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January 1

Maths - discovered or invented?

To me its amazing that inferances from the rules of mathematics can predict what happens in the real world. As a second question, have there ever been empirical studies that have led to the rules of mathematics being altered? 78.146.210.81 (talk) 11:43, 1 January 2010 (UTC)

Isn't it amazing the way the real world acts as a model for mathematics ;-) Naive set theory is a good example of some maths that was found to be wrong after a bit of study and had to be amended, see the section on paradoxes at the end. Dmcq (talk) 11:59, 1 January 2010 (UTC)

An important aspect of mathematics lies within the relationships between different branches. For instance, one could use algebra in topology, without delving deep into the subject (but of course, a deeper form of algebra may shed new light on topology, if applied). Likewise, although contradictions were found in naïve set theory approximately 100 years ago, it continues to be applied to this day. In fact, many branches of mathematics could function without reference to Russel's paradox or any deep form of axiomatic set theory. On the other hand, as I pointed out, fields such as set-theoretic topology require a very deep form of set theory; naïve set theory alone does not suffice. --PST 12:29, 1 January 2010 (UTC)

The OP might not be that interested, and there is an overlap between who helps at the two help desks, but he or she might ask something close to what is being asked here at the Science desk and shuttle back and forth when it seems interesting if a more complex answer is desired. Part of the issue is what the 'real world' is from a scientific standpoint. There are some who will argue that it is a mathematical object when all is said and done. I wouldn't hold out too much hope for a serious reply at the Computer Science desk, but you might actually get the most interesting one there. Ultimately, the world consists of objects with rules for their behavior (it seems), and the fact that an abstraction from the real world (mathematics) would be able to feed back predictions on the behavior of the real world isn't what surprises anyone. It's that purposeless questions internal to the abstractions can be taken far from their natural sources and still find applications that is often surprising.Julzes (talk) 12:56, 1 January 2010 (UTC)

Of course there are religious/philosophical questions like whether you support the Axiom of choice or the Axiom of determinacy or go for a third way like projective determinacy. I think I read a science fiction story once where the laws of nature kept getting more complicated to cover up problems as more and more flaws were shown up in the previous laws. So relativity and quantum mechanics only became true recently as problems and paradoxes were shown in the previous simpler laws. Eventually it'll all become so complicated we'll never be able to show something is actually wrong or it might all disappear as just too paradoxical. :) Dmcq (talk) 14:09, 1 January 2010 (UTC)

Nobody has answered the first question. Does that mean the answer is "Nobody knows"? 92.24.69.222 (talk) 18:54, 1 January 2010 (UTC)

What was the first question. All I see there is "To me it is amazing that..." etc. That's not a question. Did you mean "Why does that work"? Or maybe "Am I the only one who's amazed?" Or did you maybe want an answer that would leave you unamazed? Michael Hardy (talk) 00:30, 2 January 2010 (UTC)

The first question is the title. -- Meni Rosenfeld (talk) 16:16, 2 January 2010 (UTC)

Mathematicians discover properties of natural numbers, shapes etc. and then invent notation, theories and structures to describe, investigate and explain what they have discovered, and go on to make more discoveries about what they invented. The answer has to be "both". Dbfirs 19:36, 1 January 2010 (UTC)

"Shape" can be formulated in the language of natural numbers, after employing various other mathematical (algebraic and analytical) ideas. --PST 00:58, 2 January 2010 (UTC)

That's true, but the discoveries were made thousands of years before this formulation was invented. Dbfirs 09:57, 2 January 2010 (UTC)

Um, I'm not quite sure what the claim is supposed to mean. On its face, to talk about shape using only natural numbers, you need second-order logic; I'm not sure why you wouldn't just bring in the real numbers from the start, which are more natural to use when discussing shapes. --Trovatore (talk) 10:12, 2 January 2010 (UTC)

Not sure what kind of answer is wanted. If we had an answer that could be codified then it would be mathematical and then we'd simply have something saying it itself is applicable using methods which work only because it is applicable. Something like that which doesn't sound very sensible when applied to the real world but worked okay for Gödel in maths. That we are able to make sense of the world is only because it is reasonable. Dmcq (talk) 22:15, 1 January 2010 (UTC)

This is a classic philosophical question about the nature of mathematics. See mathematical realism for the "discovered" viewpoint, and see several alternative views on that page. Staecker (talk) 22:31, 1 January 2010 (UTC)

January 2

Original Basic Discovery or No?

I have been wondering if the following factorization has a history, if there is any reasonably simple way to answer this. Oddly, if it is new, it was discovered to be factorable by one person (me) and actually factored by someone else (User:PrimeHunter) who knew the quick way to do it. My own reaction is to be agnostic on the question of its originality. It's not far from the subject of cyclotomic polynomials, but I can't see any particular way the question is likely to have come up other than by simply experimenting as I was (as a new user of PARI/GP). I also haven't a clue as to whether its factorability has any more than educational value (in regards to small-number coincidences?), though I am looking into it on paper. Here is the result:

$x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+20=(x^{4}-x^{3}+x^{2}-3x+4)(x^{4}+2x^{3}+2x^{2}+4x+5)\,$

I'm virtually positive that the following is a re-discovery, but I'd be interested in its history as well:

$x^{4}+x^{3}+x^{2}+x+12=(x^{2}-2x+3)(x^{2}+3x+4)\,$

Julzes (talk) 00:26, 2 January 2010 (UTC)

These sorts of factorizations are often discovered (in more general situations), so it is unlikely that it has not been stumbled upon before, as you point out. However, I have not thought about this one, or an analogue (recently) in particular, so I cannot guarantee anything about it. You might like to read the books "Cyclotomic fields" and "Cyclotomic fields II", by Serge Lang, in the GTM series. --PST 07:41, 2 January 2010 (UTC)

That's a good suggestion. I think of the subject as more fundamental and simple than it surely is and another couple of Lang's books is not going to hurt me.Julzes (talk) 09:49, 2 January 2010 (UTC)

How to factorise

It seems by an off chance ive forgotten how to factorise the $x^{3}-27$ in ${\lim _{x\to 3}{\frac {x^{3}-27}{(x-3)}}}$, its been a while since ive did this so any help would be appreciated. —Preceding unsigned comment added by 121.214.27.148 (talk) 04:16, 2 January 2010 (UTC)

Two things you should know:

$a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})\,$

and a basic fact from algebra: If you plug the number 3 into a polynomial and get 0, then (x − 3) is a factor of that polynomial; if you plug in 42 and get 0, then (x − 42) is a factor, etc.

This looks like another typical case of someone learning calculus without really knowing the algebra prerequisites, which means you'll probably pass but you'll find it unpleasant. Michael Hardy (talk) 04:43, 2 January 2010 (UTC)

I think that one cannot generalize such situations without sufficient evidence regarding the circumstances. For instance, I was self-taught in calculus; theoretical ideas such as "maxima" and "minima" evoked interest in me when I was first presented a calculus book. But I never "liked" elementary algebra at the time, and thus never really thought about these sorts of factorizations (for instance) when doing calculus, even when I saw them arise in the theory of limits. But once I actually noticed that I was taking these (algebraic) ideas for granted in my intuitive visualization of limits, I thought it worthwhile to develop some intuition about them, and in doing so, generalize them to higher degree polynomials (and other sorts of factorizations). Thus it was not a question of me not being able to understand basic algebra; rather, I did not see its purpose (or interest) prior to studying calculus and did not attempt to understand it. And this was perfectly appropriate (at least in my view) because I saw its need in consolidating my intuition when doing calculus, and subsequently sought to think about it in greater depth. That said, I was not "dumb" at algebra prior to studying calculus; I was reasonably competent, though there were some minor aspects which although I knew, I did not examine in close depth. Succinctly, my point is that it may be more effective for someone to realize the importance of algebra in calculus with time, rather than be forced to do it (algebra) without having a clue about where they will use it in mathematics (not every mathematician would have enjoyed doing elementary algebra (and by this I do not mean original discoveries; just mindless application of identities) without reference to calculus (for instance), for a long period of time, although I could be wrong). But of course, I am assuming one is also not being forced to do calculus; unless calculus is found interesting at the time of its study, it is not worthwhile to do anything earlier in depth anyhow. --PST 07:34, 2 January 2010 (UTC)

....and notice that if you get something other than 0 in the numerator, then the question of what to do with the limit becomes trivial. Michael Hardy (talk) 04:44, 2 January 2010 (UTC)

As Michael explained, factorizing a polynomial f(x) amounts to solving the equation f(x) = 0. But this requires tedious computations if you do it by hand. So why not use your computer? In the J (programming language) the expression

     p.(_27,0,0,1)

immediately evaluates to

┌─┬────────────────────────────┐
│1│3 _1.5j2.59808 _1.5j_2.59808│
└─┴────────────────────────────┘

showing that

−27·x⁰ + 0·x¹ + 0·x² + 1·x³ = 1 · (x−3) · (x−(−1.5+2.59808i)) · (x−(−1.5−2.59808i))

Bo Jacoby (talk) 17:11, 2 January 2010 (UTC).

This suggestion is silly. One finds the factorization

$x^{3}-27=(x-3)(x^{2}+3x+9)\,$

and there is no occasion to look for a further factorization since that quadratic polynomial is not 0 when x = 3. Michael Hardy (talk) 19:36, 3 January 2010 (UTC)

Because that takes longer than just observing that 3 is a root, doing some quick polynomial division, cancelling the x-3 terms and evaluating the remaining polynomial at 3. Computers are good for difficult numerical problems, but they are usually slower for the easy ones. --Tango (talk) 17:25, 2 January 2010 (UTC)

Of course one can always look up Factorization. I don't think I'll suggest Lenstra–Lenstra–Lovász lattice basis reduction algorithm except via an algebra package ;) [[User:D[[]]mcq|Dmcq]] (talk) 17:41, 2 January 2010 (UTC)

I was merely answering the question 'how to factorise the $x^{3}-27$ '. Bo Jacoby (talk) 08:04, 3 January 2010 (UTC).

Pardon? What's that about? Dmcq (talk) 08:12, 3 January 2010 (UTC)

About Tango's objection to my factoring $x^{3}-27$ rather 'than just observing that 3 is a root'. Bo Jacoby (talk) 13:14, 3 January 2010 (UTC).

