Math Class

I've been doing math my whole life - I love the stuff. When I get a problem I can't do, I keep working on until I run out of mathematical rules to try against it. Like the Collatz conjecture. I love the Collatz conjecture - I've got a whole folder of programs I've written to try billions and billions of numbers, hoping against hope to find a counterexample. The version I wrote in C can calculate the sequences for about a range of half a trillion numbers, mainly because it doesn't look at even numbers - it can easily be proven that no even number the lowest counterexample to the Collatz conjecture. The same is true for numbers equal to 1 mod 4, 3 mod 16, 11 or 23 mod 32, and several others. But that's all very boring stuff to people who have no idea what I'm talking about. This is the internet, there's no room for intelligent discussions of heavy mathematics. The people want entertainment! The people demand cats playing piano!

Anyway, math class. Here's a problem even the best of calculus teachers would have trouble with:

Question: Find the line A given by the equation y = mx + b such that A is tangent to these two curves:

f(x) = 2x² + 3

g(x) = (-1/2)x² - 2

Answer: I have to admit, I got different answers every time I did this problem. I eventually got it right, and checked it on my graphing calculator.

Let one such tangent line intersect y = 2x²+3 at (a,b)
in the first quadrant and intersect y = (-1/2)x² - 2 at (c,d)
in the third quadrant.

Taking derivatives gives the slope at (a,b) to be y' = 4x = 4a,
and at (c,d) the slope is y' = -x = -c. So 4a = -c and c = -4a.

Substitution gives b = 2a² + 3
and d = (-1/2)c² - 2 = (-1/2)(-4a)² - 2 = -8a² - 2.

The slope of the line also equals (b-d)/(a-c),
or (2a²+3 +8a²+2)/(a-(-4a)) or (10a²+5)/5a or 2a+1/a

Thus equating values for the slope gives 4a = 2a + 1/a,
or 2a = 1/a, or 2a² = 1, or a = (1/2) * sqrt(2).

This means b = 2a² + 3 = 2((1/2) * sqrt(2))² + 3 = 2 * (1/2) + 3 = 4.

Also c = -4a = -2 * sqrt(2).
And d = (-1/2)(-2 * sqrt(2))² - 2 = -6.

The equation of this line through ((1/2)sqrt(2),4)
with slope = 4a = 2 * sqrt(2) is

y = (2 * sqrt(2))x + b' (b' is the y-intercept).

But when x = a = (1/2)sqrt(2), y = b = 4,
so b' = 4 - (2 * sqrt(2) * 1/2 * sqrt(2)) = 4 - 2 = 2.

So the line is

y = (2 * sqrt(2))x + 2.

The other symmetric tangent line has the same y-intercept
but the negative of this slope, so:

y = (-2 * sqrt(2))x + 2.

If you graph all this, it looks like so:

You like that? Made that in Grapher. Pretty sweet, if I must say so myself.

Anyway, you only get one good one for Math Class today. But wow, what a problem. I mean, damn.