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.

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Comments

3 Comments so far. Leave a comment below.
  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. Neat stuff, keep up the good work. Those of us who have given up the fight and retired can still find hope for the future with teachers like you out there.

  3. 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.

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