"Just observing that 3 is a root" was only the first step in the procedure Tango suggested for factoring $x^{3}-27$ (over the reals). What he objected to was the use of a computer program. -- Meni Rosenfeld (talk) 13:26, 3 January 2010 (UTC)

For computing the limit you do not need to factorize. Just use l'Hôpital's rule

${\lim _{x\to 3}{\frac {x^{3}-27}{x-3}}}={\lim _{x\to 3}{\frac {d(x^{3}-27)}{d(x-3)}}}={\lim _{x\to 3}{\frac {3x^{2}dx}{dx}}}={\lim _{x\to 3}{3x^{2}}}=3\cdot 3^{2}=27$

But the OP asked 'how to factorise'. Bo Jacoby (talk) 13:46, 3 January 2010 (UTC).

L'Hopital's Rule will involve you in logical circularity if the limit arises as part of an explanation of how to find derivatives. Michael Hardy (talk) 19:34, 3 January 2010 (UTC)

And Tango explained how to factorize. -- Meni Rosenfeld (talk) 14:42, 3 January 2010 (UTC)

'"Just observing that a is a root" is not a method for finding a root. Bo Jacoby (talk) 16:15, 3 January 2010 (UTC).

Not a very efficient method for finding a root indeed! There is a small ambiguity in the OP's request: the headers ask how to factorize x³-27, but the aim is computing a limit of a fraction where the denominator x-3 is given. For that limit, the relevant operation was not to factorize the numerator, but rather to divide it by x-3, had it a remainder or not. I think this is what Tango meant: the OP had first to divide x³-27 by x-3, and in consequence (s)he should observe that 3 is a root, so there is no remainder and the fraction could be simplified. --pma (talk) 17:27, 3 January 2010 (UTC)

Bo Jacoby's suggestions are silly. One has

$\lim _{x\to 3}{x^{3}-27 \over x-3}.$

Since the numerator and denominator are both 0 when x = 3, one seeks to write it like this:

$\lim _{x\to 3}{(x-3)(\cdot \cdots ) \over x-3}.$

So one divides (x³ − 27) by x − 3, getting

$\lim _{x\to 3}{(x-3)(x^{2}+3x+9) \over x-3}.$

Further factoring beyond that point is not appropriate since

$x^{2}+3x+9\,$

is not 0 when x = 3. Michael Hardy (talk) 19:41, 3 January 2010 (UTC)

• Also, see the article removable singularity. ~~ Dr Dec (Talk) ~~ 19:46, 3 January 2010 (UTC)

For completeness sake: see also the definition of derivative of x³ at x=3. --pma (talk) 20:54, 3 January 2010 (UTC)

Geometric algebra question - involutions

What are "the" (or "some") geometric interpretations of the three main geometric algebra involutions (reversion, grade involution, and Clifford conjugation)? I believe that grade involution corresponds to reflection through the origin, but I don't know about the other two.

ALSO, when representing a geometric algebra using a faithful matrix representation, where ordinary matrix multiplication corresponds to a geometric product, are there analogous matrix involutions to these geometric algebra involutions. When using a sensible basis, the transpose is equal to the reverse, but I've found nothing for the other two.--Leon (talk) 13:09, 2 January 2010 (UTC)

I know I've already tried to answer this but I'll have another go as no-one else has, and besides it's an interesting topic. For the first it depends what your geometric interpretation of the algebra is, as there are a few. The most common is associating the even sub-algebra (e.g. the bivectors, or normed even elements) with rotations in space. There I think reversion gets you the inverse transformation, e.g. e.g. in 2D for unit vectors a and b the product ab is simply the even element that rotates from b to a. Reversing them generates the opposite direction. It gets more complex in higher dimensions but with care you can build all rotations the same way, so I think the same will be true.

Grade inversion I think straightforwardly corresponds to inversion in the origin, i.e. the same as the matrix with -1 along its main diagonal. As for Clifford conjugation, which is the combination of inversion and reversion, if both of the above are correct then I don't see it has any simple interpretation, though it may have one if geometric algebra is interpreted differently.

On matrices it depends what you mean by "faithful matrix representation". The matrix representations usually given for the Geometric algebras, e.g. from here, are quite varied with matrices and sums of matrices in ℝ, ℂ and ℍ. As ℂ and ℍ can be conjugated, and all the matrices can be inverted/transposed, there may be correspondences but I suspect they will be different for different dimensions. --John Blackburne (words ‡ deeds) 15:20, 3 January 2010 (UTC)

Okay...that's about as far as I understood it. Thanks anyway.

Further question: regarding interpretations of the geometric algebra, I'm aware of both the projective model (4 basis vectors for 3-dimensional space) and the conformal model (5 basis vectors for 3-dimensional space). What is the name given to the model that contains as many vectors as there are spatial dimensions, with the corresponding metric signature.--Leon (talk) 10:09, 4 January 2010 (UTC)

I'm not aware of one. It's the standard model if you like, i.e. it's the standard interpretation of GA, qualified with the dimension and metric (especially if non-Euclidean). E.g. the [geometric] algebra over ℝ³, the [geometric] algebra over ℝ^3,1 etc.. --John Blackburne (words ‡ deeds) 12:42, 4 January 2010 (UTC)

Just noticed there's some detail on matrices at Classification of Clifford algebras#Center. Over my head mostly but it's got more detail than the link I gave and wikilinks to relevant other articles. --John Blackburne (words ‡ deeds) 18:50, 8 January 2010 (UTC)

Proof of continuity

How to prove the continuity of f(x)=ln(x)? --84.62.205.233 (talk) 13:12, 2 January 2010 (UTC)

I assume you wish to prove the continuity of f on the domain $D=(0,\infty )$, in which case, you would need to show that $\lim _{h\rightarrow 0}\ln(x+h)-\ln(x)=0$ for all $x\in D$. However, we may compute as follows:

$\lim _{h\rightarrow 0}\ln(x+h)-\ln(x)=\lim _{h\rightarrow 0}\ln({\frac {x+h}{x}})$

Since $\lim _{h\rightarrow 0}{\frac {x+h}{x}}=1$, if f is continuous at 1, $\lim _{h\rightarrow 0}\ln({\frac {x+h}{x}})=\ln(\lim _{h\rightarrow 0}{\frac {x+h}{x}})=\ln(1)=0$ thus establishing continuity of f. I will leave the check that f is continuous at 1 to you. However, note that since every differentiable function is continuous (Proof: If g is differentiable at a, $\lim _{h\rightarrow 0}g(a+h)-g(a)=\lim _{h\rightarrow 0}{\frac {g(a+h)-g(a)}{h}}\cdot h=\lim _{h\rightarrow 0}{\frac {g(a+h)-g(a)}{h}}\cdot \lim _{h\rightarrow 0}h=g'(a)\cdot 0=0$ and thus $\lim _{h\rightarrow 0}g(a+h)=g(a)$ establishing the continuity of g at a), the continuity of f in its domain follows from its differentiability (and the differentiability of f may be a familiar fact to you). Hope this helps. --PST 13:41, 2 January 2010 (UTC)

Note: a precise answer depends on how you define the natural logarithm $\log \,x.$ A reasonable choice is to define it as the inverse function of the exponential function, in which case the continuity is a consequence of $\exp$ being continuous and strictly increasing. The exponential function also can be defined in several ways. So first of all, choose you definition of $\log$ and of $\exp .$ Here is a nice way though a bit out of fashion: prove that for all x>0 the limit

$\lim _{n\to \infty }n(x^{\frac {1}{n}}-1)$

exists, name it $\log(x)$, and prove that it's a continuous inverse to $\exp .$ --pma (talk) 20:08, 3 January 2010 (UTC)

What's a mathematical coincidence?

A post above by User:Julzes (and in fact, others strange mathematical facts recently reported by the same user) made me ask: what is a mathematical coincidence? In fact, do mathematical coincidences exist? (For instance, is it a coincidence that the twelfth Fibonacci number is 144?) At least, a mathematical coincidence is a remarkable mathematical fact of whom there are not yet satisfactory explanations, and that doesn't fit into a general pattern. But this is quite a relative notion. Does anybody have a more absolute definition of what a "mathematical coincidence" is? Is there a way of giving a measure of how remarkable is a coincidence in mathematics? --pma (talk) 15:18, 2 January 2010 (UTC)

See mathematical coincidence for an article that I've always had mixed feelings about. Staecker (talk) 15:24, 2 January 2010 (UTC)

Mathematical coincidence is an article requiring some major writing effort and organizational work (as well as some content changes to the list). I'll say something about this at its discussion page later today. Incidentally, I failed to mention a while back that 3360633 is not only the first number which is palindromic of length seven digits in three different bases, but that it is eighth on an obscure list (palindromes--in base ten--which are sums of all composite numbers up to a certain point) that also has 33633 sixth on the list. You can find that at OEIS.Julzes (talk) 08:43, 3 January 2010 (UTC)

The argument I used there is a complexity one, if the number of keys you have to press on a calculator to get the 'mathematical coincidence' is larger than the number of digits accuracy then it's a pretty poor coincidence. Which would remove most of them if applied strictly and can be used on people's original research when they discover new 'coincidences'. The ones there should only be ones people have noted in literature as coincidences. Dmcq (talk) 15:31, 2 January 2010 (UTC)

In fact using that definition if only one could find a way of easily elimating trivial cases like 1+10^-45 is equal to 1 to 45 digits it should be possible to list all the mathematical coincidences up to say 12 button presses and say how good they are. Above that I think one wold have to start getting a little more picky, how often would one really find an expression with 45 symbols in it giving something to 50 places all that surprising or interesting. What I would consider rally surprising is if an expression gave something with twice the number of figures accuracy which I think should actually happen occasionally. Dmcq (talk) 17:33, 2 January 2010 (UTC)

OH I hadn't tried searching it. Thanks to both. Wikipedia contains everything... Nice article. --pma (talk) 23:55, 2 January 2010 (UTC)

When computing the coefficients of low-degree cyclotomic polynomials, one notices that all the nonvanishing coefficients are either 1 or -1. In fact, and somewhat surprisingly, this holds for all cyclotomic polynomials of degree at most 104, but fails for degree 105! Why the "104" and "105" appear is somewhat perplexing at first. But a result due to Migotti asserts than in order for a cyclotomic polynomial to have a coefficient other than 0, 1 or -1, it is necessary that its degree is divisible by at least three different odd primes (and the smallest such integer having this property is 105). In fact, this is probably one of the most well-known "conjectures" in number theory that holds for consecutive integers up to a large value, but subsequently fails (I do not know of any other as well-known as this, so if someone does I would like to hear it). Migotti's result perhaps dispels any idea that it is a mathematical coincidence, but it is interesting nonetheless. --PST 00:42, 3 January 2010 (UTC)

I remember reading about that a while back, the Cyclotomic polynomial article doesn't have that result in, it should definitely be included I think if you could stick it in. ... Sorry I see the result is in but not attributed to Migotti, I'll find a ref and stick it in. Dmcq (talk) 01:02, 3 January 2010 (UTC)

The 105th cyclotomic polynomial is example 27 in Richard K. Guy's The Strong Law of Small Numbers (in Guy's paper, not the Wikipedia article, but that could use an example...). PrimeHunter (talk) 02:09, 3 January 2010 (UTC)

The article Root_of_unity#Cyclotomic_polynomials talks about the cyclotomic polynomial of degree 105. Bo Jacoby (talk) 07:57, 3 January 2010 (UTC).

