thanks, 85.194.245.82 (talk) 20:47, 14 March 2008 (UTC)

(original poster): Still be not-overly-confident of how well I explained my confusion, here's an addendum I just thought of: from the perspective of the paper as it is twisting into three-dimensions, both of its (indistinguishable) diagonals curve into circular arc-segments, one "up", the other "down" -- the puzzle is what logic does the paper use in conspiring with three-dimensional space to decide which of these diagonals goes up and which goes down?! 85.194.245.82 (talk) 20:53, 14 March 2008 (UTC)

I must admit I had no idea what you were talking about until I actually made one of these. Do you have one on hand? If so, try this: grab two opposite sides of the square piece of paper (which are two opposite edges of the tetrahedron) and twist in such a way as to flatten out the square. Then keep twisting in the same direction far past that point. Try to do it in one smooth motion. If you do it right, the thing will flip inside out and become an identical shape, but this time the diagonals go the opposite directions. So really it's not that a symmetrical pattern leads to an asymmetrical shape, it's that a symmetrical pattern leads to two possible stable shapes which are opposites, and they're symmetrical if you consider them as a pair. In between them is an unstable equilibrium which is a corrugated flat square (it's unstable because you effectively give the paper a negative Gaussian curvature, so it doesn't like to be flat anymore). Neat. —Keenan Pepper 21:41, 14 March 2008 (UTC)

Technical nitpick with one of your statements: The shape it makes is not chiral. It has point group $D_{2d}$, which is a proper subgroup of the point group of the flat square you start out with ($D_{4h}$), so it does have "less symmetry" in a specific sense, but it still has mirror symmetry, so it's not chiral. —Keenan Pepper 21:51, 14 March 2008 (UTC)

Hah, just realized I used many "arcane technical terms". Sorry about that. I'll be happy to explain them to you. —Keenan Pepper 21:53, 14 March 2008 (UTC)

Aye, empirical experimentation shortly after framing the question lead me to the "the paper does it" conclusion you point to, which at first made me think, "awww, reality's boring", as I was hoping for some kind of crazy metaphysical implication about the non-independence of spacial dimensions relative to one another (I have an overactive imagination that way). But then, I thought about it some more and had a little discussion on IRC, and decided it was still a very interesting phenomenon despite the "choice" of axis-curving resulting from environmental factors. Interesting because a purely geometrical consideration lead to an amplification of information. That is to say, although really only one bit of information is added to the system in choosing the subset of the point group, the effect is to cause a large difference in eventual topology. Someone said this was an example, or at least analogous, to the theoretical-physical concept of spontaneous symmetry breaking (I personally think the term "spontaneous" is a terrible misnomer, as it implies that the information arrives randomly, ex nihilo, from nowhere, which is not how mathematics or the universe works, no matter how hard we try to pretend otherwise). 85.194.245.82 (talk) 22:09, 14 March 2008 (UTC)

(thanks for the terminology info, btw. I had a feeling chirality wasn't right, but couldn't think of any other asymmetry. Is there a specific term for it? Will look at the article you linked to.) 85.194.245.82 (talk) 22:09, 14 March 2008 (UTC)

The phenomenon is an instance of spontaneous symmetry breaking. An even much simpler example is when you put a small ball (as from a ball bearing) exactly in the middle on top of a larger ball. Perfect symmetry. Turn around and the little ball will have dropped off, in an asymmetric way (since all ways of dropping off are asymmetric). If it does not drop off instantaneously, you may have to wait for a butterfly or Heisenberg to assist the process.  --Lambiam 23:43, 14 March 2008 (UTC)

March 15

A measurable set?

Let $(X,\mu )$ and $(Y,\nu )$ be measure spaces and $\{A_{k}\},\{B_{k}\}$ sequences of sets of finite measure in X and Y respectively. Let the "rectangles" $R_{k}=A_{k}\times B_{k}$, and assume that

$\sum _{k=1}^{\infty }(\mu \otimes \nu )(R_{k})<\infty .$

Let $R_{k,x}=\{y\ |\ (x,y)\in R_{k}\}$ and

$T_{n}=\{x\ |\ 1/n\leq \sum _{k}\nu (R_{k,x})\}.$

Why is it obvious that T_n is measurable?  — merge 17:01, 15 March 2008 (UTC)

Well, $R_{k,x}$ is just B_k if x is in A_k, and empty otherwise. So $\sum _{k}\nu (R_{k,x})$ is the some of the measures of the B_k such that x is in A_k. Thus whether x is in T_n is determined by which of the A_ks x is in, and T_n is a union of intersections of the A_ks. Algebraist 17:45, 15 March 2008 (UTC)

Oh, I think I see how it works out. If $\{f_{k}\}$ is a sequence of nonnegative measurable real-valued functions and α is a real number, the sets

$V_{k}=\{x\ |\ f_{1}(x)+\cdots +f_{k}(x)>\alpha \}$

are measurable, and so are

$W_{\alpha }=\bigcup _{k}V_{k}=\{x\ |\ \sum _{k=1}^{\infty }f_{k}(x)>\alpha \}$

and

$\bigcap _{j}W_{\alpha -1/j}=\{x\ |\ \sum _{k=1}^{\infty }f_{k}(x)\geq \alpha \}$.

In this case $f_{k}(x)=\nu (R_{k,x})=\chi _{A_{k}}(x)\nu (B_{k})$ and $\alpha =1/n$.  — merge 22:30, 16 March 2008 (UTC)

March 16

grasping math to an intermediate level

hi, I like to get to an intermediate level of understanding of different subjects and I want to get a decent grasp of mathematics, but I'm having a hard time with it. I have a BS in math and an MS in statistics. I feel as if I'm wandering blindly in a mathematical universe, feeling the obvious, stumbling over the less-obvious, but without confidence in my travels. When I started to learn typography, say, I had that blind feeling at first but with study soon found the fundamentals and understood the basic ideas and questions in that field.

I want to get to that level of confidence with math, but I'm not sure how to get there. Obviously it's a big world out there, but is there a field that I should concentrate study on (analysis maybe?) to get an intermediate grasp (i.e., not expert and not blind beginner) ? thanks 72.150.136.11 (talk) 14:24, 16 March 2008 (UTC)

In all honesty, it's pretty much impossible to get an intermediate grasp of everything, since there is so much. I mean, I am doing an MMath, and I'd say that by doing so I was intermediate in the subjects I do, but there are others I simply have no clue about - topology, fluid mechanics, relativity, anything related to statistics... maths is so vast, and a lot of it is so remote from anything you do at school as to be an entirely different subject. Group theory is probably a useful thing to try and understand, as is basic vector calculus and mathematical analysis. I personally like complex analysis and combinatorics. Of course, combinatorial game theory is also fun to try. -mattbuck (Talk) 15:00, 16 March 2008 (UTC)

Linear algebra, while rather boring, is also a useful subject to have a working knowledge of - it seems to pop up all over the place. However, if you've got a BS in Maths I would expect you've already covered the basics of all these areas. If you want to do more Maths, you need to specialise, there's no way to have a better than undergraduate understanding of more than one or two areas. --Tango (talk) 15:19, 16 March 2008 (UTC)

I disagree with the last statement, there are many researchers who are active in several quite distinct areas. It's not easy, of course.

Mathematical logic is very important to know the basics of (say, up to the level of understanding Godel's theorems). Likewise for topology, which at the very least one needs to know to the level of feeling the difference between metric features and topological ones. I recall that when I have been applying for a math MSc, virtually all schools frowned upon me not having taken a topology course in my BA. Fortunately I had learnt the material independently - my peers who hadn't had some trouble with many of the courses.

