I still want to run a negative split, but upping my tempo in discrete steps is such a slog. Instead, my next plan is to continuously increase my pace in the following way:
At the beginning of the race, I’ll start with a 24-minute pace. Then, wherever I am on the race course, I’m always running 10 percent faster than I was when I was twice as far from the finish line. Also, my speed should always be increasing in a continuous and smooth fashion. Using this strategy, how long will it take me to complete the 5K? Once again, I’m hoping it’s faster than my steady 23-minute pace, even though I started out slower.
Let's again assume that the race start at x=0 and finish at x=5, but we will need to understand some sort of history, since otherwise when I am one step into the race, I would need to know how fast I was going about 2 steps closer than 5 kilometers before the beginning of the race. So here let's define x_n = 5(1 - 2^{-n}) for n = \mathbf{-1}, 0, 1, 2, \dots be breakpoints where we will be interested in defining our speed. Let v(x) denote the velocity you are running (in kmph), when you are at some x \in (-5,5). Again, here we are given that v(0) = 12.5 and that at any point x \in (0,5) we have v(x) = 1.1 v(2x-5). Since x_0 = 0, and 2x_n-5 = x_{n-1} for each n = 0, 1, \dots, we can use the recursive formula to find the values of v_n = v(x_n) = 12.5 \cdot 1.1^n, \,\, n = -1, 0, 1, 2, \dots
Now let's tackle what smooth means here. The link provided in the prompt to Wolfram suggests that we are talking about some sort of C^n(-5,5) space of functions on (-5,5) that are continuous and have continuous derivatives up to order n for some n. We could of course overachieve and go with C^\infty(-5,5), which we will start with here, but then see if we can relax this for some more Zeno's 5K fun. So without further adieu, let's start with the C^\infty(-5,5) solution:
First, by design and analogy, we are likely to have a velocity function v that blows up to \infty as x \to 5^-. Let's try something from the parametric family v_\alpha(x) = \frac{12.5}{(1-\frac{x}{5})^\alpha} for \alpha \gt 0, which satisfy v_\alpha \in C^\infty(-5,5), v_\alpha(0) = 12.5 and v_\alpha^\prime (x) \geq 0 for all x \in (-5,5). The only thing left to do is to decide which value of \alpha satisfied the recursion formula. Note that v_\alpha(2x-5) = \frac{12.5}{\left(1 - \frac{2x-5}{5}\right)^\alpha} = \frac{v_\alpha(x)}{2^\alpha}, for all x \in (0, 5) so in order to satisfy the recursion formula v(x) = 1.1 v(2x-5) we must have \alpha = \frac{\ln 1.1}{\ln 2}. So we have a C^\infty (-5,5) solution for velocity at position x \in (0,5), v^*(x) = 12.5 \left( 1 - \frac{x}{5} \right)^{-\frac{\ln 1.1}{\ln 2}}, which gives a completion time for the C^\infty extra credit Zeno's 5K of \begin{align*}T = \int_0^5 \, \frac{dx}{v(x)}& = \int_0^5 \frac{ \left(1 - \frac{x}{5} \right)^{\frac{\ln 1.1}{\ln 2}} }{12.5} \,dx\\ &= \frac{5}{12.5} \int_0^1 u^{\frac{\ln 1.1}{\ln 2}} \,du \\ &= \frac{5}{12.5} \frac{1}{1 + \frac{\ln 1.1}{\ln 2}} = 0.351647262315\dots \,\,\,\text{hours},\end{align*} or 21.0988357389\dots minutes.
Obviously, C^\infty qualifies as smooth, but what if we are willing to go with only C^1(-5,5) solutions. Obviously, since C^\infty \subsetneq C^1, we still have the solution v^* from above, but what if we use our points \mathcal{X} = \{x_{-1}, x_0, x_1, \dots \} as knots to define a C^1 quadratic spline. Can we still get a solution that is C^1 and satisfies the recursion formula? It turns out, \dots yes!