The conjecture li(n) ≥ π(n) holds for all natural numbers up to an insane value estimated as 1.397 × 10³¹⁶, but then it fails infinitely many times. — Emil J. 13:40, 5 January 2010 (UTC)

January 3

Generalized Heat Equation

Studying for a PDE exam, I came across the following practice problem

$T_{t}=(k(x)T_{x})_{x}+F(x),\,0<x<L,\,t>0$

where F(x)>0, k(x)>0, T(x,0)=f(x) with T(0,t)=T(L,t)=0. My question is how to even get started on this? Can someone please at least point me in the correct direction? The domain in x is finite so I am thinking separation of variables or the method of eigenfunction expansion and I have no idea what k(x) is. In fact, part of the problem asks for the conditions on F(x),f(x), and k(x) that are needed. I can't find it in any of my books either. Thanks! 75.171.178.10 (talk) 08:34, 3 January 2010 (UTC)

Start by studying a simple special case such as L=1, k(x)=1, F(x)=0. Bo Jacoby (talk) 13:30, 3 January 2010 (UTC).

But what's exactly your question? Yours is a second order parabolic equation. There are several approach to it (what PDE course was yours? Classical PDE's? Distributions? Semigroups? I think you are not supposed to rewrite the whole theory by yourself. If you are not given a precise problem, I suggest to go and look any textbook on PDE (a tame starting point is e.g. Evans, Partial Differential Equations, Chapt.7). --pma (talk) 16:59, 3 January 2010 (UTC)

The question is "For the boundary conditions given, find the solution T(x,t) in terms of suitable functions and stipulate any conditions on F(x), k(x), and f(x) that are needed". Specific cases I can do. For the past couple of weeks I have doing a plethora of such problems with all their variations on the forcing terms and the damping terms and the boundary conditions in different coordinate systems. And I know that sometimes even if a function (like the forcing term in $u_{t}=u_{xx}+F(x,t)$ with u(0,t)=u(L,t)=0) is not explicitly given you can still write down the solution (in this case using the method of eigenfunction expansion) in terms of F. But in the problem I mentioned above, k(x) is inside the partial derivative. So I have two unknown functions and even if I expand the partial, it looks just as formidable. So what method would I use on this to solve this PDE? This is just a classical PDEs course. Thanks!75.171.178.10 (talk) 00:20, 4 January 2010 (UTC)

"So I have two unknown functions"? No! T(x,t) is one unknown function. The functions F(x), f(x), and k(x) are supposed to be known, and so is the derivative k'(x). What is the PDE for G(x,t)=k(x)T(x,t)? Bo Jacoby (talk) 14:38, 4 January 2010 (UTC).

Diffusion Equation

Hi. I need some help with a homework question but I'll show you what I've done so far.

"The function $\theta (x,t)$ obeys the diffusion equation ${\frac {\partial ^{2}\theta }{\partial x^{2}}}={\frac {\partial \theta }{\partial t}}$. Find, by substitution, solutions of the form $\theta (x,t)=f(t)\exp {\frac {-(x+a)^{2}}{4(t+b)}}$, where a and b are arbitrary constants and the function f is to be determined. Hence find a solution that satisfies the initial condition $\theta (x,0)=exp(-(x+2)^{2})-exp(-(x-2)^{2})$ and sketch its behaviour for t≥0."

So, after some tedious algebra, I get $f(t)=k(t+b)^{-0.5}$ for some constant k. But I can't see how to get the initial condition part for the entire function - it seems to me that you would need to consider two separate cases, when a=2 and a=-2, to get it but surely you want to solve this for one particular a value not two. Help anyone? Thanks 92.0.129.48 (talk) 17:18, 3 January 2010 (UTC)

Hint: the heat equation u_t=u_xx being linear, if you are able to solve the initial value problems $v(x,0)=\exp(-(x+2)^{2})$ and $w(x,0)=\exp(-(x-2)^{2}),$ then you solve the given one with the function $\theta (x,t):=v(x,t)-w(x,t).$ Remark: on the first part (that you have already done) you could have considered just the case $a=b=0$ and found $\scriptstyle f(t)=k/{\sqrt {t}}.$ Then it is clear that if $u(x,t)$ solves the heat equation so does $u(x+a,\,t+b).$ This remark is useful for the previous hint too: you just have to solve the heat equation with initial condition $u(x,0)=\exp(-x^{2}),$ after all. Further remark: yours is the initial value problem for the homogeneous heat equation, whose solution admits an integral representation. You do not need it here of course, but you may notice that the general integral formula is based on the same idea of your exercise: finding solutions as (limits of) linear combinations of translations of the "fundamental solution" (that is, convolutions). --pma (talk) 18:29, 3 January 2010 (UTC) Was I clear? You should have found

$\scriptscriptstyle \theta (x,t)={\frac {1}{\sqrt {4t+1}}}\left(\exp \left({\frac {-(x+2)^{2}}{4t+1}}\right)-\exp \left({\frac {-(x-2)^{2}}{4t+1}}\right)\right)$--pma (talk) 21:32, 3 January 2010 (UTC)

Yes, thank you, I got that. Your comments were very helpful. 92.0.129.48 (talk) 23:13, 3 January 2010 (UTC)

Formal sums for quotients of primorials

What are the formal sums, with lower limit m and upper limit n, for (P(j)-1)#/(P(j))# and for (P(j+1)-1)#/(P(j))#, where P(j) means the j'th prime?

For (P(j-1)-1)#/(P(j))# I obtained the sum as -(P(j-1)-1)#/(P(j-1))#, but those other two have me stumped.

(Assuming that this answer for (P(j-1)-1)#/(P(j))# is correct, then, e.g., instead of (P(j+1)-1)#/(P(j))#, the answer for P(j)*(P(j)-1)#/(P(j))#, or (P(j)-1)#/(P(j-1))#, should be just as useful.)

I tried summation by parts and Abel's transformation, to no avail (but then again maybe I didn't apply them correctly).

FWIW: Last week I traveled a few dozen miles to the library of a university that awards graduate degrees in mathematics, and looked for books on primorials. CWOT. Yes, they have books on number theory and primes and finite differences and summation, but nothing on primorials.ImJustAsking (talk) 22:43, 3 January 2010 (UTC)

January 4

conversions

How many gallons is 60,000 liters? What fahrenheit is 500 °C?24.90.204.234 (talk) 01:56, 4 January 2010 (UTC)

We have a template, {{convert}}, for this kind of question. It returns results in the form Original number Original unit (Converted number target unit). For example, {{convert|60000|l|gal}} produces 60,000 litres (16,000 US gal), and {{convert|500|C|F}} produces 500 °C (932 °F). Intelligentsium 02:00, 4 January 2010 (UTC)

Thanks a lot.24.90.204.234 (talk) 02:36, 4 January 2010 (UTC)

Between gallons and liters is somewhat messy. In the USA, an inch is defined as 2.54 centimeters, and a gallon is 231 cubic inches. A liter is 1000 cubic centimeters.

${\begin{aligned}1{\text{ liter}}&=1000{\text{ cm}}^{3}=1000{\text{ cm}}^{3}\cdot \left({\frac {1{\text{ in}}}{2.54{\text{ cm}}}}\right)^{3}={\frac {1000}{2.54^{3}}}{\text{ in}}^{3}\\[8pt]&={\frac {1000}{2.54^{3}}}{\text{ in}}^{3}\cdot {\frac {1{\text{ gallon}}}{231{\text{ in}}^{3}}}={\frac {1000}{2.54^{3}\cdot 231}}{\text{ gallons}}.\end{aligned}}$

Now multiply that by 60000 and you've got it. Do not round until after you've crunched all the numbers. Rounding should be the last step. I'm getting about 60000 liters = about 15850.3 gallons. A British or "imperial" gallon is different, I think, and maybe the British actually have inches of a different size. Michael Hardy (talk) 02:42, 4 January 2010 (UTC)

Well, actually I'm referring to American gallons.24.90.204.234 (talk) 03:40, 4 January 2010 (UTC)

Everything above is in American gallons and American inches. Michael Hardy (talk) 07:04, 4 January 2010 (UTC)

Google is your friend: [1], [2]. Of course, knowing how to do the conversions yourself is very important. -- Meni Rosenfeld (talk) 09:13, 4 January 2010 (UTC)

American and British inches are exactly the same (unless you want the obscure American Survey Inch, if it exists?). The British (Imperial) gallon is one of the few things that are bigger (and better) in England ;) One Imperial gallon = 4.54609188 litres. (Yes, I know we don't have an empire any more!) and one US "dry gallon" is 4.40488377086 liters.Dbfirs 10:22, 4 January 2010 (UTC)

Cantor's Ordinal Numbers

Last night I was reading a book about some of Cantor's work. After reading the articles here on wikipedia, I had a question. Assuming the axiom of choice, what is the cardinality of the set of all transfinite cardinal numbers? I am thinking it must aleph-zero because we can write them in a sequence

$0,1,2,3,\cdots ,n,\cdots ;\aleph _{0},\aleph _{1},\aleph _{2},\cdots ,\aleph _{\alpha },\cdots .$
So there are countably infinite of these transfinite cardinal numbers, right? And also just to clarify,

$\aleph _{0}=card(\mathbb {N} )$

$\aleph _{1}=card(P(\mathbb {N} ))$

$\aleph _{2}=card(P(P(\mathbb {N} )))$ and so on, right?-Looking for Wisdom and Insight! (talk) 03:35, 4 January 2010 (UTC)

Not exactly, in answer to your latter question. Have you read the article on the continuum hypothesis? The article cardinal numbers helps answer your questions. You read it, but did not understand it?Julzes (talk) 04:03, 4 January 2010 (UTC)

