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!


(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.

4 thoughts on “Math that made you go whoa!

  1. All the Cantor stuff is ‘whoa-some’ ;-), but I’ve always found “Cantor’s dust” more mind-blowing than the diagonalisation proof. Several paradoxes and fractal results are also mind-expanding, including ‘Gabriel’s horn’!

  2. One that I always loved was a bit of a cheat on trisecting an angle: that if you make two marks on the straightedge, you can easily trisect an arbitrary angle. I believe I encountered it from Courant and Robbins. But as a demonstration of something neat to do and of just how the exact rules of the operation affect the possibilities it’s great.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s