Of course, there's the obvious stuff, like analysis (real, complex, functional, differential geometry...), algebra (groups, rings, fields, Galois theory...), discrete (combinatorics, graphs, ...), and of course computer science (computational complexity, information theory, machine learning, ...) and applied math (numerical methods, mathematical physics, game theory...). Disclaimer: This categorization is based on my own personal view and the subjects are not equally divided. -- Meni Rosenfeld (talk) 17:37, 16 March 2008 (UTC)

Well, I guess you run into trouble defining an "area" of maths. There are plenty of people that do research in what would usually be considered distinct areas, but they're usually working in some kind of overlap between them - pretty much all branches of maths overlap somewhere. Are there many people that have a solid understanding of the whole of several areas? I would say most understand just the parts of those areas that are relevant to their research. It does depend very much on how large you consider an area to be (is algebra a single area or is ring theory a distinct area to Galois theory? - obviously, you can understand both ring theory and Galois theory to a high level, but understanding the whole of algebra to a high level is pretty challenging without even trying to start on any non-algebraic areas). Put simply: I think we're both right for appropriate definitions of "area". --Tango (talk) 18:12, 16 March 2008 (UTC)

Thanks for these answers. I thought it might be impossible for a not-particularly-gifted person to walk with confidence in the math spaces. I just hate that feeling of flailing around blindly, holding only to those proofs I really really understand. Sometimes I feel like I'm missing a sense that some others have; which I suppose is true for everyone on some level. I'll explore mathematical logic and topology and see how that goes. thanks everyone 72.150.136.11 (talk) 19:00, 16 March 2008 (UTC)

While we all strive for perfection, nobody gets there. However far you go, there's always going to be more you don't understand, so you'll probably always be walking blindly, you'll just be walking in more advanced parts of the world of maths. I think the important thing is to enjoy it - study the areas of maths that interest you and that you find fun. There's little point studying anything else. --Tango (talk) 19:31, 16 March 2008 (UTC)

From what you say, I'm not sure that what you're looking for is so much a subject as a teaching style. You need some reintroductions to subjects you already know, from a different viewpoint. At the moment, I don't know of any better way to find good sources than to read things randomly and hope to hit gold (as I have occasionally), but someone else on the board might. If you want to grasp the material, try some popularizers, avoiding as best you can those who dumb it down instead of actually making it clearer. I was just reading Indra's Pearls, for instance, which gives some solid visual meaning to some pretty abstract techniques from several fields (analysis, group theory, linear algebra, complex analysis, maybe some others). If you want to understand the motivation, that is the goals, behind something, look into its history. Even the original papers on the subject, if you can get them. For instance, I've heard over and over again this same fable about complex numbers being the solution to the equation x^2+1=0, which ties in nicely to some other equations, like x^2-2=0 and x+1=0. It wasn't until fairly recently that I found out that complex numbers were motivated for the first time, after thousands of years of people intentionally ignoring their effects on quadratic equations, by the solution to the cubic equation. There are a lot of nice books that give the story behind that, not least the Ars Magna itself. Also, something that will always help you grasp a subject is to use it. Set your sights on a number of small, random goals, and see what you can do with them. Abstraction is bad on an empty stomach - you need a lot of bulk to soak it up. Once you can navigate real situations with confidence, the patterns you find among them will motivate and support abstraction. The exercizes in textbooks are a start, but don't rely on them for originality, follow your own curiosity. I'm trying to keep my studies broad-based, since I like them that way, but if you want to focus on a single subject for awhile, there's certainly plenty to choose from. It doesn't sound like you plan to use this study immediately, so focus your decision on how much you can learn about learning. It may be best to relearn something concrete as an introduction. Euclid's geometry, for instance, moving through some historical works like the Arithmetica Infinitorum or Newton's papers, then up through the crisis a century ago with Cauchy and Weierstrass et al, to get to analysis. An interesting stop you might take on the way is in Algebraland - most of calculus, as it applies to polynomials, can be derived without hardly touching limits. Black Carrot (talk) 02:28, 17 March 2008 (UTC)

I would have to agree with Black Carrot there, for me, my confidence in my mathematical ability comes from my thorough understanding of the basics (that is HS level and some college level) mathematics, I'm so comfortable with these rules, that I have little trouble knowing whether or not they apply in a situation I've never encountered in a text. Memorization of the basic rules to a very exacting degree helps with that a lot. And curiosity is definitely the best motivator, I have learned more from my own curiosity than from those who have taught math to me. (Mainly do to the fact that I wasn't challenged in HS math class even AP) 69.54.143.177 (talk) 00:22, 19 March 2008 (UTC)

I wouldn't recommend memorising the basic rules, it's much better to make sure you fully understand them and why they work. Once you've done that, you'll find you never forget them without having to make any actually effort to memorise them. (I frequently make mistakes in basic integration [as you may have seen lower down!], and that's because I didn't understand it when I first learnt it so tried to just memorise the methods, and ended up memorising them incorrectly and now get confused between my incorrect memory and my correct understanding and end up making mistakes - had I just taken the time to work through it and understand it, I wouldn't have the problems I do now.) --Tango (talk) 00:37, 19 March 2008 (UTC)

Just out of curiosity (which we have already established to be important), care to elaborate what is it that you have incorrectly memorized? -- Meni Rosenfeld (talk) 16:55, 19 March 2008 (UTC)

Local Sidereal Time Conversion

How do I convert local sidereal time to something that is more commonly used? Is this actually a measure of time? I read the article about sidereal time and it appeared to be an angle measurement, a measure of place. So when someone tells me "do this action at 13:20" local sidereal time, is this something that makes sense, or is it just astrological mumbo jumbo? And if it makes sense, how do I convert it to Eastern Standard Time or Hawaii Standard Time or GMT or anything else that is more commonly used? —Preceding unsigned comment added by RastaNancy (talk • contribs) 20:45, 16 March 2008 (UTC)

Sidereal time is measure of time, but it's not easy to convert to normal time. Regular time is determined by watching the Sun move across the celestial sphere, sidereal time is determined by watching the stars move - the stars move only due to Earth's rotation on it's axis, the Sun moves because of that and because of Earth orbiting it. Therefore, a solar day differs from a sidereal day (by about 4 minutes). This means that to convert from solar to sidereal time you need to know the date in order to work out how much they differ by. The easiest way to convert would be to find a converter online, I'm sure there are plenty. --Tango (talk) 21:42, 16 March 2008 (UTC)

Possibility of solving this equation

I have the following equation:

${\it {\lambda _{i}}}=c\int _{0}^{L}\!\left|-1+R\theta \ '\left(s\right)\cos \left({\it {\phi _{i}}}\right)\right|{ds}$

I can choose any values for ${\it {\phi _{i}}}$, and get the corresponding values of $\lambda _{i}$ through an experiment. The aim is to find the function $\theta \left(s\right)$ as accurately as possible. I am wondering if it is possible at all to do this. I was trying to expand $\theta \left(s\right)$ in a Fourier series with n terms (and n undetermined coefficients) and then using n different values of ${\it {\phi _{i}}}$ to get a system of n equations. However, it turns out that all these equations are linearly dependent. Note that ${\it {\phi _{i}}}$ is not a function of s. Any ideas on how best we can find $\theta \left(s\right)$ will be highly appreciated. Regards, deeptrivia (talk) 21:40, 16 March 2008 (UTC)

Do you have any additional knowledge about $\theta '$? If it is known to be monotonic then I have an idea which might work in some cases. The essence is to start with $\phi \approx \pi /2$, for which the absolute value can be solved; then decrease $\phi$ until you see a deviation from normal. The rate of deviation should give you the ratio $\theta '/\theta ''$ at the x-intercept. This might then be solvable. -- Meni Rosenfeld (talk) 22:56, 16 March 2008 (UTC)

Thanks, Meni. Although I can choose any values for ${\it {\phi _{i}}}$, I can't choose too many. Practically, n is two or three, at most six. $\theta '$ is smooth but not monotonous. Also, I don't need to find $\theta$ exactly, but just as best as I can. How best do you think can I find it by finding $\lambda _{i}$ for, say, n wisely chosen values of ${\it {\phi _{i}}}$ ? Regards, deeptrivia (talk) 23:13, 16 March 2008 (UTC)

PS: I think in my case it is safe to assume that

${\it {\lambda _{i}}}=c\int _{0}^{L}\!\left(1-R\theta \ '\left(s\right)\cos \left({\it {\phi _{i}}}\right)\right){ds}$

Then, since R and ${\it {\phi _{i}}}$ are not functions of s, we can directly integrate and get:

${\it {\lambda _{i}}}=c\left(L-R\cos \left({\it {\phi _{i}}}\right)\left(\theta \left(L\right)-\theta \left(0\right)\right)\right)$

Consequently, it appears that no matter how we vary φ, all we can find about θ is the difference between its values at 0 and L. Now, if we remove the assumption I just made regarding the absolute value, we can still break the integral up into a countable number of parts that can be directly integrated, and using a similar logic, we can argue that we cannot know anything more than θ(L) - θ(0) by varying φ. Am I right? Thanks, deeptrivia (talk) 23:33, 16 March 2008 (UTC)

It's certainly very hard to deduce information about $\theta$ in the general case, but it's not theoretically impossible. If you break up the interval, in some segments you will have $\theta (a)-\theta (b)$ and in some $\theta (b)-\theta (a)$, and when you add them up they don't cancel. Since changing $\theta$ can change the integral, knowledge of the integral can at least eliminate some possibilities. -- Meni Rosenfeld (talk) 00:15, 17 March 2008 (UTC)

Thanks. Am I at least right for the case where there is no sign change, becaause $R\theta \ '\left(s\right)\cos \left({\it {\phi _{i}}}\right)\ll 1$ and no need to break the interval? Regards, deeptrivia (talk) 00:47, 17 March 2008 (UTC)

Yes, indeed. That is what I called the "normal" case, where the integral is reduced to a multiple of $\theta (L)-\theta (0)$. Deviations from this tells you something about $\theta$, but how to extract anything useful from it, especially with limited observations, is a mystery. -- Meni Rosenfeld (talk) 06:32, 17 March 2008 (UTC)

Am I missing something? If $\theta \left(s\right)$ solves the equation, then so does $\theta \left(s\right)+C$, so you can only determine solutions up to an additive constant.  --Lambiam 17:17, 17 March 2008 (UTC)

Certainly, but it appears that it's rather difficult, if not impossible, to even determine it that far. --Tango (talk) 17:28, 17 March 2008 (UTC)

March 17

is mathematics subjective?