I thought I had it but I guess not. The article says $2^{\aleph _{0}}=\aleph _{1}$ and $2^{\aleph _{\alpha }}=\aleph _{\alpha +1}$ and I thought that for a given set A, $card(P(A))=2^{card(A)}$. For sure, I know that this is true for finite sets A but I thought that $2^{\aleph _{\alpha }}=\aleph _{\alpha +1}$ was motivated by this. So if $\aleph _{0}=card(\mathbb {N} )$, then $card(P(\mathbb {N} ))=2^{card(\mathbb {N} )}=2^{\aleph _{0}}=\aleph _{1}$. Is this not right?-Looking for Wisdom and Insight! (talk) 04:17, 4 January 2010 (UTC)

Sir (or madam), is this a mathematician's idea of a joke? Surely you are contradicted by the fact that the article does not say what you claim.Julzes (talk) 04:23, 4 January 2010 (UTC)

I'll assume you breezed over the article for a moment. The power set of one set is proven to have larger cardinality than the set itself, but the generalized continuum hypothesis is known to be independent of ZFC (and the truth or falsity's status is of disputable value for applied mathematics). What this means is that the equality you state depends on the assumption of an hypothesis that can neither be proved true nor false inside or outside of mathematics.Julzes (talk) 04:31, 4 January 2010 (UTC)

Well, I wasn't joking and that is why I posted this question because I didn't understand it. I posted it here to see if someone can clarify it a bit. I didn't do it to waste mine or anyone else's time. So you are saying that my second question cannot be answered? What about the first? Is that at least correct?-Looking for Wisdom and Insight! (talk) 04:40, 4 January 2010 (UTC)

Cardinals are weird. Correct, either set theory is uselessly broken or it is independent of ZFC that ℵ₁ is the cardinality of the set of subsets of the integers. Some people do prefer the cardinality of the set of subsets of the integers to be a different aleph number. For every cardinal X, there is a set of cardinals Y such that |Y| is strictly larger than X. In particular, in ZFC there is no "set of all cardinals". You can construct Y fairly easily, Y = { ℵ_k : k is an ordinal such that |k| ≤ |P(X)| }. Note that the aleph numbers do not "stop" at the positive integers, after ℵ₁, ℵ₂, ..., you get ℵ_ω where ω is the least ordinal that is infinite; ω = {0,1,2,...} is the set of non-negative integers (finite ordinals) under the obvious ordering in von Neumann's model of ordinals. The problem with your idea is that the "sequence" you wrote down is transfinite, not countable. JackSchmidt (talk) 04:55, 4 January 2010 (UTC)

It looks right to me. That appears to me to be a statement about the sizes of sets that Cantor proved prior to a lot of controversy during which the concept of a set was clarified.Julzes (talk) 04:51, 4 January 2010 (UTC)

(e/c) To be clear, the first question is not correct. There is no set of all transfinite cardinals, and there are sets of transfinite cardinals that are uncountably infinite. JackSchmidt (talk) 04:59, 4 January 2010 (UTC)

The funny thing is that you can run into Russell's paradox when you try to write a decent sentence answering your first question.Julzes (talk) 05:00, 4 January 2010 (UTC)

Thanks, JackSchmidt.Julzes (talk) 05:02, 4 January 2010 (UTC)

Thanks and sorry if I annoyed anyone with my slowness.-Looking for Wisdom and Insight! (talk) 05:35, 4 January 2010 (UTC)

No problem. To clarify one thing Jack Schmidt seems to say, there is mathematics done under assumptions of where in the aleph scheme the cardinality of the continuum fits (the cardinality of the continuum is 'c' written in Fraktur script and equals the cardinality of the set of subsets of the integers). To clarify an earlier parenthetical by me; in terms of real world applications of mathematics, it so far has not been decided if it matters what one assumes on the subject--and perhaps different assumptions will each find some use. Most mathematicians get along just fine without any of the mathematics around that particular question, with the following quote made by Gerald Folland in his Real Analysis: Modern Techniques and Their Applications probably being typical:

"My own feeling, subject to revision in the event of a major breakthrough in set theory, is that if the answer to one's question turns out to depend on the continuum hypothesis, one should give up and ask a different question."

I'm not sure one also shouldn't also relate what such a question is to those who specialize in the subject area, but you should get the idea.Julzes (talk) 07:24, 4 January 2010 (UTC)

I have to object to the subtext here, that CH or not CH is a matter of what we assume. There are very serious set theorists trying to find out whether or not it's true. The fact that it can neither be proved nor refuted from ZFC has little bearing on that, though it does certainly indicate that any successful approach to the question will have to go outside the sorts of argument that are familiar to mathematicians.

Folland's attitude is highly practical, given that as he was not doing research in set theory, he was unlikely (though you never know!) to stumble on anything shedding much light on CH's truth or falsity. But you'll notice the subject to revision clause, which is also very appropriate. Some people see the work of W. Hugh Woodin on Ω-logic as the possible start of Folland's "breakthrough", though few if any are bold enough to claim the question is settled. --Trovatore (talk) 07:45, 4 January 2010 (UTC)

Well, wouldn't you say that the question of whether it is "true" essentially depends on its usefulness? ZFC and ZFC+CH are either both internally consistent or both not. So far, we should probably say that at least they are equally useful and withhold judgement about whether the truth or falsity of CH has any meaning at all without appealing to what it means to what mathematics ultimately models. I personally find the possibility that both "choices" of the truth value have application in some way highly unlikely, but even that can't be determined at the present state of our knowledge, can it? I ask this as a relative ignoramus, so don't take offense. I'll look into the article on omega logic.Julzes (talk) 08:05, 4 January 2010 (UTC)

No, I wouldn't say that whether it's true depends on its usefulness. The truth or falsity of the statement the apple is red does not depend on whether it's useful for the apple to be red; it depends on whether the apple is red.

Similarly, from a realist perspective, the truth of the statement there exists a wellordering of the real numbers whose every proper initial segment is countable (aka CH) does not depend on whether such a wellordering is useful, but just on whether there is one. --Trovatore (talk) 08:09, 4 January 2010 (UTC)

Are you saying the real numbers cannot be formed from either ZFC+~CH or ZFC+CH? I'm misunderstanding something in this subject, perhaps. Pardon me if I seem a little tired, but I was simply here originally because a questioner hadn't done some simple basic research very carefully. Perhaps I'm doing the same thing.Julzes (talk) 08:42, 4 January 2010 (UTC)

I've answered my own question. Second order logic versus first-order logic. New fundamental concepts I hadn't been given any reason to study in the past. No, ZFC requires augmentation with some second order logic to get the reals because of the supremum axiom. Sorry to be so uninformed about your specialty, Trovatore, but it looks like the truth of the continuum hypothesis may be contained in the reals themselves regardless of the axiom scheme, as you said.Julzes (talk) 09:23, 4 January 2010 (UTC)

Trovatore, when you said "There are very serious set theorists trying to find out whether or not it's true", do you mean it in a realist's perspective? I.e is CH true for the platonic real numbers? I've read that people are trying to figure out new "intuitive" axioms that can be used to derive CH or its negation, for example Freiling's axiom of symmetry. I just got my books for logic and set theory today and hopefully I will understand the von Neumann universe soon. Wish me luck! Money is tight (talk) 17:50, 4 January 2010 (UTC)

It may be like the P versus NP problem. It may be that an actual proof is found using techniques we haven't dreamed of yet. It may be people just accept it as an axiom because the evidence becomes so great. Or it may just be left for ever as a problem. Dmcq (talk) 18:26, 4 January 2010 (UTC)

Term by Term Integration of a Fourier Series

On an different note, I understand that if you have a sum of functions converging uniformly to the limiting function, then that series can be (Riemann) integrated term by term. In context of Fourier series, how can you tell when a series can be integrated term by term? Because as I understand it, uniform convergence is just sufficient for term by term integration. It is not necessary. We have examples of series which can be integrated term by term even though they are not uniformly convergent. So IF we have uniform convergence, then we know we can integrate term by term but what if we don't have uniform convergence, then how can I tell if the Fourier series in question can be integrated term by term or not? I can always just find the Fourier series representation of the integral of the function and then see if it matches the term by term integration of the original Fourier series but is there any other (quicker perhaps) way?-Looking for Wisdom and Insight! (talk) 03:43, 4 January 2010 (UTC)s

There are several convergence theorem in integration theory to do that, of course. However note that Riemann integrability is unsatisfactory, for the reason that it is a quite unstable property with respect to convergence. A limit of a sequence of Riemann integrable functions in a convergence weaker than uniform may fail to be Riemann integrable. This is the main reason why at a certain moment a more general integration theory became needed (in fact, the main problem was exactly the convergence of Fourier series). Here's an elementary result about integration and series: if a series of integrable functions $\scriptstyle \sum _{k=0}^{\infty }f_{k}$satisfies $\scriptstyle \sum _{k=0}^{\infty }\|f_{k}\|_{1}<\infty$ then it converges almost everywhere and in the $L^{1}$ norm (in particular, you can integrate term by term). Related facs: you may see the exchange sum/integration as a particular case of a change of order in iterated abstract integration and apply the Fubini and Tonelli theorems. --pma (talk) 11:44, 4 January 2010 (UTC)

Sums of squares

Hi. I found a spreadsheet on my hard drive that I don't remember making. It must have been a while ago. In it, I've got a column with the natural numbers, a second column with their squares, a third column with partial sums of squares (1, 5, 14, 30, 55, etc.), and a fourth column with the square roots of those partial sums. The only two integers in the fourth column seem to be 1 and 70, hence:

$\sum _{k=1}^{1}k^{2}=1^{2}$ and $\sum _{k=1}^{24}k^{2}=70^{2}$

I wonder: Are those the only two integers I'll ever see in column 4? I've only got 1024 4096 rows so far... Thanks in advance for any insight. (Also, any ideas as to what I was thinking when I made these computations would be welcome. o_0) -GTBacchus^(talk) 07:56, 4 January 2010 (UTC)

There are no more integers in the first million rows. It seems likely that there are no more integers at all, but I don't know how to prove that.

In case you didn't already know, the formula $\sum _{k=1}^{n}k^{2}={\frac {n(n+1)(2n+1)}{6}}$ may be useful. -- Meni Rosenfeld (talk) 09:27, 4 January 2010 (UTC)

Actually, the proof should be easy. For large enough n, ${\frac {n(n+1)(2n+1)}{6}}\approx {\frac {2n+3}{6}}n^{2}$, and ${\sqrt {\frac {2n+3}{6}}}$ is irrational. The rest is left as an exercise. -- Meni Rosenfeld (talk) 09:41, 4 January 2010 (UTC)

The only numbers that are simultaneously square and pyramidal (the cannonball problem) are $P_{1}=1$ and $P_{24}=4900$, corresponding to $S_{1}=1$ and $S_{70}=4900$ (Ball and Coxeter 1987, p. 59; Ogilvy 1988; Dickson 2005, p. 25), as conjectured by Lucas (1875, 1876) and proved by Watson (1918).

