Aaaanyways, though, that gives $V_0\approx 113.6$. Now the moment of truth, when I compare that to Mathematica and see if I've done a silly algebra error myself :/
So if you solve that for the collision time $t$, you get $$t=\frac{-(9-V_0)\pm\sqrt{(9-V_0)^2-4(16)(171)}}{2(16)}$$
There are then three possibilities. If what's inside the square root is negative, then you don't get any real solutions.
If it's positive, you get two real solutions.
If it's zero, you get exactly one zero i.e. the quadratic has a double root.
The condition for that is that $(9-V_0)^2=4(16)(171)$. If we assume that $V_0>9$ (it has to for this to make any sense) then taking the appropriate square root and solving gives $V_0=9+\sqrt{4(16)(171)}=9+48\sqrt{19}$.
This is assuming I did the algebra right, though. Lemme make sure.
Just read the number of Hopf-Galois structures on L/K of type Gal(L/K)=Zp^n is asymptotically p^((2/27)n^3), which is coincidentally asymptotically the number of groups of order p^n up to isomorphism.
If you think about the picture I gave earlier, this is telling you that 1) If the velocity is sufficiently high, then the ball would pass Colleen on the way up and then again on the way down, 2) if it's too small, the ball never reaches Colleen at all, 3) if the velocity is just right, there's only one time when it reaches Colleen.
@WillNjundong The upshot is that it is probably faster to do it via the quadratic equation and reasoning about when you'd get a double root (and why you want one)
OEIS calls them Numbers whose square is a nontrivial concatenation of other squares, and calls the sequence of their squares Squares which are a decimal concatenation of two or more squares. One of the references from that latter sequence calls the squares Smarandache square-partial-digital [numb...
If $n \in \mathbb{N}$, then $(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16}) \dotsb (1-\frac{1}{2^n}) \ge \frac{1}{4}+\frac{1}{2^{n+1}}$
Anyone has any idea how to prove this by induction? I checked the base case when $n=1$ and got $\frac{1}{2} \ge \frac{1}{2}$. Now I need to show that $(1-\frac{1}{2})(1-\frac{1}{4}) \dotsb (1-\frac{1}{2^{k+1}}) \ge \frac{1}{4}+\frac{1}{2^{k+1}}$
Right. More precisely, if you multiply the inductive hypothesis on both sides by $1-1/2^{k+1}$, then you'll have the left-hand side of the desired result.
Okay, so we have $(1-\frac{1}{2})(1-\frac{1}{4}) \dotsb (1-\frac{1}{2^{k}})(1-\frac{1}{2^{k+1}}) \ge (\frac{1}{4}+\frac{1}{2^{k+1}})(1-\frac{1}{2^{k+1}})$
$(\frac{1}{4}+\frac{1}{2^{k+1}})(1-\frac{1}{2^{k+1}}) \ge (\frac{1}{4}+\frac{1}{2^{k+1}})(1-\frac{1}{2^{k+1}})$ by the inductive hypothesis (inductive step)
Or does that not make sense. I think I just went in a nonsense loop