So my understanding is that math isn't falsifiable, because it doesn't make physical predictions. Is it then subjective?

For example, whether "something has been proved", is that a question of viewpoint, a subjective thing, or is it objective? —Preceding unsigned comment added by 79.122.42.52 (talk) 01:41, 17 March 2008 (UTC)

I can't speak for all of mathematics but probability is subjective.

The probability of an event depends on what the observer/questioner knows. So two different observers can assign two different probability to the same event if they have different knowledge. 202.168.50.40 (talk) 02:26, 17 March 2008 (UTC)

Mathematics is objective in the sense that all proofs are derived from a set of axioms, so unlike the physical sciences where different experiments can lead to different results, whenever you start from the same axioms you'll get the same result for your proof. That said, the choice of axioms is somewhat subjective, since there is nothing to say that axiom set A is any "better" than axiom set B, although set A may result in something more familiar such as the natural numbers. Confusing Manifestation(Say hi!) 03:26, 17 March 2008 (UTC)

Sets of axioms are necessarily completely different, according to Gödel's incompleteness theorems. Because there are an infinite number of sets, subjectivity can not exist. Mac Davis (talk) 05:06, 17 March 2008 (UTC)

Mathematics is falsifiable, in that guesses can be checked. That's what falsifiability, as a doctrine, is for in the sciences - make sure your clever guesses aren't dead wrong. In mathematics, what's called proof serves roughly the same purpose. We can check our guesses. Whether we can check the system we use to check our guesses is open to philosophical debate, as the system of empirical validation used in the sciences is. Maybe it works, but maybe we live in god's big toe and he's playing games with us. That's a different level of fact-checking, and one that even falsifiable theories can't pass. So, to answer your question, mathematics is subjective to roughly the same extent that science is. Black Carrot (talk) 06:14, 17 March 2008 (UTC)

Mathematics differs in subjectivity from other sciences in a quantitative, not qualitative, way. Mathematics certainly has a higher standard of derivation - while in science, repeating an experiment several times passes as "proof" for its result, in mathematics we prefer to formulate our statements in an abstract way and derive them in a way we consider "logical". But don't be fooled into thinking that mathematics is completely rigorous, objective and devoid of human weaknesses. Not so much because of the choice of axioms (since it is understood that establishing a theorem doesn't mean that it's absolute, but rather that it follows from the specified axioms), but because of the choice of phrasing and logical inference. We have rules like "if we know A and we know A->B then we can deduce B", but there is nothing objectively "true" about this rule, other than that humanity's experience over the eons seems to support it. To this extent, mathematics relies on empirical evidence the same way as other sciences, the only difference being the amount of expected evidence. -- Meni Rosenfeld (talk) 06:49, 17 March 2008 (UTC)

The axioms of the logic in use are, to me, just as much part of the "choice of axioms" as the more mathematical axioms. It turns out that these are deeply intertwined anyway; a lovely result I saw once was that DeMorgan's Laws as logical axioms are equivalent to axiomatically stating that maximal ideals of commutative rings are necessarily prime -- one statement appears purely logical, the other most definitely in the realm of math not logic, yet in practice they are equivalent axioms. What I'm getting at with this little digression is that I don't really see the distinction you make between "choice of axioms" and axioms of logic that we assume. -- Leland McInnes (talk) 14:19, 17 March 2008 (UTC)

As I understand it - something that is 'subjective' depends on who is examining it. So a good maths proof should not be subjective.

Really though 'subjective' and 'objective' have no meaning here ie all mathematical knowledge should be 'objective'. Subjective examples of mathematical things include: imperfect models of phyical behaviour, best estimates of statisical behaviour/probabilities - eg things that are (educated) guesse.

I suggest reading about 'subjective and objective' and forming your own opinion. Picked these definitions 'at random' from the net:

Subjective: characteristic of or belonging to reality as perceived rather than as independent of mind : phenomenal

Objective: of, relating to, or being an object, phenomenon, or condition in the realm of sensible experience independent of individual thought and perceptible by all observers : having reality independent of the minds objective reality

Emphasis mine.87.102.13.144 (talk) 11:59, 17 March 2008 (UTC)

Answer - when something has been proved mathematically that should be objective, and not subjective knowledge. (It's work noting that peoples perception of what the words 'subjective' and 'objective' mean can be a bit subjective...)87.102.13.144 (talk) 14:07, 17 March 2008 (UTC)

Yes, something that has been proved should be objective, in the sense that we would want it to be. Unfortunately, it is not, per the responses above. -- Meni Rosenfeld (talk) 15:06, 17 March 2008 (UTC)

Our choice of what to try and prove is very subjective, but the actual proof is objective. We don't prove XYZ, we prove that our choice of axioms implies XYZ. That choice of axioms is very subjective, but the implication is entirely objective (give or take Godel, anyway - even then, I don't think it's subjective, it's just not wholly reliable). So, I guess the answer to the question "Is maths subjective?" is that maths is a subjective collection of objective facts (there is more to maths than a collection of facts, of course, but for the sake of this discussion I think that definition will do). --Tango (talk) 15:59, 17 March 2008 (UTC)

I disagree. First, nobody proves theorems in a completely formal language. People use natural language to describe their proofs, and hope that others will interpret it as they meant and agree that each of their implications is indeed valid. Sometimes it works, but there is still a lot variance in people's interpretation of arguments in a language such as used in articles. Second, even if the proof was written in a completely formal language, it still relies on the person's ability to validate each step. This validation is also composed of steps that can only be described in a natural language ("look at this line, see what's written here, compare it to what you see there..."). Especially for a long and complicated proof (like most proofs will be when written in a formal language), the possibility of human error is huge. It's also not unthinkable for the same person to make the same error repeatedly because of a certain mindset he is in. Thus different people will report different views on the correctness of a proof. In fact, when given a certain proof, there is no way for us to tell if it is truly valid (whatever that means) or is it just that every person who has ever examined it happened to mistakenly accept it. Thus I reiterate my claim that everything in mathematics is subjective, but to a much lesser extent than some other sciences. -- Meni Rosenfeld (talk) 16:15, 17 March 2008 (UTC)

What you say is certainly true, but I'm not sure if that's really a form of subjectivity - if it is, then nothing is truly objective and the question is moot. --Tango (talk) 17:16, 17 March 2008 (UTC)

Personally, I do believe there is an "objective truth", but that anything a human can ever discover is only a crude approximation of it, and thus subjective. The original question is not moot, as a priori one might think that modern mathematics has, at least in some parts, achieved the holy grail of true objectiveness - my argument is that this is not the case. It's probably as close as we'll be for a while, though. -- Meni Rosenfeld (talk) 18:38, 17 March 2008 (UTC)

Simultaneous Equation

how to solve this simultaneous equation 4x+y+8=x²+x-y=2 —Preceding unsigned comment added by 60.48.198.225 (talk) 07:32, 17 March 2008 (UTC)

You have the two equations $4x+y+8=2\;\!$ and $x^{2}+x-y=2\;\!$. There is a simple operation you can take to get rid of the y's and have an equation with only x. Can you see it? -- Meni Rosenfeld (talk) 07:35, 17 March 2008 (UTC)

Infinitely Many Factors

Consider the ring of formal power series over something like the integers. Is there any power series that can be split into infinitely many (non-unit) factors? That is, is there one where the process of pulling off factors will never stop with some number of irreducible factors? Black Carrot (talk) 10:02, 17 March 2008 (UTC)