— Weisstein, Eric W., Square Number., From MathWorld

--132.230.101.16 (talk) 09:47, 4 January 2010 (UTC)

I guess it's not as easy as I thought. -- Meni Rosenfeld (talk) 11:10, 4 January 2010 (UTC)

By the way, no need to go as far as (the obsolescent, IMO) MathWorld. The information is available locally, at Square pyramidal number. -- Meni Rosenfeld (talk) 12:28, 4 January 2010 (UTC)

Square pyramidal numbers that are also square numbers; that's it. Thanks to all. -GTBacchus^(talk) 17:31, 4 January 2010 (UTC)

Regarding the other question... Does this spreadsheet look like something that you would have written, but you just don't recall the act of having done so, or does it look completely alien? If the latter, it was probably written by someone else. -- Meni Rosenfeld (talk) 12:28, 4 January 2010 (UTC)

I don't think anybody but me has used my copy of OpenOffice Calc for number crunching. I'm sure I wrote it, and it may have been after reading something about figurate numbers. The formulas are all written they way I'd write them, but there's not a lot of creativity involved in typing "=sum(b$2..b15)" or something. Hard to say... -GTBacchus^(talk) 17:31, 4 January 2010 (UTC)

It's a bit off-topic, but that's a tricky age-dependent question involving human memory, and if the user feels certain (s)he neither allowed someone else to use the computer nor did the spreadsheet him- or herself, I would lean toward being mistaken on the latter. Perhaps when you were 13 or 14 years old you wanted to check something you read on the subject, GTBacchus? Or it could have been independent work (People are always rediscovering things). It could have been related to a school assignment also. Unless there is some jarring of your memory (or it occurs to you who else might have been working on the computer), we probably can't help with that question.Julzes (talk) 13:01, 4 January 2010 (UTC)

I wasn't 13 or 14, because I didn't own a computer then. I've had this one since '06, which is the year I turned 30. It was probably just some random late-night number theory doodling, I suppose. That sort of thing happens around here. Thanks again. :) -GTBacchus^(talk) 17:31, 4 January 2010 (UTC)

Actually this question has also been posed one year ago; check Wikipedia:Reference_desk/Archives/Mathematics/2009_February_25#Simultaneously_pyramidal_and_square and the links. --pma (talk) 17:26, 4 January 2010 (UTC)

One of those links triggered the memory! I was thinking about Squaring the square, no question. I was wondering whether one could have square tiles with edges 1, 2,..., x, and arrange them to cover a square region. Apparently, the only set where you'd have a chance would be if x=24 and the playing board is 70X70. It appears from our Squaring the square article that there's no solution using that set, but that's what I was after. I think it came up when I was teaching high school in Seattle, and cutting out a lot of shapes with my geometry students. I'm glad I remembered that. -GTBacchus^(talk) 17:38, 4 January 2010 (UTC)

If anyone is able to edit SVG images, File:Squaring the square.svg could do with having the "112 x 112" in the bottom right brought properly into view. (BTW there are also other size PNGs derived from that - is that automatic?) -- SGBailey (talk) 08:20, 5 January 2010 (UTC)

The original SVG was correct, but it apparently hit a bug in the PNG rendering library (a weird one, I couldn't even reproduce it with a local copy of rsvg). I replaced all the css with regular attributes, which seems to have fixed the problem. — Emil J. 13:16, 5 January 2010 (UTC)

Rates of decline

Hi, This should be quite a simple problem, but I am having problems figuring out the quickest way to solve it... Basically a business made $10 million in the last per year, and this profit is declining at a rate of 10% per year. Estimate the total sum of profit made by the business over the next 5.7 years....

While it would be possible to solve this problem by working this out year by year (although the 0.7 of a year might pose problems) i was wondering if there was a quicker and easier way to solve it? I am sure there is!!

Many Thanks 80.47.236.247 (talk) 08:10, 4 January 2010 (UTC)

When you say it is declining at 10% per year, are you talking about an instantaneous rate or an "effective" rate? I.e. will the profit per year one year from now be 90% of what it is now (so it's an effective rate), or is the current instantaneous rate of decline such that that's what it would be if it continued at the same rate (so it's an instantaneous rate)? Michael Hardy (talk) 16:34, 4 January 2010 (UTC)

Do you know calculus? You could solve it as an integral, assuming the rate of decline is continuous. -GTBacchus^(talk) 08:15, 4 January 2010 (UTC)

I don't really know calculus very well, however yes the rate of decline is continuous... i.e. 10% down from $10m in the first year = $9m, and then $8.1m in the second year and so on.... could u explain with use of a formula how this therefore might be solved? It would be much appreciated! :) 80.47.137.127 (talk) 08:56, 4 January 2010 (UTC)

Actually what you've described is not what we consider continuous. In the discrete case, which is what you've described, you can use the formula for the sum of a geometric series. -- Meni Rosenfeld (talk) 09:32, 4 January 2010 (UTC)

Oh right.. so put simply, could you explain how to compute a 10% per year decline on $10m over 5.7 years? I am still having difficulties getting my head around this! thanks. 80.47.137.127 (talk) 10:02, 4 January 2010 (UTC)

Just looking at the 5.7 years, it raises the issue, do you earn in .7 years (0.7 * what you would have earned in the whole year) or is the 10% a gradual change over the course of the year? If you do, then you can use the formula half way down Geometric progression, k=0, n=4, r=0.9, a=10 million. Then, work out the sixth term, ar^5, multiply by 0.7 and add for your final answer. - Jarry1250 ^{[Humorous? Discuss.]} 10:31, 4 January 2010 (UTC)

Of course, it would be natural to look for a continuous decline that matches up with the discrete decline. For an article, you might look at compound interest, but it won't be quite right for your problem. One problem with trying to answer this honestly is that you are simply saying what the problem is and then asking people to work it out for you in detail. Let's get to the bottom of whether the decline is or is not continuous before trying to answer this further.Julzes (talk) 11:42, 4 January 2010 (UTC)

I think we need to know more about the background to the problem, then we can work out whether it should be modelled as a continuous or discrete decline and what ought to be done about the incomplete year. --Tango (talk) 17:12, 4 January 2010 (UTC)

jection

Bijection, injection, surjection. What is a jection? Wiktionary has nothing on the etymology. -- SGBailey (talk) 14:00, 4 January 2010 (UTC)

oed says both bijection and surjection arise from injection, which comes from the latin verb injicere Tinfoilcat (talk) 14:21, 4 January 2010 (UTC)

Usually spelled inicere, and itself derived from the verb iacere, which gives you the actual root the three -jections share. — Emil J. 15:15, 4 January 2010 (UTC)

Also related to trajectory, I think. Michael Hardy (talk) 16:23, 4 January 2010 (UTC)

And I'd conjecture that some other compound of iacio used in maths is here around. We may add "projection" if you have no objection (why should we reject it?) --pma (talk) 17:35, 4 January 2010 (UTC)

So it's "two-way throwing", "Throwing in" and "Throwing on"? -- SGBailey (talk) 19:25, 4 January 2010 (UTC)

Sounds right. -GTBacchus^(talk) 03:19, 5 January 2010 (UTC)

Adjective, conjecture, ejection, subject. Bo Jacoby (talk) 09:43, 5 January 2010 (UTC).

"pma"'s puns will be lost on some people, I suspect.... Michael Hardy (talk) 19:29, 5 January 2010 (UTC)

Well, some people have trouble parsing anything more complex than a simple interjection... -GTBacchus^(talk) 03:50, 6 January 2010 (UTC)/me ducks the inevitable rotten fruit, hopes he's not ejected from the board for it all

BTW, to make the list complete, I wonder what kind of a map would be a dejection in maths. The Italian word kept a meaning which is somehow closer to the concreteness of mathematical objects, compared with the English meaning. --pma 11:04, 6 January 2010 (UTC)

A quotient? A covering map? -GTBacchus^(talk) 18:07, 6 January 2010 (UTC)

"Covering map" sounds perfect --especially fitting with the Italian term "deiezione" ;-) pma 14:19, 7 January 2010 (UTC)

Of course, all these -jections have got adjective forms as well... -GTBacchus^(talk) 18:15, 6 January 2010 (UTC)

January 5

Convergence

I am working through a proof and I am at a part that might be pretty simple but I am a bit confused. It's probably stuff I understood well a few months ago and haven't worked on since and now I feel silly. Here is the expression I am working with, and what the proof says right after it:

$\sum _{n=1}^{\infty }{\frac {1}{n^{2+2\epsilon }}}+\sum _{m=1}^{\infty }\sum _{n=-\infty }^{\infty }{\bigg [}{\frac {1}{(mz+n)^{2}|mz+n|^{2\epsilon }}}-\int _{n}^{n+1}{\frac {dt}{(mz+t)^{2}|mz+t|^{2\epsilon }}}{\bigg ]},\qquad \epsilon >0,z\in \mathbb {H} {\text{(upper half-plane)}}.$

Both sums on the right converge absolutely and locally uniformly for $\epsilon >-{\frac {1}{2}}$ (the second one because the expression in square brackets is $O(|mz+n|^{-3-2\epsilon })$ by the mean-value theorem, which tells us that $f(t)-f(n)$ for any differentiable function $f$ is bounded in $n\leq t\leq n+1$ by $\max _{n\leq u\leq n+1}|f'(u)|$), so the limit of the expression on the right as $\epsilon \to 0$ exists and can be obtained simply by putting $\epsilon =0$ in each term.

So, I get the first sum, that's easy. The second one confuses me. I think I understand partially. I believe I understand the mean value theorem part as we would have a $c$ such that $f'(c)={\frac {f(t)-f(n)}{t-n}}$ and $t-n\leq 1$. So, for that $c$, we have $f(t)-f(n)\leq (t-n)f'(c)\leq f'(c)$. Then, the derivative of the integral is 0 and so the derivative of the thing in square brackets is just the derivative of the other term. I guess I definitely don't understand why showing the limit exists tells us we can just plug in $\epsilon =0$. I also don't understand if the double sum screws things up. Thanks. StatisticsMan (talk) 03:00, 5 January 2010 (UTC)

Maybe a simpler way to say it is $\left|f(n)-\int _{n}^{n+1}f(t)dt\right|\leq \int _{n}^{n+1}|f(n)-f(t)|dt\leq \sup _{n\leq u\leq n+1}|f(n)-f(u)|\leq \sup _{n\leq u\leq n+1}|f'(u)|.$

The fact that there's a double sum does come into play, but you can show that $\sum _{a}\sum _{b}{\frac {1}{|a+b|^{s}}}$ converges for s > 2, so you're still fine there with s = 3 + 2ε.

