Day: December 15, 2015

3 levels of learning math – all surprisingly similar

Had an interesting morning with my son working through a challenging problem that required him to pull together some ideas from geometry and from algebra. I’m slightly annoyed with myself because we were running a little behind schedule and ended up choosing to walk him through the solution rather than finding a better path. Still though, it was a good example of how hard it can be to pull together ideas from different parts of math.

But, I also saw to other pieces about learning math today – one was a post about slightly younger kids, and one was a post about math professors struggling to understand a complicated proof. The similarities in the struggles were fascinating to me.

(1) Young kids – from Michael Pershan:

Dissent of the Day by Michael Pershan

“While walking around, I noticed some kids getting lost in their calculations. Lots of great ideas, but constantly losing the thread.”

Not losing the thread is truly one of the biggest challenges in math.

(2) slightly older kid this morning – the struggle is to find the right path through all of the algebra and geometry:

 

(3) Now skip ahead a few years: professional mathematicians struggling to understand a difficult proof.

Notes on the Oxford IUT Workshop by Brian Conrad on Mathbabe.org

I think if you are interested in communicating / teaching math at any level, this article is an important read. For example:

“There was substantial audience frustration in the final 2 days. Here is an example.

We kept being told many variations of “consider two objects that are isomorphic,” or even something as vacuous-sounding as “consider two copies of the category D, but label them differently.” Despite repeated requests with mounting degrees of exasperation, we were never told a compelling example of an interesting situation of such things with evident relevance to the goal.”

Sort of amazing to have run across all three of these examples on the same day.

Math that made you go whoa!

Saw this tweet from Dan Anderson a few days ago:

I had a 7 hour round trip drive yesterday and spent a little time thinking about the math ideas that really grabbed me in high school. Three really stuck out in my mind:

(A) The Extended law of sines:

We learned in our trigonometry class that for a triangle with sides A, B, and C, and corresponding angles a, b, and c that:

\frac{A}{Sin(a)} = \frac{B}{Sin(b)} = \frac{C}{Sin(c)}

But it turns out that these ratios are equal to 2R where R is the radius of the circumscribed circle. I learned this idea from the wonderful book Geometry Revisited by Coxeter:

Book 2

In fact, the first math movie I uploaded to youtube was about this idea:

This identity made me think that there was a lot more going on in geometry that met the eye. One neat particularly neat thing that the identity shows is that the area of a triangle with side lengths A, B, and C is equal \frac{ABC}{4R}. Beautiful!

(B) 1 + 1/4 + 1/9 + . . . . = \frac{\pi^2}{6}

Mr. Waterman used the idea that the coefficients of a polynomial were symmetric functions of the roots to prove this sum. It blew me away. (yes, this is a non-rigorous proof, but it is what captured my attention)

In general, for a polynomial of degree n, x^n + c_{n-1}x^{n-1} + \ldots + c_1 x + c_0, sum of the reciprocals of the roots is given by -c_1 / c_0.

We know that Sin(x) = x - x^3 / 3! + x^5 / 5! + \ldots. Factoring out an x we are left with a polynomial whose roots are \pm \pi, \pm 2\pi, \pm 3\pi, \ldots, namely:

\frac{Sin(x)}{x} = 1 - x^2 / 3! + x^4 / 5! + \ldots

making the substitution u = x^2, we see that the polynomial

1 - u/3! + u^2 / 5! + \ldots

are \pi^2, 4\pi^2, 9\pi^2 \ldots

By the “sum of the reciprocals of the roots” formula above, we see that

\frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \ldots = \frac{1}{6}.

Multiplying both sides by \pi^2 gives us the result.

This result showed me that there was more going on with the integers than I realized! How could they be connected to \pi? A few years later I’d see this identity in a complex analysis class and see that \pi and e were connected in a strange way, too!

Formula

(3) A formula for the Fibonacci numbers

I think it was my sophomore year in high school when a former student, Anita Barnes, came back to lecture to Mr. Waterman’s Enrichment Math class. Her talk showed a way to find closed form solutions for simple recurrence relations like the one for the Fibonacci numbers:

F_{n+1} = F_n + F_{n - 1}

The idea seemed incredibly simple – for the Fibonacci numbers just assume the solution took the form F_n = x^n and solve for x. Solving the recurrence relation for the Fibonacci numbers was reduced to solving the quadratic equation x^2 = x + 1. From there it was not hard at all to show that the Fibonacci numbers were connected to the Golden ratio. If we let \phi = \frac{1 + \sqrt{5}}{2}, then

F_n = ( \phi^n - (-\phi)^{-n}) / \sqrt{5}

That just blew me away – there was a simple formula for the Fibonacci numbers (and any simple recurrence relation). You could calculate the 100th Fibonacci number by just knowing the first 2 plus the recurrence relation. I think this was the first idea from advanced math that totally blew my mind.