Maybe I'm missing some subtlety in the question, but isn't

$\sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots =x\prod _{n=-\infty }^{\infty }\left(1-{\frac {x}{\pi n}}\right)$

an example of a power series with an infinite number of irreducible factors ? Gandalf61 (talk) 10:44, 17 March 2008 (UTC)

Another example (unless I, too, misunderstand the question) is $\prod _{n=0}^{\infty }\left(1+2^{-n}x\right)=\sum _{n=0}^{\infty }{\frac {2^{n}}{\prod _{k=1}^{n}(2^{k}-1)}}x^{n}$. -- Meni Rosenfeld (talk) 10:50, 17 March 2008 (UTC)

Those are units: (1-ax)(1+ax+a^2*x^2+a^3*x^3...) = 1. You can always pull off infinitely many unit factors, even from something irreducible. Black Carrot (talk) 11:12, 17 March 2008 (UTC)

Oh, I see. No idea then. -- Meni Rosenfeld (talk) 11:22, 17 March 2008 (UTC)

Let's see if I have got this straight. You want an infinite number of factors each of which has, say, integer coefficients, is irreducible and is not a unit when considered as a formal power series over the integers ? Let's take out the largest power of x that we can, so we have an initial factor of x^k for some k (possibly 0). Doesn't this mean that the constant term of every other factor cannot be 0 (otherwise it is reducible) or +/-1 (otherwise it is a unit when considered as a formal power series) ? So you have x^k times an infinite number of factors each of which has constant term with absolute value greater than or equal to 2 - which seems to imply that the coefficient of the lowest power of x in the product is not finite, so the product is not a power series. Gandalf61 (talk) 11:29, 17 March 2008 (UTC)

What is meant by "non unit factor"?87.102.13.144 (talk) 11:47, 17 March 2008 (UTC)

In this context, a unit is something you can divide by without leaving the number system. For instance, in the whole numbers 1 is a unit, but anything bigger isn't. That's why 1 isn't considered prime - every number would have infinitely many prime factorizations, eg 5 = 5*1 = 5*1*1 = 5*1*1*1... You're right Gandalf, I hadn't thought of looking at it like that. So, does that mean that power series over the rationals also have finitely many factors? Since everything with a constant term would be a unit, and everything without could be divided by x. Black Carrot (talk) 12:04, 17 March 2008 (UTC)

Meni, should that be 2^n at the top of your last formula? Black Carrot (talk) 12:09, 17 March 2008 (UTC)

Yes, indeed. Fixed. -- Meni Rosenfeld (talk) 14:55, 17 March 2008 (UTC)

Was the first example given not an example? eg a power series over rational fractions expressed, but made of factors which are irrational (contain 1/pi)?87.102.13.144 (talk) 14:12, 17 March 2008 (UTC)

The original question needs to be made more precise. What do you mean by "something like the integers"? For example, are algebraic integers something like the integers? And wat about the complex numbers? Should the factors be over the same ring as the power series? Can the factors again be formal power series, or should they be polynomials?  --Lambiam 14:49, 17 March 2008 (UTC)

Gandalf has solved one form of the question: I believe his proof shows that if R is a Noetherian integral domain, then any element of R[[X]] can be expressed as a product of finitely many irreducibles in R[[X]]. Algebraist 16:09, 17 March 2008 (UTC)

math: polynomial as the difference of two squares

How do you determine if a polynomial is the difference of two squares? —Preceding unsigned comment added by 72.94.230.129 (talk) 14:08, 17 March 2008 (UTC)

Assuming that by "square" you mean the square of a polynomial, and that the coefficients are from a number system that is closed under division by 2, such as the rationals and the reals, any polynomial P can be rewritten as:

$\left({\frac {P+1}{2}}\right)^{2}-\left({\frac {P-1}{2}}\right)^{2}.$

 --Lambiam 14:33, 17 March 2008 (UTC)

I'm not sure that's what the OP means. We studied polynomials last year in algebra I. We learned the difference of two squares as $(x+S)(x-S)\,\!$, in which S is any number. This then takes the form $x^{2}-S^{2}\,\!$. So, to determine if a polynomial is the difference of two squares, just look for two square terms (such as $x^{2}-64$). The solution to a polynomial of the form $(x+S)(x-S)$ is $x=\pm {S}$. So the solution to $x^{2}-64$ would be $x=\pm {8}$.

Disclaimer: I learned these things last year so they are subject to possible error.

Zrs 12 (talk) 19:48, 17 March 2008 (UTC)

That's right, but the principle can be used for more complicated polynomials, as well. For example $x^{2}-y^{2}+8x+4y+12=x^{2}+8x+16-y^{2}+4y-4=(x+4)^{2}-(y-2)^{2}=(x+4+y-2)(x+4-y+2)=(x+y+2)(x-y+6)$. If you want everything to have integer coefficients, then I'm not sure there is a general algorithm to find out if you can write a polynomial as the difference of two squares, you just have to fiddle around with it and see what happens (you can try completing the square as a starting point). --Tango (talk) 20:08, 17 March 2008 (UTC)

We also have an article on this. See difference of two squares. Zrs 12 (talk) 20:06, 17 March 2008 (UTC)

Tango: A polynomial (over Z, say) is the difference of two squares iff it can be factorised with the sum of the (two) factors being divisible by 2 (i.e. corresponding coefficients in the factors have to be of the same parity). Thus a poly is a difference of two squares iff its reduction mod 2 is a square. Algebraist 23:25, 17 March 2008 (UTC)

Or rather not. We get an algorithm, anyway. There's an algorithm that factors a polynomial over Z into its irreducible factors, so we can just test all possible factorizations. I've no idea if this can be improved on. Algebraist 23:28, 17 March 2008 (UTC)

need a more indepth determination

Bold textHow do you determine if a polynomial is the difference of two roots? Thats how the question was brought to me.I don't quite get it either. —Preceding unsigned comment added by 72.94.241.142 (talk) 19:36, 17 March 2008 (UTC)

The difference of two roots? Well, assuming you mean square root, any number is a square root because it is able to be squared. Just go down a number line and mulitiply each term by itself. For example $S$ is the square root of $S^{2}$; 4 is the square root of 16. However, every number is not a square because not every number has an integer square root. I think the answer to your previous question above may be what you seek. Zrs 12 (talk) 19:55, 17 March 2008 (UTC)

Solve for y: x = y - 1/y

How can you solve the equation $x=y-{\frac {1}{y}}$ for y in terms of x? I know that the solution is $y={\frac {x\pm {\sqrt {x^{2}+4}}}{2}}$, and this works when you substitute it back into the original equation, but how can this be found analytically? (I used a calculator). --BrainInAVat (talk) 20:59, 17 March 2008 (UTC)

• Try multiplying both sides by $y$... you should then have a quadratic in terms of $y$ which can be solved using the "usual method". --Kinu ^t/_c 21:14, 17 March 2008 (UTC)
  • I'm at a loss, sorry. You get $xy=y^{2}-1$, but then what?--BrainInAVat (talk) 01:29, 18 March 2008 (UTC)

Take it all over to one side and you have a quadratic in y (just treat x like a number), you can then solve it using the quadratic formula and you'll get the answer you gave above. --Tango (talk) 01:53, 18 March 2008 (UTC)

If you are not familiar with it, you can take a look at out Quadratic equation article. -- Meni Rosenfeld (talk) 07:19, 18 March 2008 (UTC)

This can be a little confusing the first time you see it - here:

x=y - 1/y

xy=y² - 1 (equation 1)

y² - xy - 1 = 0

Now complete the square

(y-x/2)² - x²/4 - 1 = 0

As you should be able to see multilpling by y gives a quadratic in y (equation 1)

The solve the quadratic by completing the square - you can see that x is a parameter of the quadratic. Hopefully from the last equation you should be able to get the result.