Lets define the velocity function u = u(x) by spline components u_n(x) = a_n + b_nx + c_nx^2 for x_{n-1} \leq x \leq x_n, for n = 0, 1, 2, \dots. In order to satisfy the recursion formula, since 2x-5 \in (x_{n-1},x_n) for all x \in (x_n, x_{n+1}), we would need \begin{align*}u_{n+1}(x) &= a_{n+1} + b_{n+1} x + c_{n+1} x^2 \\ &= 1.1 u_{n}(2x-5)\\ &= 1.1 \left( a_n + b_n (2x-5) + c_n (4x^2 - 20x + 25) \right)\\ &= 1.1 (a_n - 5 b_n + 25 c_n) + 1.1 (2b_n - 20c_n) x + 4.4 c_n x^2,\end{align*} so we must have the recursion formulae \begin{align*} a_{n+1} &= 1.1 \left( a_n - 5b_n + 25 c_n \right) \\ b_{n+1} &= 2.2 \left( b_n - 10 c_n \right) \\ c_{n+1} &= 4.4 c_n \end{align*} for all n = 0, 1, \dots.
Using this recursion formula we can define each quadratic spline component in terms of the parameters (a_0, b_0, c_0). We need to confirm that we can choose appropriate values of (a_0, b_0, c_0) such that the resulting spline u = u(x) \in C^1(-5,5), that u^\prime(x) \geq 0 for all x \in (-5,5) and the u(0) = 12.5. Since x_0 = 0 is on both the u_0 and u_1 components of the spline and we want both u and u^\prime continuous at x_0 we must have u_0(0) = a_0 = u_1(0) = a_1 = 1.1 (a_0 - 5b_0 + 25 c_0) = 12.5 and u^\prime_0(0) = b_0 = u^\prime_1(0) = b_1 = 2.2 (b_0 - 10c_0). So we can group these conditions into the system of three equations and three unknowns \begin{align*}a_0 &= 12.5 \\ 0.1 a_0 - 5.5 b_0 + 27.5 c_0 &= 0 \\ 1.2 b_0 - 22 c_0 &= 0,\end{align*} which can be solved by (a_0,b_0,c_0) = (12.5, \frac{5}{16}, \frac{3}{176}).
From the choice of (a_0,b_0,c_0) we have already established the base case u_0(x_0) = u_1(x_0) = v_0 = 12.5. Suppose that for some n we have u_n(x_{n-1}) = u_{n-1}(x_{n-1}) = v_{n-1}. By construction, we have u_n(x_n) = 1.1u_{n-1}(2x_n-5) = 1.1u_{n-1}(x_{n-1}) = v_n. Similarly, by construction, we have u_{n+1}(x_n) = 1.1u_n(2x_n-5) = 1.1u_n(x_{n-1}) = v_n. Therefore, for each n = 0, 1, \dots, by induction we have u_n(x_n) = u_{n+1}(x_n) = v_n, which proves that u \in C(-5,5). Similarly, we have from the choice of (a_0,b_0,c_0) already established that u_0^\prime(x_0) = u_1^\prime(x_0). Assume that for some n we have u_n^\prime(x_n) = u_{n+1}^\prime(x_n). Since u_{n+1}(x) = 1.1u_n(2x-5) we have u^\prime_{n+1}(x) = 2.2u_n^\prime(2x-5), so in particular u^\prime_{n+1}(x_{n+1})=2.2u_n^\prime(2x_{n+1}-5) = 2.2u_n^\prime(x_n). Similarly, we have u^\prime_{n+2}(x) = 2.2u_{n+1}^\prime (2x-5), so in particular u^\prime_{n+2}(x_{n+1}) = 2.2u^\prime_{n+1}(2x_{n+1}-5) = 2.2u_{n+1}^\prime(x_n) = 2.2u_n^\prime(x_n) = u^\prime_{n+1}(x_{n+1}), so we conclude that u_n^\prime(x_n) = u_{n+1}^\prime (x_n), for all n = 0, 1, 2, \dots, so u \in C^1(-5,5). From the solution of (a_0,b_0,c_0) we have u_0^\prime(x) \geq 0 for all x \in (x_{-1}, x_0), and since u^\prime_{n}(x) = 2.2^n u_0(2^nx - 5(2^n-1)) \geq 0 for each n = 1, 2, \dots, we have u^\prime(x) \geq 0 for all x \in (-5,5). So the C^1 monotonic increasing quadratic spline u is another C^1 solution to the extra credit Zeno 5K. This quadratic spline solution provides a slightly slower C^1 completion time of T = \int_0^5 \,\frac{dx}{u(x)} \approx 0.354718139\dots \,\,\text{hours} or 21.28308834\dots minutes. Ultimately, it would appear that for any n there is a C^n solution comprised of a monotonically increasing (n+1)th order polynomial spline that satisfies the recursion formula of v(x) = 1.1v(2x-5) for all x \in (0,5).