Month: May 2018

Incredibly difficult math puzzle

For math/puzzle enthusiasts (as well for as my own future reference): This was one of the most diabolically difficult puzzles that I’ve ever seen. The object: use the numbers 1-9 exactly once in each row and column while ensuring that the given arithmetical operation in each cage is also correct. Here it is. Fair warning: while most MathDoku+ puzzles take me 20-40 minutes to solve, this one took me over 3 hours (spread out over 5 days).

 

 

5 Ways to go Beyond Recitation

Most students will encounter recitation in a math class during their academic career. How can math professors make the experience more meaningful? MAA Teaching Tidbits blog has 5 ways educators can enhance the student experience during recitation.

  1. Focus on getting students to do the work instead of doing it for them.
  2. Incorporate group work into your sessions.
  3. Get students to communicate what they understand to each other and to the class.
  4. Have students relate mathematics to their own experiences.
  5. Cultivate an environment where failure is ok and experimentation is encouraged.

Full article: http://maateachingtidbits.blogspot.com/2017/09/5-ways-to-go-beyond-recitation.html

My Favorite One-Liners: Part 114

In this series, I’m compiling some of the quips and one-liners that I’ll use with my students to hopefully make my lessons more memorable for them.

I’ll use today’s one-liner whena step that’s usually necessary in a calculation isn’t needed for a particular example. For example, consider the following problem from probability:

Let X be uniformly distributed on \{-1,0,1\}. Find \hbox{Cov}(X,X^2).

The first step is to write \hbox{Cov}(X,X^2) = E(X \cdot X^2) - E(X) E(X^2) = E(X^3) - E(X) E(X^2). Then we start computing the expectations. To begin,

E(X) = (-1) \cdot \displaystyle \frac{1}{3} + 0 \cdot \displaystyle \frac{1}{3} + 1 \cdot \displaystyle \frac{1}{3} = 0.

Ordinarily, the next step would be computing E(X^2). However, this computation is unnecessary since E(X^2) will be multiplied by E(X), which we just showed was equal to 0. While I might calculate E(X^2) if I thought my class needed the extra practice with computing expectations, the answer will not ultimately affect the final answer. Hence my one-liner:

To paraphrase the great philosopher The Rock, it doesn’t matter what E(X^2) is.

P.S. This example illustrates that the covariance of two dependent random variables (X and X^2) can be zero. If two random variables are independent, then the covariance must be zero. But the reverse implication is false.