Once you've got it, keep an eye out for equations of this type as it's a useful thing to remember.87.102.74.53 (talk) 11:35, 18 March 2008 (UTC)

March 18

Proof of Chain Rule

Let

$y=f(u)$ (1)

$u=g(x)$ (2)

where $f(u)$ and $g(x)$ are both differentiable functions. Then

${\frac {dy}{dx}}$ (3)

$=\lim _{\Delta x\rightarrow 0}{\frac {f(g(x+\Delta x))-f(g(x))}{\Delta x}}$ (4)

$=\lim _{\Delta x\rightarrow 0}\left[{\frac {f(g(x+\Delta x))-f(g(x))}{g(x+\Delta x)-g(x)}}\cdot {\frac {g(x+\Delta x)-g(x)}{\Delta x}}\right]$ (5)

$=\lim _{\Delta x\rightarrow 0}{\frac {f(g(x+\Delta x))-f(g(x))}{g(x+\Delta x)-g(x)}}\cdot \lim _{\Delta x\rightarrow 0}{\frac {g(x+\Delta x)-g(x)}{\Delta x}}$ (6)

Treat the arrow $\rightarrow$ as equal sign $=$. We can do the same operation on both sides of the arrow without changing the relationship

$\Delta x\rightarrow 0$ (7)

$x+\Delta x\rightarrow x$ (8)

Function $g(x)$ is continuous since it is differentiable. Apply $g(\cdot )$ to both sides of (8)

$g(x+\Delta x)\rightarrow g(x)$ (9)

$g(x+\Delta x)-g(x)\rightarrow 0$ (10)

Let

$\Delta u=g(x+\Delta x)-g(x)$ (11)

Replace (11) into (10)

$\Delta u\rightarrow 0$ (12)

Therefore

$\lim _{\Delta x\rightarrow 0}\cdot$  implies $\lim _{\Delta u\rightarrow 0}\cdot$ (13)

Replace (2) into (11)

$g(x+\Delta x)=u+\Delta u$ (14)

Replace (2), (11), (13) and (14) into (6)

${\frac {dy}{dx}}=\lim _{\Delta u\rightarrow 0}{\frac {f(u+\Delta u)-f(u)}{\Delta u}}\cdot \lim _{\Delta x\rightarrow 0}{\frac {g(x+\Delta x)-g(x)}{\Delta x}}={\frac {dy}{du}}{\frac {du}{dx}}$ (15)

Q.E.D.

Is the proof of chain rule above correct and rigorous? - Justin545 (talk) 06:25, 18 March 2008 (UTC)

There are some questionable details. First, if we want a proof we can consider "rigorous", we would want to avoid treating functions as quantities (e.g., u instead of $g(x)$) and using Leibniz notation (${\tfrac {du}{dx}}$). So as a first step you should try formulating the proof without using u or y, only f, g and their composition $h=f\circ g$ (equivalently, $h(x)=f(g(x))$. Second, the limit notation, $\lim _{x\to x_{0}}f(x)=L$, is one unit. You shouldn't take out the $x\to x_{0}$ and treat it as something that stands on its own. This would be acceptable for a handwaving proof, but not for a rigorous one. -- Meni Rosenfeld (talk) 07:37, 18 March 2008 (UTC)

>> "the limit notation, $\lim _{x\to x_{0}}f(x)=L$, is one unit. You shouldn't take out the $x\to x_{0}$ and treat it as something that stands on its own."

I think you mean the result of (13) is incorrect or not rigorous. Does it mean the whole proof should be re-derived in a completely different way or we can somehow fix the problem so that we don't have to re-derive the whole proof? If (13) is not rigorous, is there any example which opposes it? Thanks! - Justin545 (talk) 09:00, 18 March 2008 (UTC)

(13) and the derivations that lead to it are "not even wrong" in the sense that in the standard framework of calculus they are pretty much meaningless - if you look at the standard rigorous definitions of limits, you will see that they do not allow a function to be used as a variable. It is "correct" in the sense that intuitively, the limit of a function "when" the variable approaches some value is equal to the limit when some othee function approaches its appropriate limit value. However, this "when" business lacks a rigorous interpretation and is haunted by Bishop Berkeley's ghosts.

I have thought about how one might amend the proof, and realized that you also have a mistake much earlier. Step (5), dividing and multiplying by $g(x+\Delta x)-g(x)$, is only valid if $g(x+\Delta x)-g(x)\neq 0$, but there is no reason to assume that should be the case. Take, for example

$g(x)=\left\{{\begin{array}{ll}x^{2}\sin {\tfrac {1}{x}}&x\neq 0\\0&x=0\end{array}}\right.$

- a perfectly differentiable function at 0, and yet $g(x)=0=g(0)$ infinitely many times in any neighborhood of 0. Thus your proof will not work for it. Those kinds of pathological counterexamples are one of the things that separates rigorous proofs from not-so-rigorous ones. -- Meni Rosenfeld (talk) 10:58, 18 March 2008 (UTC)

>> "Step (5), dividing and multiplying by $g(x+\Delta x)-g(x)$, is only valid if $g(x+\Delta x)-g(x)\neq 0$, but there is no reason to assume that should be the case. Take, for example..."

I think $g(x+\Delta x)-g(x)$ will never be zero since $\Delta x$ "is not zero", $\Delta x$ is just a value that "very close to zero". Thus, $g(x+\Delta x)-g(x)$ will only close to zero but $g(x+\Delta x)-g(x)$ will not be zero, and I believe the step (5) would be still correct. As for your example, we may first need to evaluate

$\lim _{\Delta x\rightarrow 0}g(0+\Delta x)$ (16)

$=\lim _{\Delta x\rightarrow 0}(0+\Delta x)^{2}\sin {\frac {1}{0+\Delta x}}$ (17)

$=\lim _{\Delta x\rightarrow 0}{\Delta x}^{2}\sin {\frac {1}{\Delta x}}$ (18)

$=\lim _{\Delta x\rightarrow 0}{\Delta x}^{2}\cdot \lim _{\Delta x\rightarrow 0}\sin {\frac {1}{\Delta x}}$ (19)

But what will $\lim _{\Delta x\rightarrow 0}\sin {\frac {1}{\Delta x}}$ evalute to? I'm not sure... - Justin545 (talk) 01:41, 19 March 2008 (UTC)

You've made two mistakes here. First, $g(x+\Delta x)-g(x)$ can be zero for arbitrarily small values of $\Delta x$. That's what Meni's example shows. Your (18)=(19) is also mistaken: it would be valid if both limits in (19) existed, but as it happens the second one doesn't. Btw, your error at step (5) is a reasonably common one: IIRC, it occurs in the first few editions of G H Hardy's A Course of Pure Mathematics. Though there are other ways round it, perhaps the best is to avoid division at all in the proof. This has the advantage that your proof immediately generalises to the multi-dimensional case. Algebraist 02:36, 19 March 2008 (UTC)

>> "First, $g(x+\Delta x)-g(x)$ can be zero for arbitrarily small values of $\Delta x$. That's what Meni's example shows."

Meni's example is not so obvious to me why $g(x+\Delta x)-g(x)=0$ where

$g(x)=\left\{{\begin{array}{ll}x^{2}\sin {\tfrac {1}{x}}&x\neq 0\\0&x=0\end{array}}\right.$ (20)

Could you provide more explanation for it? Or could you tell what theorem supports that $g(x+\Delta x)-g(x)$ could be exactly zero?

>> "perhaps the best is to avoid division at all in the proof."

Division could be avoided at all, but it is "intuitive" since the definition of derivative involves division. Besides, even this proof involves division I think. If it does involve division, the proof would be considered non-rigorous. - Justin545 (talk) 03:08, 19 March 2008 (UTC)

>> "(13) and the derivations that lead to it are "not even wrong" in the sense that in the standard framework of calculus they are pretty much meaningless - if you look at the standard rigorous definitions of limits, you will see that they do not allow a function to be used as a variable."

I'm afraid I don't get it that "rigorous definitions of limits do not allow a function to be used as a variable" and why the derivations lead to (13) is meaningless. - Justin545 (talk) 03:37, 19 March 2008 (UTC)

A better question is how are they not meaningless. Where in your textbook did anyone mention taking the $\Delta x\to 0$ notation, treating it as a formula on its own, and doing manipulations on it? -- Meni Rosenfeld (talk) 16:48, 19 March 2008 (UTC)

If you want a similar proof that really works, one way would be to apply the mean value theorem to f at (4). This allows you to replace $f(g(x+h))-f(g(x))$ 163.1.148.158 (talk) 12:54, 18 March 2008 (UTC)

The mean value theorem I found is

$f'(c)={\frac {f(b)-f(a)}{b-a}}$

where $c\in (a,b)$. But I have no idea how to apply it to f at (4) and why it's needed to replace $f(g(x+h))-f(g(x))$? Thanks! - Justin545 (talk) 02:38, 19 March 2008 (UTC)

To the OP: it is not necessary to avoid division to make the proof rigorous, but it is one way of doing it. I meant division specifically by values of the domain or codomain of f and g (since these are the things that become vectors when you generalise), but I see I failed to say it. Apologies. The definition of the derivative need not involve such division (the one lectured to me didn't, for example), and one could argue that it shouldn't. Not sure if one would be right, mind. To your specific question, Meni's function is zero whenever x is 1/(nπ) (n a non-zero integer). Thus we have g(x)=0 for arbitrarily small x. Algebraist 03:34, 19 March 2008 (UTC)

>> "The definition of the derivative need not involve such division (the one lectured to me didn't, for example), and one could argue that it shouldn't."

