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 $x \in (0,5).$
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