Wikipedia:Reference desk/Archives/Mathematics/2010 January 2

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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)