The familiar definition of derivative is

${\frac {dy}{dx}}=f'(x)=\lim _{\Delta x\rightarrow 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}$ (21)

It seems you was saying that (21) is not a "rigorous" definition. It sounds pretty odd to me. I thought (21) is the only way of defining derivative. There are many lemmas or theorems about derivative in my textbook are originated from (21). It's not easy to imagine there other definition without division. - Justin545 (talk) 05:13, 19 March 2008 (UTC)

No, that's not what he was saying. He said that you can define the derivative without division, not that you should. Definition (21) (at least the $f'(x)=\lim _{\Delta x\rightarrow 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}$ part) is rigorous and is indeed the standard definition. There is nothing wrong with division, except for division by zero. The main flaw in your proof is dividing by $g(x+\Delta x)-g(x)$ which may be zero. Just because $\Delta x\neq 0$ doesn't mean that $g(x+\Delta x)\neq g(x)$. This is just common sense, you don't need my complicated example for that. -- Meni Rosenfeld (talk) 16:48, 19 March 2008 (UTC)

>> "To your specific question, Meni's function is zero whenever x is 1/(nπ) (n a non-zero integer). Thus we have g(x)=0 for arbitrarily small x."

I'm afraid I'm not able to proof (20) is zero when $x\in \left\{{\frac {1}{n\pi }}{\Bigg |}n\in \mathbb {Z} \land n\neq 0\right\}$. But I think $g(x+\Delta x)-g(x)$ will be zero when $g(x)=a$ where $a$ is any fixed constant. (Edit: which means I was ridiculously wrong. Apologies.) - Justin545 (talk) 05:42, 19 March 2008 (UTC)

x²sin(1/x) is zero whenever sin(1/x) is zero, which happens whenever 1/x is a multiple of pi, which happens whenever x = 1/npi for some integer n. You know, you're not really all that wrong. You have the right idea, you just don't have the tools to implement it. Here's roughly how my analysis textbook solves the problem. First, you define a new function h(y). I'll skip the details about intervals and mappings, and just say that it's focused on f and ignoring g, and assumes some interesting value c has been chosen. Let h(y) = (f(y)-f(g(c)))/(y-g(c)) if y does not equal g(c), and let h(y) = f'(y) if y=g(c). All that should be possible by assumption. Since g is differentiable at c, g is continuous at c, so h of g is continuous at c, so lim x->c (hog)(x)=h(g(c))=f'(g(c)). By the definition of h, f(y)-f(g(c))=h(y)(y-g(c)) for all y, so ((fog)(x)-(fog)(c)) = (hog(x))(g(x)-g(c)), so for x not equal to c we have ((fog)(x)-(fog)(c))/(x-c) = (hog(x))(g(x)-g(c))/(x-c). Taking the limit of both sides as x->c, then (fog)'(c)=lim x->c ((fog)(x)-(fog)(c))/(x-c) = (lim x->c hog(x))(lim x->c (g(x)-g(c))/(x-c)) = f'(g(c))g'(c). Black Carrot (talk) 06:36, 19 March 2008 (UTC)

>> "x²sin(1/x) is zero whenever sin(1/x) is zero, which happens whenever 1/x is a multiple of pi, which happens whenever x = 1/npi for some integer n."

Thanks! Now I understand it.

>> "You know, you're not really all that wrong. You have the right idea, ..."

Excuse my rewiring of your response for readability:

Here's roughly how my analysis textbook solves the problem. First, you define a new function $h(y)$. I'll skip the details about intervals and mappings, and just say that it's focused on $f$ and ignoring $g$, and assumes some interesting value $c$ has been chosen. Let

$h(y)={\begin{cases}{\frac {f(y)-f(g(c))}{y-g(c)}},&y\neq g(c)\\f'(y),&y=g(c)\end{cases}}$

All that should be possible by assumption. Since $g$ is differentiable at $c$, $g$ is continuous at $c$, so $h$ of $g$ is continuous at $c$, so

$\lim _{x\rightarrow c}h(g(x))=h(g(c))=f'(g(c))$.

By the definition of $h$,

$f(y)-f(g(c))=h(y)[y-g(c)]$, $\forall y\neq g(c)$,

so

$f(g(x))-f(g(c))=h(g(x))[g(x)-g(c)]$,

so for $x$ not equal to $c$ we have

${\frac {f(g(x))-f(g(c))}{x-c}}={\frac {h(g(x))[g(x)-g(c)]}{x-c}}$.

Taking the limit of both sides as $x\rightarrow c$, then

$(f\circ g)'(c)=\lim _{x\rightarrow c}{\frac {f(g(x))-f(g(c))}{x-c}}=\left[\lim _{x\rightarrow c}h(g(x))\right]\left[\lim _{x\rightarrow c}{\frac {g(x)-g(c)}{x-c}}\right]=f'(g(c))g'(c)$.

Did I misunderstand your response? Thanks! - Justin545 (talk) 09:01, 19 March 2008 (UTC)

After "By the definition of h", it should be for all y. If y=g(c), both sides are equal to zero, and the equality still holds. That one line is pretty much the goal of the whole thing, finding a way to get that conclusion without dividing by zero anywhere. Black Carrot (talk) 16:02, 19 March 2008 (UTC)

Calculating inflation by consumer price index statistics

Hello,

I'm currently working on HMAS Melbourne (R21), which is currently at Featured Article Candidates. One of the reviewers has requested that more recent financial figures be provided for the various sums of money (listed at User talk:Saberwyn/HMAS Melbourne (R21)) mentioned in the history of the ship. I've been pointed to Australian consumer price index statistics (available at [5] from 1969 to 2007, and [6] from 1949 to 1997 - Both open directly as Excel spreadsheets).

I've looked at them, and realise I don't have the first idea as to using these statistics to convert, for example, a AU$1.4 million figure from 1985 to a AU$ figure in 2007. Any help, at least a point in the right direction to an online tutorial or something, would be muchly appreciated. -- saberwyn 10:32, 18 March 2008 (UTC)

Well, the table says that AU$69.7 in 06/1985 are equivalent to AU$157.5 in 06/2007 in terms of overall purchasing power, so 1.4m in 1985 AU$ would roughly amount to 1.4m*157.5/69.7 = 3.2m in 2007 AU$. Bikasuishin (talk) 12:21, 18 March 2008 (UTC)

You calculate percentage of inflation from Year 1 to Year 2 by using the formula $Inflation={\frac {(CPI\ in\ Year\ 2)-(CPI\ in\ Year\ 1)}{(CPI\ in\ Year\ 1)}}\times 100$

For example, according to the table CPI in March 1988 was 87 and in March 1998 it was 120.3. Thus, by using the above formula, we find that prices in this period have inflated by ${\frac {120.3-87}{87}}\times 100\approx 38.28\%$. Therefore, AU$ 1,000 in March 1988 was equivalent to AU$ $1000+1000\times 0.3828=1382.8$ in March 1998. Cheers,  ARTYOM  17:14, 18 March 2008 (UTC)

Thank you for your assistance, it has been a great help!! -- saberwyn 00:12, 19 March 2008 (UTC)

Area Between Curves

I'm having a little trouble with an area between curves problem. The functions are f(x)=x^3-x and g(x)=3x. What I'm trying to figure out is if I should add the two areas (the one bellow and the one above the curve) or if i should subtract the area from underneath the X axis from the one above it. Thanks RedStateV (talk) 14:28, 18 March 2008 (UTC)

The distance between two points is the absolute value of one minus the other. So long as you keep your signs straight, it doesn't matter where either curve is in relation to the x-axis or to the other. — Lomn 14:42, 18 March 2008 (UTC)

So should I be getting a answer of 8 then? RedStateV (talk) 14:52, 18 March 2008 (UTC)

That's not what I get, but either one of us could have got the arithmetic wrong. I've done it twice and got the same answer both times, do you want to try doing it again and see if you still get 8? --Tango (talk) 15:17, 18 March 2008 (UTC)

My silicon master confirms that the answer is indeed 8. -- Meni Rosenfeld (talk) 15:35, 18 March 2008 (UTC)