For arguing that thing is continuous in ε near zero, I'm pretty sure there must be some sort of theorem about the continuity of a series of continuous functions that are bounded by an absolutely convergent series or something like that. I'm not really sure exactly what it is, but it shouldn't be too hard to show if you don't have anything like that. The sum of all but finitely many terms is small and a finite sum of continuous functions is continuous. Probably someone will come along with a better informed answer to that. Rckrone (talk) 06:10, 5 January 2010 (UTC)

Yes, you are talking of the so called Weierstrass M-test. --pma (talk) 13:48, 5 January 2010 (UTC)

Does the fact that our function is complex-valued and not real-valued affect the mean value theorem? StatisticsMan (talk) 14:58, 6 January 2010 (UTC)

No, because the mean value theorem in the form of the inequality $\|f(b)-f(a)\|\leq (b-a)\sup _{a<t<b}\|f^{\,\prime }(t)\|$ holds true for any normed space valued function f continuous on the interval [a,b] and differentiable on (a,b) (BTW even continuous and differentiable up to a countable set is OK, and even absolutely continuous and differentiable a.e. is OK). What is no longer true for vector valued functions, even for E = R², is that there is a a<t<b such that f(b)-f(a)=(b-a)f'(t). --pma 00:18, 7 January 2010 (UTC)

Question on combinations

I suspect there's an easy formula for figuring this out, but I can't figure out what it is:

Given n teams in a league (assume n is even), how many possible opening day match-ups are there? The order of each match-up determines which is the home team, so Team-A vs. Team-B is different from Team-B vs. Team-A. However, it doesn't matter in which order the matches are listed: A v B and C v D is the same as C v D and A v B.

It's not a simple combination, nor a permutation that I can figure. Any ideas? –RHolton≡– 17:40, 5 January 2010 (UTC)

The first team has n-1 choices for who to play and then 2 choices for who is the home team. Once that's decided, the next team on the list that isn't already scheduled for a game has n-3 choices for who to play and 2 choices for who is the home team. The next has n-5 choices, etc. So there are $\prod _{k=1}^{n/2}2(2k-1)$ possibilities assuming everyone plays. Rckrone (talk) 17:57, 5 January 2010 (UTC)

First find the number of (unordered) partitions of a set of size n into sets of size 2. Then you can multiply that by 2^n/2 (2 choices for each of n/2 pairs) to get the number you're looking for. To be continued..... Michael Hardy (talk) 19:23, 5 January 2010 (UTC)

...and here is something possibly somewhat relevant.

But I think Rckone's answer may be enough for your purpose. Michael Hardy (talk) 19:26, 5 January 2010 (UTC)

Also, if N is such a number, N(n/2)!=n! (permuting the n/2 pairs in each of the N sets of pairs you get every permutation of the n teams, each once).--pma (talk) 19:54, 5 January 2010 (UTC)

Also you may first choose the subset of n/2 home teams, and then select a bijection with its complement (there are of course as many such bijections as there are (n/2)-permutations); and of course the result is again Rckrone's formula.--pma (talk) 20:03, 5 January 2010 (UTC)

I like Maple's answer for Rckrone's formula. Since n is even, let's assume that n = 2m where m is a positive integer. Then Maple gives

$\prod _{k=1}^{m}2(2k-1)={\frac {4^{m}}{\sqrt {\pi }}}\,\Gamma \!\left(m+{\frac {1}{2}}\right),$

where Γ is Euler's Gamma function. More beautiful, but far less applicable. ~~ Dr Dec (Talk) ~~ 20:15, 5 January 2010 (UTC)

Möbius Maps

Hi. I am trying to find the group G of Möbius transformations that map the set {0,1,$\infty$} onto itself. Now, I have looked through my notes (this is a course on group theory incidentally) and have found the general function to be $g(z)={\frac {z-z_{0}}{z-z_{\infty }}}{\frac {z_{1}-z_{\infty }}{z_{1}-z_{0}}}$, where you choose your $z_{i}$ accordingly. Problems obviously arise though when you pick one z to be $\infty$; indeed, what meaning does $g(z)={\frac {z-z_{0}}{z-{\infty }}}{\frac {z_{1}-{\infty }}{z_{1}-z_{0}}}$? So, to solve this problem my lecturer told us that, in the case $z_{\infty }=\infty$ for example, rewrite your function as $g_{1}(z)={\frac {z-z_{0}}{z_{1}-z_{0}}}$, which obviously takes $z_{0}$ to 0 and $z_{1}$ to 1. What I don't understand is how this is, in general, supposed to take $z_{\infty }$ to ${\infty }$. How does this work when $z_{0}=1$ and $z_{1}=0$? By my logic, for this case $g(z_{\infty })=-\infty$, most definitely not a result I want. Can anyone help me out here? Thanks 92.0.129.48 (talk) 20:17, 5 January 2010 (UTC)

Check this. Does it give you a clue? Also note that $\infty$ here is the point that compactifies the complex plane to get the Riemann sphere. There's no $-\infty$ (or if you like, it coincides with $\infty ,$ and is one of the two fixed points of the involution $z\mapsto -z$).--pma (talk) 20:38, 5 January 2010 (UTC)

If $z_{\infty }=\infty$ then you are restricting yourself to the sub-set of Möbius maps that fix $\infty$ - these are the affine maps $g(z)=az+b$. To map $z_{0}$ to 0 and $z_{1}$ to 1 then $a={\frac {1}{z_{1}-z_{0}}}$ and $b={\frac {z_{0}}{z_{0}-z_{1}}}$, so you have $g(z)={\frac {z-z_{0}}{z_{1}-z_{0}}}$, as you said. In particular, if $z_{0}=1$ and $z_{1}=0$ then $g(z)=1-z$. Geometrically, in the complex plane, this is a rotation through 180 degrees about the point $z={\frac {1}{2}}$. Gandalf61 (talk) 08:27, 6 January 2010 (UTC)

A maybe-obvious-but-maybe-worth-to-recall remark. Since a Moebius map is fixed by the image of three points, it should be clear that the subgroup G is isomorphic to the symmetric group S₃. It is generated e.g. by the maps 1-z and 1/z, the map 1-z corresponding to a transposition (0,1)(∞), and 1/z corresponding to (0,∞)(1). You may enjoy finding the other 4 maps by means of compositions of these, together with the associated permutations of {0,1,∞}. --pma 10:28, 6 January 2010 (UTC)

Thank you both for your help. On a related note though, what exactly is the order of a Möbius map f(z)? Is it just how many times must you apply f to reach the identity? Thanks 92.0.129.48 (talk) 19:45, 6 January 2010 (UTC)

Yepp. Note that order has a lot of meanings in maths; but here the group theoretic one (that is the one you mention) is I think the only reasonable one. --pma 22:34, 6 January 2010 (UTC)

NP and NPC

Just wondering, because it is not specifically stated in the NP and NP-complete articles... Does a solution to an NP problem imply that all NPC is solvable? If it is proven that there is no solution to an NP problem, does it imply that there is absolutely no solution to all NPC problems? -- kainaw™ 21:50, 5 January 2010 (UTC)

I think you have the wrong idea about what NP means. All P problems are automatically NP. Problems that are not NP are harder than NP problems, not easier. --Trovatore (talk) 21:59, 5 January 2010 (UTC)

I wasn't wondering about NP and P. I was wondering if the following statement is reversible... Solving (in polynomial time) a single NP-complete problem proves that there is a polynomial-time solution to all NP problems. By "reversible", I mean is the following true: Proving that there is no polynomial time solution to an NP problem proves that there is no polynomial-time solution to any NP-complete problem. -- kainaw™ 22:09, 5 January 2010 (UTC)

Sure. That's just the contrapositive of the original statement. You don't need to know anything about P and NP for that. --Trovatore (talk) 22:13, 5 January 2010 (UTC)

That is what I thought, but I wasn't certain that there wasn't some rarely mentioned snag in the whole thing that made the contrapositive incorrect. -- kainaw™ 22:23, 5 January 2010 (UTC)

For what it's worth, your statement is a more precise fit for "NP-hard" than for "NP-complete"; NP-complete is just the intersection of NP-hard and NP. (In case it's confusing not to state it explicitly, NP-hard is not a subset of NP.) --Tardis (talk) 22:30, 5 January 2010 (UTC)

January 6

Predicting number of items in a set (statistics)

I remember reading about a method to solve this type of problem, but I don't remember what it's called.

There is a set of n integers. I pick 10 randomly, one at a time, and with replacement.

10

11

14

0

9

11

7

13

14

0

Using this list, how would I predict n? The actual numbers pulled are not important, but repetition is. Can someone explain how to analyze this data? (n=15 in this example)

The general problem is:

1. Using the output list of k items, predict n. k can be greater than n.
2. What is the probability that the predicted n is accurate (predicted n = actual n)?

--Yanwen (talk) 00:33, 6 January 2010 (UTC)

I believe (although did not check rigorously) that maximum likelihood estimator of n would be maximum of your set of numbers. I'd leave calculating probabilities to others (Igny (talk) 02:25, 6 January 2010 (UTC))

We have an article about this: German tank problem. The maximum likelihood estimator in this case is the maximum observed value, but generally that's not likely to be the right value; it's quite biased. The best unbiased estimator is (k + 1)m/k where m is the maximum observed value. If I recall correctly, there's another multiple of m that gives a biased estimator with a slightly smaller mean squared error than that of the bset unbiased estimator. Michael Hardy (talk) 03:35, 6 January 2010 (UTC)

PS: You haven't said explicitly that you take the set to be {1, 2, 3, ..., n}, but I've assumed that's what you meant. But If we drop that assumption, I think we can still deal with that problem, but I'd have to think that one through. Michael Hardy (talk) 03:37, 6 January 2010 (UTC)

You are correct. I did not specify that the list was {1, 2, 3, ..., n} because I did not mean for it to be. In fact, I am attempting to solve this problem where a server randomly outputs a message. And I need to estimate the number of distinct messages so I can make sure I find all of them without wasting resources by simply sending a billion requests. The simple numbering is to make the problem easier, and then generalize it. --Yanwen (talk) 04:26, 6 January 2010 (UTC)

If your observations are drawn from a population that is not numbered consecutively, then the analysis in mark and recapture may be useful. Gandalf61 (talk) 08:05, 6 January 2010 (UTC)

Mark and recapture seems to give underestimates. If you divide the list in two and pretend that 5 were on the first capture, and another 5 were on the second capture, you get.

