Motivation

While working on a project in analog complexity theory, I encountered a simple but elegant question: how quickly does the function

converge to $\pi/4$ as $t \to \infty$?

This function arises naturally in the study of chemical reaction networks (CRNs) as computational devices. A CRN can compute a real number $\alpha$ by having a designated species whose concentration converges to $\alpha$ over time. The speed of this convergence determines the time complexity of the computation.

The function $f(t) = \arctan(1 - e^{-t})$ appears in a construction that proves $\pi$ is computable by a CRN in real time — meaning the concentration reaches $r$ bits of precision in $O(r)$ time. The key step is showing that $f(t) \to \pi/4$ at an exponential rate.

The Taylor expansion approach

One natural approach is to expand $\arctan(x)$ in a Taylor series around $x = 1$:

Substituting $x = 1 - e^{-t}$, so that $x - 1 = -e^{-t}$, we get:

This immediately gives $|f(t) - \pi/4| = \Theta(e^{-t})$.

But wait — is this Taylor expansion legitimate? We are, after all, dealing with a composite function. Might we be “misusing” Taylor’s theorem?

The answer is: the expansion is perfectly valid. Here’s why.

The function $\arctan(x)$ is real-analytic on all of $\mathbb{R}$. Its Taylor series at $x = 1$ converges for all $|x - 1| < R$, where $R$ is the distance from $1$ to the nearest singularity of $\arctan$ in the complex plane. The function $\arctan(z) = \frac{1}{2i}\log\frac{1+iz}{1-iz}$ has branch points at $z = \pm i$, so:

Since $|u(t) - 1| = e^{-t} \leq 1 < \sqrt{2}$ for all $t \geq 0$, the substitution $x = 1 - e^{-t}$ stays within the radius of convergence for all $t$. The Taylor expansion, and hence the convergence estimate, is valid.

We are not expanding the composite function $\arctan(1 - e^{-t})$ directly in powers of $t$. Rather, we first expand $\arctan(x)$ at $x = 1$ (which is standard), and then substitute $x = u(t)$, which is valid as long as $u(t)$ stays in the disk of convergence.

A simpler proof via the Mean Value Theorem

If the Taylor expansion approach feels like overkill, there is a much more elementary proof.

This gives us a tight, two-sided bound showing that $|f(t) - \pi/4| = \Theta(e^{-t})$, with no Taylor series needed at all.

What does this mean computationally?

In the framework of computable analysis, the time complexity of computing a real number $\alpha$ is the function $\mu(r)$ giving the time needed to achieve $r$ bits of precision (i.e., error $< e^{-r}$).

For $f(t) = \arctan(1 - e^{-t})$:

• Error $< e^{-r}$ requires $e^{-t} < e^{-r}$, i.e., $t > r$.
• So $\mu(r) = \Theta(r)$: linear time.

In the ODE/dynamical systems world, this kind of convergence is called “exponential convergence” (error decays as $e^{-t}$). But in computable analysis, it corresponds to the fastest possible rate for computing a constant — linear time.

This terminological inversion is worth noting:

What the ODE community calls “fast” (exponential convergence) is merely “linear time” in computational terms. And what they call “slow” (polynomial convergence) is actually “exponential time” — a fundamentally harder computation!

Takeaway

The proof that $\arctan(1 - e^{-t}) \to \pi/4$ exponentially quickly is correct, whether via Taylor expansion or the Mean Value Theorem. The MVT approach is arguably more elementary and gives tight two-sided bounds.

This small verification exercise highlights a beautiful connection between classical analysis, analog computation, and complexity theory: the same convergence estimate that an analyst might view as routine has deep implications for the computational complexity of the number $\pi$ in the world of chemical reaction networks.