Yeah, I can't integrate, it's definitely 8... sorry. --Tango (talk) 17:31, 18 March 2008 (UTC)

Are you going from x=-2 to x=2 ? If so, you can take advantage of the symmetry of the curves - take the area between the curves from x=0 to x=2 and then double it. I also get 8. Gandalf61 (talk) 15:49, 18 March 2008 (UTC)

Ok, so the value of the area bellow the X axis is not negative then. Good, Thanks. RedStateV (talk) 16:03, 18 March 2008 (UTC)

Area is, in fact, always positive (at least in the usual settings). What would negative area (or length, or volume) even mean? — Lomn 16:11, 18 March 2008 (UTC)

To answer this question we need to go back to the source. Lengths, areas etc. are measures given to sets in, say, an Euclidean space. To this extent they cannot be negative. But we can generalize to a sort of multiset in which elements can appear several times or even a negative number of times, and define measures on these in the natural way. Then the area of a set which only has negative points will be negative. In analytic geometry, there are formulae for area that can give negative results if an absolute value is not taken explicitly, such as $S={\frac {1}{2}}\det {\begin{pmatrix}x_{A}&x_{B}&x_{C}\\y_{A}&y_{B}&y_{C}\\1&1&1\end{pmatrix}}$ for a triangle. The sign is traditionally taken to represent the orientation, with only the absolute value representing the "true" area. But the signed number can be taken to be the area, if a negatively oriented triangle is seen as containing its points negatively. The same goes for calculating length by simply subtracting two coordinates, or area under a function by evaluating its integral. -- Meni Rosenfeld (talk) 17:21, 18 March 2008 (UTC)

If in doubt - plot the graph. Always check for places where the line crosses y=0 because if you don't notice this can mess up your areas if you are using integration.87.102.47.176 (talk) 16:56, 18 March 2008 (UTC)

If you're calculating the area between a curve and the x-axis, then the curve crossing the axis is significant, however this question is about the area between 2 curves, so it's where the curves cross each other that's significant, not where they cross the x-axis. --Tango (talk) 15:45, 19 March 2008 (UTC)

Rational function conjecture

If $f:\mathbb {R} \rightarrow \mathbb {R}$ is an analytic function such that $x\in \mathbb {Q}$ implies $f(x)\in \mathbb {Q}$, must $f$ be a rational function? I expect it's false, so what's a counterexample? —Keenan Pepper 19:02, 18 March 2008 (UTC)

How about f(x)=2^(lg(x+1))?, where lg is log base 2. GromXXVII (talk) 20:08, 18 March 2008 (UTC)

Silly me...bad domain. GromXXVII (talk) 20:10, 18 March 2008 (UTC)

And it equals x+1, which is rational. Nice try, though! I can't think of a counterexample. It wouldn't surprise me if it were true, but I can't prove it. --Tango (talk) 20:13, 18 March 2008 (UTC)

Any ideas? Or search keywords? —Keenan Pepper 15:26, 19 March 2008 (UTC)

March 19

Angle bisector proof

In triangle ABC, draw angle bisectors AD and CE, where D is on BC and E is on AB. If angle B is 60 degrees, show that AC=CD+AE.

I've figure out that if the intersection of AD and CE is F, then <CFA and <EFD are 60 degrees and <DFC and <EFA are 30 degrees, but I'm not sure what's next. Thanks, 76.248.244.196 (talk) 01:06, 19 March 2008 (UTC)

Your four angles around F are all the angles there are around F and you have them adding up to 60+60+30+30 = 180. Surely this should be 360, so some of your figuring has gone astray. Present what working you have done so far and maybe it will become clearer... -- SGBailey (talk) 07:24, 19 March 2008 (UTC)

Algebra question

Given any three natural numbers, show that there are two of them, a and b, that a^3b-b^3a is divisible by thirty.

I have no clue how to prove it, please help. Thanks, 76.248.244.196 (talk) 01:09, 19 March 2008 (UTC)

Note that "divisible by thirty" is equivalent to "divisible by two, three, and five". I would first factor the polynomial as $a^{3}b-b^{3}a=ab(a+b)(a-b)$. If either a or b is even, then this product is even, and if they are both odd, then (a+b) is even so the product is still even. Therefore $a^{3}b-b^{3}a$ is even for every pair of natural numbers. A similar argument can be used to show that it is also a multiple of three (work modulo three). So the polynomials for all three of the pairs are multiples of six, and you can use the pigeonhole principle to show that at the polynomial for at least one of the pairs is also a multiple of five (work modulo guess-what). —Keenan Pepper 01:29, 19 March 2008 (UTC)

Divisibility curiosity

If n is a positive integer that is not divisible by 6, (n^2-1) seems to be always a multiple of 24. Why? Imagine Reason (talk) 02:54, 19 March 2008 (UTC)

It only works if neither 2 nor 3 divides n. Difference of two squares says a^2-b^2 = (a-b)*(a+b). Can you use that to rewrite n^2-1 and make arguments about where 2 and 3 must divide? PrimeHunter (talk) 03:18, 19 March 2008 (UTC)

You're right, it's about 2 and 3. So n^2-1 = (n-1)(n+1), which when n is not a multiple of 2 equals a multiple of 4. And since n isn't a multiple of 3, (n-1)(n+1) must be, but that's as far as I got. Imagine Reason (talk) 19:05, 19 March 2008 (UTC)

Gödel's Incompleteness Theorems: Is The Math Reliable?

Many sciences depend on the math to prove something and use it for rigorous study. But Gödel's incompleteness theorems states:

For any consistent formal, computably enumerable theory that proves basic arithmetical truths, an arithmetical statement that is true, but not provable in the theory, can be constructed.1 That is, any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete.

Therefore, I would like to know are all the theories we use (for biology, chemistry, physics, medicine, computer science, etc.) considered to be consistent theories themself? And are all of maths we learn from elementary school to university considered to be reliable and don't contradict each other? - Justin545 (talk) 07:00, 19 March 2008 (UTC)

What do you mean by "reliable"? I would say the mathematics underlying biology, chemistry, etc is far less likely to be in error than the biology and chemistry themselves. But if you're looking for apodeictic certainty -- the sort of thing that, by its nature, cannot be wrong -- well, sorry, we don't have any of that. In my humble opinion, anyway. We'll settle for being right; we don't have to be completely certain.

Or as the Eagles put it -- "I could be wrong, but I'm not". --Trovatore (talk) 07:18, 19 March 2008 (UTC)

Math is used as a tool for studying many sciences. If the tool itself is "problematic" or "questionable", the consequences of employing it are very likey to be wrong! "reliable" means "consistent" and "don't contradict". Incompleteness theorems, in other words, states: if every arithmetical statement that is true and is provable in the theory, the theory is inconsistent but it is complete. So what I want to know is: the math we use is either

(1) consistent but not complete, or

(2) complete but not consistent - Justin545 (talk) 07:48, 19 March 2008 (UTC)

Well, we don't know for certain, but the general view is that we are in the consistent but not complete case, which is really not as bad as it sounds at first. If you know any group theory, consider that there are plenty of facts about groups that cannot be deduced from the axioms for a group alone -- the theory of groups, as given by the most basic group axioms, is not complete. In some sense this is because there are different models, different groups, that all meet those basic axioms, and thus have truths that are not derivable from just those axioms. You can think of arithmetic as being similar, with different models, just with the proviso that, unlike groups, we haven't found models that disagree on any arithmetic facts you or I would generally care about. -- Leland McInnes (talk) 12:11, 19 March 2008 (UTC)

Could you give an example of what you mean, there? I've done quite a bit of group theory and have never come across something that's true but can't be proven from the axioms of a group (together with ZF). --Tango (talk) 13:39, 19 March 2008 (UTC)

What I mean is that the group axioms don't uniquely define the group, but rather a whole slew of possible objects each of which satisfies the axioms of being a group. Thus there isn't a unique model of "group" specified by the axioms, but rather each and every different group is a different model that satisfies the basic group axioms. There are things that are true of particular groups that you can't deduce from just the group axioms -- you need more information (more axioms in essence) to pin down which group (or class of groups) you are talking about. Thus there are truths that occur in systems that fulfill the group axioms that are not provable from the group axioms alone. Does that make more sense? -- Leland McInnes (talk) 17:26, 19 March 2008 (UTC)