10

11

14

0

9

11

7

13

14

0

$N\approx {\frac {MC}{R}}={\frac {5\cdot 5}{3}}=8.{\bar {3}}$ --Yanwen (talk) 15:57, 6 January 2010 (UTC)

The number of times each integer appears in your sample sequence is distributed multinomially with k trials and n equally probable outcomes (the notation is reversed from that in the article). A logical thing to do is to choose n for which the distribution most closely matches the observed counts.

One way to do this is choose some statistic of the counts and pick n for which the expected statistic is equal to the observed one. For example, you can take the sum of squares of counts. The expected value is $\mathbb {E} [S]={\frac {k^{2}}{n}}+k\left(1-{\frac {1}{n}}\right)$ so you can estimate $n\approx {\frac {k^{2}-k}{S-k}}$. In your example, the count vector is (2,1,1,1,2,1,2), the sum of squares is 16 and so $n\approx 15$.

This method is not rigorous enough to answer your second question. For this, you need to choose a prior distribution on n and update it using Bayesian inference. -- Meni Rosenfeld (talk) 09:58, 6 January 2010 (UTC)

I thought that in second question OP asked for $P({\hat {n}}=n|n)$ where n is fixed and the estimate ${\hat {n}}$ is random. But if one treats n as random, one needs the prior distribution indeed. (Igny (talk) 13:38, 6 January 2010 (UTC))

That's one way to interpret it. But since the sample is known and n is not, it is more useful to know the posterior distribution (in the subjective probability sense) of n. -- Meni Rosenfeld (talk) 13:59, 6 January 2010 (UTC)

That seems like an accurate method Meni Rosenfeld. Just one thing, how did you derive the formula for expected value? --Yanwen (talk) 15:57, 6 January 2010 (UTC)

Each of the counts (call it $X_{i}$) is distributed binomially with k trials of probability $1/n$ (they are dependent, but it doesn't matter). We have $\mathbb {E} [X_{i}^{2}]=\mathbb {E} [X_{i}]^{2}+\mathbb {V} [X_{i}]=(k/n)^{2}+k(1/n)(1-1/n)$. Then $\mathbb {E} [S]=\mathbb {E} [\Sigma X_{i}^{2}]=\Sigma \mathbb {E} [X_{i}^{2}]=n\mathbb {E} [X_{i}^{2}]$. -- Meni Rosenfeld (talk) 19:37, 6 January 2010 (UTC)

The article on Multiset may be useful for your problem. Bo Jacoby (talk) 23:01, 6 January 2010 (UTC).

Try a simpler problem first: "There is a set of n objects. I pick k=3 objects randomly, one at a time, and with replacement". The n³ outcomes are classified into 3 possible observations: "three of a kind", "a pair", and "singletons". For n=1 the odds for these three observations are as 1 : 0 : 0. For n=2 the odds are as 2 : 6 : 0. For n=3 the odds are as 3 : 18 : 6. For n=4 the odds are as 4 : 36 : 24. Generally the odds are as n : 3n(n−1) : n(n−1)(n−2). The probabilities are obtained by dividing odds by n³, giving n⁻² : 3n⁻¹(1−1/n) : (1−1/n)(1−2/n). Each of these probabilities, considered as a function of n, is a likelihood function containing the information of the observation. The likelihood function n⁻² can be normalized into a probability distribution function (see Basel problem) with infinite mean value and infinite standard deviation. The likelihood functions 3n⁻¹(1−1/n) and (1−1/n)(1−2/n) cannot be normalized. It now remains to generalize this approach to higher values of k. Bo Jacoby (talk) 16:56, 7 January 2010 (UTC).

The following seems relevant:

• Inferences from Multinomial Data: Learning about a Bag of Marbles
• Author(s): Peter Walley
• Source: Journal of the Royal Statistical Society. Series B (Methodological), Vol. 58, No. 1 (1996), pp. 3-57
• Published by: Blackwell Publishing for the Royal Statistical Society
• Stable URL: http://www.jstor.org/stable/2346164

Abstract
A new method is proposed for making inferences from multinomial data in cases where there is no prior information. A paradigm is the problem of predicting the colour of the next marble to be drawn from a bag whose contents are (initially) completely unknown. In such problems we may be unable to formulate a sample space because we do not know what outcomes are possible. This suggests an invariance principle: inferences based on observations should not depend on the sample space in which the observations and future events of interest are represented. Objective Bayesian methods do not satisfy this principle. This paper describes a statistical model, called the imprecise Dirichlet model, for drawing coherent inferences from multinomial data. Inferences are expressed in terms of posterior upper and lower probabilities. The probabilities are initially vacuous, reflecting prior ignorance, but they become more precise as the number of observations increases. This model does satisfy the invariance principle. Two sets of data are analysed in detail. In the first example one red marble is observed in six drawings from a bag. Inferences from the imprecise Dirichlet model are compared with objective Bayesian and frequentist inferences. The second example is an analysis of data from medical trials which compared two treatments for cardiorespiratory failure in newborn babies. There are two problems: to draw conclusions about which treatment is more effective and to decide when the randomized trials should be terminated. This example shows how the imprecise Dirichlet model can be used to analyse data in the form of a contingency table.

Michael Hardy (talk) 19:55, 7 January 2010 (UTC)

division by zero

I have some thing posted in my blog about the zero division and logarithm of zero i need to check it right or not please some one help me please visit here...

http://www.unfoldallmysteries.blogspot.com/ —Preceding unsigned comment added by Ajithrajnair (talk • contribs) 13:00, 6 January 2010 (UTC)

No, that is wrong. $\log 0\;\!$ is definitely not 0. If anything, it's $-\infty$.

See also division by zero. -- Meni Rosenfeld (talk) 13:11, 6 January 2010 (UTC)

I see that you have also provided a "proof" that $\log 0=0$. Your error is that $2x=x\implies x=0$ only holds if x is a (real, or complex, or likewise) number. It does not hold in more general settings. -- Meni Rosenfeld (talk) 13:15, 6 January 2010 (UTC)

Reading further, I see that you also tried to find $\log(-1)$. The correct answer is $\log _{e}(-1)=(2k+1)\pi i\;\!$ where $i={\sqrt {-1}}$ (see also Complex number).

Perhaps you should study a bit more math before trying to invent things of your own? -- Meni Rosenfeld (talk) 13:21, 6 January 2010 (UTC)

our calculators and teachers will not explain what log0 is

they says it is a mathematical error but it is our duty for all Indians we must find out the value of log0

it is because zero is discovered by our grand fathers

I am quoting the above comment from your blog. Although you certainly seem like a nice person, you must understand that the evaluation of "log0" is based on mere convention, and has little mathematical value. Furthermore, I think a more modest duty for all Indians would be to do research in areas such as modern algebra (Indians are excellent at this, as well as at number theory, but of course this is not to say that Indians are not excellent at every branch of mathematics); one could argue that the invention of 0 ultimately served to create this field. Do you not agree? If I were feeling a little more bored at present, I would also note that a "calculator" can do as much as a "human" (in fact, humans can do much more than a calculator or even a supercomputer; the research in which a mathematician engages subsumes ideas to which a calculator cannot be associated); if a calculator cannot "explain log0" neither can any human, whoever his grandfather(s) may be. In any case, log0 is mathematically not very interesting; one could argue for it endlessly, but in the end it serves little purpose. (And to answer your question, if $\ln(0)=x$, $e^{x}=0$ by "definition". However, this is impossible, since the graph of the exponential function never intersects the x-axis. Perhaps a slighly more intellectual question would be to formally prove this, using theoretical ideas?). --PST 14:07, 6 January 2010 (UTC)

And let me also add that this elementary mathematics does not require a genius to crack, partly because it is not real mathematics, and partly because it does not subsume any mathematical intuition. In fact, even if someone could not "solve it", I would not dispel the idea that he/she is a mathematical genius; in fact the two circumstances are entirely independent. But perhaps I am being slightly extravagent in these remarks. --PST 14:20, 6 January 2010 (UTC)

I beg to differ. log0 is very interesting, and while a calculator cannot explain it, I certainly can: The function $\exp :\mathbb {R} \to (0,+\infty ),x\mapsto \sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}$ can be extended uniquely to a continuous function $\exp :[-\infty ,+\infty ]\to [0,+\infty ]$, where the domain is the extended real number line. The extended function is strictly increasing and satisfies $\exp(-\infty )=0$ and $\exp(+\infty )=+\infty$. It has an inverse function $\log :[0,+\infty ]\to [-\infty ,+\infty ]$, which satisfies $\log 0=-\infty$. A similar argument follows for the definition $\log :(0,+\infty )\to \mathbb {R} ,x\mapsto \int _{1}^{x}{\frac {dt}{t}}$. -- Meni Rosenfeld (talk) 14:36, 6 January 2010 (UTC)

OK. But what are the non-trivial consequences of this particular evaluation of "log0"? The fact that the exponential function can be uniquely extended to a larger domain is of course interesting, but I cannot see furthur than the fact that it is a special case of the (general) question of "extending continuous functions" (or maybe I am simply not thinking properly; if you could contradict me, and show me how this leads to something truly non-trivial, I would be curious). Of course, one could note that it (the exponential function) cannot be extended to a continuous function defined on the one-point compactification of the space of real numbers (since that would define an unbounded continuous function on a compact space). This, and that which you mentioned, are both quite interesting when compared. But I think we are losing the OP at this stage... --PST 15:17, 6 January 2010 (UTC)

This extension of log is indeed a special case of a much more general phenomenon - it is worth noting simply because log is such an important function, so how this general question applies to it specifically is of interest.

It probably doesn't have any far-reaching consequences, but it can certainly pop up occasionally and simplify things.

Hopefully the OP will get a feeling for how there's more to these things than some handwavy calculations. -- Meni Rosenfeld (talk) 15:41, 6 January 2010 (UTC)

A first nontrivial consequence is that this way one reduces the notion of limit, including all cases at infinity, to the notion of continuity, which is more elementary and handable (just compare the simplicity of the composition property of continuity with the weird proofs of rules of compositions of limits, that still one finds in some sadistic highschool textbooks).--pma 19:48, 6 January 2010 (UTC)

Quotients and products of topological spaces.

