Acceleration of convergence

Let .

1. Evaluate , , and .
2. Evaluate .
3. Can you explain why gives a better approximation to the limit than does?

Concept classification via Google page counts

Two years ago, I explored the possibility of using Google page counts to measure similarity between concepts. The results are contained in this article. Although the results were not as strong as I had hoped, they may serve as a starting point for further exploration.

My favorite word problem

My “favorite” word problem is from Beginning Algebra, 6th edition by Lial, Miller, and Hornsby. I had the misfortune of using this textbook to teach a course in beginning algebra.

The distance between Singapore and Tokyo is 3300 miles. On a certain wall map, this distance is represented by 11 inches. The actual distance between Mexico City and Cairo is 7700 miles. How far apart are they on the same map? [#53, p. 117]

Now, this would be an excellent problem… IF THE WORLD WERE FLAT!

Introduction to Differentials

I gave an exam to my calculus students last week, and I was disappointed to find that my students did not understand differentials very well. So, I have written a tutorial on differentials to supplement the discussion in Stewart. I have posted the document in PDF format, as well as the LaTeX source, in hopes that it is useful for other calculus instructors or students. Constructive comments are welcome.

Semidirect Products II

In today’s class we continued to discuss semidirect products.

A homomorphism f : G -> Q is called a split epimorphism if there exists a homomorphism s : Q -> G so that the composition fs is the identity on Q. We proved that there exists a split epimorphism f : G -> Q if and only if G is a semidirect product of K and Q1, where Q1 is isomorphic to Q. If this occurs, then we may take K = ker(f) and Q1 = sQ.

We also learned that if G is the semidirect product of K and Q (where K is normal) then Q acts on K by conjugation. In other words, there is a homomorphism from Q to Aut(K) defined by

q -> (k -> q^-1 k q)

It turns out that the semidirect product is uniquely determined up to isomorphism by the groups K and Q and the action of Q on K.

Semidirect products

I am auditing a group theory course this semester, and I intend to post my notes to this blog. The first class meeting was today. The professor spent a great deal of time discussing the syllabus, and there was not much time left to present material, but he did introduce semidirect products.

Let K and Q be subgroups of G. We say that G is the internal semidirect product of K and Q if the following conditions hold:

1. K is a normal subgroup of G,
2. G = KQ, and
3. the intersection of K and Q is the identity.

Notation: G = K Q.

We recall that, for any two subgroups U and V of G, UV is defined as {uv : u in U, v in V}. In general UV is not a subgroup, but it is a subgroup if at least one of the factors is normal. Also, UV is a subgroup of G if and only if UV = VU.

If K and Q are both normal, then G is the internal direct product of K and Q. In that case it is easily shown that the elements of K commute with the elements of Q.

Examples of semidirect products:

1. S_n = A_n C₂
2. D_2n = C_n C₂
3. S₄ = V S₃

S_n denotes the symmetric group on n letters. A_n is the alternating group, consisting of the even permutations of S_n. D_2n is the dihedral group of order 2n, i.e. the symmetry group of a regular n-sided polygon. C_n is the cyclic group of order n, and V is the Klein 4-group.

44th Mersenne Prime discovered!

http://mathworld.wolfram.com/news/2006-09-04/mersenne-44/

Proof of Burnside’s Lemma using Multiorbits

Burnside’s Lemma is one of my favorite theorems. It says that if a finite group G acts on a finite set X, then the number of orbits is equal to the average number of elements fixed by a group element. The standard proof uses the orbit-stabilizer theorem. The proof is elegant and easy to understand, but it left me unsatisfied. Fortunately, I found a different proof that I find more intuitive. The proof is due to Kenneth Bogart (1943 – 2005).

If g is an element of G, then let Fix(g) denote the number of elements of X so that g*x = x. Likewise, if x is an element of X, then let Fix(x) denote the number of elements of G so that g*x = x. Our first observation is that

sum (g in G) Fix(g) = sum (x in X) Fix(x).

This holds because both sums enumerate the pairs (g,x) so that g*x = x.

The next ingredient is the idea of a multiorbit. The multiorbit of x, denoted G(x), is the multiset [gx : g in G]. The size of G(x) is equal to |G|, and the multiplicity of x in G(x) is equal to Fix(x).

Now, the union of the distinct multiorbits has size n * |G|, where n is the number of orbits. The multiplicity of x in this union is equal to Fix(x). Since the size of a multiset is equal to the sum of the multiplicities of its distinct elements, we conclude that

sum (g in G) Fix(g) = sum (x in X) Fix(x) = n * |G|.

It follows that the number of orbits (n) is

1/|G| * sum (g in G) Fix(g)

Reference: Bogart, Kenneth P. An obvious proof of Burnside’s lemma. Amer. Math. Monthly 98 (1991), no. 10, 927–928.

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