Right, but arithmetic (and set theory) are quite a different case from group theory. Arithmetic is not the study of models of arithmetic; it's the study of numbers. All models of arithmetic have (copies of) all the true natural numbers, but some of them also have fake natural numbers. The one true Platonic intended model of arithmetic has only the true ones, and none of the fake ones, and is unique up to a canonical isomorphism. There's a limit to what we can find out about the behavior of the true natural numbers from a fixed set of axioms and first-order logic alone. That doesn't mean we have to stop there. --Trovatore (talk) 17:49, 19 March 2008 (UTC)

Sure, there are plenty of things that can't be proven using just the axioms of a group, but those things aren't true. $gh=hg\quad \forall g,h\in G$ can't be proven just from the group axioms, because it isn't true in general. That's not incompleteness, it's just a false statement. If you want it to be true, you have to add an additional assumption (that the group G be cyclic, say). If the statement can be stated in terms of only the group axioms, and is true, then it can be proven using only the group axioms. If it can't be stated using only those axioms, then it being impossible to prove isn't a case of incompleteness. A framework is incomplete if there are unprovable true statements within that framework. --Tango (talk) 18:06, 19 March 2008 (UTC)

Tango, I think you have not thought these things through terribly well. At least it isn't clear to me what you mean by a framework, or unprovable but true within a framework. Is a framework a first-order theory, or a model, or what exactly? --Trovatore (talk) 18:19, 19 March 2008 (UTC)

Let me be a little less Socratic and hopefully more constructive (took me more time to figure out how to say this than it did to ask a question). Let's take a specific example. Peano arithmetic neither proves nor (we suppose) refutes the claim "Peano arithmetic is consistent" (the claim is usually abbreviated Con(PA). Therefore there are models of PA in which Con(PA) is true, and there are models of PA in which Con(PA) is false. So we can make an analogy with your example statement "multiplication is commutative": There are models of group theory (that is, groups) in which "multiplication is commutative" is true, and there are other models of group theory in which it's false.

Here's the big difference: There's no such thing as "the intended group", the group that defines the truth value of "multiplication is commutative in group theory". We're interested in Abelian groups, and we're also interested in non-Abelian groups, and you just have to specify which ones you're talking about.

But Peano arithmetic (we suppose) really is consistent. The models of PA that think otherwise are wrong about that. That's not to say they're not interesting (people devote whole careers to them), but merely by their opinion on this one issue, they prove that they are not the intended model. --Trovatore (talk) 18:35, 19 March 2008 (UTC)

Ok, I think I understand what you're saying now. I'm not sure I agree, though. Group theory is defined in terms of set theory. Once you've determined a model of set theory, your model of group theory is completely determined (a group is simply a set together with a function - both concepts defined outside of group theory). Is there a (reasonable) model of set theory in which all groups are abelian? --Tango (talk) 18:56, 19 March 2008 (UTC)

Whoah, we have to be careful here -- the phrase "group theory" is being used in two different ways (my fault, probably). When I say "model of group theory==group", I'm using "group theory" to mean the first-order theory defined by the three axioms (identity existence, existence of two-sided inverses, associativity). That's different of course from "group theory" as in "the study of groups", which is not a formal first-order theory at all. Please re-read my remarks keeping this clarification in mind -- they won't have made any sense at all if you were thinking of "model of group theory" as meaning "model of the study of groups(?)". --Trovatore (talk) 19:02, 19 March 2008 (UTC)

Ok, but I think my point still stands. Group theory, in that sense, is still built on set theory. Any model of group theory must be a model of set theory, since it has to satisfy ZF plus the 3 axioms of a group. Can you have such model of set theory in which all groups are abelian? For example, set theory provides all kinds of methods of combining sets to produce sets - those method can be used to combine groups and produce other groups. Is there a model in which all such possible combinations are abelian? --Tango (talk) 19:22, 19 March 2008 (UTC)

Theories in physics (thought to be the trunk of the science "tree") are not necessarily consistent. Helpfulness of established theories begin and end with orders of magnitude. This is why we have semiclassical physics, and the mesoscopic scale, and why we differentiate "Physics in the Classical Limit," Relativity, and quantum theory. Mac Davis (talk) 08:01, 19 March 2008 (UTC)

Did I misunderstand Gödel's Incompleteness Theorems or the Incompleteness Theorems is really about distinguishing between classical physics and modern physics? I thought Incompleteness Theorems is just all about the math but not the physics. And Incompleteness Theorems should be able to be applied to all kinds of science, not just physics. I'm not offending, just hope someone can clarify the concept. - Justin545 (talk) 09:23, 19 March 2008 (UTC)

The incompleteness theorems don't apply particularly well to the kind of math you're probably familiar with. That is, they aren't relevant. They claim that a specific very sensible, very general way of justifying things doesn't work very well in certain contexts. That doesn't mean that what we were trying to justify is wrong, just that we'll have to look somewhere else for confidence in it. It also throws essentially no doubt on actual arithmetic, which deals only with fairly small numbers and can be justified by direct experience and some common sense. Black Carrot (talk) 08:10, 19 March 2008 (UTC)

Theorems are proved based on axioms. Experience in proving theorems made mathematicians conjecture that every true statement could eventually be proved. This conjecture turned out to be naive. The incompletenes theorem states that the conjecture is not true: the fact that some statement cannot be proved does not imply that the statement is false. The incompleteness theorem does not threaten the reliabilty of mathematics. Bo Jacoby (talk) 11:06, 19 March 2008 (UTC).

In general, mathematicians believe that mathematics (however we may choose to define that term) is consistent. This is mainly because we have not found an inconsistency (a statement P such that both P and not-P can be proved). We can even express this "conjecture" as a (humungously complex) arithmetical statement. Problem is that we also know, thanks to Gödel, that we cannot prove this statement - at least, not without stepping up to some more powerful axiom system, which then leads a "turtles all the way down" type of regression. Bottom line is, most mathematicians say "that's interesting and slightly weird" but they don't lose sleep worrying that mathematics might be inconsistent. On a scale of rational evidence-based confidence, you can put the consistency of mathematics right up at the 99.99% mark. Gandalf61 (talk) 12:21, 19 March 2008 (UTC)

Poincare Video

I tried to access the video at [7], but it doesn't work. Does anyone know where I can find a good copy of it? Black Carrot (talk) 08:14, 19 March 2008 (UTC)

Diophantine equation

I've got the equation n(2n+1)=a(a-1), both n and a to be positive integers. The solutions (10,15) and (348,493) came by searching, but nothing higher has been found. I'm wondering if the absence of further solutions can be shown from the fact that as a and n increase, their ratio will tend to the irrational root 2?—81.132.237.15 (talk) 13:16, 19 March 2008 (UTC)

Next two solutions are (11830, 16731) and (401880, 568345). As you conjecture, there is indeed a connection with rational approximations to the square root of 2. Gandalf61 (talk) 15:50, 19 March 2008 (UTC)

So there will be an infinite number of solutions?—81.132.237.15 (talk) 19:26, 19 March 2008 (UTC)

Complex Analysis

Determine, with motivation, all funtions that is analytic on {z/ |z| < 4}, with f( 0 ) = i and |f( z ) <= 1 on {z/ |z| < 4} —Preceding unsigned comment added by 163.187.240.51 (talk) 13:33, 19 March 2008 (UTC)

Simple tip: When trying to get other people to do your homework, don't include phrases like with motivation. Also, you'd have better chances if you explained specifically what you've tried so far, and what you're having trouble with. —Keenan Pepper 15:23, 19 March 2008 (UTC)

Oh, what the heck. I'm feeling nice, so I'll practically give you the answer: Maximum modulus principle. —Keenan Pepper 15:28, 19 March 2008 (UTC)

How do you explain the fractional root of a number?

I can't find an answer to this because I'm not sure if I'm using the right terminology. What I mean is, how does one explain what the "1/2"th root of a number is? Or the "2/3"rd root? I understand that the nth root of A is that number x which when multiplied by itself n times yields A, but how do I explain a fractional root in similar terms? Thanks in advance, Narxysus (talk) 17:22, 19 March 2008 (UTC)

the n th root of a is in fact a^1/n eg sqrt(10) = 10^1/2

Therefor the a/b th root of n is n^b/a. That's a consistent intepretation. So the 1/2 root of x is in fact x squared.83.100.183.180 (talk) 18:24, 19 March 2008 (UTC)

It looks like you are trying to do two conceptual steps at once here. First you should be asking what is the fractional power of a number, such as power of 1/2, and we can try to answer that question if you do. Only then should you ask about fractional roots, and the answer is that the 1/2th root of A is that number x such that $x^{1/2}=A$. -- Meni Rosenfeld (talk) 18:29, 19 March 2008 (UTC)