Let X and Y be topological spaces and let ${\overline {Y}}$ be a quotient of Y with quotient map $q\colon Y\to {\overline {Y}}$. (1) Is $1\times q\colon X\times Y\to X\times {\overline {Y}}$ a quotient map? (2) If not, is it true if we further assume that $Y=Z\times [0,1]$ and ${\overline {Y}}=(Z\times [0,1])/(Z\times 0)$. (3) If not, is it true if we further assume that $Z=S^{n}$.

I'm asking because this question often arises when trying to determine if a map with domain $X\times {\overline {Y}}$ is continuous. Case (3) above is the case I'm looking at at the moment. Aenar (talk) 17:50, 6 January 2010 (UTC)

Do you know the definition of a quotient map? Answering your first question should be a simple matter of applying it. --Tango (talk) 22:01, 6 January 2010 (UTC)

I can't remember clearly I had to return my topology books. There's a theorem about maps of the form $1\times q\colon X\times Y\to X\times {\overline {Y}}$; I think it's a quotient map if X is locally compact hausdorff. The restriction must be placed on X not Y. If your trying to prove continuity I suggest trying something else because quotient maps are quite special continuous maps (the topology on the range is the finest for which the function is continuous), so it's probably not a quotient map. Plus it's generally VERY difficult to prove a map is quotient. Hope that helps Money is tight (talk) 10:38, 7 January 2010 (UTC)

Thanks. I was speculating that I might need compactness or local compactness at some place. Do you have a reference for the result that $1\times q$ is a quotient map when X is locally compact Hausdorff? I found a reference for the result. Aenar (talk) 11:37, 7 January 2010 (UTC)

Limits

$\lim _{t\to -3}{\frac {t^{2}-9}{2t^{2}+7t+3}}$ will equal 0 if you apply the usual limit laws and consider the limit as t tends to -3 in the numerator.

Factorisation of the numerator and denominator gives us ${\frac {t-3}{2t+1}}$, which gives us the correct limit ${\frac {6}{5}}$. However, I only know this because I looked at the graph.

It seems arbitrary that a simplification should produce a different result, if the expression is equal. How can I possibly tell whether a function has a limit at a particular point unless I do the "correct" manipulations? There must be some strict method I can apply to every function to ensure I know I am taking limits correctly. —Preceding unsigned comment added by 94.171.225.236 (talk) 19:09, 6 January 2010 (UTC)

You are incorrect about the first assumption. You have to put -3 into both the numerator AND denominator. You will get 0/0, which is nonsense. So, you know that you must reduce in some way to get a valid answer since 0/0 is not valid. -- kainaw™ 19:16, 6 January 2010 (UTC)

Also, L'Hôpital's rule is very useful for computing such limits. -- Meni Rosenfeld (talk) 19:44, 6 January 2010 (UTC)

Try writing the original expression as ${\frac {(t+3)(t-3)}{(t+3)(2t+1)}}$, then plug in numbers very close to -3. You'll see the first factor of the numerator and denominator canceling and leaving you with ${\frac {t-3}{2t+1}}$. This happens at every t value except for -3, where you get ${\frac {0}{0}}$ as noted above. Thus, the unreduced fraction and the reduced fraction are identical as functions, except that -3 is in the domain of the second one, and not in the domain of the first. Thus, you just can't plug in -3 at all until you switch to the reduced expression. -GTBacchus^(talk) 20:08, 6 January 2010 (UTC)

Intuitively, what is $\lim _{t\to -3}{\frac {t^{2}-9}{2t^{2}+7t+3}}$? It is the value ${\frac {t^{2}-9}{2t^{2}+7t+3}}$ as t approaches −3 (while still maintaining a distance from this number). Thus, it is not necessarily the value of ${\frac {t^{2}-9}{2t^{2}+7t+3}}$ at $t=-3$. Furthermore, if one considers t sufficiently close to (but not equal to) −3, one can divide both numerator and denominator by $t+3$, since $t+3\neq 0$ when t is close to, but not equal to, −3. Therefore, $\lim _{t\to -3}{\frac {t^{2}-9}{2t^{2}+7t+3}}=\lim _{t\to -3}{\frac {(t-3)(t+3)}{(2t+1)(t+3)}}=\lim _{t\to -3}{\frac {t-3}{2t+1}}={\frac {-6}{-5}}={\frac {6}{5}}$, as desired. Since $\lim _{t\to -3}{\frac {t^{2}-9}{2t^{2}+7t+3}}$ exists, but ${\frac {(-3)^{2}-9}{2(-3)^{2}+7(-3)+3}}$ does not, we say that ${\frac {t^{2}-9}{2t^{2}+7t+3}}$ has a removable discontinuity at −3. Hope this helps. --PST 03:00, 7 January 2010 (UTC)

Extension fields

Hi. I'm preparing for an algebra exam, and I've been chewing on this problem, from a past exam. I'd appreciate a nudge in the right direction, if someone can give me a hint.

Let $f(x)\in \mathbb {Q} \left[x\right]$ be irreducible over $\mathbb {Q}$, let the degree of f be odd, and let $\alpha \in \mathbb {R}$ be a zero of f. Prove or find a counter-example: $\mathbb {Q} \left(\alpha \right)=\mathbb {Q} \left(\alpha ^{2}\right)$.

My hunch is that the claim holds for, say, $f(x)=x^{3}-2$, but not for $f(x)=x^{5}-2$. I don't know how to show that, though. Thanks in advance. -GTBacchus^(talk) 19:56, 6 January 2010 (UTC)

For α a root of x³-2, α⁴/2 = α and for α a root of x⁵-2, α⁶/2 = α so the hunch is not right. It might be helpful to think about it in terms of the degrees of the extensions. In Q(α²), α is a root of x² - α² so [Q(α) : Q(α²)] divides 2. On the other hand [Q(α) : Q] is odd by assumption, which tells you [Q(α²) : Q]. Rckrone (talk) 22:09, 6 January 2010 (UTC)

Thank you very much. -GTBacchus^(talk) 22:12, 6 January 2010 (UTC)

(ec) Write f(x)=xg(x²)-h(x²) with g and h in Q[x], and deg(g)≥deg(h). Then we have to show that g(α²)≠0, which implies that α=h(α²)/g(α²), which is in the field Q[α²]. Certainly g(α²)=0 implies h(α²)=0, and it would follow that f(x) is not the minimal polynomial of α (there is a polynomial of less degree of the form x(g(x)-a(x)h(x)). I wrote that without thinking; my algebraic head is somewhere in the hayloft you know; I'd be glad of any comment.--pma 22:23, 6 January 2010 (UTC)

How to solve for B

I swear this is not a homework question as I am in my late 20's and not in school, but I was helping my niece with her homework last month and I never got around to ask her how she solved it. I don't want to look stupid and ask her but this was bugging me for some time:

(A+B) x (B+C) = D

I know the answer is staring at me in the face, but I have gone completely blank. How do you solve for B? --Reticuli88 (talk) 21:39, 6 January 2010 (UTC)

Expand the brackets on the L.H.S What do you get? Theresa Knott | token threats 21:43, 6 January 2010 (UTC)

(edit conflict.) (Assuming these are all real numbers.) Multiplying out the left hand side gives you the equation $B^{2}+(A+C)B+AC=D$. This is a quadratic equation -- the article explains how to solve such an equation. Aenar (talk) 21:49, 6 January 2010 (UTC)

Note that $D=(A+B)\times (B+C)=A\times B+A\times C+B\times B+B\times C=AB+AC+B^{2}+BC=B^{2}+(A+C)B+AC$, so that we obtain the equation $B^{2}+(A+C)B+(AC-D)=0$. Make the substituition $X=B$ and we obtain the equation $(1)x^{2}+(A+C)x+(AC-D)=0$. Now, by the quadratic formula (see quadratic equation for more details), we have:

$B=x={\frac {-(A+C)\pm {\sqrt {(A+C)^{2}-4(1)(AC-D)}}}{2(1)}}$

Is this clear? --PST 02:40, 7 January 2010 (UTC)

how did you make it equal to zero? --Reticuli88 (talk) 14:57, 7 January 2010 (UTC)

Subtract D from both sides. -- Meni Rosenfeld (talk) 15:10, 7 January 2010 (UTC)

He subtracted D from both sides of the equation, which moves the D from the left-hand side of the equals sign to the right-hand side. asyndeton talk 15:10, 7 January 2010 (UTC)

Logarithms and order of operations

Is $log_{b}x^{5}$ considered $log_{b}(x^{5})$ or $(log_{b}x)^{5}$? According to [3], logarithms are on the same level as square roots and exponents, so then with the above question, as one goes right-to-left with equivalent-level operations, is $(log_{b}x)^{5}$ thus correct? Thanks, Spencer^{T♦Nominate!} 22:51, 6 January 2010 (UTC)

With log x (regardless of the base), log x^a = log (x^a). To raise the function to a power, log^a x = (log x)^a. Of course, you can use it however you like as long as you make it clear what you mean. -- kainaw™ 00:35, 7 January 2010 (UTC)

Caution: some people also write $\log ^{a}x=\underbrace {\log \log \dots \log } _{a{\text{ times}}}x$. — Emil J. 11:47, 7 January 2010 (UTC)

Note that, in the context of the question, $\log _{b}(x^{5})=5\log _{b}(x)\neq (\log _{b}(x))^{5}$. For instance, if we consider the natural logarithm, $\ln e^{5}=\ln(e^{5})=5\ln(e)=5\cdot 1=5$ but $(\ln(e))^{5}=1^{5}=1$ (and of course, 5 is not equal to 1). Thus, in general, no: $\log _{b}x^{5}=\log _{b}(x^{5})\neq (\log _{b}x)^{5}$. Hope this helps. --PST 02:31, 7 January 2010 (UTC)

Okay, thanks. This answers my question. Spencer^{T♦Nominate!} 02:54, 7 January 2010 (UTC)

January 7

TI 84 plus calculator program

Can somone please make me a program for factoring? Thanks! Accdude92 (talk to me!) (sign) 17:58, 7 January 2010 (UTC)

Have you tried a google search with "TI 84 plus calculator program factoring"? Dmcq (talk) 18:26, 7 January 2010 (UTC)

Are you talking about factoring integers, or polynomials? -GTBacchus^(talk) 18:31, 7 January 2010 (UTC)